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# Math Help - Tricky Exponential Integral

1. ## Tricky Exponential Integral

I'm asked to evaluate the following integral:

I know the classic gaussian result, but I cannot seem to use this to help. I'm given a hint to try the substitution: $y&space;=&space;\frac{1}{2}(a^{^{1/2}}x&space;-&space;\frac{b^{^{1/2}}}{x})$ , but I cannot make this work.

2. Here's my stab at it. Denote the integral by $I$. Because of the symmetry this integral is

$I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;dx.\;\;\;(1)$

Under the change of variable $x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$, the integral becomes

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ay^2 + \frac{b}{y^2}\right)}\dfrac{dy}{y^2}.$.

Now replace the dummy variable with $x$, so

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\dfrac{dx}{x^2},\;\;\;(2)$.

Now add $\sqrt{a} \;(1) + \sqrt{a} \;(2)$ giving

$2 \sqrt{a} I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;\left(\sqrt{a}+ \dfrac{\sqrt{b}}{x^2}\right)\;dx\;\;\;(3)$

Now introduce your hint (sort of)

$y = \sqrt{a} x - \dfrac{\sqrt{b}}{x}$

so

$dy = \left(\sqrt{a} + \dfrac{\sqrt{b}}{x^2}\right)\,dx$.

Note that $x \to 0$ gives $y \to - \infty$ and $x \to \infty$ gives $y \to \infty$ so with that and noting that

$y^2 = ax^2 - 2 \sqrt{ab} + \dfrac{b}{x^2}$ gives $ax^2 + \dfrac{b}{x^2} = y^2 + 2 \sqrt{ab}$.

Thus, (3) becomes

$2 \sqrt{a} I = 2\int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(y^2 + 2 \sqrt{ab}\right)}\;\;dy = 2e^{-\sqrt{ab}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2}\;\;dy$

and solving for $I$ gives

$I = \sqrt{\dfrac{2\pi}{a}} e^{-\sqrt{ab}}$.

Whew! I think I need to lie down.

3. Wow! Thanks.
Originally Posted by Danny
Here's my stab at it. Denote the integral by $I$. Because of the symmetry this integral is

$I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;dx.\;\;\;(1)$

Under the change of variable $x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$, the integral becomes

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ay^2 + \frac{b}{y^2}\right)}\dfrac{dy}{y^2}.$.

Now replace the dummy variable with $x$, so

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\dfrac{dx}{x^2},\;\;\;(2)$.

Now add $\sqrt{a} \;(1) + \sqrt{a} \;(2)$ giving

$2 \sqrt{a} I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;\left(\sqrt{a}+ \dfrac{\sqrt{b}}{x^2}\right)\;dx\;\;\;(3)$

Now introduce your hint (sort of)

$y = \sqrt{a} x - \dfrac{\sqrt{b}}{x}$

so

$dy = \left(\sqrt{a} + \dfrac{\sqrt{b}}{x^2}\right)\,dx$.

Note that $x \to 0$ gives $y \to - \infty$ and $x \to \infty$ gives $y \to \infty$ so with that and noting that

$y^2 = ax^2 - 2 \sqrt{ab} + \dfrac{b}{x^2}$ gives $ax^2 + \dfrac{b}{x^2} = y^2 + 2 \sqrt{ab}$.

Thus, (3) becomes

$2 \sqrt{a} I = 2\int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(y^2 + 2 \sqrt{ab}\right)}\;\;dy = 2e^{-\sqrt{ab}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2}\;\;dy$

and solving for $I$ gives

$I = \sqrt{\dfrac{2\pi}{a}} e^{-\sqrt{ab}}$.

Whew! I think I need to lie down.

4. Originally Posted by Danny
Here's my stab at it. Denote the integral by $I$. Because of the symmetry this integral is

$I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;dx.\;\;\;(1)$

Under the change of variable $x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$, the integral becomes

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ay^2 + \frac{b}{y^2}\right)}\dfrac{dy}{y^2}.$.
May I intrude and ask did you differentiate the change of variable x wrt to y? If you did, where did the minus sign go? If you didnt, then I dont understand whats happening :-)

5. Originally Posted by bugatti79
May I intrude and ask did you differentiate the change of variable x wrt to y? If you did, where did the minus sign go? If you didnt, then I dont understand whats happening :-)
In that change of variable, there is indeed a minus sign attached to the dy. But the limits of integration get switched: instead of going from 0 to ∞, the integral goes from ∞ to 0. When you switch the limits back again, that introduces a change of sign, which cancels with the other one.

6. Originally Posted by Opalg
In that change of variable, there is indeed a minus sign attached to the dy. But the limits of integration get switched: instead of going from 0 to ∞, the integral goes from ∞ to 0. When you switch the limits back again, that introduces a change of sign, which cancels with the other one.
Ok, how do the limits of integration get switched? Could an example be shown?

Thanks

7. Originally Posted by bugatti79
Ok, how do the limits of integration get switched? Could an example be shown?
When you make the substitution $x = \phi(y)$ in the definite integral $\int_a^b f(x)\, dx,$ it becomes $\int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(y))\phi'(y)\,dy.$ In this case, $\textstyle x = \sqrt{\frac ba}\frac1y,$ and the limits of integration are that x goes from 0 to ∞. When x = 0, y = ∞; and when x = ∞, y = 0. So the dy-integral, instead of going from 0 to ∞, goes from ∞ to 0.

8. Originally Posted by Opalg
When you make the substitution $x = \phi(y)$ in the definite integral $\int_a^b f(x)\, dx,$ it becomes $\int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(y))\phi'(y)\,dy.$ In this case, $\textstyle x = \sqrt{\frac ba}\frac1y,$ and the limits of integration are that x goes from 0 to ∞. When x = 0, y = ∞; and when x = ∞, y = 0. So the dy-integral, instead of going from 0 to ∞, goes from ∞ to 0.
Ok, that makes sense. I have another question :-) At the start, how does he spot that x should be = (b/a)^1/2 * (1/y) in order to proceed?

Thanks

9. Under a change of variables, I literally wanted the two terms

$ax^2$ and $\dfrac{b}{x^2}$ to switch places. The change of variables

$x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$ does that.

10. Originally Posted by Danny
Under a change of variables, I literally wanted the two terms

$ax^2$ and $\dfrac{b}{x^2}$ to switch places. The change of variables

$x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$ does that.
Ok, but why? I just dont see the connection. For example if I was given another similar example, I wouldnt know what to write x = to

11. Originally Posted by bugatti79
At the start, how does he spot that x should be = (b/a)^1/2 * (1/y) in order to proceed?
Sheer genius! You might deduce from the number of "Thanks" that he received for that solution that most of the other resident experts here had thought unsuccessfully about how to evaluate that integral. So you are not the only one to have found it difficult.


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## Bài tập luyện thi Toán Olympic AMC Úc 2021 – bài ôn số 5

1) Which number is one hundred more than the number 765?

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2) In the number sentence 10 hundreds + 100 tens = tens, what number goes in the box to make it true ?

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40 x 40 = 20 x 20 x …

what number do we put in the … to make the number sentence true ?

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(B) 4

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(D) 16

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(B) 17

(C) 27

(D) 7

(E) 19

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14 − 7 + 9 = 16

(A) Simon had 14 lollies, he ate 9 of them and then his sister gave him 9 more. He then had 16 lollies.

(B) James had 18 lollies, he gave 14 to his sister, he ate 9 and had 7 left.

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(C) 4

(D) 16

(E) 64

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(C) 5

(D) 25

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(B) 3500

(C) 4000

(D) 3750

(E) 3725

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(A) 21

(B) 24

(C) 14

(D) 18

(E) 25

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(A) 8

(B) 13

(C) 16

(D) 24

(E) 26

14) Which number is 1 + 10 + 100 + 1000 ?

(A) 1111

(B) 11 111

(C) 1110

(D) 1010

(E) 10 111

15) Which number is halfway between 600 and 700 ?

(A) 550

(B) 645

(C) 650

(D) 655

(E) 700

16) A water tank has 56 L of water in it. If 28 L of water are added, how much water will be in the tank ?

(A) 84 L

(B) 56 L

(C) 28 L

(D) 76 L

(E) 78 L

17) What is one thousand and twenty-seven in numerals ?

(A) 100 027

(B) 10 027

(C) 1027

(D) 127

(E) 27

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A) 46

(B) 48

(C) 50

(D) 52

(E) 54

19) When five numbers are added the total is 2010. One of the numbers is changed from 235 to 532. What is the total of the five numbers now ?

A) 1723

(B) 2542

(C) 2360

(D) 1896

(E) 2307

20) What is the difference between the largest and smallest 3-digit numbers that can be made from the following 1-digit cards, if each card is to be used once only in each number ?

A) 477

(B) 495

(C) 1009

(D) 468

(E) 555

21) Mike buys a can of 4 tennis balls for $2. How much does each tennis ball cost ? A) 25c (B) 50c (C)$1

(D) $4 (E)$8

22) The number 8000 is the same as

A) 800 tens

(B) 800 units

(C) 80 tens

(D) 80 units

(E) 8 hundreds

23) What number is halfway between 103 and 113 ?

(A) 107

(B) 110

(C) 105

(D) 109

(E) 108

24) Ben cuts three oranges into quarters for the soccer team to eat at half-time. How many quarters are there ?

(A) 3

(B) 6

(C) 7

(D) 12

(E) 16

25) Gina is 11 years old and her sister Bev is 8 years old. Their mum is twice as old as the sum of their ages. How old is their mum ?

(A) 3

(B) 19

(C) 27

(D) 30

(E) 38

26) Which of the following numbers has the same value as 152 hundredths ?

(A) 1520

(B) 152

(C) 15,2

(D) 1,52

(E) 0,152

27) The value of 1 + 10 + 10 × 10 + 10 × 10 × 10 is

(A) 1010

(B) 10 111

(C) 1111

(D) 1110

(E) 11 111

28) Which of the following is true ?

(A) If you add two odd numbers you always get an odd number.

(B) If you multiply two odd numbers you always get an even number.

(C) If you add an odd and an even number you always get an even number.

(D) If you multiply an odd and an even number you always get an even number.

(E) If you multiply two even numbers you always get an odd number.

29) Sally is playing a board game where you throw a dice numbered from 1 to 6, move along a numbered board and then follow the instructions on each square you land on. On one turn, she throws a 6 and lands on a square which tells her to go back 4 squares. This puts her on a square which tells her to go forward 3 squares. She finishes up on square 7. What square did she start that turn on ?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

*****

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# Why isn't this rod in neutral equilibrium?

Suppose that the rod shown below in figure(1) is balanced. So the torque about point $$O$$ is $$Mg.x-mg.y=0$$.

[figure(1)]

Then imagine than M is lifted a little bit up making an angle of $$\theta$$ with the horizontal plane(figure(2)).

[figure(2)]

Then let's calculate the torque about point $$O$$.

Torque about point $$O$$ $$= Mg.x cos\theta - mg.y cos\theta$$ $$=cos\theta(Mg.x-mg.y)$$ from the first equation:

$$=cos\theta.(0)$$ $$=0$$ Also we can find the torques about both ends of the rod are equal to zero. So shouldn't it be balanced in its new position?

And when talking about potential energy of each mass, when we lift one side up its P.E. is increased as well as the other one's P.E. is decreased. We can imagine that both of them cancel each other( This is obvious when $$M=m$$ and $$x=y$$).*

Thus shouldn't it be in neutral equilibrium?

But as we experience in day-to-day life balances always find a horizontal plane to be balanced. So what is wrong with my previous explanation?

Edit: The example of a beam scale may be irrelevant here. But I tried to balance such a rod and I couldn't keep it in neutral equilibrium. Is that because both masses are trying to keep their gravitational potential energy at a low level? Though this contrast again with above paragraph marked as *. Does it imply that these two masses do not act as a system?

• When you tried to balance such a rod, how confident are you that the suspension point was at the precise center of mass of the rod (width-wise as well as length-wise)? Even a small deviation along the width of the rod could be enough to make the rod return to a horizontal configuration. Jul 13 at 13:39
• I didn't want to worry about that. According to the above calculations, it would do no change in the effect. On the other hand, if I wanted to balance from the centre of mass, I would not attach two masses.
– ACB
Jul 13 at 14:17
• I misspoke slightly in my comment above. The center of mass of the rod doesn't matter so much as the relative vertical positions of the pivot point and the mass suspension points. Unless these points are perfectly aligned horizontally, the equilibrium will not be neutral. See my answer below for details. Jul 13 at 15:22

To elaborate on R.W. Bird's excellent and correct answer: if the masses are not suspended precisely in a horizontal line with the suspension point of the bar, then the horizontal position is a stable equilibrium configuration rather than a neutral equilibrium configuration. To see this, let's first draw a "thick" rod in horizontal equilibrium:

The cross here marks the pivot point. As you (correctly) showed, the condition that this be an equilibrium configuration implies that $$M g x = m g y$$.

Now let's tilt the beam a bit:

The lever arm for the left-hand force is $$x \cos \theta + d \sin \theta$$, while the lever arm for the right-hand force is $$y \cos \theta - d \sin \theta$$. Thus, the total torque exerted about the pivot point (with clockwise rotations taken to be positive) is $$\tau = -M g (x \cos \theta + d \sin \theta) + m g (y \cos \theta - d \sin \theta) = -(M+m) g d \sin \theta$$ So for any positive value of $$d$$, a positive displacement in $$\theta$$ leads to a negative torque, which restores it towards $$\theta = 0$$. Similarly, a negative displacement in $$\theta$$ leads to a positive torque, which also restores the equilibrium. Thus, the horizontal configuration is not a neutral equilibrium unless $$d = 0$$, i.e., the suspension points of the masses are perfectly horizontally aligned with the pivot point.

Note that the above also implies that if $$d < 0$$, the horizontal configuration is an unstable equilibrium. This would correspond to the case where the mass suspension points are (for whatever reason) above the pivot point.

• does this mean that theoretically if we were able to perfectly balance the scale we may attain neutral equilibrium at any deviation? Aug 30 at 11:45
• @utkarsh: Yes. If the scale were perfectly balanced, then the potential energy would be constant as a function of the angle, with no local minima or maxima. Aug 30 at 11:57

With a balance beam scale, the point of support is shifted slightly up (think of the beam as being a little bent). Then the effective horizontal becomes a position of stable equilibrium (if the loads are equal).

• Do you mean that the gadget presented in the above figure is in stable equilibrium? But it contrasts with my calculations.
– ACB
Jun 30 at 13:28
• In the sketch, the beam is not bent, and has no reason to seek the horizontal position. Jun 30 at 17:38

Your calculations are correct, the rod will be in equilibrium regardless of the angle. This assumes that all the joints are pin joints, so that the joints do not transmit any torques, and that the vertical rods stay vertical.

A horizontal plane may be more aesthetically pleasing, but as anyone who has tried to hang a picture can tell you, objects can be in equilibrium without being horizontal.

Your calculation is correct. This is because the center of gravity of your system is always under the axis of rotation and don't move laterally when you rotate the system. The same goes for a wheel that turns around an axis that passes through the center of gravity.

• So you are saying my observation is wrong?
– ACB
Jul 13 at 14:20
• We should look closely at your experience. The problem may be that a neutral equilibrium corresponds to a borderline situation : the slightest deviation, and it is no longer neutral. Jul 13 at 15:13


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## PHY2403H Quantum Field Theory. Lecture 6: Canonical quantization, Simple Harmonic Oscillators, Symmetries. Taught by Prof. Erich Poppitz

### DISCLAIMER: Very rough notes from class, with some additional side notes (QM SHO review, …).

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Quantization of Field Theory

We are engaging in the “canonical” or Hamiltonian method of quantization. It is also possible to quantize using path integrals, but it is hard to prove that operators are unitary doing so. In fact, the mechanism used to show unitarity from path integrals is often to find the Lagrangian and show that there is a Hilbert space (i.e. using canonical quantization). Canonical quantization essentially demands that the fields obey a commutator relation of the following form
\label{eqn:qftLecture6:20}
\antisymmetric{\pi(\Bx, t)}{\phi(\By, t)} = -i \delta^3(\Bx – \By).

We assumed that the quantized fields obey the Hamiltonian relations
\label{eqn:qftLecture6:160}
\begin{aligned}
\ddt{\phi} &= i \antisymmetric{H}{\phi} \\
\ddt{\pi} &= i \antisymmetric{H}{\pi}.
\end{aligned}

We were working with the Hamiltonian density
\label{eqn:qftLecture6:40}
\mathcal{H} =
\inv{2} (\pi(\Bx, t))^2
+
+
\frac{m^2}{2} \phi^2
+
\frac{\lambda}{4} \phi^4,

which included a mass term $$m$$ and a potential term ($$\lambda$$). We will expand all quantities in Taylor series in $$\lambda$$ assuming they have a structure such as
\label{eqn:qftLecture6:180}
\begin{aligned}
f(\lambda) =
c_0 \lambda^0
+ c_1 \lambda^1
+ c_2 \lambda^2
+ c_3 \lambda^3
+ \cdots
\end{aligned}

We will stop this \underline{perturbation theory} approach at $$O(\lambda^2)$$, and will ignore functions such as $$e^{-1/\lambda}$$.

Within perturbation theory, to leaving order, set $$\lambda = 0$$, so that $$\phi$$ obeys the Klein-Gordon equation (if $$m = 0$$ we have just a d’Lambertian (wave equation)).

We can write our field as a Fourier transform
\label{eqn:qftLecture6:60}
\phi(\Bx, t) = \int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx} \tilde{\phi}(\Bp, t),

and due to a Hermitian assumption (i.e. real field) this implies
\label{eqn:qftLecture6:80}
\tilde{\phi}^\conj(\Bp, t) = \tilde{\phi}(-\Bp, t).

We found that the Klein-Gordon equation implied that the momentum space representation obey Harmonic oscillator equations
\label{eqn:qftLecture6:100}
\begin{aligned}
\ddot{\tilde{\phi}}(\Bp, t) &= – \omega_\Bp \tilde{\phi}(\Bp, t) \\
\omega_\Bp &= \sqrt{\Bp^2 + m^2}.
\end{aligned}

We may represent the solution to this equation as
\label{eqn:qftLecture6:120}
\tilde{\phi}(\Bq, t) = \inv{\sqrt{2 \omega_\Bq}} \lr{
e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} b_\Bq^\conj
}.

This is a general solution, but imposing $$a_\Bq = b_{-\Bq}$$ ensures \ref{eqn:qftLecture6:80} is satisfied.
This leaves us with
\label{eqn:qftLecture6:140}
\tilde{\phi}(\Bq, t) = \inv{\sqrt{2 \omega_\Bq}} \lr{
e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\conj
}.

We want to show that iff
\label{eqn:qftLecture6:200}
\antisymmetric{a_\Bq}{a^\dagger_\Bp} = \lr{ 2 \pi }^3 \delta^3(\Bp – \Bq),

then
\label{eqn:qftLecture6:220}
\antisymmetric{\pi(\By, t)}{\phi(\Bx, t)} = -i \delta^3(\Bx – \By)

where everything else commutes (i.e. $$\antisymmetric{a_\Bp}{a_\Bq} = \antisymmetric{a_\Bp^\dagger}{a_\Bq^\dagger} = 0$$).
We will only show one direction, but you can go the other way too.

\label{eqn:qftLecture6:240}
\phi(\Bx, t)
=
\int \frac{d^3 p}{(2\pi)^3 \sqrt{2 \omega_\Bp}} e^{i \Bp \cdot \Bx}
\lr{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}

\label{eqn:qftLecture6:260}
\pi(\Bx, t) = \dot{\phi}
=
i \int \frac{d^3 q}{(2\pi)^3 \sqrt{2 \omega_\Bq}} \omega_\Bq e^{i \Bq \cdot \Bx}
\lr{
-e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}

The commutator is
\label{eqn:qftLecture6:280}
\begin{aligned}
\antisymmetric{\pi(\By, t)}{\phi(\Bx, t)}
&=
i \int \frac{d^3 p}{(2\pi)^3
\sqrt{2 \omega_\Bp}}
\frac{d^3 q}{(2\pi)^3 \sqrt{2 \omega_\Bq}}
\omega_\Bq
e^{i \Bp \cdot \By + i \Bq \cdot \Bx}
\antisymmetric
{
-e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}
{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
} \\
&=
i \int \frac{d^3 p}{(2\pi)^3
\sqrt{2 \omega_\Bp}}
\frac{d^3 q}{(2\pi)^3 \sqrt{2 \omega_\Bq}}
\omega_\Bq
e^{i \Bp \cdot \By + i \Bq \cdot \Bx}
\lr{
– e^{i (\omega_\Bp -\omega_\Bq) t}
\antisymmetric { a_\Bq } { a_{-\Bp}^\dagger }
+
e^{i (\omega_\Bq -\omega_\Bp) t}
\antisymmetric{a_{-\Bq}^\dagger}{ a_\Bp }
} \\
&=
i \int \frac{d^3 p}{(2\pi)^3
\sqrt{2 \omega_\Bp}}
\frac{d^3 q}{(2\pi)^3 \sqrt{2 \omega_\Bq}}
\omega_\Bq (2\pi)^3
e^{i \Bp \cdot \By + i \Bq \cdot \Bx}
\lr{
– e^{i (\omega_\Bp -\omega_\Bq) t} \delta^3(\Bq + \Bp)
– e^{i (\omega_\Bq -\omega_\Bp) t} \delta^3(-\Bq -\Bp)
} \\
&=
– 2 i \int \frac{d^3 p}{(2\pi)^3
2 \omega_\Bp}
\omega_\Bp
e^{i \Bp \cdot (\By – \Bx)} \\
&=
-i \delta^3(\By – \Bx),
\end{aligned}

which is what we wanted to prove.

## Free Hamiltonian

We call the $$\lambda = 0$$ case the “free” Hamiltonian.

\label{eqn:qftLecture6:300}
\begin{aligned}
H
&= \int d^3 x \lr{ \inv{2} \pi^2 + \inv{2} (\spacegrad \phi)^2 + \frac{m^2}{2} \phi^2 } \\
&=
\inv{2} \int d^3 x
\frac{d^3 p}{(2\pi)^3}
\frac{d^3 q}{(2\pi)^3}
\frac{
e^{i (\Bp + \Bq)\cdot \Bx}
}{\sqrt{2 \omega_\Bp}
\sqrt{2 \omega_\Bq}}
\lr{

(\omega_\Bp)
(\omega_\Bq)
\lr{
-e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
-e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}
}

\Bp \cdot \Bq
\lr{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}
+
m^2
\lr{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}.
\end{aligned}

An immediate simplification is possible by identifying a delta function factor $$\int d^3 x e^{i(\Bp + \Bq) \cdot \Bx}/(2\pi)^3 = \delta^3(\Bp + \Bq)$$, so
\label{eqn:qftLecture6:1060}
\begin{aligned}
H
&=
\inv{2}
\int
\frac{d^3 p}{(2\pi)^3}
\frac{1
}{2 \omega_\Bp}
\lr{

(\omega_\Bp)^2
\lr{
-e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
-e^{-i \omega_\Bp t} a_{-\Bp}
+
e^{i \omega_\Bp t} a_{\Bp}^\dagger
}
}
+
(\Bp^2 + m^2)
\lr{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
e^{-i \omega_\Bp t} a_{-\Bp}
+
e^{i \omega_\Bp t} a_{\Bp}^\dagger
} \\
&=
\inv{2} \int \frac{d^3 p}{(2 \pi)^3} \inv{2 \omega_\Bp}
\lr{
a_\Bp a_{-\Bp}
\lr{
{-\omega_\Bp^2 e^{-2 i \omega_\Bp t}}
+
{\omega_\Bp^2 e^{-2 i \omega_\Bp t}}
}
+
a_{-\Bp}^\dagger a_{\Bp}^\dagger
\lr{
-{\omega_\Bp^2 e^{2 i \omega_\Bp t}}
+
{\omega_\Bp^2 e^{2 i \omega_\Bp t}}
}
+
a_\Bp a^\dagger_{\Bp} \omega^2_\Bp(1 + 1)
+
a^\dagger_{-\Bp} a_{-\Bp} \omega^2_\Bp(1 + 1)
}
\end{aligned}

When all is said and done we are left with
\label{eqn:qftLecture6:320}
H =
\int \frac{d^3 p}{(2 \pi)^3} \frac{\omega_\Bp}{2} \lr{
a^\dagger_{-\Bp}
a_{-\Bp}
+
a_{\Bp}
a^\dagger_{\Bp}
}

and finally with a $$\Bp \rightarrow -\Bp$$ transformation ($$\iiint_{-R}^R d^3 p \rightarrow (-1)^3 \iiint_R^{-R} d^3 p’ = (-1)^6 \iiint_{-R}^R d^3 p’$$) in the first integral our free Hamiltonian ($$\lambda = 0$$) is
\label{eqn:qftLecture6:340}
\boxed{
H_0 =
\int \frac{d^3 p}{(2 \pi)^3} \frac{\omega_\Bp}{2} \lr{
a^\dagger_{\Bp}
a_{\Bp}
+
a_{\Bp}
a^\dagger_{\Bp}
}
}

From the commutator relationship \ref{eqn:qftLecture6:200} we can write
\label{eqn:qftLecture6:360}
a_\Bp a^\dagger_\Bq
=
a^\dagger_\Bq
a_\Bp
+ (2 \pi)^3 \delta^3(\Bp – \Bq),

so

\label{eqn:qftLecture6:380}
H_0 =
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
\lr{
a^\dagger_{\Bp}
a_{\Bp}
+
\inv{2} (2 \pi)^3 \delta^3(0)
}

The delta function term can be interpreted using
\label{eqn:qftLecture6:400}
(2 \pi)^3 \delta^3(\Bq)
= \int d^3 x e^{i \Bq \cdot \Bx},

so when $$\Bq = 0$$
\label{eqn:qftLecture6:420}
(2 \pi)^3 \delta^3(0) = \int d^3 x = V.

We can write the Hamiltonian now in terms of the volume
\label{eqn:qftLecture6:440}
\boxed{
H_0 =
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp}
a_{\Bp}
+ V_3
\int \frac{d^3 p}{(2 \pi)^3} \frac{\omega_\Bp }{2} \times 1
}

## QM SHO review

In units with $$m = 1$$ the non-relativistic QM SHO has the Hamiltonian
\label{eqn:qftLecture6:460}
H
= \inv{2} p^2 + \frac{\omega^2}{2} q^2.

If we define a position operator with a time-domain Fourier representation given by
\label{eqn:qftLecture6:480}
q = \inv{\sqrt{2\omega}} \lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} },

where the Fourier coefficients $$a, a^\dagger$$ are operator valued, then the momentum operator is
\label{eqn:qftLecture6:500}
p = \dot{q} =
\frac{i\omega}{\sqrt{2\omega}} \lr{ -a e^{-i \omega t} + a^\dagger e^{i \omega t} },

or inverting for $$a, a^\dagger$$
\label{eqn:qftLecture6:1080}
\begin{aligned}
a &= \sqrt{\frac{\omega}{2}} \lr{ q – \inv{i \omega} p } e^{-i \omega t} \\
a^\dagger &= \sqrt{\frac{\omega}{2}} \lr{ q + \inv{i \omega} p } e^{i \omega t}.
\end{aligned}

By inspection it is apparent that the product $$a^\dagger a$$ will be related to the Hamiltonian (i.e. a difference of squares). That product is
\label{eqn:qftLecture6:1120}
\begin{aligned}
a^\dagger a
&=
\frac{\omega}{2}
\lr{ q + \inv{i \omega} p }
\lr{ q – \inv{i \omega} p } \\
&=
\frac{\omega}{2} \lr{ q^2 + \inv{\omega^2} p^2 – \inv{i \omega} \antisymmetric{q}{p} } \\
&= \inv{2 \omega} \lr{ p^2 + \omega^2 q^2 – \omega },
\end{aligned}

or
\label{eqn:qftLecture6:1140}
H = \omega \lr{ a^\dagger a + \inv{2} }.

We can glean some of the properties of $$a, a^\dagger$$ by computing the commutator of $$p, q$$, since that has a well known value
\label{eqn:qftLecture6:520}
\begin{aligned}
i
&= \antisymmetric{q}{p} \\
&=
\frac{i\omega}{2 \omega} \antisymmetric
{ a e^{-i \omega t} + a^\dagger e^{i \omega t} }
{ -a e^{-i \omega t} + a^\dagger e^{i \omega t} } \\
&=
\frac{i}{2} \lr{
\antisymmetric{a}{a^\dagger} –
\antisymmetric{a^\dagger}{a} } \\
&=
i
\antisymmetric{a}{a^\dagger},
\end{aligned}

so
\label{eqn:qftLecture6:1100}
\antisymmetric{a}{a^\dagger} = 1.

The operator $$a^\dagger a$$ is the workhorse of the Hamiltonian and worth studying independently. In particular, assume that we have a set of states $$\ket{n}$$ that are eigenstates of $$a^\dagger a$$ with eigenvalues $$\lambda_n$$, that is
\label{eqn:qftLecture6:1160}
a^\dagger a \ket{n} = \lambda_n \ket{n}.

The action of $$a^\dagger a$$ on $$a^\dagger \ket{n}$$ is easy to compute
\label{eqn:qftLecture6:1180}
\begin{aligned}
a^\dagger a a^\dagger \ket{n}
&=
a^\dagger \lr{ a^\dagger a + 1 } \ket{n} \\
&=
\lr{ \lambda_n + 1 } a^\dagger \ket{n},
\end{aligned}

so $$\lambda_n + 1$$ is an eigenvalue of $$a^\dagger \ket{n}$$. The state $$a^\dagger \ket{n}$$ has an energy eigenstate that is one unit of energy larger than $$\ket{n}$$. For this reason we called $$a^\dagger$$ the raising (or creation) operator.
Similarly,
\label{eqn:qftLecture6:1200}
\begin{aligned}
a^\dagger a a \ket{n}
&=
\lr{ a a^\dagger – 1 } a \ket{n} \\
&=
(\lambda_n – 1) a \ket{n},
\end{aligned}

so $$\lambda_n – 1$$ is the energy eigenvalue of $$a \ket{n}$$, having one less unit of energy than $$\ket{n}$$.
We call $$a$$ the annihilation (or lowering) operator.
If we argue that there is a lowest energy state, perhaps designated as $$\ket{0}$$ then we must have
\label{eqn:qftLecture6:560}
a\ket{0} = 0,

by the assumption that there are no energy eigenstates with less energy than $$\ket{0}$$.
We can think of higher order states being constructed from the ground state from using the raising operator $$a^\dagger$$
\label{eqn:qftLecture6:580}
\ket{n} = \frac{(a^\dagger)^n}{\sqrt{n!}} \ket{0}

## Discussion

We’ve diagonalized in the Fourier representation for the momentum space fields. For every value of momentum $$\Bp$$ we have a quantum SHO.

For our field space we call our space the Fock vacuum and
\label{eqn:qftLecture6:620}
a_\Bp\ket{0} = 0,

and call $$a_\Bp$$ the “annihilation operator”, and
call $$a^\dagger_\Bp$$ the “creation operator”.
We say that $$a^\dagger_\Bp \ket{0}$$ is the creation of a state of a single particle of momentum $$\Bp$$ by $$a^\dagger_\Bp$$.

We are discarding the volume term, a procedure called “normal ordering”. We define
\label{eqn:qftLecture6:640}
\frac{a^\dagger a + a a^\dagger}{2}
:\equiv
a^\dagger a

We are essentially forgetting the vacuum energy as some sort of unobservable quantity, leaving us with the free Hamiltonian of
\label{eqn:qftLecture6:660}
H_0 =
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp}
a_{\Bp}

Consider
\label{eqn:qftLecture6:680}
\begin{aligned}
H_0
a^\dagger_\Bq \ket{0}
&=
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp}
a_{\Bp}
a^\dagger_\Bq
\ket{0} \\
&=
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp}
\lr{
a^\dagger_\Bq a_\Bp + (2 \pi)^3 \delta^3(\Bp – \Bq)
}
\ket{0} \\
&=
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp} \lr{
a^\dagger_\Bq {a_\Bp \ket{0}}
+ (2 \pi)^3 \delta^3(\Bp – \Bq) \ket{0}
} \\
&=
\omega_\Bq a^\dagger_\Bq \ket{0}.
\end{aligned}

### Question:

Is it possible to modify the Lagrangian or Hamiltonian that we start with so that this vacuum ground state is eliminated? Answer: Only by imposing super-symmetric constraints (that pairs this (Bosonic) Hamiltonian to a Fermonic system in a way that there is exact cancellation).

We will see that the momentum operator has the form
\label{eqn:qftLecture6:700}
\mathcal{P}
=
\int \frac{d^3 p}{(2 \pi)^3} \Bp a^\dagger_\Bp a_\Bp.

We say that $$a^\dagger_\Bp a^\dagger_\Bq \ket{0}$$ a two particle space with energy $$\omega_\Bp + \omega_q$$, and $$(a^\dagger_\Bp)^m (a^\dagger_\Bq)^n \ket{0} \equiv (a^\dagger_\Bp)^m \ket{0} \otimes (a^\dagger_\Bq)^n \ket{0}$$, a $$m + n$$ particle space.

There is a connection to statistical mechanics that is of interest

\label{eqn:qftLecture6:720}
\begin{aligned}
\expectation{E}
&= \inv{Z} \sum_n E_n e^{-E_n/\kB T} \\
&= \inv{Z} \sum_n \bra{n} e^{-\hat{H}/\kB T} \hat{H} \ket{n},
\end{aligned}

so for a SHO Hamiltonian system
\label{eqn:qftLecture6:740}
\begin{aligned}
\expectation{E}
&= \inv{Z} \sum_n e^{-E_n/\kB T} \bra{n} \hat{H} \ket{n} \\
&= \inv{Z} \sum_n e^{-E_n/\kB T} \bra{n} \omega a^\dagger a \ket{n} \\
&= \frac{\omega}{e^{\omega/\kB T} – 1 } \\ \\
&= \expectation{ \omega a^\dagger a }_{\kB T}
\end{aligned}

the $$\kB T$$ ensemble average energy for a SHO system. Note that this sum was evaluated by noting that $$\bra{n} a^\dagger a \ket{n} = n$$ which leaves sums of the form
\label{eqn:qftLecture6:1220}
\begin{aligned}
\frac{\sum_{n = 0}^\infty n a^n }{\sum_{n = 0}^\infty a^n}
&=
a \frac{\sum_{n = 1}^\infty n a^{n-1} }{\sum_{n = 0}^\infty a^n} \\
&=
a (1 – a) \frac{d}{da} \lr{ \inv{1 – a} } \\
&=
\frac{a}{1 – a}.
\end{aligned}

If we consider a real scalar field of mass $$m$$ we have $$\omega_\Bp = \sqrt{ \Bp^2 + m^2 }$$, but for a Maxwell field $$\BE, \BB$$ where $$m = 0$$, our dispersion relation is $$\omega_\Bp = \Abs{\Bp}$$.

We will see that for a free Maxwell field (no charges or currents) the Hamiltonian is
\label{eqn:qftLecture6:760}
H_{\text{Maxwell}} =
\sum_{i = 1}^2
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp {a^i}^\dagger_\Bp {a^i}_\Bp,

where $$i$$ is a polarization index.

We expect that we can evaluate an average such as \ref{eqn:qftLecture6:740} for our field, and operate using the analogy

\label{eqn:qftLecture6:780}
\begin{aligned}
a a^\dagger &= a^\dagger a + 1 \\
a_\Bp a_\Bp^\dagger &= a_\Bp^\dagger a_\Bp + V_3.
\end{aligned}

so if we rescale by $$\sqrt{V_3}$$
\label{eqn:qftLecture6:800}
a_\Bp = \sqrt{V_3} \tilde{a}_\Bp

Then we have commutator relations like standard QM
\label{eqn:qftLecture6:820}
\tilde{a} \tilde{a}^\dagger = \tilde{a}^\dagger \tilde{a} + 1.

So we can immediately evaluate the energy expectation for our quantized fields
\label{eqn:qftLecture6:840}
\begin{aligned}
\expectation{H_0}
&=
\expectation{
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp a_\Bp^\dagger a_\Bp
} \\
&=
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp V_3 \expectation{ \tilde{a}^\dagger_\Bp a_\Bp } \\
&=
V_3
\int \frac{d^3 p}{(2 \pi)^3} \frac{\omega_\Bp}{e^{\omega_\Bp/\kB T} – 1}.
\end{aligned}

Using this with the Maxwell field, we have a factor of two from polarization
\label{eqn:qftLecture6:860}
U^{\text{Maxwell}} = 2 V_3
\int \frac{d^3 p}{(2 \pi)^3} \frac{\Abs{\Bp}}{e^{\omega_\Bp/\kB T} – 1},

which is Planck’s law describing the blackbody energy spectrum.

## Switching gears: Symmetries.

The question is how to apply the CCR results to moving frames, which is done using Lorentz transformations. Just like we know that the exponential of the Hamiltonian (times time) represents time translations, we will examine symmetries that relate results in different frames.

### Examples.

For scalar field(s) with action
\label{eqn:qftLecture6:880}
S = \int d^d x \LL(\phi^i, \partial_\mu \phi^i).

For example, we’ve been using our massive (Boson) real scalar field with Lagrangian density
\label{eqn:qftLecture6:900}
\LL = \inv{2} \partial_\mu \phi\partial^\mu \phi – \frac{m^2}{2} \phi^2 – V(\phi).

Internal symmetry example

\label{eqn:qftLecture6:920}
H = J \sum_{\expectation{n, n’}} \BS_n \cdot \BS_{n’},

where the sum means the sum over neighbouring indexes $$n, n’$$ as sketched in

fig. 1. Neighbouring spin cells.

Such a Hamiltonian is left invariant by the transformation $$\BS_n \rightarrow -\BS_n$$ since the Hamiltonian is quadratic.

Suppose that $$\phi \rightarrow -\phi$$ is a symmetry (it leaves the Lagrangian unchanged). Example

\label{eqn:qftLecture6:940}
\phi =
\begin{bmatrix}
\phi^1 \\
\phi^2 \\
\vdots
\phi^n \\
\end{bmatrix}

the Lagrangian
\label{eqn:qftLecture6:960}
\LL = \inv{2} \partial_\mu \phi^\T \partial^\mu \phi – \frac{m^2}{2} \phi^\T \phi – V(\phi^\T \phi).

If $$O$$ is any $$n \times n$$ orthogonal matrix, then it is symmetry since
\label{eqn:qftLecture6:980}
\phi^\T \phi \rightarrow \phi^\T O^\T O \phi = \phi^\T \phi.

O(2) model, HW, problem 2. Example for complex $$\phi$$
\label{eqn:qftLecture6:1000}
\phi \rightarrow e^{i \phi} \phi,

\label{eqn:qftLecture6:1020}
\phi = \frac{\psi_1 + i \psi_2}{\sqrt{2}}

\label{eqn:qftLecture6:1040}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
\rightarrow
\begin{bmatrix}
\cos\alpha & \sin\alpha \\
-\sin\alpha & \cos\alpha
\end{bmatrix}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}

## PHY2403H Quantum Field Theory. Lecture 5: Klein-Gordon equation, Hamilton’s equations, SHOs, momentum space representation, raising and lowering operators. Taught by Prof. Erich Poppitz

### DISCLAIMER: Very rough notes from class. Some additional side notes, but otherwise barely edited.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Canonical quantization

Last time we introduced a Lagrangian density associated with the Klein-Gordon equation (with a quadratic potential coupling)
\label{eqn:qftLecture5:20}
L = \int d^3 x
\lr{
\inv{2} \lr{\partial_0 \phi}^2 – \inv{2} \lr{\spacegrad \phi}^2 – \frac{m^2}{2} \phi^2 – \frac{\lambda}{4} \phi^4
}.

This Lagrangian density was related to the action by
\label{eqn:qftLecture5:40}
S = \int dt L = \int dt d^3 x \LL,

with momentum canonically conjugate to the field $$\phi$$ defined as
\label{eqn:qftLecture5:60}
\Pi(\Bx, t) = \frac{\delta \LL}{\delta \phidot(\Bx, t) } = \PD{\phidot(\Bx, t)}{\LL}

The Hamiltonian defined as
\label{eqn:qftLecture5:80}
H = \int d^3 x \lr{ \Pi(\Bx, t) \phidot(\Bx, t) – \LL },

led to
\label{eqn:qftLecture5:680}
H
= \int d^3 x
\lr{ \inv{2} \Pi^2 + (\spacegrad \phi)^2 + \inv{2} m^2 \phi^2 + \frac{\lambda}{4} \phi^4 }.

Like the Lagrangian density, we may introduce a Hamiltonian density $$\mathcal{H}$$ as
\label{eqn:qftLecture5:100}
H = \int d^3 x \mathcal{H}(\Bx, t).

For our Klein-Gordon system, this is
\label{eqn:qftLecture5:120}
\mathcal{H}(\Bx, t) =
\inv{2} \Pi^2 + (\spacegrad \phi)^2 + \inv{2} m^2 \phi^2 + \frac{\lambda}{4} \phi^4.

### Canonical Commutation Relations (CCR)

:

We quantize the system by promoting our fields to Heisenberg-Picture (HP) operators, and imposing commutation relations
\label{eqn:qftLecture5:140}
\antisymmetric{\hat{\Pi}(\Bx, t)}{\hat{\phi}(\By, t)} = -i \delta^3 (\Bx – \By)

This is in analogy to
\label{eqn:qftLecture5:160}
\antisymmetric{\hat{p}_i}{\hat{q}_j} = -i \delta_{ij},

To choose a representation, we may map the $$\Psi$$ of QM $$\rightarrow$$ to a wave functional $$\Psi[\phi]$$
\label{eqn:qftLecture5:180}
\hat{\phi}(\By, t) \Psi[\phi] = \phi(\By, t) \Psi[\phi]

This is similar to the QM wave functions
\label{eqn:qftLecture5:200}
\begin{aligned}
\hat{q}_i \Psi(\setlr{q}) &= q_i \Psi(q) \\
\hat{p}_i \Psi(\setlr{q}) &= -i \PD{q_i}{} \Psi(p)
\end{aligned}

Our momentum operator is quantized by expressing it in terms of a variational derivative
\label{eqn:qftLecture5:220}
\hat{\Pi}(\Bx, t) = -i \frac{\delta}{\delta \phi(\Bx, t)}.

(Fixme: I’m not really sure exactly what is meant by using the variation derivative $$\delta$$ notation here), and to
quantize the Hamiltonian we just add hats, assuming that our fields are all now HP operators
\label{eqn:qftLecture5:240}
\hat{\mathcal{H}}(\Bx, t)
=
\inv{2} \hat{\Pi}^2 + (\spacegrad \hat{\phi})^2 + \inv{2} m^2 \hat{\phi}^2 + \frac{\lambda}{4} \hat{\phi}^4.

### QM SHO review

Recall the QM SHO had a Hamiltonian
\label{eqn:qftLecture5:260}
\hat{H} = \inv{2} \hat{p}^2 + \inv{2} \omega^2 \hat{q}^2,

where
\label{eqn:qftLecture5:280}
\antisymmetric{\hat{p}}{\hat{q}} = -i,

and that
HP time evolution operators $$O$$ satisfied
\label{eqn:qftLecture5:700}
\ddt{\hatO} = i \antisymmetric{\hatH}{\hatO}.

In particular
\label{eqn:qftLecture5:300}
\begin{aligned}
\ddt{\hat{p}}
&= i \antisymmetric{\hat{H}}{\hatp} \\
&= i \frac{\omega^2}{2} \antisymmetric{\hat{q}^2}{\hatp} \\
&= i \frac{\omega^2}{2} (2 i \hat{q}) \\
&= -i \omega^2 \hat{q},
\end{aligned}

and
\label{eqn:qftLecture5:320}
\begin{aligned}
\ddt{\hat{q}}
&= i \antisymmetric{\hat{H}}{\hat{q}} \\
&= i \inv{2} \antisymmetric{\hatp^2}{\hat{q}} \\
&= \frac{i}{2}(-2 i \hatp ) \\
&= \hatp.
\end{aligned}

Applying the time evolution operator twice, we find
\label{eqn:qftLecture5:340}
\frac{d^2}{dt^2}{\hat{q}}
= \ddt{\hat{p}}
= – \omega^2 \hat{q}.

We see that the Heisenberg operators obey the classical equations of motion.

Now we want to try this with the quantized QFT fields we’ve promoted to operators
\label{eqn:qftLecture5:360}
\begin{aligned}
\ddt{\hat{\Pi}}(\Bx, t)
&= i \antisymmetric{\hatH}{\hat{\Pi}(\Bx, t)} \\
&=
i \int d^3 y \inv{2} \antisymmetric{ \lr{\spacegrad \phihat(\By) }^2 }{\hat{\Pi}(\Bx) }
+
i \int d^3 y \frac{m^2}{2} \antisymmetric{ \phihat(\By)^2 }{\hat{\Pi}(\Bx) }
+
i \frac{\lambda}{4} \int d^3 \antisymmetric{ \phihat(\By)^4 }{\hat{\Pi}(\Bx) }
\end{aligned}

Starting with the non-gradient commutators, and utilizing the HP field analogues of the relations $$\antisymmetric{\hat{q}^n}{\hatp} = n i \hat{q}^{n-1}$$, we find
\label{eqn:qftLecture5:780}
\int d^3 y \antisymmetric{ \lr{ \phihat(\By) }^2 }{\hat{\Pi}(\Bx) }
=
\int d^3 y 2 i \phihat(\By) \delta^3(\Bx – \By)
= 2 i \phihat(\Bx).

\label{eqn:qftLecture5:740}
\int d^3 y \antisymmetric{ \lr{ \phihat(\By) }^4 }{\hat{\Pi}(\Bx) }
=
\int d^3 y 4 i \phihat(\By)^3 \delta^3(\Bx – \By)
= 4 i \phihat(\Bx)^3.

For the gradient commutators, we have more work. Prof Poppitz blitzed through that, just calling it integration by parts. I had trouble seeing what he was doing, so here’s a more explicit dumb expansion required to calculate the commutator
\label{eqn:qftLecture5:720}
\begin{aligned}
\int d^3 y (\spacegrad \phihat(\By))^2 \hat{\Pi}(\Bx)
&=
\int d^3 y
&=
\int d^3 y
\lr{ \spacegrad (\phihat(\By) \hat{\Pi}(\Bx)) } \\
&=
\int d^3 y
\lr{ \spacegrad (\hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By)) } \\
&=
\int d^3 y
\Biglr{
+ i
} \\
&=
\int d^3 y
\Biglr{
\spacegrad \lr{ \hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By) } \cdot \spacegrad \phihat(\By)
+ i
} \\
&=
\int d^3 y
\hat{\Pi}(\Bx)
\lr{
}
+ 2 i
\int d^3 y
&=
\int d^3 y
+
2 i
\int d^3 y

2 i
\int d^3 y
\delta^3(\Bx – \By) \spacegrad^2 \phihat(\By) \\
&=
\int d^3 y
+
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)

\end{aligned}

Here we take advantage of the fact that the derivative operators $$\spacegrad = \spacegrad_\By$$ commute with $$\hat{\Pi}(\Bx)$$, and use the identity
$$\spacegrad \cdot (a \spacegrad b) = (\spacegrad a) \cdot (\spacegrad b) + a \spacegrad^2 b$$, so the commutator is
\label{eqn:qftLecture5:800}
\begin{aligned}
&=
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)

&=

\end{aligned}

where the boundary integral is presumed to be zero (without enough justification.) All the pieces can now be put back together
\label{eqn:qftLecture5:820}
\ddt{} \hat{\Pi}(\Bx, t)
=

m^2 \phihat(\Bx, t)

\lambda \phihat^3(\Bx, t).

Now, for the $$\phihat$$ time evolution, which is much easier
\label{eqn:qftLecture5:380}
\begin{aligned}
\ddt{\hat{\phi}}(\Bx, t)
&= i \antisymmetric{\hatH}{\hat{\phi}(\Bx, t)} \\
&= i \inv{2} \int d^3 y \antisymmetric{\hat{\Pi}^2(\By)}{\hat{\phi}(\Bx)} \\
&= i \inv{2} \int d^3 y (-2 i) \hat{\Pi}(\By, t) \delta^3(\Bx – \By) \\
&= \hat{\Pi}(\Bx, t)
\end{aligned}

\label{eqn:qftLecture5:400}
\frac{d^2}{dt^2}{\hat{\phi}}(\Bx, t)
=
-m^2 \phi – \lambda \phihat^3.

That is
\label{eqn:qftLecture5:420}
\ddot{\phihat} – \spacegrad^2 \phihat + m^2 \phihat + \lambda \phihat^3 = 0,

which is the classical Euler-Lagrange equation, also obeyed by the Heisenberg operator $$\phi(\Bx, t)$$. When $$\lambda = 0$$ this is the Klein-Gordon equation.

## Momentum space representation.

Dropping hats, we now consider the momentum space representation of our operators, as determined by Fourier transform pairs
\label{eqn:qftLecture5:440}
\begin{aligned}
\phi(\Bx, t) &= \int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx} \tilde{\phi}(\Bp, t) \\
\tilde{\phi}(\Bp, t) &= \int d^3 x e^{-i \Bp \cdot \Bx} \phi(\Bx, t)
\end{aligned}

We can discover a representation of the delta function by applying these both in turn
\label{eqn:qftLecture5:480}
\tilde{\phi}(\Bp, t)
= \int d^3 x e^{-i \Bp \cdot \Bx} \int \frac{d^3 q}{(2 \pi)^3} e^{i \Bq \cdot \Bx} \tilde{\phi}(\Bq, t)

so
\label{eqn:qftLecture5:500}
\boxed{
\int d^3 x e^{i \BA \cdot \Bx} = (2 \pi)^3 \delta^3(\BA)
}

Also observe that $$\phi^\conj(\Bx, t) = \phi(\Bx, t)$$ iff $$\tilde{\phi}(\Bp, t) = \tilde{\phi}^\conj(-\Bp, t)$$.

We want the EOM for $$\tilde{\phi}(\Bp, t)$$ where the operator obeys the KG equation
\label{eqn:qftLecture5:520}
\lr{ \partial_t^2 – \spacegrad^2 + m^2 } \phi(\Bx, t) = 0

Inserting the transform relation \ref{eqn:qftLecture5:440} we get
\label{eqn:qftLecture5:540}
\int \frac{d^3 p}{(2 \pi)^3} e^{i \Bp \cdot \Bx}
\lr{
\ddot{\tilde{\phi}}(\Bp, t) + \lr{ \Bp^2 + m^2 }
\tilde{\phi}(\Bp, t)
}
= 0,

or
\label{eqn:qftLecture5:580}
\boxed{
\ddot{\tilde{\phi}}(\Bp, t) = – \omega_\Bp^2 \,\tilde{\phi}(\Bp, t),
}

where
\label{eqn:qftLecture5:560}
\omega_\Bp = \sqrt{ \Bp^2 + m^2 }.

The Fourier components of the HP operators are SHOs!

As we have SHO’s and know how to deal with these in QM, we use the same strategy, introducing raising and lowering operators
\label{eqn:qftLecture5:600}
\tilde{\phi}(\Bp, t) = \inv{\sqrt{2 \omega_\Bp}} \lr{ e^{-i \omega_\Bp t } a_\Bp + e^{i \omega_\Bp t} a^\dagger_{-\Bp}
}

Observe that
\label{eqn:qftLecture5:840}
\begin{aligned}
\tilde{\phi}^\dagger(-\Bp, t)
&= \inv{\sqrt{2 \omega_\Bp}} \lr{ e^{i \omega_\Bp t } a^\dagger_{-\Bp} + e^{-i \omega_\Bp t} a_{\Bp} } \\
&=
\tilde{\phi}(\Bp, t),
\end{aligned}

or
\label{eqn:qftLecture5:620}
\tilde{\phi}^\dagger(\Bp, t) = \tilde{\phi}(-\Bp, t),

so $$\phi(\Bp, t)$$ has a real representation in terms of $$a_\Bp$$.

We will find (Wednesday) that
\label{eqn:qftLecture5:640}
\antisymmetric{a_\Bq}{a^+_\Bp} = \delta^3(\Bp – \Bq) (2 \pi)^3.

These are equivalent to
\label{eqn:qftLecture5:660}
\antisymmetric{\hat{\Pi}(\By, t)}{\tilde{\phi}(\Bx, t)} = -i \delta^3(\Bx – \By)

## PHY2403H Quantum Field Theory. Lecture 4: Scalar action, least action principle, Euler-Lagrange equations for a field, canonical quantization. Taught by Prof. Erich Poppitz

### DISCLAIMER: Very rough notes from class. May have some additional side notes, but otherwise probably barely edited.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Principles (cont.)

• Lorentz (Poincar\’e : Lorentz and spacetime translations)
• locality
• dimensional analysis
• gauge invariance

These are the requirements for an action. We postulated an action that had the form
\label{eqn:qftLecture4:20}
\int d^d x \partial_\mu \phi \partial^\mu \phi,

called the “Kinetic term”, which mimics $$\int dt \dot{q}^2$$ that we’d see in quantum or classical mechanics. In principle there exists an infinite number of local Poincar\’e invariant terms that we can write. Examples:

• $$\partial_\mu \phi \partial^\mu \phi$$
• $$\partial_\mu \phi \partial_\nu \partial^\nu \partial^\mu \phi$$
• $$\lr{\partial_\mu \phi \partial^\mu \phi}^2$$
• $$f(\phi) \partial_\mu \phi \partial^\mu \phi$$
• $$f(\phi, \partial_\mu \phi \partial^\mu \phi)$$
• $$V(\phi)$$

It turns out that nature (i.e. three spatial dimensions and one time dimension) is described by a finite number of terms. We will now utilize dimensional analysis to determine some of the allowed forms of the action for scalar field theories in $$d = 2, 3, 4, 5$$ dimensions. Even though the real world is only $$d = 4$$, some of the $$d < 4$$ theories are relevant in condensed matter studies, and $$d = 5$$ is just for fun (but also applies to string theories.)

With $$[x] \sim \inv{M}$$ in natural units, we must define $$[\phi]$$ such that the kinetic term is dimensionless in d spacetime dimensions

\label{eqn:qftLecture4:40}
\begin{aligned}
[d^d x] &\sim \inv{M^d} \\
[\partial_\mu] &\sim M
\end{aligned}

so it must be that
\label{eqn:qftLecture4:60}
[\phi] = M^{(d-2)/2}

It will be easier to characterize the dimensionality of any given term by the power of the mass units, that is

\label{eqn:qftLecture4:80}
\begin{aligned}
[\text{mass}] &= 1 \\
[d^d x] &= -d \\
[\partial_\mu] &= 1 \\
[\phi] &= (d-2)/2 \\
[S] &= 0.
\end{aligned}

Since the action is
\label{eqn:qftLecture4:100}
S = \int d^d x \lr{ \LL(\phi, \partial_\mu \phi) },

and because action had dimensions of $$\Hbar$$, so in natural units, it must be dimensionless, the Lagrangian density dimensions must be $$[d]$$. We will abuse language in QFT and call the Lagrangian density the Lagrangian.

## $$d = 2$$

Because $$[\partial_\mu \phi \partial^\mu \phi ] = 2$$, the scalar field must be dimension zero, or in symbols
\label{eqn:qftLecture4:120}
[\phi] = 0.

This means that introducing any function $$f(\phi) = 1 + a \phi + b\phi^2 + c \phi^3 + \cdots$$ is also dimensionless, and
\label{eqn:qftLecture4:140}
[f(\phi) \partial_\mu \phi \partial^\mu \phi ] = 2,

for any $$f(\phi)$$. Another implication of this is that the a potential term in the Lagrangian $$[V(\phi)] = 0$$ needs a coupling constant of dimension 2. Letting $$\mu$$ have mass dimensions, our Lagrangian must have the form
\label{eqn:qftLecture4:160}
f(\phi) \partial_\mu \phi \partial^\mu \phi + \mu^2 V(\phi).

An infinite number of coupling constants of positive mass dimensions for $$V(\phi)$$ are also allowed. If we have higher order derivative terms, then we need to compensate for the negative mass dimensions. Example (still for $$d = 2$$).
\label{eqn:qftLecture4:180}
\LL =
f(\phi) \partial_\mu \phi \partial^\mu \phi + \mu^2 V(\phi) + \inv{{\mu’}^2}\partial_\mu \phi \partial_\nu \partial^\nu \partial^\mu \phi + \lr{ \partial_\mu \phi \partial^\mu \phi }^2 \inv{\tilde{\mu}^2}.

The last two terms, called \underline{couplings} (i.e. any non-kinetic term), are examples of terms with negative mass dimension. There is an infinite number of those in any theory in any dimension.

### Definitions

• Couplings that are dimensionless are called (classically) marginal.
• Couplings that have positive mass dimension are called (classically) relevant.
• Couplings that have negative mass dimension are called (classically) irrelevant.

In QFT we are generally interested in the couplings that are measurable at long distances for some given energy. Classically irrelevant theories are generally not interesting in $$d > 2$$, so we are very lucky that we don’t live in three dimensional space. This means that we can get away with a finite number of classically marginal and relevant couplings in 3 or 4 dimensions. This was mentioned in the Wilczek’s article referenced in the class forum [1]\footnote{There’s currently more in that article that I don’t understand than I do, so it is hard to find it terribly illuminating.}

Long distance physics in any dimension is described by the marginal and relevant couplings. The irrelevant couplings die off at low energy. In two dimensions, a priori, an infinite number of marginal and relevant couplings are possible. 2D is a bad place to live!

## $$d = 3$$

Now we have
\label{eqn:qftLecture4:200}
[\phi] = \inv{2}

so that
\label{eqn:qftLecture4:220}
[\partial_\mu \phi \partial^\mu \phi] = 3.

A 3D Lagrangian could have local terms such as
\label{eqn:qftLecture4:240}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \mu^{3/2} \phi^3 + \mu’ \phi^4
+ \lr{\mu”}{1/2} \phi^5
+ \lambda \phi^6.

where $$m, \mu, \mu”$$ all have mass dimensions, and $$\lambda$$ is dimensionless. i.e. $$m, \mu, \mu”$$ are relevant, and $$\lambda$$ marginal. We stop at the sixth power, since any power after that will be irrelevant.

## $$d = 4$$

Now we have
\label{eqn:qftLecture4:260}
[\phi] = 1

so that
\label{eqn:qftLecture4:280}
[\partial_\mu \phi \partial^\mu \phi] = 4.

In this number of dimensions $$\phi^k \partial_\mu \phi \partial^\mu$$ is an irrelevant coupling.

A 4D Lagrangian could have local terms such as
\label{eqn:qftLecture4:300}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \mu \phi^3 + \lambda \phi^4.

where $$m, \mu$$ have mass dimensions, and $$\lambda$$ is dimensionless. i.e. $$m, \mu$$ are relevant, and $$\lambda$$ is marginal.

## $$d = 5$$

Now we have
\label{eqn:qftLecture4:320}
[\phi] = \frac{3}{2},

so that
\label{eqn:qftLecture4:340}
[\partial_\mu \phi \partial^\mu \phi] = 5.

A 5D Lagrangian could have local terms such as
\label{eqn:qftLecture4:360}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \sqrt{\mu} \phi^3 + \inv{\mu’} \phi^4.

where $$m, \mu, \mu’$$ all have mass dimensions. In 5D there are no marginal couplings. Dimension 4 is the last dimension where marginal couplings exist. In condensed matter physics 4D is called the “upper critical dimension”.

From the point of view of particle physics, all the terms in the Lagrangian must be the ones that are relevant at long distances.

## Least action principle (classical field theory).

Now we want to study 4D scalar theories. We have some action
\label{eqn:qftLecture4:380}
S[\phi] = \int d^4 x \LL(\phi, \partial_\mu \phi).

Let’s keep an example such as the following in mind
\label{eqn:qftLecture4:400}
\LL = \underbrace{\inv{2} \partial_\mu \phi \partial^\mu \phi}_{\text{Kinetic term}} – \underbrace{m^2 \phi – \lambda \phi^4}_{\text{all relevant and marginal couplings}}.

The even powers can be justified by assuming there is some symmetry that kills the odd powered terms.

fig. 1. Cylindrical spacetime boundary.

We will be integrating over a space time region such as that depicted in fig. 1, where a cylindrical spatial cross section is depicted that we allow to tend towards infinity. We demand that the field is fixed on the infinite spatial boundaries. The easiest way to demand that the field dies off on the spatial boundaries, that is
\label{eqn:qftLecture4:420}
\lim_{\Abs{\Bx} \rightarrow \infty} \phi(\Bx) \rightarrow 0.

The functional $$\phi(\Bx, t)$$ that obeys the boundary condition as stated extremizes $$S[\phi]$$.

Extremizing the action means that we seek $$\phi(\Bx, t)$$
\label{eqn:qftLecture4:440}
\delta S[\phi] = 0 = S[\phi + \delta \phi] – S[\phi].

How do we compute the variation?
\label{eqn:qftLecture4:460}
\begin{aligned}
\delta S
&= \int d^d x \lr{ \LL(\phi + \delta \phi, \partial_\mu \phi + \partial_\mu \delta \phi) – \LL(\phi, \partial_\mu \phi) } \\
&= \int d^d x \lr{ \PD{\phi}{\LL} \delta \phi + \PD{(\partial_mu \phi)}{\LL} (\partial_\mu \delta \phi) } \\
&= \int d^d x \lr{ \PD{\phi}{\LL} \delta \phi
+ \partial_\mu \lr{ \PD{(\partial_mu \phi)}{\LL} \delta \phi}
– \lr{ \partial_\mu \PD{(\partial_mu \phi)}{\LL} } \delta \phi
} \\
&=
\int d^d x
\delta \phi
\lr{ \PD{\phi}{\LL}
– \partial_\mu \PD{(\partial_mu \phi)}{\LL} }
+ \int d^3 \sigma_\mu \lr{ \PD{(\partial_\mu \phi)}{\LL} \delta \phi }
\end{aligned}

If we are explicit about the boundary term, we write it as
\label{eqn:qftLecture4:480}
\int dt d^3 \Bx \partial_t \lr{ \PD{(\partial_t \phi)}{\LL} \delta \phi }
=
\int d^3 \Bx \evalrange{ \PD{(\partial_t \phi)}{\LL} \delta \phi }{t = -T}{t = T}
– \int dt d^2 \BS \cdot \lr{ \PD{(\spacegrad \phi)}{\LL} \delta \phi }.

but $$\delta \phi = 0$$ at $$t = \pm T$$ and also at the spatial boundaries of the integration region.

This leaves
\label{eqn:qftLecture4:500}
\delta S[\phi] = \int d^d x \delta \phi
\lr{ \PD{\phi}{\LL} – \partial_\mu \PD{(\partial_mu \phi)}{\LL} } = 0 \forall \delta \phi.

That is

\label{eqn:qftLecture4:540}
\boxed{
\PD{\phi}{\LL} – \partial_\mu \PD{(\partial_mu \phi)}{\LL} = 0.
}

This are the Euler-Lagrange equations for a single scalar field.

Returning to our sample scalar Lagrangian
\label{eqn:qftLecture4:560}
\LL = \inv{2} \partial_\mu \phi \partial^\mu \phi – \inv{2} m^2 \phi^2 – \frac{\lambda}{4} \phi^4.

This example is related to the Ising model which has a $$\phi \rightarrow -\phi$$ symmetry. Applying the Euler-Lagrange equations, we have
\label{eqn:qftLecture4:580}
\PD{\phi}{\LL} = -m^2 \phi – \lambda \phi^3,

and
\label{eqn:qftLecture4:600}
\begin{aligned}
\PD{(\partial_\mu \phi)}{\LL}
&=
\PD{(\partial_\mu \phi)}{} \lr{
\inv{2} \partial_\nu \phi \partial^\nu \phi } \\
&=
\inv{2} \partial^\nu \phi
\PD{(\partial_\mu \phi)}{}
\partial_\nu \phi
+
\inv{2} \partial_\nu \phi
\PD{(\partial_\mu \phi)}{}
\partial_\alpha \phi g^{\nu\alpha} \\
&=
\inv{2} \partial^\mu \phi
+
\inv{2} \partial_\nu \phi g^{\nu\mu} \\
&=
\partial^\mu \phi
\end{aligned}

so we have
\label{eqn:qftLecture4:620}
\begin{aligned}
0
&=
\PD{\phi}{\LL} -\partial_\mu
\PD{(\partial_\mu \phi)}{\LL} \\
&=
-m^2 \phi – \lambda \phi^3 – \partial_\mu \partial^\mu \phi.
\end{aligned}

For $$\lambda = 0$$, the free field theory limit, this is just
\label{eqn:qftLecture4:640}
\partial_\mu \partial^\mu \phi + m^2 \phi = 0.

Written out from the observer frame, this is
\label{eqn:qftLecture4:660}
(\partial_t)^2 \phi – \spacegrad^2 \phi + m^2 \phi = 0.

With a non-zero mass term
\label{eqn:qftLecture4:680}
\lr{ \partial_t^2 – \spacegrad^2 + m^2 } \phi = 0,

is called the Klein-Gordan equation.

If we also had $$m = 0$$ we’d have
\label{eqn:qftLecture4:700}
\lr{ \partial_t^2 – \spacegrad^2 } \phi = 0,

which is the wave equation (for a massless free field). This is also called the D’Alembert equation, which is familiar from electromagnetism where we have
\label{eqn:qftLecture4:720}
\begin{aligned}
\lr{ \partial_t^2 – \spacegrad^2 } \BE &= 0 \\
\lr{ \partial_t^2 – \spacegrad^2 } \BB &= 0,
\end{aligned}

in a source free region.

## Canonical quantization.

\label{eqn:qftLecture4:740}
\LL = \inv{2} \dot{q} – \frac{\omega^2}{2} q^2

This has solution $$\ddot{q} = – \omega^2 q$$.

Let
\label{eqn:qftLecture4:760}
p = \PD{\dot{q}}{\LL} = \dot{q}

\label{eqn:qftLecture4:780}
H(p,q) = \evalbar{p \dot{q} – \LL}{\dot{q}(p, q)}
= p p – \inv{2} p^2 + \frac{\omega^2}{2} q^2 = \frac{p^2}{2} + \frac{\omega^2}{2} q^2

In QM we quantize by mapping Poisson brackets to commutators.
\label{eqn:qftLecture4:800}
\antisymmetric{\hatp}{\hat{q}} = -i

One way to represent is to say that states are $$\Psi(\hat{q})$$, a wave function, $$\hat{q}$$ acts by $$q$$
\label{eqn:qftLecture4:820}
\hat{q} \Psi = q \Psi(q)

With
\label{eqn:qftLecture4:840}
\hatp = -i \PD{q}{},

so
\label{eqn:qftLecture4:860}
\antisymmetric{ -i \PD{q}{} } { q} = -i

Let’s introduce an explicit space time split. We’ll write
\label{eqn:qftLecture4:880}
L = \int d^3 x \lr{
\inv{2} (\partial_0 \phi(\Bx, t))^2 – \inv{2} \lr{ \spacegrad \phi(\Bx, t) }^2 – \frac{m^2}{2} \phi
},

so that the action is
\label{eqn:qftLecture4:900}
S = \int dt L.

The dynamical variables are $$\phi(\Bx)$$. We define
\label{eqn:qftLecture4:920}
\begin{aligned}
\pi(\Bx, t) = \frac{\delta L}{\delta (\partial_0 \phi(\Bx, t))}
&=
\partial_0 \phi(\Bx, t) \\
&=
\dot{\phi}(\Bx, t),
\end{aligned}

called the canonical momentum, or the momentum conjugate to $$\phi(\Bx, t)$$. Why $$\delta$$? Has to do with an implicit Dirac function to eliminate the integral?

\label{eqn:qftLecture4:940}
\begin{aligned}
H
&= \int d^3 x \evalbar{\lr{ \pi(\bar{\Bx}, t) \dot{\phi}(\bar{\Bx}, t) – L }}{\dot{\phi}(\bar{\Bx}, t) = \pi(x, t) } \\
&= \int d^3 x \lr{ (\pi(\Bx, t))^2 – \inv{2} (\pi(\Bx, t))^2 + \inv{2} (\spacegrad \phi)^2 + \frac{m}{2} \phi^2 },
\end{aligned}

or
\label{eqn:qftLecture4:960}
H
= \int d^3 x \lr{ \inv{2} (\pi(\Bx, t))^2 + \inv{2} (\spacegrad \phi(\Bx, t))^2 + \frac{m}{2} (\phi(\Bx, t))^2 }

In analogy to the momentum, position commutator in QM
\label{eqn:qftLecture4:1000}
\antisymmetric{\hat{p}_i}{\hat{q}_j} = -i \delta_{ij},

we “quantize” the scalar field theory by promoting $$\pi, \phi$$ to operators and insisting that they also obey a commutator relationship
\label{eqn:qftLecture4:980}
\antisymmetric{\pi(\Bx, t)}{\phi(\By, t)} = -i \delta^3(\Bx – \By).

# References

[1] Frank Wilczek. Fundamental constants. arXiv preprint arXiv:0708.4361, 2007. URL https://arxiv.org/abs/0708.4361.


text:
CalGraph Bob's Calculus

## The Cauchy-Schwartz Inequality

What is it?

The Cauchy-Schwartz  Inequality  |<U,W>|      ||U||∙ ||W||      is an example of a lemma.

"Lemma" is Latin for "unimportant theorem".

( Not exactly:  a lemma is a result which is needed to prove a more general or harder theorem .)

Where did it come from?

We can visualize it and explore its origin at the same time.

Let's start with the assumption that the shortest distance between two points on a flat (See Note 1) surface is a straight line.

In the diagram, it's a shorter walk from O to B, than from O to A and then from A to B.

(By the way, the diagram was generated using Calgraph, a Graphing/Statistics/Calculus Calculator created by Kishore Anand. You can download it at http://math-interactive.com/download.html and run it on any Windows laptop or desktop.)

Let's call the vector joining O to A:  U = (u1  , u2).

Similarly, after reaching the point A, we walk parallel to the vector  W = (w1 , w2).                             (Line 1)

If we walked directly from  O to  B, we would be at the point with coordinates:  (u1 + w1 , u2 + w2).

So, translating our statement about the shortest distance between two points being a straight line into vector algebra, ||U + W|| ||U|| + ||W||

Since all the quantities are non-negative, the preceding inequality is still true if we square it. (The reason to square it is to avoid having to deal with square roots.)

|| U + W ||2   ( ||U|| + ||W||)2

||(u1 + w1 , u2 + w2)||2  ||U||2+ ||W||2  +  2||U|| ∙||W||

(u1 + w1)2+( u2 + w2)2  ||( u1 ,  u2)||2+|| (w1  , w2)||2 + 2 ||U||∙ ||W||

+ + 2 u1 w1  +  +  +2 u2 w2   ||( u1 , u2)||2+|| (w1  , w2)||2 + 2||U||∙ ||W||

Rearranging the terms on the left side:

(+ ) + (  +) + 2 u1 w1 + 2 u2 w2   ||( u1 , u2)||2+|| (w1 , w2)||2 +2 ||U||∙ ||W||

On the left,  the terms in parentheses  are  exactly ||( u1 , u2)||2 and ||(w1 , w2)||2 respectively. Subtracting them from both sides and then dividing everything by 2 we have:

u1 w1+ u2 w2    ||U|| ∙||W||

But the left is exactly the dot or  inner product < U, W > of the two vectors  U and W , so we have:

|< U, W >|    ||U|| ∙||W||                                                                                (Line 2)

(If you are unsatisfied, see Note 2.)

Who needs it?

The following  three applications can be proved in one of two ways:  directly or by showing that the relations  live in a normed vector space, so that they derive from the general Cauchy-Schwartz Inequality.   However, either type of proof demands a knowledge of subjects traditionally taught after Vector Algebra. (Quantum Mechanics,  Integration and Infinite Series).

1   Physics: Arguably, the most important tenet of Quantum Mechanics is a direct consequence of the Cauchy-Schwartz Inequality:

The Heisenberg Uncertainty Principle: There is a fundamental limit to the precision to which certain pairs of physical quantities of a particle can simultaneously be measured.

e.g.  A particle's position x and momentum p:

∆x ∙∆p

For example, the more accurately the position is measured (∆x small), the greater will be the uncertainty in the momentum (∆p), as the product of the two errors must be at least  .

An elegant, fairly simple proof on the Uncertainty Principle appears here:

http://www.phys.ufl.edu/courses/phy4604/fall18/uncertaintyproof.pdf

2  If f and g are integrable functiona over domain E: (f* is the complex conjugate of f)

|∫E  f*g dx|      ( E  | f |2 dx)1/2 ( E  |g|2 dx)1/2

3  (

Taking the limits of both sides as  →∞, this inequality can be used in convergence proofs.

(

Here's the skeleton of a proof by Mathematical Induction:

(The n=1 case is trivial and the Algebra for the n=2 case is instructive for the Induction Step.)

We will need: 0  =  +  - 2    2     +             (Line 3)

(=  =  +2

+ = (  + (  + =

(The inequality comes from replacing x by  and y by  in  (Line 3).

The Induction Step can be accomplished with analogous algebra to the last few lines.

How do you know it's valid?

The general proof is constructive, in other words,  a cute trick.

U and W are vectors in an Inner Product Space over a field F.    (See Note 3)

Consider  U- α W (where α ε  F)

We will assume that W 0 as that case is trivial.

(Line 4)

0 ≤   < U- α W, U- α W >     (true for the inner product of any vector with itself, equality only for 0)

Expanding the left hand side:

<U, U- α W > -  α <W, U- α W > = <U, U> - α *<U, W> - α <W, U> + α∙ α* <W, W>

Set  α  =

(No worries about a 0 denominator as W was assumed to be a non-zero vector.)

0      < U- α W, U- α W >  =

<U, U> <U, W> -  <W, U> + <W, W>                                       (Line 4)

Simplifying the preceding line:

0        <U, U> -

A little Algebra and the realization that   and  <U, U> =

yields

or simply:  |<U, W>|   ||U||||W||                                                                                ( Line 5)

Incidentally, tracing back to close to the beginning of the proof  ( Line 4), we have equality if and only if  U = α W,  so equality occurs only when  U and W are linear multiples.

i.e. In  or  equality occurs in the Cauchy-Schwartz Inequality (Line 5) only if U and W are parallel vectors.

Notes:

1) By normal, flat 2-dimensional space, we mean the surface of a piece of paper as opposed to the curved 2-dimensional surface of, for example,  a sphere or a bagel.

In real two dimensional real space, the inner product is the same as the dot product of two vectors.

So if U = (u1  , u2) and W = (w1 , w2)

<U, W> = U  W = (u1  , u2)  ∙ (w1 , w2)  = 2 u1 w1  + 2 u2 w2

2) The eagle-eyed reader might have noticed a hole in the proof.  The absolute value in Line 2 wasn't justified. However, if, after Line 1,  point B had been taken as (u1 -  w1 , u2 -  w2), then  proceeding analogously,  the left side of Inequality I would have had been multiplied by -1 . Thus,  the absolute value at Line 2  can be justified.

3) Definition of an Inner Product Space.

An inner product space is a vector space V over a field F (F = R or C) together with an operation <. , .>

<. , .>: VxV -> F

(The idea is that the inner product is a function with two vectors as the input and a single real or complex number as the output.)

·         Conjugate symmetry:

<U, W> = <U, W>C

·        Linearity in the first argument:

< αU,w> = α <U,W>

<U + V,W> = <U,W> + <V,W>

·        Positive Definite:  <U,U>   0   with equality only if U  is the 0 vector.

4) Sometimes, Mathematicians develop abstract systems for their own amusement and are surprised or even miffed when their work proves useful. For example, Group Theory was unknown outside pure Mathematics from its inception circa 1855 until 1920 when Weyl rolled it into Quantum Mechanics. Inner product spaces are generalizations of our familiar R2 and, as one can see from the applications of just this Inequality, have proved to be extremely useful.


text:
Slanted numbers in running header using Times

How do I get slanted numbers in the running page header when using Times? For CM they are slanted. TIA.

\documentclass{book}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{newtxtext,newtxmath}
\begin{document}
\section{Introduction}
Some text
\end{document}

Is changing the font to another Times variant an option? If so, I have a couple of alternatives for you.

Caveat

Thanks to campa's excellent comments, I have learned that tgtermes uses italics for the header, not a slanted variant (since there apparently is none).

If that doesn't bother you, have a look at the following example:

\documentclass{book}
\usepackage[T1]{fontenc}

% Loading tgtermes and newtxmath (in that order!)
% works by inserting italics instead of slanted text
% \usepackage{tgtermes}
% \usepackage{newtxmath}

% txfonts also works
\usepackage{txfonts}

\begin{document}
\section{Introduction}
Some text

$E=mc^2$

\end{document}

• Sorry, I have in part misunderstood the answer: you meant loading tgtermes and newtxmath together, right? But the problem is the same: tgtermes has no slanted font, so it is replacing it with the italic one. Look at the log, you'll find the corresponding warning. So yes, it seems to work, but one might question whether this is good. I'd rather patch the book class to use italic text in headings. But that might be a matter of opinion. – campa Aug 11 '20 at 12:53
• @campa, yes, you are right, of course! So loading txfonts would be the only preferred package? – phil-elkabat Aug 11 '20 at 12:54
• Also, if you want to use tgterms and newtxmath you should load tgtermes first. Look at the numbers in math mode otherwise. – campa Aug 11 '20 at 12:54
• txfonts is a possibility, or the old mathptmx. Or you can cheat: tex.stackexchange.com/a/248981/82917 – campa Aug 11 '20 at 12:55
• @campa, excellent, thank you! I will edit my answer to make it transparent that some font choices will replace the slanted font with true italics. – phil-elkabat Aug 11 '20 at 12:56


text: Microsoft Word Symbols List Codes; Microsoft Word Symbol For Section; Microsoft Word Symbol Check Mark; By default, the Symbol dialog box shows emoji at first when it's opened. equation, which looks a lot like a graphic in your document. stream Complete table â¦ All the shortcuts are checked on Microsoft Word 2010/2013/2016/Office 365 on Windows 7/8/10 platforms, hence some of the codes may not work on other software like Excel and PowerPoint. Download icons in all formats or edit them for your designs. Excel Insert Character experience has been, too, that every time I try to edit the fonts, I mess it Available symbol sets. Assigning shortcut keys to symbols. you shouldn't miss if you use Word a lot. Try to live with them the way they are. choose Microsoft Equation 3.0. The most important thing to understand when inserting symbols, fractions, special characters, or international characters is that the font you use is critical. Quick access symbols. To insert a Microsoft Word symbol, follow these steps: 1. If you use the symbol a lot, you may Your symbol may not be large enough or bold enough to be easily visible in Word. What are Paragraph Symbols or Marks in Word. To create another, you can copy the This is one of the easiest methods. Also press space after typing every closing parenthesis ")", which will adjust both the opening and closing parentheses size to fit the group's contents. In Microsoft word, you are not limited to the default hotkeys. Full list of Alt codes. ��h宇�4-R�����c謹��+��X2G ｻ�h慟����Pc�-_BG�ng*Nr庫��}#�����+�2� \�}貊�2)��e�d�� あ�Jh�tGX�X I will show you how to add clickable and non-clickable checkmarks to your list of options in Microsoft Word. dialog created by Jon Peltier, Microsoft Excel MVP and owner of Reveal Codes 3 in Word - Is there life after âReveal Codesâ? chevrons, opening and closing Microsoft term, seldom used, especially in documentation. To use it, hit Start Run Symbol AltCode Symbol AltCode Symbol AltCode â 190: ²: 253: Â: 0194 â 191 254: Ã: 0195 â 192: spc: 255: Ä: 0196 â´ 193 â¬ 0128: Å: 0197 â¬ 194 â 0130: Æ: 0198 â 195: Æ: 0131: Ç: 0199 â 196 â 0132: È: 0200 â¼ 197 â¦ 0133: É: 0201 â 198 â  0134: Ê: 0202 â 199 â¡ 0135: Ë: 0203 â 200: Ë: 0136: Ì: 0204 â 201 â° 0137: Í: 0205 â© 202: Å : 0138: Î: 0206 â¦ 203 â¹ 0139: Ï: 0207 â  204: Å: 0140: Ð: 0208 â 205: Å½: â¦ Begin by typing all the options in a list. If you can't find your symbol or character using the Symbol Microsoft Word includes two types of special characters: (1) text characters that donât appear on standard keyboards and (2) formatting characters for text layout. Then I explain how to get summation and integration, how to put one thing above another, and, finally, how to make fractions, for MS-Word. Because the 1/2 fraction is is â¦ You can create an It won't work using the regular keypad. One of the most commonly looked for symbols here is the Paragraph return, also called a Pilcrow. Word's Symbol dialog also has a Special Characters tab that you shouldn't miss if you use Word a lot. To see the Symbol menu in Microsoft Word, go to Insert > Symbols on the ribbon and click the Symbol button (or Insert > Advanced Symbol > Symbols in the menu system in Word for Mac). 2 0 obj This page is intended to supply a list of some useful symbols separated by topic so they can be found quickly without the need to search in the Unicode reference tables. Without any further ado, letâs get started. To see the Symbol menu in Microsoft Word, go to Insert > Symbols on the ribbon and click the Symbol button (or Insert > Advanced Symbol > Symbols in the menu system in Word for Mac). The following mathematical symbol sets are available in the Symbols group in Word. How to Create a List of Abbreviations in Microsoft Word. want to memorize the shortcut key for it, or create a shortcut key or Alt key codes give you access to hundreds of special ASCII characters, from accent marks to tiny icons. Ho to insert Non-Clickable checkmark in Word. If you special symbols showing up at the end of each paragraph, they are called formatting symbols (also called formatting marks). Word 2013 and Word 2010 offer similar special character options. Contains both 1 - 256 and 0128 - 0256 code ranges. endobj This is Word's Symbol dialog. As mentioned, each and every character or symbol in Microsoft Word has a character code that you can use on the keyboard. AltGr Combination Keys and the Euro Symbol. © 2005 - 2020 by MrExcel Publishing. nature, or for creating Arithmetic problems, use the Equation Editor. Access®, Excel®, FrontPage®, Outlook®, PowerPoint®, Word® are registered trademarks of the Microsoft Corporation. To access the Equation Editor, hit Insert Object, We show you how to use the Alt codes and also Word's Symbol library. â¦ The Formulas, Functions and Visual Basic procedures on this web site are provided However, you can also use a combination of Microsoft Word and Excel to quickly generate a list of abbreviations. For example, the number 4 key often has a Euro symbol (â¬), while the <> key (next to the number 1 key on the 2ndrow) also has the ¦ symbol. MS-Word File with Mathematical Symbols First I give a list of symbols for both MS-Word and Powerpoint. <> AutoCorrect entry for it. Unicode/List of useful symbols. Special for Excel. Microsoft Word Symbols with the Symbol Dialog Box You can use the Symbol dialog box to insert any symbol of your choice. Microsoft Word 2013 Symbols 10. �-b]�9F�c�1F辿�����&2���濡\$�葫����\l� � ��b93y0 蟶�a�+�0��bh�n��凡 D'��#9z�F癒@X�#��)��^���j��b��5J2l� �K-��R��霰�檸jr C%7&���B7��隠|雎�RbR?B�;��G/-!�� �a�S�u芙v粨c r4殷U8柧FQ���4�w�w�憖�KS��倨��,kh卞z�4���&ｚ��Ux@S��L>�T虫0fv&�xd）Z����]�m��xe���_=O4� 散/t∞琲��G��Z���G��X��]�'s9V徴��(輪簀��Yr筐�p礬�賂���;艶y�g;�D畸j%院�w#q��.�拱jnV -w'v&�なP欠哈隠H�S+�Q��S+跟����_衡妊�M. Microsoft Word has many features that format documents for optimal readability. Don't Todayâs post explains all of the special characters in Microsoft Word for Office 365, Word 2019, and Word 2016. and type "charmap" (without the quotes) and hit Enter. This is maybe easier, since you can see which symbol you want, point-and-click to insert it. Numbers you can type on your keyboard's Num Pad to get special symbols. Get free icons of Microsoft word in iOS, Material, Windows and other design styles for web, mobile, and graphic design projects. In this case you can just copy and paste the symbols. Thanks very much, John! All rights reserved. 4 0 obj For Powerpoint, the latter things donât seem to work. You can insert a special character or symbol in your document in a couple of ways: The Symbol [â¦] @�GBξ�b��� .��3���vm@�UY怛�惆eYk� t�比�TKl檳戉Ns}�k�r�eR�[時�觝�y��9弖I�f慨効xoiT羚���簸Gw拌蟹堪邪�\寥SYYW�l�\��'K��2����筆���島������C9Ol�]��8(B逐賦�i��%{�-������2�hp出�>cHu;�顱>�膜6}Ls&�Xc��'���g{測��1F>��8�0�.��H悄��篋���FH����M+�1��X�a�咬]R感zf[}一��~l�qｄ]ay32}5�-V0[�o��Y!.C� Once you dialog, try the Character Map, which comes with Windows. numbers on the number keypad. What are they? For instance, the expression below can be obtained with (1/2(x+1)): ( 1 2 ( x + 1 ) ) {\displaystyle \left({\frac {1}{2}}(x+1)\right)} Be careful to press space after the "2" to render the fraction, otherwise Word might put "x+1" in the denominator. - http://sbarnhill.mvps.org/WordFAQs/RevealCodes.htm. characters into their documents, so I hope to offer every method here. This brings up the toolbar. This will open a quick access menu of 20 frequently used symbols to pick from. My x��Z[���~���z4[!�uR�J&�lvgN�;N�!�篤6����?�-��09�J�c�壙�松Ip�鴎�U襖搏舸/�:Yl�%�|=/w_�嶸�C歌�の歩���P礇瀲d�V/^�W�o惓�綸���E吟甥褞�3��煕p)�b���|筑��l戎�p�?�^^|�1�o6��鰲 微vy驥�J This will open a quick access menu of 20 frequently used symbols to pick from. type the last number, your symbol or character should appear. 3 0 obj These are provided by the Segoe Color Emoji font, although some applications continue to show black and white emojis from an earlier release.. Support for Emoji 13.0 is coming soon.Now availalable for Windows Insider channels. These symbols include "(), {}, [], ||". Jump to navigation Jump to search. "as is" and we do not guarantee that they can be used in all situations. Microsoft. http://www.nexstara.com IT Consultants show users how to use the Symbol Dialog Box and Autocorrect and Shortcuts features with Symbols in Microsoft Word 2010. List of Formatting Symbols in Microsoft Word Sometimes I see special symbols are showing up at the end of each paragraph in my Microsoft Word document. %亀亀 To edit the Symbols and characters | Fast symbol and emoticon search with convenient auto-lookup This Office Add-in simplifies the use of special characters like the German Umlauts (äüöÜÄÖß), diacritic characters (diacritical marks) or other symbols like check marks, arrows, stars, boxes, emoticons, mathematical symbols, currency symbols and many more. expect to jump right in and become an expert. It seems a lot of people can't find how to get certain symbols or <>/ExtGState<>/XObject<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/Annots[ 17 0 R 20 0 R 22 0 R 24 0 R 25 0 R 27 0 R 28 0 R 30 0 R 33 0 R 34 0 R 36 0 R 38 0 R 39 0 R 41 0 R] /MediaBox[ 0 0 595.32 842.04] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>> But you can expand it to see other character sets: Scroll to the top and click the toggle button in the right corner: Reveal Codes2- Checking Formatting - Word's Reveal Codes http://www.addbalance.com/usersguide/basic_formatting.htm#RevealCodes. For example, Word provides foreign language letters and symbols â all sorts of fun stuff. Word 2010 lets you sprinkle characters beyond the keyboardâs 26 letters of the alphabet, numbers, a smattering of symbols, and punctuation thingies. Word for Microsoft 365 Word 2019 Word 2016 Word 2013 Word 2010 Word Starter 2010 More... Less. Excel lovers can download a special Both the above methods may show keystrokes to use, such as the Word's Symbol dialog also has a Special Characters tab that one above displays: To use this keystroke, hold your Alt key while typing the The free images are pixel perfect to fit your design and available in both png and vector. You can assign any keyboard shortcut key to any command or symbol to fit your needs. Your keyboard may not have many characters available but you want to use those characters in your document; in such situations, you have the option to insert Special Symbols the way we will further understand in this chapter.. To insert symbols that are occasionally used, follow the steps in this section. Most modern keyboards show more than one character on some keys. These features include bullets, numbered lists, page breaks, margins, columns, and more. Microsoft Word includes several options for formatting a document, one of them, as we said, is to â¦ In this chapter, we will discuss the use of special symbols in Word 2010. From Wikibooks, open books for an open world < Unicode. endobj <>>> OfficeArticles.com provides examples of Formulas, Functions and Visual Basic procedures for illustration only, It takes a bit of time. first one, and paste it, then double-click it to change it. Hereâs Insert | Symbol in Word 2002 (XP) that looks little different from Word 2013âs version of the same feature. without warranty either expressed or implied, including but not limited to the implied warranties of merchantability The Symbol font contains Times New Roman Greek capitals and lowercase, figures and basic punctuation together with a collection of mathematical signs and general purpose Pi characters. and/or fitness for a particular purpose. Quick access symbols. After clicking the More arrow, click the menu at the top of the symbols list to see each grouping of symbols. Grouping symbols will automatically size to the appropriate size. equation later, just double-click it. One of the most commonly looked for Insert your symbol into a Microsoft Word document. <> Our venture into the new Unicode symbols and symbol fonts highlighted how much work Microsoft needs to do on this part of Office. If you want to place the symbol in a Word document, click the symbol to select it, click Select, click Copy, and then open the Microsoft Word document and paste in the symbol by pressing Ctrl+V.. OfficeArticles.com debuted on May 26, 2005. PeltierTech. %PDF-1.5 This is an excellent article describing all of the native Word tools for discovering formatting. You can create a list of abbreviations manually by going through your work and noting each one down, then adding them all to a list at the start of your document. Emojis from Microsoft are supported on PCs and tablets running Windows 10. You can easily insert a special character, fraction, or other symbol in your Word documents. symbols here is the Paragraph return, also called a Pilcrow. But things can get tricky when trying to make a document look a particular way. up even more. Obey the following instructions to add this symbol to a list of options in Microsoft Word. The symbols that Microsoft now calls âformatting marksâ were previously referred to as ânonprinting characters.â By either name, they are characters that take up space or have a formatting function but do not appear on the printed page: spaces, tab characters, paragraph breaks, and the like. 1 0 obj Insert | Symbol has changed for a long time and many versions of Office. endobj For even more symbols, usually of a scientific or mathematic 03/11/2020; 2 minutes to read; a; P; W; m; v; In this article Overview. Symbol font family. Emojis from Microsoft are supported on PCs and tablets running Windows 10 and type  charmap '' ( without quotes... Can also use a combination of Microsoft Word symbol, follow these steps:.. One character on some keys the end of each Paragraph, they are called formatting marks ) used! Code that you can see which symbol you want, point-and-click to insert a special Characters in Microsoft Word,! Tab that you can also use a combination of Microsoft Word and to! In your document character Map, which comes with Windows a combination of Microsoft Word to see grouping. Quotes ) and hit Enter First I give a list of options in Microsoft Word Paragraph! For both ms-word and Powerpoint format documents for optimal readability case you can see which symbol you want point-and-click. The Microsoft Corporation lists, page breaks, margins, columns, and paste the symbols codes http: #! Mathematical symbol sets are available in the symbols group in Word or mathematic nature, or creating... End of each Paragraph, they are called formatting symbols ( also formatting. Word provides foreign language letters and symbols â all sorts of fun stuff each and every character or symbol fit... Life after âReveal Codesâ may not be large enough or bold enough to be easily visible in Word - there... Starter 2010 more... Less the last number, your symbol may not be large enough or bold to! Of Office especially in documentation download icons in all formats or edit them for your designs easier, since can. Download a special character, fraction, or for creating Arithmetic problems, use the codes..., each and every character or symbol to fit your needs too, that every time try! Default hotkeys should appear following Mathematical symbol sets are available in the symbols in..., just double-click it to change it 2019, and paste the symbols group in 2002. || '' a combination of Microsoft Word to use the Equation Editor mathematic. If you use Word a lot, open books for an open world < Unicode case you can copy First. Characters tab that you should n't miss if you use Word a.... Is maybe easier, since you can copy the First one, and paste the list. Excel MVP and owner of PeltierTech of abbreviations tablets running Windows 10 and an! And non-clickable checkmarks to your list of symbols is an excellent article describing of! Grouping of symbols for both ms-word and Powerpoint, we will discuss the use of special showing... End of each Paragraph, they are Microsoft Corporation on the keyboard will discuss use... Start Run and type  charmap '' ( without the quotes ) and hit.... First one, and more for discovering formatting is an excellent article describing all the. Highlighted how much work Microsoft needs to do on this part of Office options in Microsoft Word a... Way they are called formatting marks ) or other symbol in Microsoft Word your needs even more or symbol Microsoft! To add this symbol to fit your design and available in the symbols group in Word describing. The alt codes and also Word 's symbol dialog, try the character,... Language letters and symbols â all sorts of fun stuff, point-and-click to insert Microsoft., hit Start Run and type  charmap '' ( without the ). | symbol in Microsoft Word symbol, follow these steps: 1 one of the same.... Generate a list copy and paste it, hit Start Run and type  charmap '' ( the. You should n't miss if you ca n't find your symbol may not large! Especially in documentation and symbols â all sorts of fun stuff and paste it, hit insert,... I mess it up even more symbols, usually of a scientific or mathematic nature, or symbol... Is the Paragraph return, also called a Pilcrow n't expect to jump right in and become an expert want... Excel to quickly generate a list of options in Microsoft Word for Office 365, Word foreign! To read ; a ; P ; W ; m ; v ; in case. Showing up at the end of each Paragraph, they are called formatting symbols ( also called a Pilcrow special. Has been, too, that every time I try to edit the Editor... Documents for optimal readability Word tools for discovering formatting a character code that you can type your... Can assign any keyboard shortcut key to any command or symbol in Word - is there life âReveal. This part of Office Word 2016 Word 2013 and Word 2010 Word Starter 2010 more... Less the free are... Them the way they are it, hit Start Run and type ` charmap '' ( without the quotes and. Can download a special Characters in Microsoft Word has many features that format documents for optimal.! Trying to make a document look a particular way Word Starter 2010 more....... After âReveal Codesâ the keyboard later, just double-click it to change it in. Key codes give you access to hundreds of special ASCII Characters, from accent marks tiny! 2010 more... Less alt key codes give you access to hundreds of special ASCII,! From Microsoft are supported on PCs and tablets running Windows 10 download icons in all or..., click the menu at the top of the symbols list to see each grouping of for... Which comes with Windows for your designs ; W ; m ; v ; in this case can. Microsoft Corporation the native Word tools for discovering formatting Word 2013âs version the! I mess it up even more should appear access to hundreds of special.! Maybe easier, since you can assign any keyboard shortcut key to any command or symbol in Word... A graphic in your document the latter things donât seem to work by! Opening and closing Microsoft term, seldom used, especially in documentation I will show you how to clickable. Visible in Word 2010 Word Starter 2010 more... Less tablets running Windows 10 2002 ( XP ) that little... To tiny icons not be large enough or bold enough to be easily visible in 2010. Look a particular way quickly generate a list ; 2 minutes to read ; a ; ;... Paragraph return, also called a Pilcrow breaks, margins, columns, and paste it, insert! Make a document look a particular way trying to make a document look a particular way or for Arithmetic! From Microsoft are supported on PCs and tablets running Windows 10 is an excellent article describing all of the Characters... A list typing all the options in a list of options in a list hundreds! Enough or bold enough to be easily visible in Word that you should n't miss if you use Word lot! To create another, you are not limited to the appropriate size abbreviations...