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 Question from amy, a parent: My son was given the following question:  The path of a large arrow fired from a non torsion catapult can be modeled by y=-0.0044x^2 + 1.68x, where x is the distance the arrow travelled (in yards) and y is the height of the arrow (in yards).  Given the height of a castle wall, find the safest distance from the wall to launch an arrow over the wall. The given height of the castle wall is 120yards. We are stumped by this.  I have started with 120=-0.0044x^2 + 1.68x But that does not compute to a regular x value. I have asked multiple math types who have given me this same formula.  I can only assume that we and they are not understanding how to do this. Any help would be most appreciated!

Amy, we have two responses for you:

Hi Amy,

I'd try finding the maximum value of the height of the arrow (that is the maximum value of y -- you can use calculus or recognize that the function given is a parabola). If my quick calculation is right, the arrow reaches a maximum height of a little over 160 yards.

The value of x at which this occurs would be the distance away from the wall to fire the arrow from so that it reaches the top of its arc just as it crosses the wall.

I don't know if this is the "safest" distance to fire from, but I would guess that is what is being asked.

I hope this helps.

Victoria

Hi Amy.

I'm not sure why you say that these don't give a "regular x value": in fact you will get two x values corresponding to where the arrow is 120 yards up. One value (the smaller one) is where the arrow is still ascending and the other is where the arrow is descending. I would guess that the safest distance to launch your arrow from is the furthest from the wall in which the arrow can still cross over the wall.

Re-arrange your equation to make it equal zero:

0.0044x2 - 1.68x + 120 = 0

And use the Quadratic Formula to find the two values of x:

 a = 0.0044, b = -1.68, c = 120


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# Simplicial model for $\Sigma^{\infty}_+ \Omega^{\infty}X$?

Let $Sp^{\Sigma}$ be the category of symmetric spectra, and let $Sp^{\Sigma}-alg$ be the associated category of augmented commutative algebras. These are simplicial categories.

A question for experts: Can anyone give me a reference for, or explain a construction of, a simplicial functor $$Sp^{\Sigma} \rightarrow Sp^{\Sigma}-alg$$ which looks like the functor $X \mapsto \Sigma^{\infty}_+ \Omega^{\infty}X$ in the homotopy category?

Nick Kuhn

• Nick, why are you adding "+"? Isn't the relevant functor given by $X \mapsto \Sigma^\infty \Omega^\infty X$? – John Klein Jan 26 '18 at 13:50
• John: To have the suspension spectrum of an $E_{\infty}$ space, or more simply, a topological monoid, be a ring spectrum one needs to add a basepoint. (Think about it.) The output is then, in fact, a ring spectrum augmented over the sphere spectrum. By the way, I have answers to my question obtained in more traditional ways. – Nicholas Kuhn Jan 28 '18 at 18:29
• Oh yeah, I should have realized that....Well, so much for MO on the flu. – John Klein Jan 30 '18 at 1:40

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# [V] denotes the greatest integer less than or equal to 'V'. 'V' is a p

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16 Jun 2019, 16:02
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[V] denotes the greatest integer less than or equal to 'V'. 'V' is a positive integer and $$[\frac{V}{5}]-[\frac{V}{7}]=1$$.
If the minimum value of 'V' is 'a' and the maximum value of 'V' is 'b'. What is the value of (a+b)?

A. 33
B. 34
C. 35
D. 40
E. 42
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[V] denotes the greatest integer less than or equal to 'V'. 'V' is a p  [#permalink]

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18 Jun 2019, 23:37
5
V is a positive integer, therefore $$[\frac{V}{5}]−[\frac{V}{7}]=1$$ at minimum value of V should be $$1-0$$

Therefore V should be at least 5, $$a=5$$

Now looking at the answer choices, the maximum value for V should be among 28, 29, 30, 35 and 37.

When V=28, $$[\frac{V}{5}]−[\frac{V}{7}]=5-4=1$$

When V=29, $$[\frac{V}{5}]−[\frac{V}{7}]=5-4=1$$

When V=30, $$[\frac{V}{5}]−[\frac{V}{7}]=6-4=2$$

We can conclude here that V can be maximum 29 and so $$b=29$$ but just to check,

When V=35, $$[\frac{V}{5}]−[\frac{V}{7}]=7-5=2$$

When V=37, $$[\frac{V}{5}]−[\frac{V}{7}]=7-5=2$$

Therefore, $$a+b=5+29=34$$

Hit Kudos if this helped!
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Re: [V] denotes the greatest integer less than or equal to 'V'. 'V' is a p  [#permalink]

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16 Jun 2019, 23:08
nick1816 wrote:
[V] denotes the greatest integer less than or equal to 'V'. 'V' is a positive integer and $$[\frac{V}{5}]-[\frac{V}{7}]=1$$.
If the minimum value of 'V' is 'a' and the maximum value of 'V' is 'b'. What is the value of (a+b)?

A. 33
B. 34
C. 35
D. 40
E. 42

$$[\frac{V}{5}]-[\frac{V}{7}]=1$$

Minimum
When $$\frac{V}{5}$$ just crosses 0.5, while $$\frac{V}{7}$$ remains below 0.5..
This means V can be 3, SO A=3

Maximum
$$[\frac{V}{5}]-[\frac{V}{7}]<2$$, as we can have the difference just below 2, but the answer still as 1, for example 4.5 and 6.4999.
Here 4.5 will become 5, while 6.4999 will come down to 6 and answer will be 1.
$$[\frac{V}{5}]-[\frac{V}{7}]<2............[V]\frac{1}{5}]-\frac{1}{7}<2..........[V]<35$$

Let us check for the values below it..
(1) 34.....$$[\frac{34}{5}]-[\frac{34}{7}]=1........[6.8]-[4.9]=7-5..NO$$
(2) 33.....$$[\frac{33}{5}]-[\frac{33}{7}]=1........[6.6]-[4.7]=7-5..NO$$
(3) 32.....$$[\frac{32}{5}]-[\frac{32}{7}]=1........[6.4]-[4.6]=6-5=1..YES$$
SO B=32

A+B=3+32=35

c
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Re: [V] denotes the greatest integer less than or equal to 'V'. 'V' is a p  [#permalink]

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05 Jul 2019, 23:17
chetan2u wrote:
nick1816 wrote:
[V] denotes the greatest integer less than or equal to 'V'. 'V' is a positive integer and $$[\frac{V}{5}]-[\frac{V}{7}]=1$$.
If the minimum value of 'V' is 'a' and the maximum value of 'V' is 'b'. What is the value of (a+b)?

A. 33
B. 34
C. 35
D. 40
E. 42

$$[\frac{V}{5}]-[\frac{V}{7}]=1$$

Minimum
When $$\frac{V}{5}$$ just crosses 0.5, while $$\frac{V}{7}$$ remains below 0.5..
This means V can be 3, SO A=3

Maximum
$$[\frac{V}{5}]-[\frac{V}{7}]<2$$, as we can have the difference just below 2, but the answer still as 1, for example 4.5 and 6.4999.
Here 4.5 will become 5, while 6.4999 will come down to 6 and answer will be 1.
$$[\frac{V}{5}]-[\frac{V}{7}]<2............[V]\frac{1}{5}]-\frac{1}{7}<2..........[V]<35$$

Let us check for the values below it..
(1) 34.....$$[\frac{34}{5}]-[\frac{34}{7}]=1........[6.8]-[4.9]=7-5..NO$$
(2) 33.....$$[\frac{33}{5}]-[\frac{33}{7}]=1........[6.6]-[4.7]=7-5..NO$$
(3) 32.....$$[\frac{32}{5}]-[\frac{32}{7}]=1........[6.4]-[4.6]=6-5=1..YES$$
SO B=32

A+B=3+32=35

c

chetan2u

Shouldn't [v] be converted to the next largest integer as per the question. So why are we following the rounding rule of largest if >=.5 and smallest if <0.5.
So if [v]=0.45 then shouldn't it be converted to 1 rather than 0 as it is the largest integer? Sorry if am asking a stupid question, but this one striked without any plausible explanation that proves the other way...!!!

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Re: [V] denotes the greatest integer less than or equal to 'V'. 'V' is a p   [#permalink] 05 Jul 2019, 23:17
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The spatial wave functions for a single particle in a spherically symmetric po-tential can be expressed in spherical polar coordinates in the form √nlm(r, µ, ')=Rnl(r)Ylm(µ, '), (1. ) 0:ψ 0 (x)=A 0 e−x2a2 1:ψ 1 A 1 xe−x2a2 A. Introduction. Enhanced IR dynamics induces nonperturbative physics, such as superconductivity and Kondo effect. March 13, 2013 Quantum anharmonic oscillator. March 10, 2013 Langevin small approximation. a) Find the energy levels for this particle. In this tutorial we study the electron energy levels of a two-dimensional parabolic confinement potential that is subject to a magnetic field. 2630 Abstract. Prove that the allowed energy eigen. Prob 5: Use the results from. Physics 2D Lecture Slides Lecture 27: Mar 8th Vivek Sharma UCSD Physics. In the previous chapter we studied stationary problems in which the system is best described as a (time-independent). CHAPTER 2 THE HARMONIC OSCILLATOR Introduct ion The accidental degeneracy and symmetries of the harmonic oscillators has been discussed by several authors. Berkeley Physics Preliminary Exam Review Problems Kevin Grosvenor August 28, 2011. 1 The transition probability P i!n is given by the time-dependent population of the state n, as all initial population resides in the state i. com - id: 12a6c3-MDcxM. Cerveró Department de Fsica Fundamental , Facultad de Ciencias, Universidad de Salamanca, Salamanca, Spain Abstract In this paper we solve exactly the problem of the spectrum and Feynman propagator of a charged particle submitted to both an anharmonic oscillator. degeneracy of a 2D harmonic oscillator results in a particular freedom of choice of the zero-order approximation for the perturbation series. Prob 4: For the 2D Harmonic oscillator with potential V(rho) = (1/2) M omega 2 rho 2 (a) Show that separation of variables in cartesian coordinates turns this into two one-dimensional oscillators, and exploit your knowledge of the latter to determine the allowed energies. Enriquez, Ph. In quantum mechanics, there is the phenomenon of accidental degeneracy when the energy eigenvalues are multiply degenerate. For the harmonic oscillator, these integrals are zero, i. Find the 3 lowest allowed energies. Linear Molecules and Normal Modes 2. The atom-atom interaction is modelled by means of a finite-range Gaussian interaction. This means that the phase space asquires 2D-lattice structure with the basic unit-cell vectors φ1 = (p1 , q1 ) and φ2 = (p2 , q2 ) obeying (54), i. Chapter 8 The Simple Harmonic Oscillator A winter rose. 3: Degeneracy (not including spin) of the lowest 10 energy levels in a quantum well, a quantum wire with square cross-section and a quantum cube with. Hence in 2D the spectrum becomes *i,j =ℏ0 0(i + j +1), where i and j range from 0 to ∞. The Dirac oscillator was initially introduced as a Dirac operator which is linear in momentum and coordinate variables. Browse By Category. Figure 1(a) shows one example of a harmonic oscillator, where a body of mass mis. 2 An array of N 1D simple harmonic oscillators is set up with an average energy per oscillator of (m+ 1 2)~!. Because an arbitrary smooth potential can usually be approximated as a harmonic potential at the vicinity of a stable equilibrium point, it is one of the most important model systems in quantum mechanics. (b)The set of rational. For high energies, the physical nature of the states is. Lifting of the Landau level degeneracy in 2D electron gas by point impurities Two-dimensional (2D) electron systems are realized on interfaces of two condensed media. Enhanced IR dynamics induces nonperturbative physics, such as superconductivity and Kondo effect. 3D Symmetric HO in Spherical Coordinates *. Inspired by old ideas from non-critical string theory & c = 1 Matrix Quantum Mechanics, we reformulate the scattering in terms of a quantum mechanical model — of waves scattering off inverted harmonic oscillator potentials — that exactly reproduces the unitary black hole S-Matrix. By consider1ng 0 = e x 2=2 nd what n is. (a) Please write down the Schrodinger equation in x and y, then solve it using the separation of variables to derive the energy spectrum. 781 m to m c 1. That is, the potential is zero inside a cube of side L and infinite outside the cube. We have already solved the problem of a 3D harmonic oscillator by separation of variables in Cartesian coordinates. Solve this problem entirely at T= 0 K. The | ’n ’s are eigenstates of H’: We therefore have: In general, Use this to make problems easier Sample Problem Complete the square: Now shift the potential: H’ is Harmonic oscillator plus energy shift Same energies, but shifted eigenstates: What are the eigenvalues and eigenstates of A rotation is a linear mapping of the position. Enriquez, Ph. Aussois, 6 octobre 2009. 6 o ers an opportunity to demonstrate the critical relationship between symmetry and degeneracy. H  y=ly 2 Il2 y=ly 2 I hbar2 lHl+1Ly=ly. The thermal rate constant of the 3D OH + H2→H2O + H reaction was computed by using the flux autocorrelation function, with a time-independent square-integrable basis set. (Russian) Sibirsk. Klauder described coherent states. The "harmonic oscillator" sometimes means a different thing, not about finite-dimensional repns of the Lie algebra $\mathfrak{sl}_2(\mathbb C)$, but about infinite-dimensional ones. What is the energy of the state with quantum numbers n x, n y? Take the spring constant to be equal in both direction,. This open question is becoming increasingly relevant as the state of the art develops to the point where such modes can be resolved [21] and utilized [22]. Note that for a ﬁxed , the energy levels En,, increase by 2"vrather than "v. Particle in a 1D well has none, particle in a 2D square well has g=2, rigid rotor has g=2J+1, 1D harmonic oscillator has none Ionization Energy Energy needed to take an electron from the ground state (n=1) to unbound state (n=∞), He+ 4x greater than H, H-like atoms with larger Z bind electrons more strongly, not related to n. Sen: Eigenspectrum properties of the confined 3D harmonic oscillator, Journal of Physics B: Atomic, Molecular and Optical Physics 41, 225002 1-6 (2008) (M21) Lj. 2) The two-dimensional harmonic oscillator with un-perturbed Hamiltonian 𝐻𝐻0= 𝑝𝑝𝑥𝑥 2 2𝑚𝑚 + 𝑝𝑝𝑦𝑦 2𝑚𝑚 + 𝐾𝐾 2 (𝑥𝑥2+ 𝑦𝑦2). 2; a particle of mass m in this potential oscillates with frequency ω. 3 Pauli exclusion principle. However the vast majority of systems in Nature cannot be solved exactly, and we need. number lcan take values 0;1;:::;n 1, as total of n. Enhanced IR dynamics induces nonperturbative physics, such as superconductivity and Kondo effect. Using the fact that the eld-free eigenstates are normalized, we obtain P i!n= jhc n(t) njc n(t) nij= jc n(t)j2: (1) The coe cients c. Eigenvalues, eigenfunctions, position expectation values, radial densities in low and high-lying states are presented in case of small, intermediate. Prob 4: For the 2D Harmonic oscillator with potential V(rho) = (1/2) M omega 2 rho 2 (a) Show that separation of variables in cartesian coordinates turns this into two one-dimensional oscillators, and exploit your knowledge of the latter to determine the allowed energies. This will be a recurring theme of the second semester QM, so it is worth seeing it in action within a simpler. Quiz 8 0 2 4 6 8 10 12 14 16 - 1D Harmonic Oscillator Æ3D Harmonic Oscillator • Keep an eye on the number Energy Spectrum & Degeneracy 12 3 22 22 2 n,n ,n 1 2 3 i 22 111 2 22 211 121 112 2 2 3 Ground State Energy E 2 6. International Journal of Theoretical Physics, Vol. 3810 23J=K AvogadronumberN A 6. Energy eigenstate degeneracy is critical to the phenomenology of our world, including semiconductors, biology, chemistry etc. Klauder described coherent states. The ability of cucurbit[6]uril (CB6) and cucurbit[7]uril (CB7) to catalyze the thermally activated 1,2-methyl shift isomerization pathway of m-xylene in vacuum is investigated using infrequent metadynamics. The harmonic oscillator is treated in Chapter 2, both in the two and in the n dimensional case. Spherical confinement in 3D harmonic, quartic and other higher oscillators of even order is studied. Here I discuss about Degeneracy and Hermonic oscillator problems in quantum mechanics also you can see https://youtu. We ﬁnd that the energies can be well ﬁtted by the expression a TFE TF+ mod N,2 where E TF is the Thomas-Fermi. In real molecules the harmonic oscillator approximation. of this group and its. What are the energies and degeneracies of the lowest three energy levels? ii. In the center of the applet, you will see the probability distribution of the particle's position. ( ) ( ) ( ) or my t ky t cy t Fnet FH FF && =− − & = +. Diatomic molecules have vibrational energy levels which are evenly spaced, just as expected for a harmonic oscillator. For example, if our spring constants are all the same: k = k = k ≡ k ⇒ ω = ω = ω ≡ ω x y z x y z ⎛ 3 ⎞. Hi! I'm trying to calculate the degeneracy of each state for 3D harmonic oscillator. The Dirac oscillator was initially introduced as a Dirac operator which is linear in momentum and coordinate variables. 7 mg to g General Chemistry - Standalone book (MindTap Course List) A solution is prepared by mixing 13. Both involve degenerate perturbation theory. consider the case of a 2D lattice where the electron with mass m is subject to infinite square well potentials in the x and y directions, and a harmonic potential in the z direction with force constant k. Michael Fowler, University of Virginia. If we measure the energy and ﬁnd it to be 2¯hω, then the state could be |nx = 1,ny = 0i or |nx = 0,ny = 1i or any linear combination. The generalized pseudospectral method is employed for accurate solution of relevant Schrödinger equation in an optimum, non-uniform radial grid. ) gives the equation m x =−kx or x +ω2x=0, where ω=k/m is the angular frequency of sinusoidal os-cillations. This means that the phase space asquires 2D-lattice structure with the basic unit-cell vectors φ1 = (p1 , q1 ) and φ2 = (p2 , q2 ) obeying (54), i. It is the number of microstates of system 1 with energy E 1, also known as 1(E 1) = e S 1(E 1)=k B: Notice that it depends on E 1. The simplest model is a mass sliding backwards and forwards on a frictionless surface, attached to a fixed wall by a spring, the rest position defined by the natural length of the spring. However, when one combines these two fundamental models and considers interacting fermions in a 1D harmonic potential, there is no known solution in general ( 1 ). In the strongly interacting regime, we show how the many-body wave function develops. Lj Stevanović 1 and K D Sen 2. View Ravindra Shinde’s profile on LinkedIn, the world's largest professional community. , 2006 Dirac points energy bands graphene: isotropic Dirac gas isotropic Dirac gas Landau levels of graphene Hofstadter-Rammal butterfly honeycomb lattice in a magnetic field low field 1 5 3 -1 -3 -5 1 5 3 -1 -3 -5 Valley degeneracy 2-fold degeneracy of LL anisotropic Dirac gas energy bands moving Dirac points Landau levels of. Chapter 3 deals with a general Hamiltonian which may be interpreted as that of a charged mass point in a plane harmonic oscillator potential and uniform magnetic field. Harmonic oscillator in 3-dimensions, eigenvalues, eigenfunctions, degeneracy. 1 Time-dependent perturbation treatment of the harmonic oscillator 1. E(v) = (v + ½) e This is usually a fairly good approximation near the bottom of the potential well, where the potential closely resembles that of a harmonic oscillator. Using the fact that the eld-free eigenstates are normalized, we obtain P i!n= jhc n(t) njc n(t) nij= jc n(t)j2: (1) The coe cients c. The One-Dimensional Damped Forced Harmonic Oscillator Revisited. Conversely, the system is quasi-2D if higher states in the tightly con ned direction are also. HARMONIC OSCILLATOR IN 2-D AND 3-D, AND IN POLAR AND SPHERICAL COORDINATES3 In two dimensions, the analysis is pretty much the same. ϕ A =1,0 2. 3 The Harmonic Oscillator Potential. Degeneracies of the 2D Harmonic Oscillator 3. 2D harmonic oscillator in Cartesian coordinates to deduce the formula for the energy levels and their degeneracy of the hydrogen atom. Non-relativistic - 2D in semiconductors Landau gauge. Ateneo de Manila University CH 47. PERTURBATION THEORY 17. Use the v=0 and v=1 harmonic oscillator wavefunctions given below which are normalized such that ⌡⌠-∞ +∞ Ψ (x) 2dx = 1. Koleksi Contoh Makalah. The wave-functions shown above are for the ground state and 1st excited state, respectively, of an electron in a harmonic oscillator potential. Quantum s = 2 and a valley degeneracy of g v = 1. Recently the accidental degeneracy of a two-dimensional (2-D) harmonic oscillator with frequency 0a0 plus an interaction proportional to the z-th projection of the angular momentum was studied [9]. Eigenstates are labelled by occupancy of 1&2 modes E(n 1. In two dimensions and the lowest Landau level, hidden symmetries control the interaction of the interacting system with light. 3 Time-Reversal Symmetry 100. #potentialg #quantummechanics #csirnetjrfphysics In this video we will discuss about 2D and 3D Harmonic Oscillator and Degeneracy in Quantum Mechanics. Vibration in the presence of rotation. Hence the allowed energy levels are 2D: *m =ℏ0. The significance of equations 26 and 32 is that we know exactly which energies correspond to which excited state of the harmonic oscillator. We want to prove this that L z and L 2 commute with each other: ˆˆ,02 z ªº¬¼LL. In fact, one can show that for bound states in 1D, one never has degeneracy; every state has its own energy. be/ZKUlzgQC_Ug https://youtu. ring – 2D Square Well in Chapter 8 – Coulomb Potential • Homework Set #11 is available per our vote last Friday – it is due Dec. 1) Suppose we put a delta-function bump in the center of the infinite square well: where α is a constant. Vibrational motion (harmonic oscillator) 5. Quantum correlations and degeneracy of identical bosons in a 2D harmonic trap. harmonic oscillator the ground-state energy is 0 Example: average energy of the quantum harmonic oscillator Continuous Spectrum, the Density of States The units of g(!): (energy)-1(length)-" where " is the dimensionality of the volume: 1D – length, 2D – area, 3D – volume. If the rotational energy levels are lying very close to one another, we can integrate similar to what we did for q trans above. By using the Wang-Uhlenbeck method. Since the energy is the integral of motion in this case the trajectory of the electron in the momentum space is a circumference of the radius P = √ 2mEtoo. Stern-Gerlach experiment, spin, Zeeman effect. The generalized pseudospectral method is employed for accurate solution of relevant Schrödinger equation in an optimum, non-uniform radial grid. 1 Bound problems. Quantum wires and quantum dots. Ultimately the source of degeneracy is symmetry in the potential. Potential energy of just 1 2D electron in a B-field m=0,-1,-2,…-10. a) Find the energy levels for this particle. The potential energy is zero everywhere in this plane, and infinite at its walls and beyond. The number of quantum states at the same energy level is called the degree of degeneracy. Two-Electron WMs. Schematic diagram of a Josephson junction connected to a bias voltage V. Schrödinger equation (SEQ) in 2D and 3D Systems: The 3D Quantum Box and 2D Harmonic Oscillator; Degeneracy (read G. Using exact diagonalisation, the lowest monopole mode frequency is shown to depend non-monotonically on the interaction strength, having a minimum in a crossover region. 15) Note that the states with N = 2 are 3-fold degenerate, with eigenstates given by (nx,n , ) = (2, 0), (1, l) , and (0, 2). – 1D Harmonic Oscillator Æ3D Harmonic Oscillator • Keep an eye on the number of different integers needed to specify system 1Æ3 (corresponding to 3 available degrees of freedom x,y,z) y z x ri=+ˆˆxjy+kˆz G. The molecular Hamiltonian 11. Laboratoire de Physique des Solides, Orsay Cargèse, july 9, 2009. The degeneracy associated with the isotropic harmonic oscillators is partially lifted so that states of a given harmonic oscillator level are no longer degenerate with respect to ℓwhile each set of states associated with a particular orbital angular momentum ℓis split as a doublet of degenerate. It occurs in the harmonic oscillator system. March 12, 2013 Grand canonical ensemble. Eigenvalues, eigenfunctions, position expectation values, radial densities in low and high-lying states are presented in case of small, intermediate. Eigenvalues, eigenfunctions, position expectation values, radial densities in low and high-lying states are presented in case of small, intermediate. (a) Show that [H;H x] = 0. In two dimensions and the lowest Landau level, hidden symmetries control the interaction of the interacting system with light. The translational modes’ degeneracy splits at λc ≈ 0. d In the isotropic case \u03c9 1 \u03c92 \u03c9 3 what is the degeneracy of the energy levels. You can see this from the following argument - suppose y is an eigenfunc-tion of H with eigenvalue l. Mathematical equivalence in 2D between rotation (harmonic oscillator) and perpendicular magnetic ﬁeld: A = m (y,x) ⌅ B ⇤⇧⇥A =2mzˆ H = L z = xp y yp x 1 2m p 2 x + p 2 y ⇥ + 1 2 m (x + y ) = 1 2m (p A)2 + L z Eigenstates: Landau levels in the effective 'magnetic' ﬁeld B. 9, September 2006 ( 2006) DOI: 10. 2 Harmonic oscillator: one dimension The harmonic oscillator potential is 2 U(x)=1kx2, familiar to us from classical mechanics where Newton's second law applied to a harmonic oscillator potential (spring, pendulum, etc. State properties of the spherical harmonics and their relation to angular momentum. The first ex-cited wave function for the simple harmonic oscillator, Express your result in terms of the classical amplitude of oscillation A. If energy EJ is degenerate with (2 J + 1) states corresponding to it, then, the Boltzmann factor e - E , / k Trot j B has to be multiplied by (2J+ 1) to account for all these states. (b) Determine the degeneracy d(n) of E n. The above equation is usual 1D harmonic oscillator, with energy eigenvalues E0= n+ 1 2 ~!. It can be proved generally that for a harmonic oscillator = (n+ 1 2) h p k and that = (n2 + 2n+ 1) 3 h2 4 k: Verify these formulas for the rst two states of the harmonic oscillator. tures with 2D electron motion within each subband [4]. Partition Function Problems And Solutions. ; Barone, F. The potential of the harmonic oscillator is included into equation by making , where is a constant. Prob 4: For the 2D Harmonic oscillator with potential V(rho) = (1/2) M omega 2 rho 2 (a) Show that separation of variables in cartesian coordinates turns this into two one-dimensional oscillators, and exploit your knowledge of the latter to determine the allowed energies. Hi! I'm trying to calculate the degeneracy of each state for 3D harmonic oscillator. Quantum Rabi model and its analytical solution. We considered the problem of degenerate perturbation theory. Concept of degeneracy in 2D BOX and case of different levels. Spherical confinement in 3D harmonic, quartic and other higher oscillators of even order is studied. In more than one dimension, there are several different types of Hooke's law forces that can arise. The quantum mechanical oscillator can be solved exactly and has even and odd eigenfunctions fn(x) with energy ω !(n + 1/2), n = 0,1,2; n = 0 is a Gaussian, the bell-shaped curve. Koleksi Contoh Makalah. If we measure the energy and ﬁnd it to be 2¯hω, then the state could be |nx = 1,ny = 0i or |nx = 0,ny = 1i or any linear combination. Note that for a ﬁxed , the energy levels En,, increase by 2"vrather than "v. March 13, 2013 Quantum anharmonic oscillator. Sen: Eigenspectrum properties of the confined 3D harmonic oscillator, Journal of Physics B: Atomic, Molecular and Optical Physics 41, 225002 1-6 (2008) (M21) Lj. 1These particles are equivalent to the quanta of the harmonic oscillator, which have energy En = (n + 1 2)~ω, where ~ = h/2π, and h is Planck’s constant. 6 o ers an opportunity to demonstrate the critical relationship between symmetry and degeneracy. The ground-state wavefunction for a particle in the harmonic oscillator potential has the form ψ(x)=Aexp(-ax. Center of mass and relative coordinates. In this paper we give a general solution to the problem of the damped harmonic oscillator under the influence of an arbitrary time-dependent external force. Fragmentation of spin-orbit-coupled spinor Bose-Einstein condensates Shu-Wei Song, 1 ,2 3 Yi-Cai Zhang, 4 Hong Zhao, 1 ,2 3 Xuan Wang, and Wu-Ming Liu 1 State Key Laboratory Breeding Base of Dielectrics Engineering, Harbin University of Science and Technology, 150080 Harbin, China. Even for 2D and 3D systems, we have different degeneracies. There are inﬁnitely many states that satisfy this equation. It occurs in the harmonic oscillator system. been done, the effect of mode degeneracy has not yet been considered. For d→0nm, the magnetic field vanishes and hence a pure 2D harmonic oscillator is retained. We also found, through our exploration via the Lz matrix, that the two states 01 +i 10 and 01-i 10 also span this 1st excited state degeneracy manifold and, moreover, that they are Lz eigenstates with e. Two-level systems 4. H  y=ly 2 Il2 y=ly 2 I hbar2 lHl+1Ly=ly. Klauder described coherent states of the. the harmonic oscillator shell model has many symmetries and provides a useful basis for the expression of collective models with a microscopic interpretation. Derive a formula for the degeneracy of the quantum state n, for spinless particles confined in this potential. 2D Quantum Harmonic Oscillator. The degeneracy in each levels increase as the number of wells is increased. In this tutorial we study the electron energy levels of a two-dimensional parabolic confinement potential that is subject to a magnetic field. Because it is a exactly solvable model, several investigations have been developed in the context of. Piéchon, J. consider the case of a 2D lattice where the electron with mass m is subject to infinite square well potentials in the x and y directions, and a harmonic potential in the z direction with force constant k. The separation between the wells determine the magnitude of their interaction. Symmetry & Degeneracy (Dana Longcope 8/24/06) The problem of the the two-dimensional harmonic oscillator treated by Libo in x8. This paper investigates physiological responses to perceptions of unfair pay. Landau levels •One obtains •This is a 1d simple harmonic oscillator with a frequency and center c = eB c cyclotron frequency magnetic length = r ~ eB 1 2m ~2 d2 dy2 +(eB)2 y ~kx eB 2! Y = Y y 0 = ~kx eB = kx2. Here we have neglected the spin of the electron, but if we include it by neglecting its all possible e ects, the degeneracy. Consider the case of a two-dimensional harmonic oscillator with the following Hamiltonian: which may be equivalently expressed in terms of the annihilation and creation operators For your reference. If we consider a particle in a 2 dimensional harmonic oscillator potential with Hamiltonian H = p2 2m + mw2r2 2 it can be shown that the energy levels are given by Enx, ny = ℏω(nx + ny + 1) = ℏω(n + 1) where n = nx + ny. Discretization and Matrix Diagonalization 4. The electron can be considered to be localized close to the origin with hard wall potential barriers at x = y =+/-a. Eigenvalues, eigenfunctions, position expectation values, radial densities in low and high-lying states are presented in case of small, intermediate. Operator solution 3. (15 points) Fermions in a two-level or three-level system with degeneracy. There are inﬁnitely many states that satisfy this equation. 1) The Harmonic Oscillator: Classical vs. Things change as we go from 1D to 2 or 3 dimensions. Solution to Statistical Physics Exam 29th June 2015 Name StudentNumber Problem1 Problem2 Problem3 Problem4 Total Percentage Mark Usefulconstants GasconstantR 8. Problem 4: Harmonic Oscillator [30 pts] Consider a 3D harmonic oscillator, described by the potential V(x,y,z)= 1 2 m!2(x2+y2+z2). In the case of the particle in a rigid, cubical box, the next-lowest energy level is three-fold degenerate. gate physics solution , csir net jrf physics. s13d It can be seen that the shift in l, in the eigenvalues of K0sa,d, due to «, is also multiplied by the factor 2"v. 3 Infinite Square-Well Potential 6. 2 Exchange operator. Radial probability distributions. 2 Charged Particle in a Magnetic Field 2. 1) Suppose we put a delta-function bump in the center of the infinite square well: where α is a constant. The two dimensional Harmonic Oscillator has the Hamiltonian n, n'>denotes the state In> of the x-oscillator and In'> of the y-oscillator. Figure 1(a) shows one example of a harmonic oscillator, where a body of mass mis. Find the 3 lowest allowed energies. 148 LECTURE 17. Bounames Received October 25, 2005; Accepted March 13, 2006 Published Online: June 27, 2006 We use the Lewis-Riesenfeld theory to determine the exact. The 3-d harmonic oscillator can be solved in rectangular coordinates by separation of variables. However in higher dimension harmonic oscillators do show degeneracy. 4 sin( 2 Pi f t ) the fee is for this reason y'(t) = a million. other than the harmonic oscillator, wher e the 2D oscillator is the simplest example of a degenerate 2D spectrum, and the next simplest example would be the particle in a 2D box. 1) The Harmonic Oscillator: Classical vs. Degeneracy is a big part of that. Furthermore, it is one of the few quantum-mechanical systems for which an exact. be/tAUwt3Uw. Consider an isotropic 2D harmonic oscillator whose Hamiltonian is given by H = h^2 nabla^2/2m + 1/2 m omega^2 (x^2 + y^2). Separation of variables: 𝜓=𝑋𝑥𝑌𝑦 −ℏ22𝑚d2𝑋d𝑥2=𝐸𝑥𝑋 −ℏ22𝑚d2𝑌d𝑦2=𝐸𝑥𝑌 𝑚 𝑦=𝐿2 𝑦=0 𝑥=0 𝑥=𝐿1. be/ZKUlzgQC_Ug https://youtu. (remember the lattice structure?) For large number of wells, it begins to show band structure. The 2D harmonic oscillator eigenfunctions are products of the of the 1D harmonic oscillator eigenfunctions. Harmonic oscillator in 3-dimensions, eigenvalues, eigenfunctions, degeneracy. Harmonic Oscillator Assuming there are no other forces acting on the system we have what is known as a Harmonic Oscillator or also known as the Spring-Mass-Dashpot. The quantum particle in the 1D box problem can be expanded to consider a particle within a higher dimensions as demonstrated elsewhere for a quantum particle in a 2D box. How can a rose bloom in December? Amazing but true, there it is, a yellow winter rose. For d→0nm, the magnetic field vanishes and hence a pure 2D harmonic oscillator is retained. The simple harmonic oscillator, a nonrelativistic particle in a potential 1 2 k x 2, is a system with wide application in both classical and quantum physics. Quantum degeneracy just means that more than one quantum states have exactly the same energy. Assuming that the two. Sen: Eigenspectrum properties of the confined 3D harmonic oscillator, Journal of Physics B: Atomic, Molecular and Optical Physics 41, 225002 1-6 (2008) (M21) Lj. 1 - Harmonic Oscillator J10Q. So, we look for ˆ ˆ ˆ ˆ 2 2 2 2 x y z L L L L. a) Expectation value of position of harmonic oscillator: hxi= Z+1 1 v (x)x v(x)dx= 1 2vv! ˇ 1=2 Z+1 1 H2 v x p xexp x2 dx Because. Superconductivity occurs no matter how weak the attraction is. Calculate the probability that a ClO molecule treated as a harmonic oscillator will be found at a classically forbidden extension or compression when v=3. Find the shift in the ground state energy of a 3D harmonic oscillator due to relativistic correction to the kinetic energy. NASA Technical Reports Server (NTRS) Isar, Aurelian. Isotropic: Having a physical property which has the same value when measured in di erent directions. The harmonic oscillator creation and destruction operators are deﬁned in terms of the position and momentum operators, aˆ = r mω 2~ xˆ+i r 1 2mω~ pˆ. 1 Bosons, fermions. The positions of the harmonic oscillator potentials y n in the y-direction are given by the wave numbers k x that satisfy the boundary condition. The Parabolic Potential Well. 2D is N5 1 2 K21 1 2 K. be/ZKUlzgQC_Ug https://youtu. We demonstrate that an undamped few-body precursor of the Higgs mode can be investigated in a harmonically trapped Fermi gas. With these degeneracies, the total number of modes is unchanged with respect to a simple harmonic oscillator potential. Akhmerova, The asymptotics of the spectrum of nonsmooth perturba-tions of a harmonic oscillator. The different terms in the Hamiltonian are the kinetic energy, a harmonic-oscillator potential with frequency $$\omega$$ (which is a first-order approximation to the nuclear mean field), the quadratic orbital and spin–orbit terms, and the residual two-nucleon interaction. If we consider the bond between them to be approximately harmonic, then there is a Hooke's law force between. σ = rotational degeneracy of given configuration Do for both reactants and transition state Yields correction to rate constant that is equal to the reaction path. PERTURBATION THEORY 17. Because an arbitrary potential can usually be approximated as a harmonic potential at the vicinity of a stable equilibrium point, it is one of the most important model systems in quantum mechanics. Quantum Rabi model and its analytical solution. Dissipation destroys the phase space described by the momentum anomaly. 2DEG quantization in Landau gauge t 2. ϕ A =1,0 2. BCS theory. Quantum numbers, energy, excitation energy number, degeneracy, and number of states with m r = 0 for the low-energy levels of a system of two noninteracting identical bosons trapped in a 2D isotropic harmonic potential. Consider a 2-dimensional harmonic oscillator, which has a Hamiltonian given by 2m This can be thought of as a superposition of two I-D oscillators, with energy eigenvalues given by where N —n +11 (1. First excited state: 2-fold degenerate, j0 j1 and j1 j0 , E1 = 2ℏ!. Harmonic oscillator 168 2 + 10 + 18 + 26 112 6 14 + 22 2 + 10 + 18 6 + 14 2 10 3s, 2d, 2p, 2s, Figure 5. Variational method and perturbation method 8. In the classical approximation, the energy is equal to 1 kT, as shown by the magenta line. 1 Classical Particle in B-ﬂeld Want to remind ourselves how to write classical Hamiltonian describing charged particle moving in E and B ﬂelds. Piéchon, J. In general, the degeneracy of a 3D isotropic. W ork has been. It can be proved generally that for a harmonic oscillator = (n+ 1 2) h p k and that = (n2 + 2n+ 1) 3 h2 4 k: Verify these formulas for the rst two states of the harmonic oscillator. Ford 2, the DOS of the har-monic oscillator is / , so we expect a vanishing density of states as ! 4tin this case. Quantum numbers, energy, excitation energy number, degeneracy, and number of states with m r = 0 for the low-energy levels of a system of two noninteracting identical bosons trapped in a 2D isotropic harmonic potential. Rotational motion and Angular momentum (rigid rotor) 6. Here I discuss about Degeneracy and Hermonic oscillator problems in quantum mechanics also you can see https://youtu. ( ) ( ) ( ) or my t ky t cy t Fnet FH FF && =− − & = +. We distinguish between. deformed oscillators play important role in modern physics, and various quantum Using deformed oscillators, corresponding models of Bose-gas are constructed. harmonic oscillator E= ¯hω N+ 1 2 (2. The vertical axis is wave-function amplitude in. Eigenstates are labelled by occupancy of 1&2 modes E(n 1. Transformation and Michelson & Morley. We have already solved the problem of a 3D harmonic oscillator by separation of variables in Cartesian coordinates. For the three-dimensional isotropic harmonic oscillator the energy eigenvalues are E = (n + 3/2)ħω, with n = n 1 + n 2 + n 3, where n 1, n 2, n 3 are the numbers of quanta associated with oscillations along the Cartesian axes. Here I discuss about Degeneracy and Hermonic oscillator problems in quantum mechanics also you can see https://youtu. complicated bounded orbits. Consider a system of(N independent fermions. Translational motion (free particle, particle in 1D/2D box) 4. See table 1. Consider the case of a two-dimensional harmonic oscillator with the following Hamiltonian: which may be equivalently expressed in terms of the annihilation and creation operators For your reference. Creation and destruction operators a†,aare familiar from the treatment of the harmonic oscil-lator. A further study of the degeneracy of the two dimensional harmonic oscillator is made, both in the isotropic and anisotropic cases. In the previous chapter we studied stationary problems in which the system is best described as a (time-independent). Considering the correspondence that exist between the states of a two-dimensional isotropic harmonic oscillator (2DIHO) and those of a Morse oscillator (MO), the matrix elements of the latter have been calculated. If this Ji * X term la included in the potential energy, the degeneracy in the various levels is removed. (a) What is the energy of the ground state of this system? What is the degeneracy of this energy? (b) Write down the wavefunction of the ground state. The time-independent Schrödinger equation for a 2D harmonic oscillator with commensurate frequencies can generally given by. This is a reasonable result according to our analysis as follow. This system was called the generalized 2-D harmonic oscillator because presents a bigger accidental degeneracy depending on. We learned from solving Schrödinger's equation for a particle in a one-dimensional box that there is a set of solutions, the stationary states, for which the time dependence is just an overall rotating phase factor, and these solutions correspond to definite values of the. The degeneracy with respect to is often described as an accidental degeneracy, but it can be explained in terms of special symmetries of the Schrödinger equation which are only valid for the hydrogen atom in which the potential energy is given by Coulomb's law. 1 Motion of Dirac points in a 2D crystal F. Spherical confinement in 3D harmonic, quartic and other higher oscillators of even order is studied. Recall the 2D isotropic Simple Harmonic Oscillator has Hamiltonian H0(x;y) = h2 2m r2 + V(x;y) = p^2 x 2m + p^2 y 2m + 1 2 mw2 x2 + y2 = H 0(x) + H0(y). It is inturn closely. Wigner distribution function and entropy of the damped harmonic oscillator within the theory of the open quantum systems. Chapter 8 The Simple Harmonic Oscillator A winter rose. That is, the potential is zero inside a cube of side L and infinite outside the cube. First excited state: 2-fold degenerate, j0 j1 and j1 j0 , E1 = 2ℏ!. a) Find the energy levels for this particle. be/ZKUlzgQC_Ug https://youtu. In fact, it’s possible to have more than threefold degeneracy for a 3D isotropic harmonic oscillator — for example, E 200 = E 020 = E 002 = E 110 = E 101 = E 011. " We are now interested in the time independent Schrödinger equation. The vibrational and rotational spectroscopy of diatomic molecules General features Interaction of electromagnetic field with atoms/molecules Spectral range: Wave number ~ (frequency ) Wavelength Radio MW IR VIS UV X-ray rot. Harmonic Oscillator and Density of States¶ Quantum Harmonic Oscillator 2D, and 3D. Uncertainty principle Commutator, complementary variables, Heisenberg's uncertainty principle. harmonic vibrational frequency or energy expressed in units of cm 1). 1These particles are equivalent to the quanta of the harmonic oscillator, which have energy En = (n + 1 2)~ω, where ~ = h/2π, and h is Planck’s constant. oscillator potential M!2r2=2. The generalized pseudospectral method is employed for accurate solution of relevant Schrödinger equation in an optimum, non-uniform radial grid. Thus the degeneracy is 0,. Électrons dans des cristaux 2D (Landau levels ∼ harmonic oscillator) :constant of motion (degeneracy) r R R degeneracy orbital LL = flux dens. For the 2D harmonic oscillator, for example, the two orthogonal states 10 and 01 span the 1st excited state degeneracy manifold. 1 Classical Particle in B-ﬂeld Want to remind ourselves how to write classical Hamiltonian describing charged particle moving in E and B ﬂelds. The wavefunction is separable in Cartesian coordinates, giving a product of three one-dimensional oscillators with total energies. 3D Symmetric HO in Spherical Coordinates *. Chapter 3 deals with a general Hamiltonian which may be interpreted as that of a charged mass point in a plane harmonic oscillator potential and uniform magnetic field. In fact, one can show that for bound states in 1D, one never has degeneracy; every state has its own energy. Convenient to choose \Landau gauge" A = Bxy^, check that B = r Note this eqn. A particle moving under the influence of such a potential is a free particle since F = — dV(x)/dx = 0. be/ZKUlzgQC_Ug https://youtu. Appendix—Degeneracies of a 2D and a 3D Simple Harmonic Oscillator First consider the 2D case. Lecture 3: Particle in a 2D box, degeneracy 5. Wigner distribution function and entropy of the damped harmonic oscillator within the theory of the open quantum systems. There there is interaction with repns of SO(n) for all n. Physics 2D Lecture Slides Lecture 27: Mar 8th Vivek Sharma UCSD Physics. 1 - Harmonic Oscillator J10Q. Separation of variables: 𝜓=𝑋𝑥𝑌𝑦 −ℏ22𝑚d2𝑋d𝑥2=𝐸𝑥𝑋 −ℏ22𝑚d2𝑌d𝑦2=𝐸𝑥𝑌 𝑚 𝑦=𝐿2 𝑦=0 𝑥=0 𝑥=𝐿1. I will sketch the more thorough reviewgivenby, e. oscillator” (2D Quantum Harmonic Oscillator) in the QuVis HTML5 collection.
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# Thread: [SOLVED] Random Variables

1. ## [SOLVED] Random Variables

The random variable T has the:
Mean 20
Standard deviation 5

It is required to scale T by using the transformation S = aT + b where a and b are constants ( a > 0 ) so that E ( S ) and Var ( S ) satisfy specified values.

Find the value of a and b in this particular case:

when E ( S ) = 0
and when Var ( S ) = 1

2. ## Try these numbers in your equations

Your original mean is 20. So subtract 20 from all the numbers to get a mean of 0.

Your original std deviation is 5, So scale by 1/5 to reduce the std deviation to 1.

These combine to a transformation of (1/5)x -20.

3. ## Random Variables

I see what you mean, but the transformation says + b, and you've used a minus (-20 ) =/

4. The standard language is such that +B means add some constant. There is no constraint that the constant be positive unless the language says "add a positive constant". So negatives are fine. In the future, you might have to add a complex number, which is neither positive nor negative.

5. ## Random Variables

that makes sense, i understandd now, thank you for the help


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# Use of guassian elimination to find the determinate of a matrix

1. Nov 21, 2013

### question dude

1. The problem statement, all variables and given/known data

I have to find that the characteristic polynomial for a 3x3 matrix A, which is effectively the same as finding the det(A - I*lamda)

matrix A is:

the problem is that when I've written out 'A - I*lamda' in matrix form, the three values at the bottom left of the matrix (where I've circled in blue in the pic above)

I know that guassian elimination involves subtracting one row (or a multiple of a row) from another, so that I end up changing those three values to zero. But I don't get how this method would work when for one of the rows, theres more than one value which I need to change to zero.

2. Nov 21, 2013

### LCKurtz

The three values what? You didn't finish the sentence.

Why don't you just write down $A -\lambda I$ for your original matrix and calculate its determinant?

3. Nov 21, 2013

### Ray Vickson

If you have two things to 'zero out' in column 1, you need to do two row-subtractions---one for row 2 and another for row 3. I am very, very surprised that you have not seen this before!

However, I would urge you to NOT start in row 1, because you will obtain rational (not polynomial) values in columns 2 and 3. (Can you see why?) Better: interchange rows 1 and 2 (say), then start with the new row 1---that is, you will be adding or subtracting multiples of 3 to make zeros.

4. Nov 21, 2013

### SteamKing

Staff Emeritus
The usual method to find the characteristic polynomial of a matrix is to first form the matrix (A - Iλ)

Because of the variable λ in this matrix, Gaussian elimination (a numerical technique) is of limited value in calculating this determinant. However, since you are dealing with a 3x3 matrix, there is an algorithm (the Rule of Sarrus) which allows one to calculate the determinant with a fair amount of algebra involved.

http://en.wikipedia.org/wiki/Rule_of_Sarrus

http://en.wikipedia.org/wiki/Characteristic_polynomial

5. Nov 21, 2013

### question dude

The three values I was referring to, are the ones in the matrix: A - I*lamda

they're located in the same positions are the three values I've circled for matrix: A

I know I have to get them to zero, right?

6. Nov 21, 2013

### question dude

but theres two values in row 3 that I need to get to zero

I can't make a subtraction that would get them both to zero, because they're not the same values

sorry I can't see why

7. Nov 21, 2013

### LCKurtz

Like I asked before, why don't you just write down $A-\lambda I$ and take its determinant?

Also, I don't see any $\lambda$'s in your matrix, so it isn't $A-\lambda I$ in the first place.

8. Nov 21, 2013

### question dude

seems like a decent method, never come across it before though

is there any significance to the values next to each of 'a' in the matrix shown in your first link?

9. Nov 21, 2013

### question dude

the matrix I posted in the pic inb my 1st post was for matrix A, not (A - I*lamda)

I've done what you've suggested, by writing down A - I*lamda, its basically the same as matrix A, except for the three values in the central diagonal, they each have '- lamda'

I'm trying to calculate the determinate of A - I*lamda by using guassian elimination

10. Nov 21, 2013

### LCKurtz

OK then. I thought from your OP you were asking about reducing $A$, which seemed irrelevant. Both methods should work on $A-\lambda I$ if your arithmetic skills are good.

11. Nov 21, 2013

### SteamKing

Staff Emeritus
The values next to the 'a' entries represent the row and column indices of that matrix element.
For instance, a21 tells you this represents the element at row 2, column 1 of the matrix.

This is basic stuff. How much do you know about matrix algebra?


text:
When abelian group is divisible?

By definition, group $G$ is divisible if for any $g\in G$ and natural number $n$ there is $h\in G$ such that $g=h^n$. Let $A$ be abelian group with no proper subgroups of finite index. How can I prove that $A$ is divisible?

-
What have you tried? –  Qiaochu Yuan Sep 9 '12 at 20:01
This is harder than I first thought. –  Makoto Kato Sep 9 '12 at 20:53
@MakotoKato: Since $\mathbb{Z}/n\mathbb{Z}$ is such a nice ring, its modules have submodule lattices that are self-dual (I believe $\operatorname{Hom}(-,\mathbb{Z}/n\mathbb{Z})$ is the duality), and so a finite subgroup does imply a finite index subgroup (here you leverage that $\mathbb{Z}/n\mathbb{Z}$ can define $n$ module-theoretically, so also “finite”). However, things are much simpler if you take $n$ prime. $\mathbb{Z}$ is not a nice ring, and so you really do need to use something special about $A/nA$ (not just that it is torsion, but that it is bounded). –  Jack Schmidt Sep 9 '12 at 23:16

1 Answer

Consider the $n$ such that $A^n := \{ a^n : a\in A \} \neq A$. In general these $A/A^n$ will be modules over the self-injective rings $\mathbb{Z}/n\mathbb{Z}$ so Prüfer/Kulikov's theorem shows you have a finite index subgroup. However if you choose a good $n$, you can just use linear algebra. Simple version under the spoiler, since a few other people were working on it.

If $A$ is not divisible then the set $S = \{ n \in \mathbb{Z} : n > 0, \{ a^n : a \in A \} \neq A \}$ has a least element (more generally, the set of ideals without the Baer property has a maximal element). Let $n$ be the least element of $S$. If $n=ab$ for $a,b > 1$, then $A \neq A^n = A^{ab} = (A^a)^b = A^b = A$, since $a,b < n$ so $a,b \notin A$. Hence $n$ is prime and $A/A^n$ is a vector space over the field $\mathbb{Z}/n\mathbb{Z}$. In particular, by the axiom of choice it has a maximal subspace $B$ and since $B$ has codimension 1, $[A:B] =n$ is a finite index.

-
I might be picky, but does it really require choice? I wouldn't be surprised if it really did, but since you invoked it expressis verbis, I thought that maybe you can answer this question. :) –  tomasz Sep 9 '12 at 21:37
@tomasz: I'm not sure if it is equivalent to AC, but there are vector spaces in ZF without AC that have no codimension 1 subspaces. mathoverflow.net/questions/80905/… (question#2) I believe that means a non-divisible abelian group without finite index subgroups also exists. However this will just be because there are vector spaces over Z/pZ without finite index subspaces. –  Jack Schmidt Sep 9 '12 at 23:08
Thanks for the reply. :) It is not obvious to me that every space occurs as a quotient, but I'm still convinced enough that we do need some choice. :) –  tomasz Sep 9 '12 at 23:34


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### 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Examples in R

statistics-lab™ 为您的留学生涯保驾护航 在代写多元统计分析Multivariate Statistical Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写多元统计分析Multivariate Statistical Analysis代写方面经验极为丰富，各种代写多元统计分析Multivariate Statistical Analysis相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Examples in R

In this section, we consider several real data examples for PCA and illustrate the use of software R for PCA.

Example 1. In this study (Johnson and Wichern, 2007), 48 individuals who had applied for a job with a large firm were interviewed and rated on 15 criteria. Individuals were rated on the form of their letter of application (FL), their appearance (APP), academic ability (AA), likability (LA), self-confidence (SC), lucidity (LC), honesty(HON), salesmanship (SMS), experience (EXP), drive (DRV), ambition (AMB), ability to grasp concepts (GSP), potential (POT), keenness to join (KJ), and their suitability (SUIT). Each criterion was evaluated on a scale ranging from 0 to 10 , with 0 being a very low and very unsatisfactory rating, and 10 being a very high rating. In this example, there are 15 variables, so the dimension of the data is 15 . So it is difficult to graphically display the data, and to check any outliers and multivariate normality of the data. We consider a PCA to reduce the dimension.

We see that all 15 variables are correlated and some are highly correlated (e.g., $\mathrm{POT}$ and GSP have a correlation of $0.9, \mathrm{LC}$ and $\mathrm{GSP}$ also have a correlation of $0.9$, etc). Since all variables here have comparable scales, we may perform PCA on the sample covariance matrix $S=\hat{\Sigma}$.

In the above results, the “Cumulative Proportion” row gives the cumulative proportions of variation explained by the first $\mathrm{PC}$, the first two $\mathrm{PCs}$, the first three $\mathrm{PCs}$, etc. We see that the first $2 \mathrm{PCs}$ explain $69 \%$ variations (the first two eigenvalues of $\hat{\Sigma}$ are $66.54$ and $18.18$ respectively): the first $\mathrm{PC}$ explains $54 \%$ variation and the second $\mathrm{PC}$ explains $15 \%$ variation. The first $3 \mathrm{PC}$ ‘s explain about $78 \%$ of total variations. Thus, the 15 -dimensional original data may be reduced to 2 or 3 dimensions!

The “new data” (i.e., the PC scores) can be obtained from the original data. To show graphical displays of the new data, we consider PC scores from the first two PCs and then we graphically display the PC scores to check normality of the new data and any possible outliers.

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|The Basic Idea

In principal components analysis (PCA), we try to reduce the dimension of multivariate data to simplify multivariate analysis. A PCA gives us some idea about the minimal number of variables which contain most information in the original set of variables. A disadvantage of PCA is that the principal components may not always have practical interpretations. That is, sometimes the principal components do not have meaningful interpretations in practice, which is a disadvantage in real data analysis. In this section, we try to determine the minimal set of variables which also have meaningful interpretations in practice. Such analysis is called factor analysis, and the factors usually have practical interpretations.

Factor analysis (FA) tries to describe the variance-covariance relationship among variables in terms of a smaller set of unobservable and uncorrelated new random variables called factors. These factors cannot be directly observed, but they have practical meanings and can be used for data analysis. The essential idea of factor analysis is to group the original variables so that all variables in the same group are highly correlated. Each group then represents a factor (a new variable) that explains the variation and correlation in the original variables in the group. The original set of variables may then be replaced by these factors in data analysis. Thus, factor analysis is closely related to PCA. A main difference is that the factors have practical meanings while the principal components may not have practical meanings.

For example, suppose that a multivariate dataset contains exam scores on mathematics, physics, computer science, English, Chinese, French, income, education, and professionals. We wish to perform a multivariate analysis on this dataset. We see that the dimension of the original data is 9 , and we hope to reduce the dimension in data analysis. Note that exam scores on mathematics $\left(x_{1}\right)$, physics $\left(x_{2}\right)$, and computer science $\left(x_{3}\right)$ may be represented by an unobservable factor called intelligence $\left(f_{1}\right)$. Exam scores on English $\left(x_{4}\right)$, Chinese $\left(x_{5}\right)$, and French $\left(x_{6}\right)$ may be represented by an unobservable factor called verbal ability $\left(f_{2}\right)$. Data on income $\left(x_{7}\right)$, education $\left(x_{8}\right)$, and professionals $\left(x_{9}\right)$ may be represented by an unobservable factor called social status $\left(f_{3}\right)$. Thus, we have grouped the original 9 variables that are highly correlated, and obtain three factors $\left(f_{1}, f_{2}, f_{3}\right)$ : intelligence, verbal ability, and social status. These factors are unobservable and uncorrelated, and they have practical meanings. Therefore, the original 9 variables may be represented by three factors, which is a big reduction of data dimension. Although these three factors may not completely represent the original 9 variables, the three factors should contain most information in the original variables or explain most variability in the original variables. Compared to $\mathrm{PCA}$, an attractive feature of factor analysis is that the factors have practical interpretation, since intelligence, verbal ability, and social status are meaningful variables. This is the idea behind factor analysis.

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|The Factor Analysis Model

The idea of factor analysis can be formally stated as follows. Let $\mathbf{x}=\left(x_{1}, \cdots, x_{p}\right)^{\mathrm{T}} \sim$ $(\mu, \Sigma)$ be the original set of variables with mean vector $\mu=\left(\mu_{1}, \mu_{2}, \cdots, \mu_{p}\right)^{\mathrm{T}}$ and covariance matrix $\Sigma=\left(\sigma_{i j}\right){p \times p}$. Note that $\mathbf{x} \sim(\boldsymbol{\mu}, \Sigma)$ means that the random vector $\mathrm{x}$ has a mean vector $\mu$ and covariance matrix $\Sigma$, without a distributional assumption. The factor analysis (FA) model can be written as $$x{i}=\mu_{i}+\lambda_{i 1} f_{1}+\cdots+\lambda_{i m} f_{m}+\eta_{i}, \quad i=1,2, \cdots, p, \quad m \leqslant p,$$
where the $f_{j}$ ‘s are random variables called factors or common factors, the quantities $\lambda_{i j}$ ‘s are called loadings, $m$ is an positive integer smaller than the original number of variables $p$, and $\eta_{i}$ ‘s are random errors.

In the factor analysis model (3.1), the original set of variables $x_{j}$ ‘s are written as linear combinations of the common factors $f_{j}$ ‘s plus random errors. In other words, the variation in the original set of variables can be partially explained by the variation in the common factors. Typically, the number of factors is less than the number of original variables, i.e., $m<p$. The factors are unobservable. The loading $\lambda_{i j}$ ‘s represent the contribution (importance) of factor $f_{j}$ to variable $x_{i}$. The random errors $\eta_{i}$ ‘s represent variations that cannot be explained by the factors.

Note that the FA model (3.1) is different from a regression model. In a regression model, both the responses and the predictors are observed (or known). In the FA model, however, the common factors $f_{j}$ ‘s are not observed (or unknown). Thus, statistical methods for FA models are different from those for regression models.
To estimate the unknown parameters and factors based on given data, we must make some assumptions for the FA model. The common assumptions for the FA model (3.1) are

• the factors $f_{j}$ ‘s are i.i.d, $\sim(0,1)$, i.e., the factors are independently and identically distributed with mean 0 and variance 1 ;
• the random errors $\eta_{j}$ ‘s $\sim\left(0, \psi_{j}\right)$, and are independent, i.e., the random errors are independent with mean 0 and variances $\psi_{j}$ ‘s;
• the factor $f_{k}$ and the random error $\eta_{j}$ are independent for any $k, j$.
These assumptions are needed for a standard factor analysis.
The FA model and its assumptions can be written in the following compact matrix form:
\begin{aligned} &\mathbf{x}=\boldsymbol{\mu}+\Lambda \mathbf{f}+\eta \ &\mathbf{f} \sim(0, I), \quad \eta \sim(0, \Psi), \quad \mathbf{f} \text { and } \eta \text { are independent, } \end{aligned}
where $\Lambda=\left(\lambda_{i j}\right){p \times m}$ is the loading matrix, $\mathbf{f}=\left(f{1}, \cdots, f_{m}\right)^{\mathrm{T}}, \eta=\left(\eta_{1}, \cdots, \eta_{p}\right)^{\mathrm{T}}$, and $\Psi=\operatorname{diag}\left(\psi_{1}, \cdots, \psi_{p}\right)$ is a diagonal matrix. This matrix form is convenient for presentation and mathematical arguments. Like $\mathrm{PCA}$, the covariance matrix $\Sigma$ of $\mathbf{x}$ plays a key role in factor analysis, since the covariance matrix measures the variation and correlation in the data. The mean vector $\mu$ simply measures the location of the data, so it does not contain any information about the variation and correlation in the data.

In both PCA and FA, the variation in the original set of variables are partially explained by the variation in a smaller set of new and uncorrelated variables (principal components and factors). However, PCA mostly focuses on dimension reduction and the principal components may or may not have meaningful practical interpretation, while in factor analysis the factors typically have meaningful practical interpretation. Often, a PCA can be used to roughly determine the number $m$ of factors needed in factor analysis.

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Examples in R

“新数据”（即PC分数）可以从原始数据中获得。为了显示新数据的图形显示，我们考虑前两台 PC 的 PC 分数，然后我们以图形方式显示 PC 分数以检查新数据的正态性和任何可能的异常值。

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|The Factor Analysis Model

X一世=μ一世+λ一世1F1+⋯+λ一世米F米+这一世,一世=1,2,⋯,p,米⩽p,

• 因素Fj是 iid，∼(0,1)，即因子独立同分布，均值为0，方差为1；
• 随机误差这j的∼(0,ψj), 并且是独立的，即随机误差独立于均值 0 和方差ψj的;
• 因素Fķ和随机误差这j是独立的任何ķ,j.
标准因子分析需要这些假设。
FA 模型及其假设可以写成以下紧凑矩阵形式：
X=μ+ΛF+这 F∼(0,我),这∼(0,Ψ),F 和 这 是独立的，
在哪里Λ=(λ一世j)p×米是加载矩阵，F=(F1,⋯,F米)吨,这=(这1,⋯,这p)吨， 和Ψ=诊断(ψ1,⋯,ψp)是对角矩阵。这种矩阵形式便于表示和数学论证。喜欢磷C一个, 协方差矩阵Σ的X在因子分析中起着关键作用，因为协方差矩阵衡量了数据中的变化和相关性。平均向量μ只是测量数据的位置，因此它不包含任何有关数据变化和相关性的信息。

## 有限元方法代写

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# Set P consists of 5 items. The standard deviation of the items in Set

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Set P consists of 5 items. The standard deviation of the items in Set  [#permalink]

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26 Sep 2016, 03:25
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Set P consists of 5 items. The standard deviation of the items in Set P is S. Is S>2?

(1) All the items in Set P are different even numbers between 1 and 19.

(2) The smallest number in the set is 10.

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Re: Set P consists of 5 items. The standard deviation of the items in Set  [#permalink]

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26 Sep 2016, 18:55
1) All the items in Set P are different even numbers between 1 and 19.
For smallest standard deviation, lets take set with numbers closest to one another.
P = {2,4,6,8,10}
mean of P = 6
Std dev = \sqrt{$$\frac{(2-6)^2+(4-6)^2 + (6-6)^2 +(8-6)^2 +(10-6)^2}{5}$$}
this is >2.
If we take any other set say P1 = {2,4,6,8,12}, P1's standard deviation will be more than that of P.

Sufficient.

(2) The smallest number in the set is 10.
P2 = {10,10,10,10,10}
Standard deviation is 0, less than 2
P3= {10,12,14,16,18}
standard deviation is same as that of P, which is greater than 2.

(2) is not sufficient

Option A.
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Re: Set P consists of 5 items. The standard deviation of the items in Set  [#permalink]

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27 Sep 2016, 06:55
Its A

Considering the lowest/ least numbers in the given range we get (2, 4, 6, 8, 10)

SD of the least set is root(8) > 2

hence for any given numbers in the range , SD is greater than 2 . Sufficient
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Set P consists of 5 items. The standard deviation of the items in Set  [#permalink]

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27 Jan 2017, 14:56
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Bunuel wrote:
Set P consists of 5 items. The standard deviation of the items in Set P is S. Is S>2?

(1) All the items in Set P are different even numbers between 1 and 19.

(2) The smallest number in the set is 10.

Expert help please. A bit of math theory question...what is the utility of standard deviation if intuitively I think SD of set (2,4,6,8,10) should be 2? When I calculate the SD as $$\sqrt{8}$$, what is this result telling me? What would be the benefit of calculating SD with this set of numbers? I'm trying to understand the concept of SD.
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Re: Set P consists of 5 items. The standard deviation of the items in Set  [#permalink]

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28 Feb 2017, 05:19
GMAT SD problems never require to apply to SD formula itself.
Actually, SD it's the average distance between set numbers and mean.
(1) tells that the leg between numbers in sequence is 2, thus it can never be less. Suff.
(2) 10 10 10 10 10 ==> SD=0 Ans. NO
10 16 28 38 48 ==> SD>2 Ans. YES Insuff.

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Re: Set P consists of 5 items. The standard deviation of the items in Set  [#permalink]

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28 Feb 2017, 05:20
GMAT SD problems never require to apply to SD formula itself.
Actually, SD it's the average distance between set numbers and mean.
(1) tells that the leg between numbers in sequence is 2, thus it can never be less. Suff.
(2) 10 10 10 10 10 ==> SD=0 Ans. NO
10 16 28 38 48 ==> SD>2 Ans. YES Insuff.

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Re: Set P consists of 5 items. The standard deviation of the items in Set  [#permalink]

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22 Aug 2017, 00:36
raeann105 wrote:
Bunuel wrote:
Set P consists of 5 items. The standard deviation of the items in Set P is S. Is S>2?

(1) All the items in Set P are different even numbers between 1 and 19.

(2) The smallest number in the set is 10.

Expert help please. A bit of math theory question...what is the utility of standard deviation if intuitively I think SD of set (2,4,6,8,10) should be 2? When I calculate the SD as $$\sqrt{8}$$, what is this result telling me? What would be the benefit of calculating SD with this set of numbers? I'm trying to understand the concept of SD.

We dont have to calculate SD here as question is asking whether SD is greater than 2 or not. Now if we know all the values in the data set, we can find the accurate SD and can say either yes the SD is greater than 2 or no SD is not greater than 2. Thus no calculation of SD or intuition about value of SD is required in this case.
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Re: Set P consists of 5 items. The standard deviation of the items in Set &nbs [#permalink] 22 Aug 2017, 00:36
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