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I was working on a quantum mechanics problem, involving the perturbation of the 3D cubical potential well:

Suppose we perturb the infinite cubical well

\begin{equation} V(x,y,z)=\begin{cases} 0, \text{ if } 0<x<a,0<y<a,0<z<a\\ \infty, \text{ otherwise.} \end{cases} \end{equation}

by the potential

\begin{equation} V'(x,y,z)=\alpha\delta(x-a/4)\delta(y-a/2)\delta(z-3a/4),\hspace{10pt} \alpha>0. \end{equation}

I found the matrix representation of the perturbation in the unperturbed basis to be:

\begin{equation} W=\frac{\alpha}{2}\left(\frac{2}{a}\right)^3\begin{pmatrix} 1 & 0 & -1\\ 0 & 0& 0\\ -1 & 0 & 1 \end{pmatrix}, \end{equation}

with eigenvalues

\begin{equation} E_1=2, E_2=E_3=0. \end{equation}

My question is as follows: Since two of the "good states" (by the way, does anybody have a good explanation on good states, or some resources? Griffiths is horrible at explaining this.) do not get an energy correction, they are still degenerate, and I think that there must be some kind of symmetry which causes the degeneracy. I know that the "good states" for those two energy corrections are

\begin{equation} |E_2\rangle=\frac{1}{\sqrt{2}}(|211\rangle+|112\rangle), \hspace{10pt}|E_3\rangle=|121\rangle, \end{equation}

where the three numbers refer to $n_x,n_y,n_z$. By looking at these states expressed in terms of sines, I see that they are zero at the position where the perturbation is located, thus yielding no energy correction. But this, at least for me, is not the symmetry of the Hamiltonian, or at least I'm afraid of calling this the symmetry.

Can anybody help me figuring this out?

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  • $\begingroup$ Didn't you already ask this question today? I can't find it anymore, but I'm positive that I read this same post earlier today. $\endgroup$ – march Apr 15 at 22:27
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    $\begingroup$ There's clearly a $\mathbb{Z}_2$ symmetry $(x,y,z) \mapsto (a-z,\ldots)$. (Leaving the rest as an exercise.) $\endgroup$ – Hans Moleman Apr 15 at 22:33
  • $\begingroup$ @HansMoleman Could you explain to me what a $\mathbb{Z}_2$ symmetry is, as my understanding of more abstract quantum mechanics is rather limited? (Note that this is a Griffiths problem, which should say a lot about my progression in this topic) $\endgroup$ – pjHart1000 Apr 15 at 22:37
  • $\begingroup$ It's an action such that if you do it twice, you're back to where you started. Mathematicians call this group $\mathbb{Z}/2\mathbb{Z}$ but in physics we just write $\mathbb{Z}_2$. $\endgroup$ – Hans Moleman Apr 15 at 22:42
  • $\begingroup$ Actually there's one more symmetry! So the theory with the $V'$ potential has at least a $V_4$ symmetry group. $\endgroup$ – Hans Moleman Apr 16 at 1:01

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