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# Algorithm for a list of best solutions to the Assignment problem

I am trying to brute force a classical substitution cipher. The problem is that there are $26!$ possible keys. So, I'd like to do frequency analysis to try likely keys first. Then, on the first $n$ tries I will use a dictionary to see if there are actual words in the decryption. That way, I don't need to try all possible keys.

So, as suggested here, I will use chi-squared testing. I was thinking of making a matrix, best illustrated with an example. Suppose in an alphabet {A,B,C} the letter frequencies are 50%, 30% and 20% respectively, and that in a ciphertext the frequencies are 10%, 50% and 40% respectively. Then the A-B cell in the matrix is $(0.1-0.3)^2=0.04$, which is the error rate when a B in plain is encrypted as an A.

   |  A   |  B   |  C                 | Use in plaintext | in ciphertext
---+------+------+------           ---+------------------+---------------
A | 0.16 | 0.04 | 0.01             A |        0.5       |       0.1
B | 0.00 | 0.04 | 0.09             B |        0.3       |       0.5
C | 0.01 | 0.01 | 0.04             C |        0.2       |       0.4

The Hungarian algorithm then gives me a frequency-analysis-wise optimal key: $A \mapsto C, B \mapsto A, C \mapsto B$. This is because the sum of the errors (0.01+0.00+0.01=0.02) is minimal. This is useful, but what I need is the $n$ best keys.

The only thing I can come up with is to run the Hungarian algorithm, and then set one of the cells in the matrix above corresponding to the key found to a high value, so that if you run the algorithm again the key you find doesn't contain the mapping. So, in the example above, you could set A-C to 1 so that when you run the Hungarian algorithm again you find a different key that doesn't contain $A\mapsto C$.

However, this isn't guaranteed to find good keys in order. What would be a better way to extend the Hungarian algorithm to find the best $n$ keys?

Epilogue: after implementing this using D.W.'s approach below, it turned out that this method doesn't perform well enough to crack small-length (at least up to 1000 letters) cipher texts, because frequency analysis alone isn't enough. Performance may be improved by taking frequent digrams or trigrams into account, but I doubt this method can be as powerful as simple hill climbing.

• 1. If you'd like to know about better methods to crack classical substitution ciphers, I encourage you to ask a new question. There are some very cool techniques out there, but they utilize very different methods. 2. The question about how to enumerate the top-$n$ solutions to the assignment problem is still interesting in its own right. I'm trying to figure out if there's a way to reduce it to the problem of finding the $n$ shortest paths in a suitable graph, but I can't quite figure out how to make this idea work. – D.W. Jan 4 '16 at 20:48

Here's one technique to enumerate the best $n$ assignments, for any instance of the assignment problem. I suspect my approach isn't optimal, but it does run in polynomial time: it uses $O(nm)$ invocations of the Hungarian algorithm, where $m$ denotes the number of agents in the problem instance. In your example, $m=26$, so my approach requires $O(n)$ invocations of the Hungarian algorithm.

Let $A_1,A_2,A_3,\dots$ denote the assignments, from best to worse. $A_1$ is the best assignment; $A_2$ is the next-best; and so on. Our goal want to enumerate $A_1,\dots,A_n$.

You can find $A_1$ by solving the original assignment problem, e.g., with the Hungarian algorithm.

How can we find $A_2$, the second-best assignment? The idea is to use a case analysis. Let $v_1,\dots,v_m$ denote the $m$ agents in the problem instance, and let $A(v)$ denote the task assigned to agent $v$ by assignment $A$. We'll break down the space $\mathcal{S}$ of possible candidates for $A_2$ (i.e., the space of all assignments other than $A_1$) into the disjoint union $\mathcal{S} = \mathcal{S}_1 \cup \dots \cup \mathcal{S}_m$, where $\mathcal{S}_i$ is the space of assignments that agree with $A_1$ for $v_1,\dots,v_{i-1}$ but disagree with $A_1$ on $v_i$. (In other words, we look at the first agent that receives a different assignment in $A_1$ vs $A_2$. Then there are $m$ possibilities for that agent; we let $i$ denote its index, i.e., the index of the first agent whose assignment in $A_1$ is different from its assignment in $A_2$. This breaks down the space $\mathcal{S}$ into subspaces $\mathcal{S}_1,\dots, \mathcal{S}_m$, as listed before.)

Now the approach will be to find the best assignment in each $\mathcal{S}_i$, separately.

• $\mathcal{S}_1$: We find the best assignment $A$ such that $A(v_1) \ne A_1(v_1)$ using one invocation of the Hungarian algorithm, by changing the cost of the edge $(v_1,A_1(v_1))$ to $\infty$ (or some very large positive number) and then re-running the Hungarian algorithm. This finds the best assignment out of all assignments that assign $v_1$ to something different than $A_1$ did.

• $\mathcal{S}_2$: We find the best assignment $A$ such that $A(v_1) = A_1(v_1)$ and $A(v_2) \ne A_1(v_2)$ using one invocation of the Hungarian algorithm: change the cost of the edge $(v_1,A_1(v_1))$ to $0$, and change the cost of the edge $(v_2,A_1(v_2))$ to $\infty$.

• $\mathcal{S}_i$: Similarly, for each $i$, we can find the best assignment $A$ such that $A(v_j) = A_1(v_j)$ for all $j=1,2,\dots,i-1$ and such that $A(v_i) \ne A_1(v_i)$, using one invocation of the Hungarian algorithm.

This gives us $m$ assignments, i.e., $m$ candidates for $A_2$. By construction, each one of these assignments is different from $A_1$. Also, by construction, this covers all the space of all assignments that are different from $A_1$. Therefore, $A_2$ will be the best of these $m$ candidates, so we can just compare these $m$ candidates and call it $A_2$.

That find the second-best assignment. How can we find $A_3$, the third-best assignment? Well, the same ideas apply: we'll use a case split, but now the case-split will be a little more involved. Suppose that $v_i$ is the first agent where $A_1$ and $A_2$ disagree (i.e., $A_1$ and $A_2$ agree on $v_1,\dots,v_{i-1}$ but disagree on $v_i$, so that $A_2 \in \mathcal{S}_i$). Then we can break down the space of possibilities for $A_3$ by looking at the first agent that receives a different assignment from $A_2$, or from $A_1$.

In particular, let $\mathcal{T}$ denote the space of possible candidates for $A_3$ (i.e., the space of all assignments other than $A_1$ or $A_2$). We can partition it into the disjoint union

$$\mathcal{T} = \mathcal{S}_1 \cup \dots \cup \mathcal{S}_{i-1} \cup (\mathcal{T}_1 \cup \dots \cup \mathcal{T}_m) \cup \mathcal{S}_{i+1} \cup \dots \cup \mathcal{S}_m.$$

In other words, since $A_2 \in S_i$ and we now want to exclude $A_2$ from the space of allowable assignments, we partition $S_i$ into $S_i = \{A_2\} \cup \mathcal{T}_1 \cup \dots \cup \mathcal{T}_m$ and remove $A_2$. Here $\mathcal{T}_j$ denotes the set of assignments that agree with $A_2$ on $v_1,\dots,v_{j-1}$ but disagrees with $A_2$ on $v_j$ (and, if $j=i$, disagrees with $A_1$ on $v_j$ as well).

Now, we use the Hungarian algorithm to find the best assignment in each of $\mathcal{S}_1, \dots, \mathcal{S}_{i-1}, \mathcal{T}_1, \dots, \mathcal{T}_m, \mathcal{S}_{i+1}, \dots, \mathcal{S}_m$. This is doable using the techniques shown above, using one invocation of the Hungarian algorithm per subspace. Finally, we let $A_3$ be best of all the solutions found.

We can continue in this way, at each step identifying the next-best by decomposing the space of remaining assignments into multiple subspaces and invoking the Hungarian algorithm on each subspace. At each step, we introduce at most $m$ new subspaces, and we can reuse the previously-obtained results for the other subspaces. Therefore, on each step we make at most $m$ invocations of the Hungarian algorithm, so the total number of invocations of the Hungarian algorithm is $O(nm)$.

There's probably a better way to do it, but if you can't find any other algorithm, this is one you could use. Note that this is a general technique for problem of enumerating the $n$ best assignments to any instance of the assignment problem. It's not specific to your substitution-cipher example.


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# Two speakers on a horizontal axis both emit 440 Hz sound waves. The two speakers are pi radians out of phase. If there is to be a maximum constructive interference what is the minimum separation distance between the two speakers?

Mar 17, 2018

0.39 meters

#### Explanation:

Because the two speakers are off by $\pi$ radians, they are off by half a cycle. To have maximum constructive interference, they must line up exactly, meaning one of them must be shifted over half a wavelength.

The equation $v = \lambda \cdot f$ represents the relationship between frequency and wavelength. The speed of sound in air is approximately 343 m/s, so we can plug that into the equation to solve for $\lambda$, the wavelength.

$343 = 440 \lambda$
$0.78 = \lambda$

Finally, we must divide the value of the wavelength by two because we want to shift them over half a cycle.

$\frac{0.78}{2} = 0.39$ meters, which is your final answer.

format: plain


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# Continuity

• March 14th 2013, 03:51 PM
raggie29
Continuity
Please, please please help me with this questions. Really struggling as i have been off uni ill for over a week now and missed alot of classes to make it worse this assesment is due in asap!! :(

Many thanks, Hanna. x
• March 14th 2013, 04:15 PM
HallsofIvy
Re: Continuity
It will certainly be impossible to do this without knowing the definition of "continuous at x= a"! And that is that $\lim_{x\to a} f(x)= f(a)$. You will also need to know that the limit exists only if the limits, from above and below, both exist and are equal.

So for the first problem, to determine whether or not $f(x)= -2x^3+ 3x^2$ if $x\le 1$, $f(x)= sin((\pi/2)x)$ if x> 1, You need to show that
$\lim_{x\to 1} -2x^3+ 3x^2$ and $\lim_{x\to 1} sin((\pi/2)x)$ are the same and equal to f(1).

(For the last one you need to think about $(x- 1)^2 sin\left(\frac{1}{x- 1}\right)= (x- 1)\frac{sin\left(\frac{1}{x-1}\right)}{\frac{1}{x-1}}$.
What do you remember about $lim_{a\to 0}\frac{sin(a)}{a}$?


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# chance on throwing a six with 6 dice

The chance to throw a 6 with one die is 1/6

And 6 times 1/6 = 1

So, if I throw with 6 dice, the chance to throw at least 1 six should be 1.

But when I throw 6 dice, I sometimes don't throw any 6 at all..

How come?

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It's a peeve of mine: "die" is singular, "dice" is plural... –  Ｊ. Ｍ. Nov 23 '10 at 1:05
Why does nobody ever consider other "dice" than D6? –  draks ... Jul 3 '12 at 13:46

With $6$ dice there are $6^6$ possible outcomes. Of these, $5^6$ don't lead to six on any of the dice. So the number of outcomes that lead to at least one six are $6^6-5^6$, so the probability of at least one six is $\frac{6^6-5^6}{6^6} \approx 0.6651$

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Thank you.. now that's why I fail at Yatzee :p –  Enrico Pallazzo Nov 23 '10 at 0:49
Doesn't Yatzee have 5 dice? :p –  Rawling Dec 21 '10 at 14:52
Yes, 5 dice. And it has an 'h'. –  Fixee Jun 3 '11 at 2:58

The expected number of dice showing $6$ is $1$ when throwing six dice. The probability to see at least one $6$ is $1 - (5/6)^6$, as explained in Timothy's answer.

-

The probability of the union of independent events (unlike disjoint events) is not the sum of the individual probabilities. If $E_i$ denotes the event "$6$ is obtained on the $i$th throw", then $E_i$ are independent events, and it does not hold ${\rm P}(E_1 \cup \cdots \cup E_6 ) = {\rm P}(E_1 ) + \cdots + {\rm P}(E_6 )$. The left-hand side probability can be found as follows. The complement of the event $E_1 \cup \cdots \cup E_6$ is $E_1^c \cap \cdots \cap E_6^c$, where $E_i^c$ is the event "$6$ is not obtained on the $i$th throw". Hence, $${\rm P}(E_1 \cup \cdots \cup E_6 ) = 1 - {\rm P}(E_1^c \cap \cdots \cap E_6^c ).$$ Now, the probability of the intersection of independent events is the product of their individual probabilities. So, ${\rm P}(E_1^c \cap \cdots \cap E_6^c ) = {\rm P}(E_1^c) \cdots {\rm P}(E_6^c) = (5/6)^6$. Hence, ${\rm P}(E_1 \cup \cdots \cup E_6 ) = 1 - (5/6)^6$.

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+1,but you didn't explain "why the probabilities of union of independent events (unlike disjoint events) is not the sum of the individual probabilities ? " –  Quixotic Dec 7 '10 at 15:01
Try to figure out why, for any two events $A$ and $B$, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. So, if $A$ and $B$ are disjoint (meaning that $A \cap B = \emptyset$), we have $P(A \cup B) = P(A) + P(B)$, but if $A$ and $B$ are independent, then $P(A \cup B) = P(A) + P(B) - P(A)P(B)$. –  Shai Covo Dec 7 '10 at 16:49
If you flip a coin, there's a 50-50 chance you get heads. If you flip two coins there's a $3/4$ chance you get at least one head, since there's a $1/2 \times 1/2 = 1/4$ chance they're both tails. So similarly, there's a $5/6$ chance each die is not $6$. The chance they're all not $6$ is $(5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) = (5/6)^6$. So the chance that at least one is a six is $1 - (5/6)^6$.
You expect to get one $6$ on average. But sometimes you get more than one $6$.
Since the expected number of $6$s is one, this means that sometimes you have to get fewer than one $6$, i.e. zero $6$s.


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# Aerospace Curve ball problem, the spin axis will change?

1. Oct 27, 2008

### zyh

curve ball problem, the spin axis will change?image updated!

HI, I encountered a problem about the curve ball (table tennis ball) flying in the air. See the pictures below.

A ball with an initial velocity V along X axis, and has a spin along Z axis. As far As I know ( see my previous post in The air drag problem... ) .
F1 is the air drag.
F2 is the gravity.
F3 is the magnus force( sometimes we call it lift force).

So, the ball will not travel directly through X axis, it will curve to left.

My question is : What about the spin axis? Will it shift from A1 to A2? ( A1 is the axis parallel to Z axis as the initial state )? If Yes, How can I estimate the angle between A1 and A2?

Any suggestion is appreciated. Thank you very much.

#### Attached Files:

• ###### ball_force2.jpg
File size:
15.2 KB
Views:
289
Last edited: Oct 27, 2008
2. Oct 27, 2008

### zyh

I'm sorry that the attachment image is destroyed(I upload the jpg file as an attachment, but it doesn't appeared). I will upload it soon!

Last edited: Oct 28, 2008
3. Oct 30, 2008

### Phrak

The way you have it drawn, the ball will curve to the right, and also lift from gravitational effects.

Last edited: Oct 30, 2008
4. Oct 30, 2008

### zyh

Thanks Phark.
To my knowledge, I think the revolution axis(or spin axis) will remain its original direction( in the picture, it is A1), because, the drag, life and gravity force all have no effects on the spin Torque.

5. Oct 31, 2008

### Phrak

My mistake about the directions of deflection. I got everything inverted.

The question is, what happens when the spin axis is at some accute angle to the velocity, like when the ball starts dropping, right?

The lift and the induced drag, for instance, will act somewhere toward the 'leading edge'. But where is the profile drag--the viscous and eddy current drag?

6. Oct 31, 2008

### zyh

Hi, Phrak, I received your message. In my case: the revolution axis is transverse with Velocity, which is more simple than your question" spin axis is some accute angle to the velocity".

I think the lift force ( magnus force ) is defined by:
$$\overrightarrow{F_{lifg}}=k1\overrightarrow{\omega}\times\overrightarrow{V}$$
Through I'm not sure this formula is right.
The profile drag(Let me guess its meaning) is defined by:
$$\overrightarrow{F_{drag}}=-k2\parallel\overrightarrow{V}\parallel\overrightarrow{V}$$
where K1 and k2 are all drag coefficients.

7. Oct 31, 2008

### Phrak

Let me make sure I have this right. You want to know if the spin axis will precess, right?

For something with little angular mass, such as a ping pong ball, it seems, it precession could be substancial at the right angle of attack. What's required for it to precess is force couple whos axis isn't parallel with the spin axis.

I'd have to get out my crayons to see how the distributed forces would do this. It's probably more intuitive to draw the spin axis as horizontal, where the axis is oriented in the airflow like the span of a wing, then give it a little yaw.

Last edited: Oct 31, 2008
8. Oct 31, 2008

### zyh

Thanks, Yes, I'm try to find whether the spin axis will precess.
Do you mean "Angular mass" is the same as "Angular momentum" ? see in
http://en.wikipedia.org/wiki/Angular_momentum
I'm my picture, I think the torque given by "distributed forces" is symmetric on the spin axis. So, the precess may not happen.

9. Nov 1, 2008

### LURCH

First of all, let me say that your direction of curve in the original illustration is correct. The ball will curve in the same direction it is spinning. The curve in the opposite direction would be true on Mars (or maybe at extremely high altitudes), but not on Earth.

Secondly, I can see no force that would cause the axis to tilt. As I understand it, precession is a phenominon that takes place when the axis of rotation is already tilted. Maybe I'm not using the right frame of reference?

Last edited: Nov 2, 2008
10. Nov 2, 2008

### zyh

Yes, I have seem an article talking about that it will curve right in Mars.

Yes, In my opinion, I think the spin axis will never tilted from it's any original condition.

11. Nov 2, 2008

### Phrak

Lurch. At the initial position of the ball, the spin axis and velocity are perpendicular. But after a while gravity will give the ball a downward velocity component so that the velocity and spin axis are no longer perpendicular. So I wonder what the forces on the ball then.

12. Nov 2, 2008

### Phrak

It's almost a given that there will be precessional forces on the ball for the spin axis no at 0 or 90 degrees to the velocity. This would be false if there is some symmetry condition or the distributed forces somehow conspired against it.

The forces on the ball in this general condtion can be resolved into a force acting through the center of the ball (gravity included) and a force couple whos axis is not likely aligned with the spin axis. Any misalignment results in a pair of precessional forces.

After breaking a few crayons trying some picturorial solutions, I've concluded wind tunnel simulation software might be better.

In another senario, in the absense of gravity, and the spin axis not parallel nor perpendicular with the initial velocity, I think the ball will follow a corkscrew path; no precession required.

Last edited: Nov 2, 2008
13. Nov 2, 2008

### zyh

HI, Phark.
Firstly, I think the gravity will not give any precession effort on the initial Angular momentum.

Secondly, If the initial angular momentum is parallel or perpendicular to the initial velocity(consider the absence of gravity condition), after a while, the axis will not change it's direction.

I don't think it will get a "corkscrew path" ( the angular momentum is not parallel nor perpendicular to the velocity), though I can't test it in wind tunnel equipment.


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## Dipole-dipole interactions in NMR: explained

Interactions between spins are fundamental for understanding magnetic resonance. One of the most important ones is the magnetic dipole-dipole interaction. Spins act as tiny magnets, thus, they can interact with each other directly through space, pretty much the same way as classical magnetic dipoles (Figure 1). In NMR, dipole-dipole interactions very often determine lineshapes of solid-state samples and relaxation rates of nuclei in the liquid state.

In many NMR textbooks, you can find the following expression for dipole-dipole (DD) Hamiltonian between two spins 1 and 2:

$\hat{{H}}_{\rm DD} = d_{12} \left( 3 \hat{I}_{1z} \hat{I}_{2z} - \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right)$

where $d_{12} = - \displaystyle\frac{\mu_0}{4 \pi} \displaystyle\frac{\gamma_1 \gamma_2 \hbar}{r^3} \frac{\left( 3 \cos^2{\theta} - 1 \right)}{2}$, $\gamma_1$ and $\gamma_2$ are gyromagnetic ratios of the spins, $r$ is the distance between them.  Meaning of the angle $\theta$ can be seen from the Figure 2. But how was this expression derived?

In physics, I rarely struggled with imagining abstract things and concepts, but rather, I was often lazy to do math thoroughly and derive equations from the beginning to the end. That’s why I decided to write here a full derivation, from the beginning to the end, for the Hamiltonian of interacting nuclear spins. We will start from a classical expression of the magnetic field produced by the dipole and finish by the analysis of a truncated dipole-dipole Hamiltonian. Usually, you don’t see such a full derivation in textbooks. Maybe, this is because textbooks have limited space… Luckily, here we do not have such limitations, thus, we can have some fun here! Let’s go then! 🙂

Classical equation describing the magnetic field $\vec{\boldsymbol{B}}_{\rm dip} (\vec{\boldsymbol{r}})$ produced by a magnetic dipole moment $\vec{\boldsymbol{\mu}}$ is

$\vec{\boldsymbol{B}}_{\rm dip} (\vec{\boldsymbol{r}}) = \displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \left( \vec{\boldsymbol{\mu}} \cdot \hat{\boldsymbol{r}} \right) \hat{\boldsymbol{r}} - \vec{\boldsymbol{\mu}} \right)$

where $\hat{\boldsymbol{r}} = \displaystyle\frac{\vec{\boldsymbol{r}}}{|\vec{\boldsymbol{r}}|}$ is a unit vector. It is easy to show that $\left( \vec{\boldsymbol{\mu}} \cdot \vec{\boldsymbol{r}} \right) \vec{\boldsymbol{r}} = \left( \vec{\boldsymbol{r}} \cdot \vec{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}$.

Indeed,

$\left( \vec{\boldsymbol{\mu}} \cdot \vec{\boldsymbol{r}} \right) \vec{\boldsymbol{r}} = \left( \mu_x r_x + \mu_y r_y + \mu_z r_z \right) \cdot \begin{pmatrix} r_x \\ r_y \\ r_z \end{pmatrix} = \begin{pmatrix} \mu_{x} r_{x} r_{x} + \mu_{y} r_{y} r_{x} + \mu_{z} r_{z} r_{x} \\ \mu_{x} r_{x} r_{y} + \mu_{y} r_{y} r_{y} + \mu_{z} r_{z} r_{y} \\ \mu_{x} r_{x} r_{z} + \mu_{y} r_{y} r_{z} + \mu_{z} r_{z} r_{z} \end{pmatrix}$

which is the same as

$\left( \vec{\boldsymbol{r}} \cdot \vec{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}} = \begin{pmatrix} r_{x} r_{x} & r_{x} r_{y} & r_{x} r_{z} \\ r_{y} r_{x} & r_{y} r_{y} & r_{y} r_{z} \\ r_{z} r_{x} & r_{z} r_{y} & r_{z} r_{z} \end{pmatrix} \cdot \begin{pmatrix} \mu_x \\ \mu_y \\ \mu_z \end{pmatrix} = \begin{pmatrix} r_{x} r_{x} \mu_{x} + r_{x} r_{y} \mu_{y} + r_{x} r_{z} \mu_{z} \\ r_{y} r_{x} \mu_{x} + r_{y} r_{y} \mu_{y} + r_{y} r_{z} \mu_{z} \\ r_{z} r_{x} \mu_{x} + r_{z} r_{y} \mu_{y} + r_{z} r_{z} \mu_{z} \end{pmatrix}$

So, we write the magnetic field produced by a magnetic dipole $\vec{\boldsymbol{\mu}}_2$ as

$\vec{\boldsymbol{B}}_{\mu_2} = \displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}_2 - \vec{\boldsymbol{\mu}}_2 \right)$

The energy of a magnetic dipole $\vec{\boldsymbol{\mu}}_1$ interacting with the magnetic field $\vec{\boldsymbol{B}}_{\mu_2}$ produced by a magnetic dipole $\vec{\boldsymbol{\mu}}_2$ (dipole-dipole interaction) is therefore

$E_{\rm DD} = - \left( \vec{\boldsymbol{\mu}}_1 \cdot \vec{\boldsymbol{B}}_{\mu_2} \right) = -\displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \cdot \vec{\boldsymbol{\mu}}_1 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}_2 - \left( \vec{\boldsymbol{\mu}}_1 \cdot \vec{\boldsymbol{\mu}}_2 \right) \right)$

The transition from classical to quantum mechanics is realized by substituting the measurable quantities by corresponding quantum mechanical operators:

$E_{\rm DD} \rightarrow \hat{H}_{\rm DD} \quad \vec{\boldsymbol{\mu}}_1 \rightarrow \gamma_1 \hbar \hat{\mathbf{I}}_1\quad \vec{\boldsymbol{\mu}}_2 \rightarrow \gamma_2 \hbar \hat{\mathbf{I}}_2$

$\label{Eq_Hdd} \hat{H}_{\rm DD} = - \displaystyle\frac{\mu_0}{4 \pi } \displaystyle\frac{\gamma_1 \gamma_1 \hbar}{r^3} \left( 3 \cdot \hat{\mathbf{I}}_1 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \hat{\mathbf{I}}_2 - \left( \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right) \right) = b_{12} \hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2$

here $b_{12}$ is a factor which depends only on the types of the nuclear spins and the distance between them, and a tensor of dipole-dipole interactions contains information about the mutual orientation of two spins:

$\hat{\mathbf{D}} = 3 \cdot \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) - \hat{1}$

here $\hat{1}$ is a unit matrix. Note that we write Hamiltonian $\hat{H}_{\rm DD}$ in units of [rad/s], that is why one $\hbar$ is missing. In spherical coordinates:

$\hat{\boldsymbol{r}} = \begin{pmatrix} \sin{\theta} \cos{\phi} \\ \sin{\theta} \sin{\phi} \\ \cos{\theta} \end{pmatrix}$

Therefore,

$\hat{\mathbf{D}} = \begin{pmatrix} 3 \sin^2{\theta} \cos^2{\phi} - 1 & 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin{\theta} \cos{\theta} \cos{\phi} \\ 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin^2{\theta} \sin^2{\phi} - 1 & 3 \sin{\theta} \cos{\theta} \sin{\phi} \\ 3 \sin{\theta} \cos{\theta} \cos{\phi} & 3 \sin{\theta} \cos{\theta} \sin{\phi} & 3 \cos^2{\theta} - 1 \end{pmatrix}$

Looks good, doesn’t it? Now, let’s evaluate the product $\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2$:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \\ \begin{pmatrix} \hat{I}_{1x} & \hat{I}_{1y} & \hat{I}_{1z} \end{pmatrix} \begin{pmatrix} 3 \sin^2{\theta} \cos^2{\phi} - 1 & 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin{\theta} \cos{\theta} \cos{\phi} \\ 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin^2{\theta} \sin^2{\phi} - 1 & 3 \sin{\theta} \cos{\theta} \sin{\phi} \\ 3 \sin{\theta} \cos{\theta} \cos{\phi} & 3 \sin{\theta} \cos{\theta} \sin{\phi} & 3 \cos^2{\theta} - 1 \end{pmatrix} \begin{pmatrix} \hat{I}_{2x} \\ \hat{I}_{2y} \\ \hat{I}_{2z} \end{pmatrix} =$

$\begin{pmatrix} \hat{I}_{1x} & \hat{I}_{1y} & \hat{I}_{1z} \end{pmatrix} \begin{pmatrix} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) + \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) \\ \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) + \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) \\ \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) \end{pmatrix} =$

$= \hat{I}_{1x} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) + \hat{I}_{1x} \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{1x} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + + \hat{I}_{1y} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{1y} \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) + \hat{I}_{1y} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + + \hat{I}_{1z} \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{1z} \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

Let’s color terms to make it easier grouping them:

$\hat{I}_{1x} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) +$ $\hat{I}_{1x} \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) +$ $\hat{I}_{1x} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) +$ $\hat{I}_{1y} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) +$ $\hat{I}_{1y} \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) +$ $\hat{I}_{1y} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) +$ $\hat{I}_{1z} \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{1z} \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) +$ $\hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

Groupling the red terms gives

$\left( \hat{I}_{1x} \hat{I}_{2x} \cos^2{\phi} + \hat{I}_{1y} \hat{I}_{2y} \sin^2{\phi} \right) 3 \sin^2{\theta} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{1y} \hat{I}_{2y} \right)$

Let’s not forget about intrinsic connections of spin angular momentum with raising and lowering operators:

$\hat{I}_{1x} \hat{I}_{2x} = \displaystyle\frac{\left( \hat{I}_{1+} + \hat{I}_{1-} \right)}{2}\displaystyle\frac{\left( \hat{I}_{2+} + \hat{I}_{2-} \right)}{2} = \displaystyle\frac{1}{4} \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right)$

$\hat{I}_{1y} \hat{I}_{2y} = \displaystyle\frac{\left( \hat{I}_{1+} - \hat{I}_{1-} \right)}{2 i}\displaystyle\frac{\left( \hat{I}_{2+} - \hat{I}_{2-} \right)}{2 i} = -\displaystyle\frac{1}{4} \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1+} \hat{I}_{2-} - \hat{I}_{1-} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right)$

Therefore, grouping the red terms gives

$\left( \hat{I}_{1x} \hat{I}_{2x} \cos^2{\phi} + \hat{I}_{1y} \hat{I}_{2y} \sin^2{\phi} \right) 3 \sin^2{\theta} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{1y} \hat{I}_{2y} \right) = \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \frac{3}{4} \sin^2{\theta} + \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right) \frac{3}{4} \sin^2{\theta} \cdot \left( \cos{ 2 \phi} \right) - \frac{1}{2} \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) = \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \frac{1}{4} \left(1 - 3 \cos^2{\theta} \right) + \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right) \frac{3}{4} \sin^2{\theta} \cdot \left( \cos{ 2 \phi} \right)$

Groupling the blue terms gives

$\left( \hat{I}_{1x} \hat{I}_{2y} + \hat{I}_{1y} \hat{I}_{2x} \right) 3 \sin^2{\theta} \cos{\phi} \sin{\phi} = \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)\displaystyle\frac{3}{4} \sin^2{\theta} \cdot \left( - i \sin{2 \phi} \right)$

where we took into consideration that

$\hat{I}_{1x} \hat{I}_{2y} =\displaystyle\frac{\left( \hat{I}_{1+} + \hat{I}_{1-} \right)}{2} \frac{\left( \hat{I}_{2+} - \hat{I}_{2-} \right)}{2 i} = \frac{1}{4 i} \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)$
$\hat{I}_{1y} \hat{I}_{2x} = \displaystyle\frac{\left( \hat{I}_{1+} - \hat{I}_{1-} \right)}{2 i} \frac{\left( \hat{I}_{2+} + \hat{I}_{2-} \right)}{2} = \frac{1}{4 i} \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1+} \hat{I}_{2-} - \hat{I}_{1-} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)$

Red and blue terms can be combined nicely to form

$\left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \displaystyle\frac{1}{4} \left(1 - 3 \cos^2{\theta} \right) + \hat{I}_{1+} \hat{I}_{2+} \left( \frac{3}{4} \sin^2{\theta} \cdot e^{- 2 i \phi} \right) + \hat{I}_{1-} \hat{I}_{2-} \left( \frac{3}{4} \sin^2{\theta} \cdot e^{+ 2 i \phi} \right)$

Now let’s focus on purple terms:

$\left( \hat{I}_{1x} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2x} \right) \left( \hat{I}_{1x} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2x} \right) \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \left( \hat{I}_{1y} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2y} \right) \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) = \left( \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} + \hat{I}_{1z}\hat{I}_{2-} \right) \cos{\phi} \right) \displaystyle\frac{3}{4} \sin{2 \theta} + \left( \left( -i \hat{I}_{1+} \hat{I}_{2z} + i \hat{I}_{1-} \hat{I}_{2z} -i \hat{I}_{1z} \hat{I}_{2+} + i \hat{I}_{1z} \hat{I}_{2-} \right) \sin{\phi} \right) \frac{3}{4} \sin{2 \theta} = \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} \right) \left( \frac{3}{4} \sin{2 \theta} \right) e^{-i \phi} + \left( \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2-} \right) \left( \frac{3}{4} \sin{2 \theta} \right) e^{+i \phi}$

Overall, we have split our dipolar Hamiltonian into 6 term, so-called “Dipolar Alphabet”:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \hat{A} + \hat{B} + \hat{C} + \hat{D} + \hat{E} + \hat{F}$

where

$\hat{A} \quad = \quad \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

$\quad\quad \hat{B} \quad = \quad \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \cdot \displaystyle\frac{\left(1 - 3 \cos^2{\theta} \right)}{4}$

$\quad\quad \hat{C} \quad = \quad \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} \right) \left( \displaystyle\frac{3}{4} \sin{2 \theta} \right) e^{-i \phi}$

$\quad\quad \hat{D} \quad = \quad \left( \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2-} \right) \left( \displaystyle\frac{3}{4} \sin{2 \theta} \right) e^{+i \phi}$

$\hat{E} \quad = \quad \hat{I}_{1+} \hat{I}_{2+} \left( \displaystyle\frac{3}{4} \sin^2{\theta} \right) e^{- 2 i \phi}$

$\quad\quad \hat{F} \quad = \quad \hat{I}_{1-} \hat{I}_{2-} \left( \displaystyle\frac{3}{4} \sin^2{\theta} \right) e^{+ 2 i \phi}$

To summarize, the Hamiltonian of two interacting spins is a $4 \times 4$ matrix composed of 6 operators. Each of the letters of the dipolar alphabet corresponds to certain matrix elements in the final Hamiltonian (Figure 3).

Without an externally imposed direction in space (for example, in the case of two equivalent spins in zero magnetic field), all of the terms of the dipole-dipole Hamiltonian need to be used for calculating an NMR spectrum. This is because all orientations in space are equivalent. However, in the presence of the external high magnetic field, the Hamiltonian can be simplified via the use of so-called “secular approximation”.

The secular approximation concerns the case where the Hamiltonian is the sum of two terms:

$\hat{H} = \hat{A} + \hat{B}$

where $\displaystyle\hat{A}$ is a “large” operator and $\hat{B}$ is a “small” operator. In our case, $\hat{A}$ can be an operator describing the interaction with the magnetic field (Zeeman Hamiltonian) and $\hat{B}$ is DD Hamiltonian. Eigenstates of the Zeeman Hamiltonian are familiar αααβ, βα, ββGenerally, $\hat{B}$ does not commute with $\hat{A}$, therefore, if written in the eigenbasis of $\hat{A}$, it has finite elements everywhere.

The secular approximation for $\hat{B}$ means that we leave only the blocks that correspond to the eigenvalue structure of the operator $\hat{A}$ (Figure 4) and disregard all other elements.

In general, we can omit a matrix element $b_{nm}$ that is much smaller than

$|b_{mn}| \ll |E_m - E_n|$

For homonuclear case (e.g., two interacting protons), this means that only the first two terms of the dipolar Alphabet will survive:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \hat{A} + \hat{B} =$

$= \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) + \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \cdot \displaystyle\frac{\left(1 - 3 \cos^2{\theta} \right)}{4} =$

$= \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{2y} \hat{I}_{2y} \right) \cdot \displaystyle\frac{\left(3 \cos^2{\theta} - 1\right)}{2} =$

$= \displaystyle\frac{\left( 3 \cos^2{\theta} - 1\right)}{2} \cdot \left( 2 \hat{I}_{1z} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2z} - \hat{I}_{1z} \hat{I}_{2z} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{2y} \hat{I}_{2y} \right) \right) =$

$= \displaystyle\frac{\left(3 \cos^2{\theta} - 1\right)}{2} \cdot \left( 3 \hat{I}_{1z} \hat{I}_{2z} - \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right)$

Overall, this is how you go from the classical description of the magnetic field of the dipole to the truncated form of the Hamiltonian in the high nagnetic field. In the next post I will show how this Hamiltonian leads to the characteristic lineshape of the NMR line for solids.

## My “Dream Research” Project

If I was asked to identify the most challenging biological question, I would answer immediately. What is the nature of memory and thought? This question always fascinated me as a child. For a long time, I thought only biologists can figure that out. It took me 10 years deeply studying physics and chemistry, becoming a specialist in nuclear magnetic resonance (NMR) and magnetic resonance imaging (MRI), to realize that actually now we are close to start answering the question which captivated my childish mind.

With all its complexity, the end result of the genetic machinery is to affect the chemistry of the body. Small molecules, metabolites, serve as fingerprints of what is happening inside us. Studying metabolites is almost like looking at someone’s apartment and coming up with a story of their recent lives: we can make guesses about a lifestyle based on what we see! And the chemistry of thinking is not an exception – our thought processes are accompanied by myriads of chemical transformations, and leftover metabolites can tell us about the process behind them.

Studying metabolites is almost like looking at someone’s apartment and coming up with a story of their recent lives: we can make guesses about a lifestyle based on what we see!

Routinely, metabolites are measured through analytical techniques like NMR and mass spectrometry (MS). But a new astonishing era is emerging. With new sensitivity enhancement techniques (signals can be increased by more than 20,000 times [1-4]), MR imaging will become a new tool to study metabolism in vivo and will move beyond morphology onto a platform to visualize molecules. Being fundamentally a quantum mechanical technique, the full potential of MRI is yet to be discovered.

I see my “dream research” project as the development of a new experimental MRI-NMR/MS platform to study metabolomics of memory and thought in living creatures. By developing novel MRI pulse sequences (which will take into account quantum-mechanical nature of molecules) and by applying state-of-the-art signal enhancement techniques, we will be able to “light up” the regions of the brain to study chemistry in them with an unprecedented level of accuracy. I believe that once all new methodologies available today are combined, it will become possible to create functional MRI for metabolomics – a tool to study instant chemical changes in the brain associated with memory and thinking. This will not only revisit the known biochemical processes at a new quantitative level, it will allow unraveling unexpected secrets of metabolism. And it is not only a fun thing to do — understanding the biochemical reasons for making decisions will bring us much closer to a society in which everyone truly enjoys living.

[1] J. H. Ardenkjær-Larsen et al. Increase in signal-to-noise ratio of >10,000 times in liquid-state NMR. Proc. Nat. Acad. Sci., 2003, 100 (18), 10158-10163.

[2] R. W. Adams et al. Reversible Interactions with parahydrogen Enhance NMR Sensitivity by Polarization Transfer. Science, 2009, 323 (5922), 1708-1711.

[3] D. A. Barskiy et al. Over 20% 15N Hyperpolarization in Under One Minute for Metronidazole, an Antibiotic and Hypoxia Probe. J. Am. Chem. Soc., 2016, 138 (26), 8080–8083.

[4] D. A. Barskiy et al. NMR Hyperpolarization Techniques of Gases. Chem. Eur. J., 2017, 23 (4), 725-751.

## Monday Morning Effect

Friday evening. While his friends had already met in the Pub on Shattuck Avenue to celebrate a happy hour, UC Berkeley’s Ph.D. student Henry Bryndza was still in the Lab. He wanted to finish preparation of his samples so that he could come over on Monday morning to focus on the NMR measurements, not worrying about sample preparations. In order to suppress chemical reactions which could have started in his samples over the weekend, Henry put them in the liquid nitrogen dewar (T=-196oC).

Henry was working in the Laboratory of Robert Bergman, a renowned UC Berkeley professor who has made a significant contribution to the organic and metallorganic chemistry. Bergman and Bryndza were studying Fischer–Tropsch reactions using exemplary Cobalt and Iridium catalysts [1].

When he came back on Monday, Henry started to observe very interesting phenomena. 1H NMR spectra of the samples he took in the morning showed very weird “negative” NMR peaks (Figure 1). Moreover, the intensity of these peaks decreased day after day during the week when Henry tried to repeat the experiments and completely disappeared by the end of the week [2]. Henry was confused and decided to repeat his measurements. Surprisingly, this phenomenon was not observed every single time but was definitely the strongest on Mondays. Bergman and Bryndza decided to jestingly call this a “Monday phenomenon”; this was the beginning of what was known later as Parahydrogen-Induced Polarization (PHIP).

Bryndza and Bergman asked for help from many NMR specialists, including NMR expert Professor Alex Pines from UC Berkeley and Professor Joachim Bargon from the University of Bonn [2]. The last one was known for the discovery of so-called chemically-induced dynamic nuclear polarization (CIDNP). The CIDNP effect is usually manifested as positive and negative NMR signals (very similar to those observed in Henry’s experiments) for the reactions taking place via radical intermediates. After contacting Bargon and other CIDNP specialists, weird results were interpreted as “pseudo-CIDNP” in hydrogenation reactions [3]. However, it was clear that CIDNP-based explanation was at least not complete, first, because of the very unusual suggestion of radical pairs in the studied hydrogenation reactions and, second, because of the lack of convincing simulations supporting the observed phenomena. Moreover, it by no means explained why the effect was the strongest on Mondays and why it was only observed in the laboratory of Robert Bergman.

This “Monday morning” puzzle remained unresolved until the International Society of Magnetic Resonance meeting in Rio de Janeiro in June 1986. There, during an evening session, Professor Daniel Weitekamp from Caltech presented his “thought experiment” of using parahydrogen (para-H2) as a source of enhancing NMR signals. The concept and the expected results were immediately published in Physical Review Letters [4]. The experimental demonstration conducted by a Weitekamp’s Ph.D. student Russ Bowers followed in July, and brilliantly supported all theoretical predictions (Figure 2) [5].

Bowers and Weitekamp called their experiment PASADENA (Parahydrogen And Synthesis Allow Dramatically Enhanced Nuclear Alignment) to glorify the location of their institute (Caltech is located in Pasadena, CA). After their publication, it immediately became obvious that PASADENA is, in fact, a correct explanation of “Monday phenomenon” of Bryndza and Bergman. Indeed, the low-temperature storage of NMR tubes over the weekend partially converted normal hydrogen into para-H2. The conversion was not complete, but it was enough to observe antiphase lines in 1H NMR spectra (Figure 1). The PASADENA effect and discovered later effect ALTADENA (Adiabatic Longitudinal Transport After Dissociation Engenders Net Alignment) were collectively given the name PHIP (Parahydrogen-Induced Polarization) [6].

Now let’s talk about physical principles of this effect. As we discussed before, due to the absence of a net nuclear magnetic moment, para-H2 itself does not produce an NMR signal. However, this single nuclear spin state implies that, in a sense, it is cold. Indeed, a comparable degree of spin ordering is obtainable at equilibrium only at temperatures of a few mK and magnetic fields of several Tesla [7]. The brilliance of Wetekamp’s idea was to introduce magnetic inequivalence to release this potential signal by connecting the singlet to the triplet states. This would require chemistry, but simple bond cleavage would not suffice. A singlet state of two protons is a relationship of one spin relative to the other and this order would be dissipated if the pair were split and mixed with an ensemble of other such products. Rather, it is necessary that the pair have a special relationship even after being distinguished by magnetic inequivalence. This is called a “pairwise” hydrogen addition and can be realized in hydrogenation reactions in the presence of homogeneous catalysts. To see how it works, let’s take as an example the simplest situation and imagine that a chemical reaction leads to the association of para-H2 with a molecule not containing magnetic nuclei.

The two-spin system of the hydrogen molecule gives rise to four nuclear spin energy levels. As we described before, three of these energy levels correspond to orthohydrogen, the state with total nuclear spin 1 (triplet state), whereas the remaining fourth energy level corresponds to parahydrogen (singlet state), the state with zero total nuclear spin (Figure 3). Transitions between singlet and triplet spin states are forbidden by symmetry and the spin 0 parahydrogen is NMR-silent.

Now, the incorporation of para-H2 into an asymmetric molecule breaks the symmetry of the singlet spin state. For simplicity, I will consider only the PASADENA experiment, the case where hydrogenation reaction is performed at a high magnetic field (wherein the chemical shift difference between the two para-H2-nascent protons is much greater than the spin-spin coupling J between them). In this situation, the population of the singlet spin state αββα (numerical factor is omitted) of para-H2 is immediately transferred to the population of spin states αβ and βα of the formed spin system.

This can be understood as follows. Because of the chemical reaction, two H atoms from para-Hsuddenly end up in a different molecular environment. This leads to a collapse of the nuclear spin wavefunction αββα into one of the two states, αβ or βα, each with 50% probability. Next, it is easy to deduce from the energy level diagram that the NMR spectrum of the produced in such a manner molecule will contain four peaks grouped in two antiphase multiplets (Figure 3), exactly what was observed in the experiments of Bryndza (Figure 1) and Bowers (Figure 2). The key requirement is that both hydrogen protons from the para-H2 molecule are added together without significant competition from exchange reactions. This is a property of many, but not all, hydrogenations.

The assignment of the peaks to particular transitions depends on the sign of the J-coupling between the para-H2-nascent hydrogens. When J-coupling is positive, PASADENA multiplets are positive-negative; if J-coupling is negative, the spectral appearance is opposite. This feature is very useful for studying hydrogenation reaction intermediates. Normally, organic molecules possess positive J-couplings between protons; and J-couplings between them are negative in case of metal hydrides. Therefore, in a complex reaction involving many intermediates, it becomes possible to distinguish low-concentration hydrides (possessing negative-positive multiplets) from organic reaction products (Figure 4).

It is also important to realize that PHIP can lead to 100% nuclear spin polarization of the reaction product. In the case of PASADENA experiment, 100% population of para-His split into just two energy levels, making transitions from these levels enhanced by orders of magnitude compared to the thermal case. Theoretically, if all para-Hmolecules are transferred to products in a pairwise manner and relaxation loses are minimized, the reaction product can acquire 100% spin polarization. This would, of course, require an additional step to transfer spin order from αβ and βα into the state αα but this can be readily realized using a simple RF pulse sequence.

Enormous NMR signal enhancements and unique spectroscopic signatures made PHIP a very useful tool in chemistry for more than 25 years to elucidate hydrogenation reaction mechanisms, study metalorganic hydride complexes, and catalysis [6]. However, PHIP can be also used in a very different context. Imagine a suitable molecular precursor which can become a naturally occurring metabolite after hydrogenation. This metabolite can be produced in seconds, with a very high level of nuclear polarization, injected into a living organism and a metabolism of that organism can be monitored by magnetic resonance spectroscopy (MRS) and magnetic resonance imaging (MRI). Today PHIP, and its sister technology SABRE (Signal Amplification By Reversible Exchange) allow to efficiently hyperpolarize dozens of biologically relevant molecules on nuclei such as 1H, 13C, 15N, 19F, 29Si, 31P, 119Sn etc. But this is a story for a separate blog post! 🙂

It is important to emphasize that only the connection between nuclear spin and rotational degrees of freedom allows this unique situation to take place. Indeed, the fact that the nuclear spin state can be overpopulated simply by cooling is a remarkable quality inherent only to the small hydrogen molecule. Indeed, even though other molecules can have the similar connection between rotational and nuclear spin states (N2, F2 etc.), larger moments of inertia will make overpopulating these states much more challenging task (because of the lower temperature requirements). Moreover, it is very challenging to keep these molecules in the gas state at low temperatures, and the simple rule of making a total wavefunction be a product of individual wavefunctions will no longer hold true. So, it is more likely that hydrogen molecule is the only example when the rules of spin statistics and Pauli’s principle can lead to the nuclear spin hyperpolarization.

What excites me about this story is how a purely thought experiment, on the one hand, and a weird experimental phenomenon, on the other hand, emerged into a new discipline and a remarkable tool to study chemical reactions. Moreover, more exciting applications of the para-H2-based hyperpolarization techniques are expected to emerge in biomedicine. I really wish there were more Monday morning effects in science! Who knows but maybe someone today has come to a lab to look at a weird result which will form a new field of study tomorrow.

References

[1] J. Bargon. Chance Discoveries of Hyperpolarization Phenomena. eMagRes, 2007.

[2] Private conversations with Robert Bergman and Alex Pines.

[3] P. F. Seidler, H. E. Bryndza, J. E. Frommer, L. S. Stuhi, R. G. Bergman. Synthesis of Trinuclear Alkylidyne Complexes from Dinuclear Alkyne Complexes and Metal Hydrides. CIDNP Evidence for Vinyl Radical Intermediates In the Hydrogenolysis of These Clusters. Organometallics, 1983, 2 (11), 1701-1705.

[4] C. R. Bowers, D. P. Weitekamp. Transformation of Symmetrization Order to Nuclear-Spin Magnetization by Chemical Reaction and Nuclear Magnetic Resonance. Phys. Rev. Lett., 1986, 57 (21), 2645-2648.

[5] C. R. Bowers, D. P. Weitekamp. Parahydrogen and Synthesis Allow Dramatically Enhanced Nuclear Alignment. J. Am. Chem. Soc., 1987, 109 (18), 5541-5542.

[6] J. Natterer, J. Bargon. Parahydrogen-Induced Polarization. Prog. Nucl. Magn. Reson. Spect. 1997, 31, 293-315.

[7] D. Weitekamp. Sensitivity Enhancement Through Spin Statistics. Encyclopedia of Magnetic Resonance, 2007.

[8] S. Colebrooke, S. Duckett, J. Lohman, R. Eisenberg. Hydrogenation studies involving halobis(phosphine)-rhodium(I) dimers: Use of parahydrogen-induced polarisation to detect species present at low concentration. Chem. Eur. J., 2004, 10, 2459–2474.

## Parahydrogen

To begin my blog, let’s introduce parahydrogen. Lately, this little molecule has been attracting a lot of attention in the magnetic resonance community due to tremendous opportunities it brings for NMR/MRI signal enhancement. I will explain a bit later how this parahydrogen-based NMR signal enhancement works. But first, let’s talk about physical origins of parahydrogen!

Parahydrogen (para-H2) is a nuclear spin isomer of a hydrogen molecule. Nuclear spin isomerism is a very special form of isomerism. Unlike “traditional” molecular isomers (molecules having the same atomic composition but different chemical structure) and isotopologues (isomers that differ only in their isotopic composition), nuclear spin isomers are chemically identical: they have exactly the same atomic (and even isotopic) structure. However, nuclear spin isomers differ in the nuclear spin state of their atoms. It turns out that this tiny change (energy difference associated with nuclear spin transitions is only ~0.1 J/mol) may lead to different thermodynamic and spectroscopic properties of molecules. So, how does this work?

Unfortunately (or fortunately), we will have to use rules of quantum mechanics and some math. In quantum mechanics, in order to describe properties of quantum systems (atoms, molecules, etc.), physicists use wavefunctions. By knowing a wavefunction one will be able to calculate probabilities to find a quantum system in different states (namely, a squared modulus of the wavefunction determines the probability to find a system in a given state). Let’s look how it works taking as an example hydrogen molecule.

Hydrogen molecule consists of two hydrogen atoms (H) and is denoted as Н2. Each atom has a nucleus – a proton which is a spin-1/2 particle. Physicists say that hydrogen molecule has several degrees of freedom: translational, rotational, vibrational, etc., and these degrees of freedom can be considered independent. In other words, rotation of the hydrogen molecule does not depend on how and where the molecule is moving and how it is vibrating. Each degree of freedom has a wavefunction associated with it. I will use different colors to describe electron and nuclear wavefunctions. A position of the molecule in space, as well as its rotation and vibrations, are determined by the position and movements of nuclei, therefore, these degrees of freedom are described by translational (ψtr), rotational (ψrot), vibrational (ψvib), and nuclear spin (ψspin) wavefunctions. Atomic nuclei are surrounded by electrons which provides the bonding between the nuclei. The wavefunction describing movements of electrons is called orbital wavefunction ψorb, and state of the electron spins is described by the electron spin wavefunction ψspin.

Since probabilities of independent events are multiplied, the total wavefunction is a product of the above-mentioned wavefunctions:

ψtot = ψtr·ψrot·ψvib·ψspin·ψorb·ψspin

However, rules of quantum mechanics are trickier than they may sound. According to Pauli’s principle, the total wavefunction of the system of spins-1/2 particles has to be antisymmetric with respect to the exchange (also called permutation) of two identical particles. What does this mean?

Let’s take for example ψspin. A system consisting of two spins-1/2 can be described as α1α2, β1β2 or combinations α1β2+β1α2, α1β2β1α2. Here α and β denote the projection of nuclear spin angular momentum along the quantization axis (more on this stuff later, for now, one can imagine the state α as a magnetic moment – spin – pointing up along the external magnetic field and the state β as a magnetic moment pointing down, opposite to the field). Indexes 1 and 2 say to which nucleus the spin belongs. For example, the state α1α2 means that both nuclear spins point along the field while the state β1β2 means that both spins point opposite to the field. The combination states α1β2+β1α2 and α1β2β1α2 are more interesting. Neither of spins points along or opposite to the field but if we take one spin and determine its orientation, the second spin will take the opposite orientation. We can see now that two spins are correlated: the state of the second spin depends on the state of the first one.

Now let’s look what happens if we exchange (permute) particles. Mathematically, permutation simply means interchange of indexes (1→2, 2→1). One can see that upon permutation of indexes the first three states do not change: α2α1= α1α2β2β1 = β1β2, (α2β1+β2α1) = (α1β2+β1α2), but the last state changes the sign: (α2β1β2α1) = –(α1β2β1α2). Therefore, the first three states are called symmetric wavefunctions and the last one – antisymmetric with respect to permutation of particles.

So, our hydrogen molecule contains four spin-1/2 particles: two electrons and two nuclei). Permutation of electrons can only affect ψorb and ψspin. The first wave function, corresponding to the electronic ground state, is symmetric with respect to the electrons, the second, the electron spin wavefunction, is antisymmetric, and the rest are independent of the electrons’ variables and, thus, symmetric. Therefore, Pauli’s principle is fulfilled for electrons: the total wavefunction is antisymmetric with respect to permutation of electrons, thanks to antisymmetric ψspin. Permutation of nuclei can affect two wavefunctions: ψspin (as we just saw above) and ψrot. A mathematical expression for ψrot is rather complicated but it is not necessary to know its full form to understand the symmetry properties.

Figure 1. Schematic energy diagram of rotational levels of the hydrogen molecule.

This is because rotating diatomic molecules possess a set of stable rotational states, which can be described by only one parameter – the rotational quantum number J. This number can take integer values 0, 1, 2, 3, … This means that molecule can be in a stable state with J = 0, J = 1, J = 2, etc. (Figure 1). It turns out that the symmetry (with respect to permutation of nuclei) of the rotational wavefunction can be described as

P12·ψrot = (-1)J·ψrot

where P12 represents the permutation operator that interchanges the nuclei’s positions (indexes). This means that the rotational wavefunction is symmetric for even rotational states (J = 0, 2, 4, …) and antisymmetric for odd rotational states (J = 1, 3, 5, ).

Coming back to Pauli’s principle, permutation of nuclei should lead to the change of sign of the total wavefunction. Since only ψspin and ψrot can change sign upon such permutation, these two wavefunctions become connected: even (symmetric) rotational wavefunctions must be combined with the antisymmetric nuclear wavefunction (α1β2β1α2), whereas each antisymmetric rotational wavefunction has to be associated with one of the three symmetric spin functions. All this is required to yield a total wavefunction being antisymmetric with respect to the exchange of the nuclei. This is where two hydrogen spin isomers come from. One is called parahydrogen (para-H2), having an antisymmetric nuclear spin wavefunction 1β2β1α2) and existing only in even rotational states, and the other called orthohydrogen (ortho-H2), having a symmetric nuclear spin wavefunction and existing only in the odd rotational states.

It follows from the Pauli’s principle that nuclear spin state and rotational state of the hydrogen molecule are strictly correlated. This is remarkable, because the notion of independence (which allowed us to write a wavefunction as a product of individual wavefunctions) has led to complete dependence of these degrees of freedom from each other!

Remarkably, parahydrogen and orthohydrogen can be seen as two individual gases because their thermodynamic properties (boiling point, heat capacity, etc.) are slightly different. This is not surprising taking into account the fact that molecules constituting these two gases are rotating differently!

Importantly, conversion between the two states occurs extremely slowly because the transition between symmetric and antisymmetric nuclear spin states are forbidden by the selection rules of quantum mechanics. Therefore, after its production parahydrogen may be stored for long periods before use in a tank as an individual gas, as the relaxation rate of the parahydrogen back to room-temperature equilibrium can be on the order of months.

However, the use of paramagnetic catalysts (i.e., activated charcoal, nickel, hydrated iron(III) oxide) promotes the establishment of Boltzmann thermodynamic equilibrium between ortho-H2/para-H2 states for a given temperature at greatly accelerated rates. This happens because paramagnetic materials can create a strong inhomogeneous magnetic field on the atomic scale. In such fields the two hydrogen atoms are no longer equivalent, thus, spin-flip transitions between ortho-H2 and para-H2 are no longer forbidden. In practice, normal hydrogen gas (i.e., equilibrium ratio of spin isomers at room temperature) consisting of 75% ortho– and 25% para-hydrogen is passed through a chamber filled with paramagnetic catalyst and maintained at cryogenic temperatures, where the equilibration to the isomer ratio governed by the Boltzmann distribution occurs. For example, a parahydrogen generator operating at 77 K (obtained conveniently by a liquid-N2 bath) yields a mixture with ~50% parahydrogen, whereas the designs based on cryo-chillers (e.g. T~20 K) yield >99% parahydrogen (Figure 2). I should note that the enrichment of hydrogen with para-isomer happens so easily only because of the big energy gap between rotational spin states. This, in turn, is due to the small mass of molecular hydrogen (in general, the energy difference between rotational spin states is inversely proportional to the moment of inertia of a rotating molecule).

The existence of nuclear spin isomers of molecular hydrogen (which was experimentally confirmed by the early 1930s) was one of the first triumphs of quantum mechanics. Indeed, the citation of the Nobel Prize awarded to Werner Heisenberg in 1932 stated that he had “created quantum mechanics, the application of which led to the discovery of the two allotropic forms of hydrogen”!

Knowledge about ortho– to para-H2 conversion is important for the storage of liquid hydrogen (especially as a rocket fuel). The difference in energy associated with the different rotational levels means that energy is released when ortho-H2 converts to para-H2, and energy is absorbed in the reverse process. This phenomenon can be thought of as a latent heat of conversion. If one quickly liquefies normal hydrogen, it will still have 3:1 ortho:para composition which will eventually lead to the heat release. This can vaporize a significant portion of hydrogen and break the impermeability of the storage container. At the dawn of industrial liquid hydrogen production, this presented a major problem. Modern hydrogen liquefying processes now ensure that the liquid hydrogen has reached equilibrium concentration at 99.8% para-H2 before being transported and stored for use.

One may ask how can para-H2 be important for NMR? Indeed, this spin isomer has a zero total nuclear spin and, thus, it does not possess 1H NMR spectrum. However, para-H2 is a pure quantum mechanical state and a highly organized spin order which is readily achievable simply by cooling. Pure state means that all para-H2 molecules are described by the same wavefunction – 1β2β1α2). For comparison, ortho-H2 is a mixture of three wavefunctions, α1α2α1β2+β1α2 and β1β2 and, thus, it is not a pure state. It turns out that once you have a quantum mechanically pure state, you can manipulate it and transfer the spin order from one form to another. This is how parahydrogen-induced polarization (PHIP) and signal amplification by reversible exchange (SABRE) work: they transfer NMR-silent singlet spin order of para-H2 into observable nuclear magnetization.

## Let’s hyperpolarize! :)

Hi all! I am very excited to launch my personal website/blog. I plan to use it as a platform to share educational content connected to my research interests and post my opinions/thoughts regarding important events (mainly focused on science). For more information about me visit About page. Ok, let’s hyperpolarize!


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# anova.mppm

0th

Percentile

##### ANOVA for Fitted Multiple Point Process Models

Performs analysis of deviance for two or more fitted multiple point process models.

Keywords
models, methods, spatial
##### Usage
## S3 method for class 'mppm':
anova(object, \dots, test=NULL, override=FALSE)
##### Arguments
object
A fitted multiple point process model (object of class "mppm").
...
One or more fitted multiple point process models.
test
Type of hypothesis test to perform. A character string, partially matching one of "Chisq", "F" or "Cp", or NULL.
override
Logical flag indicating whether to proceed even when there is no statistical theory to support the calculation.
##### Details

This is a method for anova for comparing several fitted multiple point process models (objects of class "mppm", usually generated by the model-fitting function mppm).

If the fitted models are all Poisson point processes, then this function performs an Analysis of Deviance of the fitted models. The output shows the deviance differences (i.e. 2 times log likelihood ratio), the difference in degrees of freedom, and (if test="Chi") the two-sided p-values for the chi-squared tests. Their interpretation is very similar to that in anova.glm.

If some of the fitted models are not Poisson point processes, then there is no statistical theory available to support a similar analysis. The function issues a warning, and (by default) returns a NULL value.

However if override=TRUE, then a kind of analysis of deviance table will be printed. The deviance' differences in this table are equal to 2 times the differences in the maximised values of the log pseudolikelihood (see mppm). At the time of writing, there is no statistical theory to support inferential interpretation of log pseudolikelihood ratios. The override option is provided for research purposes only!

The argument test determines which hypothesis test, if any, will be performed to compare the models. The argument test should be a character string, partially matching one of "Chisq", "F" or "Cp", or NULL. The first option "Chisq" gives the likelihood ratio test based on the asymptotic chi-squared distribution of the deviance difference. The meaning of the other options is explained in anova.glm. For random effects models, only "Chisq" is available, and again gives the likelihood ratio test.

##### Value

• An object of class "anova", or NULL.

mppm

• anova.mppm
##### Examples
H <- hyperframe(X=waterstriders)
mod0 <- mppm(X~1, H, Poisson())
modx <- mppm(X~x, H, Poisson())
anova.mppm(mod0, modx, test="Chi")`
Documentation reproduced from package spatstat, version 1.41-1, License: GPL (>= 2)

### Community examples

Looks like there are no examples yet.


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## General Chemistry (4th Edition)

$0.599\%$
The average density of mercury at $35^{\ \circ}C$ based on the measurement results is: $$\frac{13.56\text{ g}\cdot\text{cm}^{-3}+13.62\text{ g}\cdot\text{cm}^{-3}+13.59\text{ g}\cdot\text{cm}^{-3}}{3}=13.59\text{ g}\cdot\text{cm}^{-3}$$ The true density of mercury at $35^{\ \circ}C$ is $13.509\text{ g}\cdot\text{cm}^{-3}$; thus: $$\text{percentage error}=\frac{13.59\text{ g}\cdot\text{cm}^{-3}-13.509\text{ g}\cdot\text{cm}^{-3}}{13.509\text{ g}\cdot\text{cm}^{-3}}\cdot 100=0.599\%$$ (Note, percent error is always a positive quantity, so if the equation gives a negative value, the value should be changed to a positive value.)


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GO TO... Contents Copyright BUY THE BOOK

Problem 14:  Sketching subsets of the plane ($✓$ $✓$) 1999 Paper I

Sketch the following subsets of the $x$-$y$ plane:

(i)
$|x|+|y|\le 1$ ;
(ii)
$|x-1|+|y-1|\le 1$ ;
(iii)
$|x-1|-|y+1|\le 1$ ;
(iv)
$|x|\phantom{\rule{0.3em}{0ex}}|y-2|\le 1$ .

Comments

Often with modulus signs, it is easiest to consider the cases separately, so for example in part (i), you would ﬁrst work out the case $x>0$ and $y>0$, then $x>0$ and $y<0$, and so on. Here there is a simple geometric understanding of the different cases: once you have worked out the ﬁrst case, the other three can be deduced by symmetry.

Another geometric idea should be in your mind when tackling this question, namely the idea of translations in the plane.

Post-mortem

There was no room for a post-mortem over the page, because the diagrams take up so much space.

Don’t read this until you have tried the question!

There are two key learning points, if you will excuse this horrible expression.

The ﬁrst is that if you want to sketch a region, it is often best to draw the curves that deﬁne the boundary of the region, then just work out whether you want the interior or exterior of the boundary by choosing one interior point and seeing whether it satisﬁes the inequalities.

The second is that quite complicated inequalities can sometimes be much simpliﬁed by translating or rotating the axes.

Solution to problem 14

The way to deal with the modulus signs in this question is to consider ﬁrst the case when the things inside the modulus signs are positive, and then get the full picture by symmetry, shifting the origin as appropriate.

For part (i), consider the ﬁrst quadrant $x\ge 0$ and $y\ge 0$. In this quadrant, the inequality is $x+y\le 1\phantom{\rule{0.3em}{0ex}}$. Draw the line $x+y=1$ and then decide which side of the line is described by the inequality. It is obviously (since $x$ has to be smaller than something) the region to the left of the line; or (since $y$ also has to be smaller than something) the region below the line, which is the same region.

Similar arguments could be used in the other quadrants, but it is easier to note that the inequality $|x|+|y|\le 1$ is unchanged when $x$ is replaced by $-x\phantom{\rule{0.3em}{0ex}}$, or $y$ is replaced by $-y\phantom{\rule{0.3em}{0ex}}$, so the sketch should have reﬂection symmetry in both axes, as shown in the diagram on the right.

Part (ii) is the same as part (i), except for a translation. The origin of (i) is now at $\left(1,1\right)$.

For part (iii), consider ﬁrst $x-y\le 1$, instead of $|x-1|-|y+1|\le 1$. For $x>0$ and $y>0\phantom{\rule{0.3em}{0ex}}$, this gives a region that is inﬁnite in extent in the positive $y$ direction, as shown in the ﬁrst diagram below. Then reﬂect this in both axes and translate one unit down the $y$ axis and 1 unit along the positive $x$ axis as shown in the second and third diagrams below.

For part (iv), consider ﬁrst the rectangular hyperbola $xy=1$ reﬂected in both axes to give four hyperbolas, as in the ﬁrst diagram below. Then translate the four hyperbolas translated 2 units up the $y$ axis as in the ﬁnal diagram. The region required is enclosed by the four hyperbolas.


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laser Illuminocity changes with cavity length

I got a HeNe tube with a constant size which has a concave mirror at one end.
Im changing the distance of the other mirror from the tube.. now, i know some distances arent stable, but appart from the Heaviside function, is there any other change in the beam's strength?
the only thing that comes to my mind is that as i move the mirror away less light is reflected back, because it catches less of the cone of light made by the concave mirror in the back.... so assuming that the shape of light beam from that mirror is a conus, the leading order is $$\frac{1}{r}$$... am i right?
are there any other factors?
 PhysOrg.com physics news on PhysOrg.com >> Cheap, color, holographic video: Better holographic video displays>> First entanglement between light and optical atomic coherence>> EUROnu project recommends building Neutrino Factory
 Recognitions: Science Advisor I need a little clarfication on some points before I can answer your question. Firstly, what do you mean by illuminocity and beam strength? I have a feeling you are either referring to the lasers output power, or its output irradiance, could you confirm this? Just to clarify, you are asking how the output beam of a laser will change as you vary the distance between the mirrors? One final question, is the mirror you are using at the output end planar or concave? Claude.

 Similar discussions for: laser Illuminocity changes with cavity length Thread Forum Replies Advanced Physics Homework 0 Classical Physics 4 Atomic, Solid State, Comp. Physics 7 Advanced Physics Homework 0 Introductory Physics Homework 0


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# MarkusGrassl

Markus Grassl

Solving the equations for a fiducial vector for a Weyl-Heisenberg SIC-PVOM in some small dimensions d ≤ 15 and d = 24, 35, 48. Furthermore, solutions for several other nice unitary error bases (groups of order d2).


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## 11.10 Reformat R source code

When you set the chunk option tidy = TRUE, the R source code will be reformatted by the tidy_source() function in the formatR package . The tidy_source() can reformat the code in several aspects, such as adding spaces around most operators, indenting the code properly, and replacing the assignment operator = with <-. The chunk option tidy.opts can be a list of arguments to be passed to formatR::tidy_source(), e.g.,

{r, tidy=TRUE, tidy.opts=list(arrow=TRUE, indent=2)}
# messy R code...
1+            1
x=1:10#some users prefer '<-' as the assignment operator
if(TRUE){
print('Hello world!') # indent by 2 spaces
}

The output:

# messy R code...
1 + 1
x <- 1:10  #some users prefer '<-' as the assignment operator
if (TRUE) {
print("Hello world!")  # indent by 2 spaces
}

In Section 5.3, we mentioned how to control the width of text output. If you want to control the width of the source code, you may try the width.cutoff argument when tidy = TRUE, e.g.,

{r, tidy=TRUE, tidy.opts=list(width.cutoff=50)}
# a long expression
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

The output:

# a long expression
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

Please read the help page ?formatR::tidy_source to know the possible arguments, and also see https://yihui.org/formatR/ for examples and limitations of this function.

Alternatively, you may use the styler package to reformat your R code if you set the chunk option tidy = 'styler'. The R code will be formatted with the function styler::style_text(). The styler package has richer features than formatR. For example, it can preserve alignment and works with tidyverse syntax such as %>%, !! or {{. The chunk option tidy.opts can also be used to pass additional arguments to styler::style_text(), e.g.,

{r, tidy='styler', tidy.opts=list(strict=FALSE)}
# align the assignment operators
a   <- 1#one variable
abc <- 2#another variable

By default, tidy = FALSE and your R code will not be reformatted.

### References

Müller, Kirill, and Lorenz Walthert. 2022. Styler: Non-Invasive Pretty Printing of r Code.
———. 2022b. formatR: Format r Code Automatically. https://github.com/yihui/formatR.


text:
# SSC 319-Compute a 3-month weighted moving average

SSC 319-Compute a 3-month weighted moving average
1) Compute a 3-month weighted moving average forecast for November.
Assume weights are 9, 5, 1, on Oct., Sep., Aug., respectively
2) Compute the single exponential smoothing forecast for November. Use smoothing constant alpha = 0.37. Assume forecast for January was 105.
3) Compute a trend projection line for the time series. Forecast orders delivered for the upcoming January. Round the answer to nearest integer.
4) Compute 2-month moving average forecasts for the time series and compute MAD.
5) Compute 2-month moving average forecasts for the time series and compute MAPD.
What is MAPD?
Month
Jan.
Feb.
March
April
May
June
July
Aug.
Sep.
Oct. Orders
Delivere
d
120
90
100
75
110
50
75
130
111
94


text:
## Recursos de colección

#### Project Euclid (Hosted at Cornell University Library) (203.669 recursos)

Annals of K-Theory

1. #### Stable $\mathbb{A}^1$-connectivity over Dedekind schemes

Schmidt, Johannes; Strunk, Florian
We show that [math] -localization decreases the stable connectivity by at most one over a Dedekind scheme with infinite residue fields. For the proof, we establish a version of Gabber’s geometric presentation lemma over a henselian discrete valuation ring with infinite residue field.

2. #### Connectedness of cup products for polynomial representations of $\mathrm{GL}_n$ and applications

Touzé, Antoine
We find conditions such that cup products induce isomorphisms in low degrees for extensions between stable polynomial representations of the general linear group. We apply this result to prove generalizations and variants of the Steinberg tensor product theorem. Our connectedness bounds for cup product maps depend on numerical invariants which seem also relevant to other problems, such as the cohomological behavior of the Schur functor.

3. #### A fixed point theorem on noncompact manifolds

Hochs, Peter; Wang, Hang
We generalise the Atiyah–Segal–Singer fixed point theorem to noncompact manifolds. Using [math] -theory, we extend the equivariant index to the noncompact setting, and obtain a fixed point formula for it. The fixed point formula is the explicit cohomological expression from Atiyah–Segal–Singer’s result. In the noncompact case, however, we show in examples that this expression yields characters of infinite-dimensional representations. In one example, we realise characters of discrete series representations on the regular elements of a maximal torus, in terms of the index we define. Further results are a fixed point formula for the index pairing between equivariant [math] -theory and...

4. #### Algebraic $K$-theory of quotient stacks

Krishna, Amalendu; Ravi, Charanya
We prove some fundamental results like localization, excision, Nisnevich descent, and the regular blow-up formula for the algebraic [math] -theory of certain stack quotients of schemes with affine group scheme actions. We show that the homotopy [math] -theory of such stacks is homotopy invariant. This implies a similar homotopy invariance property of the algebraic [math] -theory with coefficients.

5. #### Algebraic $K$-theory and a semifinite Fuglede–Kadison determinant

Hochs, Peter; Kaad, Jens; Schemaitat, André
In this paper we apply algebraic [math] -theory techniques to construct a Fuglede–Kadison type determinant for a semifinite von Neumann algebra equipped with a fixed trace. Our construction is based on the approach to determinants for Banach algebras developed by Skandalis and de la Harpe. This approach can be extended to the semifinite case since the first topological [math] -group of the trace ideal in a semifinite von Neumann algebra is trivial. Along the way we also improve the methods of Skandalis and de la Harpe by considering relative [math] -groups with respect to an ideal instead of the usual...

6. #### An explicit basis for the rational higher Chow groups of abelian number fields

Kerr, Matthew; Yang, Yu
We review and simplify A. Beĭlinson’s construction of a basis for the motivic cohomology of a point over a cyclotomic field, then promote the basis elements to higher Chow cycles and evaluate the KLM regulator map on them.

7. #### Cohomologie non ramifiée de degré 3 : variétés cellulaires et surfaces de del Pezzo de degré au moins 5

Cao, Yang
Dans cet article, où le corps de base est un corps de caractéristique zéro quelconque, pour [math] une variété géométriquement cellulaire, on étudie le quotient du troisième groupe de cohomologie non ramifiée [math] par sa partie constante. Pour [math] une compactification lisse d’un torseur universel sur une surface géométriquement rationnelle, on montre que ce quotient est fini. Pour [math] une surface de del Pezzo de degré [math] , on montre que ce quotient est trivial, sauf si [math] est une surface de del Pezzo de degré 8 d’un type particulier. ¶ We consider geometrically cellular varieties [math] over an arbitrary field...

8. #### Equivariant noncommutative motives

Given a finite group [math] , we develop a theory of [math] -equivariant noncommutative motives. This theory provides a well-adapted framework for the study of [math] -schemes, Picard groups of schemes, [math] -algebras, [math] -cocycles, [math] -equivariant algebraic [math] -theory, etc. Among other results, we relate our theory with its commutative counterpart as well as with Panin’s theory. As a first application, we extend Panin’s computations, concerning twisted projective homogeneous varieties, to a large class of invariants. As a second application, we prove that whenever the category of perfect complexes of a [math] -scheme [math] admits a full exceptional collection of...

9. #### Abstract tilting theory for quivers and related categories

Groth, Moritz; Šťovíček, Jan
We generalize the construction of reflection functors from classical representation theory of quivers to arbitrary small categories with freely attached sinks or sources. These reflection morphisms are shown to induce equivalences between the corresponding representation theories with values in arbitrary stable homotopy theories, including representations over fields, rings or schemes as well as differential-graded and spectral representations. ¶ Specializing to representations over a field and to specific shapes, this recovers derived equivalences of Happel for finite, acyclic quivers. However, even over a field our main result leads to new derived equivalences, for example, for not necessarily finite or acyclic quivers. ¶ Our...

10. #### Suslin's moving lemma with modulus

Kai, Wataru; Miyazaki, Hiroyasu
The moving lemma of Suslin (also known as the generic equidimensionality theorem) states that a cycle on [math] meeting all faces properly can be moved so that it becomes equidimensional over [math] . This leads to an isomorphism of motivic Borel–Moore homology and higher Chow groups. ¶ In this short paper we formulate and prove a variant of this. It leads to a modulus version of the isomorphism, in an appropriate pro setting.

11. #### Localization, Whitehead groups and the Atiyah conjecture

Lück, Wolfgang; Linnell, Peter
Let [math] be the [math] -group of square matrices over [math] which are not necessarily invertible but induce weak isomorphisms after passing to Hilbert space completions. Let [math] be the division closure of [math] in the algebra [math] of operators affiliated to the group von Neumann algebra. Let [math] be the smallest class of groups which contains all free groups and is closed under directed unions and extensions with elementary amenable quotients. Let [math] be a torsionfree group which belongs to [math] . Then we prove that [math] is isomorphic to [math] . Furthermore we show that [math] is a...

12. #### Hochschild homology, lax codescent, and duplicial structure

Garner, Richard; Lack, Stephen; Slevin, Paul
We study the duplicial objects of Dwyer and Kan, which generalize the cyclic objects of Connes. We describe duplicial objects in terms of the decalage comonads, and we give a conceptual account of the construction of duplicial objects due to Böhm and Ştefan. This is done in terms of a 2-categorical generalization of Hochschild homology. We also study duplicial structure on nerves of categories, bicategories, and monoidal categories.

13. #### A plethora of inertial products

Edidin, Dan; Jarvis, Tyler; Kimura, Takashi
For a smooth Deligne–Mumford stack [math] , we describe a large number of inertial products on [math] and [math] and inertial Chern characters. We do this by developing a theory of inertial pairs. Each inertial pair determines an inertial product on [math] and an inertial product on [math] and Chern character ring homomorphisms between them. We show that there are many inertial pairs; indeed, every vector bundle [math] on [math] defines two new inertial pairs. We recover, as special cases, the orbifold products considered by Chen and Ruan (2004), Abramovich, Graber and Vistoli (2002), Fantechi and Göttsche (2003), Jarvis, Kaufmann...

14. #### The joint spectral flow and localization of the indices of elliptic operators

Kubota, Yosuke
We introduce the notion of the joint spectral flow, which is a generalization of the spectral flow, by using Segal’s model of the connective [math] -theory spectrum. We apply it for some localization results of indices motivated by Witten’s deformation of Dirac operators, and rephrase some analytic techniques in terms of topology.

15. #### On some negative motivic homology groups

Kohrita, Tohru
For an arbitrary separated scheme [math] of finite type over a finite field [math] and a negative integer [math] , we prove, under the assumption of resolution of singularities, that [math] is canonically isomorphic to [math] if [math] or [math] , and [math] vanishes if [math] and [math] . As the group [math] is explicitly known, this gives a explicit calculation of motivic homology of degree [math] and weight [math] or [math] of an arbitrary scheme over a finite field.

16. #### On the Deligne–Beilinson cohomology sheaves

Barbieri-Viale, Luca
We prove that the Deligne–Beilinson cohomology sheaves [math] are torsion-free as a consequence of the Bloch–Kato conjectures as proven by Rost and Voevodsky. This implies that [math] if [math] is unirational. For a surface [math] with [math] we show that the Albanese kernel, identified with [math] , can be characterized using the integral part of the sheaves associated to the Hodge filtration.

17. #### Stable operations and cooperations in derived Witt theory with rational coefficients

Ananyevskiy, Alexey
The algebras of stable operations and cooperations in derived Witt theory with rational coefficients are computed and an additive description of cooperations in derived Witt theory is given. The answer is parallel to the well-known case of K-theory of real vector bundles in topology. In particular, we show that stable operations in derived Witt theory with rational coefficients are given by the values on the powers of the Bott element.

18. #### Rational mixed Tate motivic graphs

Agarwala, Susama; Patashnick, Owen
We study the combinatorics of a subcomplex of the Bloch–Kriz cycle complex that was used to construct the category of mixed Tate motives. The algebraic cycles we consider properly contain the subalgebra of cycles that correspond to multiple logarithms (as defined by Gangl, Goncharov and Levin). We associate an algebra of graphs to our subalgebra of algebraic cycles. We give a purely combinatorial criterion for admissibility. We show that sums of bivalent graphs correspond to coboundary elements of the algebraic cycle complex. Finally, we compute the Hodge realization for an infinite family of algebraic cycles represented by sums of graphs...

19. #### Chow groups of some generically twisted flag varieties

Karpenko, Nikita
We classify the split simple affine algebraic groups [math] of types A and C over a field with the property that the Chow group of the quotient variety [math] is torsion-free, where [math] is a special parabolic subgroup (e.g., a Borel subgroup) and [math] is a generic [math] -torsor (over a field extension of the base field). Examples of [math] include the adjoint groups of type A. Examples of [math] include the Severi–Brauer varieties of generic central simple algebras.

20. #### $K\mkern-2mu$-theory of derivators revisited

Muro, Fernando; Raptis, Georgios
We define a [math] -theory for pointed right derivators and show that it agrees with Waldhausen [math] -theory in the case where the derivator arises from a good Waldhausen category. This [math] -theory is not invariant under general equivalences of derivators, but only under a stronger notion of equivalence that is defined by considering a simplicial enrichment of the category of derivators. We show that derivator [math] -theory, as originally defined, is the best approximation to Waldhausen [math] -theory by a functor that is invariant under equivalences of derivators.

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# Can two waves interfere head on?

Can two waves (like sound or electromagnetic waves) interfere head on? If yes, and suppose they are out of phase with each other and thus interfere destructively, where does the energy of the waves go?

• Jul 21, 2019 at 23:39
• Yes but it is only temporary, waves reemerge after the collision as if they have just passed thru each other. 2 water waves (one up and one down) show a flat line surface when they meet, the energy is stored in the elasticity of the water temporarily. Jul 22, 2019 at 0:18
• @PhysicsDave Is water that elastic? Jul 22, 2019 at 3:37
• Water is about 80 times more compressible that steel, water hammer is another phenomenon of water elasticity. Jul 22, 2019 at 10:54
• @PhysicsDave True... but I don't see how that means when waves destructively interfere that means that the energy is stored in the elasticity of the water. Jul 22, 2019 at 17:27

Can two waves (like sound or electromagnetic waves) interfere head-on?

Yes. When waves add in a superposition it is called interference. Two waves heading towards each other with have interference.

suppose they are out of phase with each other and thus interfere destructively, where does the energy of the waves go?

It depends on what you mean by "interfere destructively". If you mean at some point in time the amplitude is $$0$$ for all points in space, then there isn't a problem. The wave equation is a second-order equation, so the wave is not only determined by its amplitude. A simple example is seen with waves on a string. Send one pulse to the right and another opposite pulse to the left on the string. When they meet the amplitude of the superposition is $$0$$ at all points on the string for that instant in time. But the various parts of the string still have a velocity, and hence the two pulses will then move past each other. No energy is lost.

If instead, you mean can we have two waves approach each other so that for all times larger than some finite time the superposition is $$0$$, then this is impossible? This assumes that we have two non-zero waves with some sort of localization in space. These waves will eventually move past each other. Even if you had a continuously oscillating source you couldn't cancel everything out. Of course, your energy conservation argument is sufficient as well in my opinion.

• Yes, moving past each other like this: commons.wikimedia.org/wiki/File:Standing_waves1.gif
– user137289
Jul 21, 2019 at 18:45
• Re. "still have a velocity, and hence the two pulses will then move past each other. No energy is lost", during the time the string displacement is zero where is the energy stored--in the momentum? Jul 21, 2019 at 19:17
• @user45664 There parts of the string still in motion. That is where the energy is. Look at Pieter's animation. Jul 21, 2019 at 21:15
• @Aaron Stevens I was thinking of two pulses of opposite polarity in two different directions approaching each other. Where is the energy when they pass each other and for an instant the net displacement is zero? Jul 22, 2019 at 1:53
• @user45664 Parts of the string have vertical velocities still... $0$ displacement doesn't mean all parts of the string are not moving. Please look at the black line in Pieter's animation. His is for a standing wave, but you can still see this happening. Jul 22, 2019 at 1:58

Two waves of the same frequency and oppposite direction of propagation will produce a standing wave.

Like this figure by LucasVB:

Edit: In a standing wave, the energy oscillates back and forth between different forms. For a mechanical wave (transverse wave on a string for example), that is elastic potential energy and kinetic energy. At instances when the string is straight (minimal potential energy), the kinetic energy is at its maximum.

• ... in which case it is important to remark that the energy doesn't "go" anywhere. The details depend on the medium, though: (i) for sound, the energy is not the square of the function being plotted, whereas (ii) for light, the electric energy density is the square of the black line, which does fluctuate -- out of sync in space and time from the magnetic energy density, so that the total energy density stays constant. Jul 21, 2019 at 18:48

Why not? Superposition is superposition. Standing waves result from interference of waves of opposite propagation direction.

For sound waves and other waves that depend on the motion of atoms and molecules, destructive interference will eventually disperse the energy to the atoms and molecules finally ending up in heat (motion of matter).

For light, which does not depend on a medium to propagate, it is more complicated. Light only superimposes, does not interact. This MIT video shows what is happening in superposition interference of a laser beam split in two and made to interfere. The energy not appearing in the pattern goes back to the lazing mechanism!

In general for electromagnetic waves, when interference light patterns are seen, the energy goes from the dark regions to the bright regions, i.e.it is a function of at least two spatial dimensions, not the one-dimensional plots usually showing amplitudes interfering. In my answer here to an almost duplicate question, I go into more detail. Note that in order to see light, it has to interact with matter, so in the end, the energy will go into heating the screen material.

• Destructive interference isn't energy dispersion. Jul 21, 2019 at 13:34
• @AaronStevens total destructive interference raises the problem of energy conservation. In light, which is composed of photons yes, total interference as the video shows needs and explanation for where the energy goes. Jul 21, 2019 at 14:24
• Yes, I agree with that Jul 21, 2019 at 14:28
• The linked by you video from MIT is so obvious, “This MIT video is instructive and a real experiment that shows that in destructive interference set up with interferometers there is a return beam, back to the source, as far as classical electromagnetic waves go. So the energy is balanced by going back to the source.” How ignorant someone could be and downvote the answer, than more doing this without explanation. Jul 21, 2019 at 19:53
• The MIT video does not explain where the energy goes because by setting up the strong interference the laser cavity actually lase less, i/e a current meter on the power supply would show a decrease. It is easy to recreate this type of interference with a laser diode for example, external mirrors can easily upset the wave function of the cavity and reduce lasing. Jul 23, 2019 at 12:46

Of course annav's answer is right. Let me add a few things, because there are two basic theories:

1. in the double slit experiment, when we interpret the photon (shot one at a time) pass through both slits as partial waves, those waves interfere. They can interfere constructively (their in phase, create a dot on the screen) or destructively (out of phase and no dot on the screen). Of course the explanation here is that the photons, and the slits are entangled, and so the energy of the photons that interfere destructively, will give their energy to the slits as they scatter inelastically, or get absorbed by the slits. Energy is conserved. If they interfere constructively, the energy is in the dot on the screen. Energy is still conserved.

2. virtual particles and antiparticles, these are of course duo to wave-particle duality acting as waves too, and when they in your case interfere head on, they cancel out, and the energy goes to where it came from. Back to the vacuum energy density. Yes, photons are their own antiparticles. When you talk about elementary particles in the SM, particles can annihilate with their own antiparticles. This is what you are asking about, because in this case, when the particle antiparticle is created, it pops in and out of existence. From where? From the vacuum energy density (virtual particles).

Pair production is the creation of a subatomic particle and its antiparticle from a neutral boson. Examples include creating an electron and a positron, a muon and an antimuon, or a proton and an antiproton. Pair production often refers specifically to a photon creating an electron–positron pair near a nucleus. For pair production to occur, the incoming energy of the interaction must be above a threshold of at least the total rest mass energy of the two particles, and the situation must conserve both energy and momentum.

https://en.wikipedia.org/wiki/Pair_production

In particle physics, annihilation is the process that occurs when a subatomic particle collides with its respective antiparticle to produce other particles, such as an electron colliding with a positron to produce two photons.[1] The total energy and momentum of the initial pair are conserved in the process and distributed among a set of other particles in the final state.

https://en.wikipedia.org/wiki/Annihilation

Now it is very important to understand that virtual particles are not real particles, and they can pop in and out of existence from the energy density of quantum vacuum. Virtual particles are a mathematical model of describing the interaction of quantum fields, in this case the poping of in and out of virtual particles.

This is not the case in pair production and annihilation. These are real particles. In the case of pair production and annihilation, and this is what you are asking about, two particles (particle antiparticle pair) is created or annihilated. When the particle and anti particle, like an electron and a positron, traveling as waves, collide head on, both cease to exist as fermions (or when the photon ceases to exist and creates a electron and positron), and the energy (momentum) transforms into a photon. So energy is conserved.

• I don't agree with the double slit interpretation that a photon is cancelling "itself" out. All photons that enter the slits emerge and appear on the screen, no energy is lost. See Feynman path interpretation on the "pattern", the word interference is historical. Jul 23, 2019 at 12:27