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Chapter 14: Problem 7

Show that the eigenvalues of the three-dimensional harmonic oscillater have the form \(\left(n+\frac{3}{2}\right) \hbar \omega\) where \(n\) is a non-negative integer. Show that the dcpeneracy of the \(n\)th eigenvalue is \(\frac{1}{2}\left(n^{2}+3 n+2\right)\). Find the corresponding eigenfunctions.

Short Answer

Expert verified
The eigenvalues of the 3D Quantum Harmonic Oscillator will take the form \((n + 3/2) ħω\). The degeneracy of any eigenvalue \(n\) of this system is given by the formula \(\frac{1}{2} (n^2 + 3n + 2)\). The eigenfunctions are a product of 1D QHO eigenfunctions for each dimension.

Step by step solution

01

Understand the 3D Quantum Harmonic Oscillator

The eigenstates of a 3D Quantum Harmonic Oscillator are product of one dimensional harmonic oscillator eigenstates for each dimension. The quantum numbers of a 3D QHO are given by three integers (n_x, n_y, n_z), each corresponding to one dimension. The eigenstates are signified by |n_x, n_y, n_z>. The energy of each state is: E = (n_x + n_y + n_z + 3/2) ħω
02

Determine the eigenvalues

Substituting \(n=n_x + n_y + n_z\) in the energy equation we get: \(E= (n + 3/2) ħω\). This indicates that eigenvalues of a 3D QHO take the form \((n + 3/2) ħω\) where \(n=n_x + n_y + n_z\) is a non-negative integer.
03

Calculate the degeneracy

Degeneracy is the number of different states corresponding to a particular energy level. For a 3D QHO, degeneracy is given by the number of different combinations of \(n_x, n_y, n_z\) that sum up to \(n\). This is given by the formula: \(\frac{1}{2} (n^2+3n+2)\). This can be shown mathematically through combinatorial principles.
04

Finding the eigenfunctions

The 3D QHO eigenfunctions are a product of the 1D eigenfunctions for each dimension. The 1D QHO eigenfunctions are given by: \(\psi_n(x) = ((mω/(πħ))^(1/4) / √2^n n!) H_n((mω/ħ)^1/2 x) e^(−mωx^2 / 2ħ)\) where H_n is the Hermite polynomial of order n.

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Most popular questions from this chapter

Verify for each direction $$ \mathbf{n}=\sin \theta \cos \phi \mathbf{e}_{x}+\sin \theta \sin \phi c_{y}+\cos \theta \boldsymbol{e}_{x} $$ the spin operator $$ \sigma_{n}=\left(\begin{array}{cc} \cos \theta & \sin \theta \mathrm{e}^{-1 \phi} \\ \sin \theta \mathrm{e}^{i \phi} & -\cos \theta \end{array}\right) $$ has cigenvalues \(\pm 1\). Show that up to phase, the cigenvectors can be expressed as $$ \left.\mid+n)=\left(\begin{array}{c} \cos \frac{1}{2} \theta \mathrm{c}^{-t \phi} \\ \sin \frac{1}{2} \theta \end{array}\right), \quad \mid-n\right)=\left(\begin{array}{c} -\sin \frac{1}{2} \theta \mathrm{e}^{-1 \phi} \\ \cos \frac{1}{2} \theta \end{array}\right) $$ and compute the expectation values for spin in the dircction of the various axes $$ \left\langle\sigma_{1}\right\rangle_{\text {tn }}=\left(\pm n\left|\sigma_{1}\right| \pm n\right) $$ For a bcam of particles in a pure state \(\mid+n\) ) show that after a measurcment of spin in the \(+x\) direction the probability that the spin is in this direction is \(\frac{1}{2}(1+\sin \theta \cos \phi)\).

Calculate the canonical partition function, mean encrgy \(U\) and entropy \(S\), for a system having just two energy levels 0 and \(E\). If \(E=E(a)\) for a paramcter \(a\), calculate the force \(A\) and verify the thermodynamic relation \(\mathrm{dS}=\frac{1}{r}(\mathrm{~d} U+A \mathrm{da})\).

Show that the time reversal of angular momentum \(\mathrm{L}=Q \times \mathbb{P}\) is \(\Theta^{*} L_{t} \Theta=-L_{i}\) and that the commutation relations \(\left[L_{i}, L_{j}\right]=a h \epsilon_{i j} L_{k}\) are only preserved if \(\Theta\) is anti-unitary:

A particle of mass \(m\) is confined by an tnfinite potental barrier to rematn within a box \(0 \leq x, y, z \leq a\), so that the wave function vanishes on the boundary of the box. Shew that the energy levels are $$ E=\frac{1}{2 m} \frac{\pi^{2} \hbar^{2}}{a^{2}}\left(n_{1}^{2}+n_{2}^{2}+n_{3}^{2}\right) $$ where \(n_{1}, n_{2}, n_{3}\) are positive intcecrs, and calculate the stationary wave functions \(\psi_{E}(\mathrm{x})\). Verify that the lowest energy state is non-degenerate, but the next highest is triply degenerate.

In the Heisenberg picture show that the time evolution of the expection value of an operator \(A\) is given by $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(A^{\prime}\right\rangle_{\psi}^{\prime}=\frac{1}{\mathrm{~s} h}\left\langle\left[A^{\prime} \cdot H^{\prime}\right]\right\rangle_{\psi}+\left\langle\frac{\partial A^{\prime}}{\partial t}\right\rangle_{\phi} $$ Convert this to an equation in the Schrödinger picture for the time evolution of \((A\rangle_{\psi}\).
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