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I'm working on an introductory qm project, hope somebody has the time to help me (despite the length of this post), it will be highly appreciated.

My goal is to determine the bound states and their energies for the potential

$V_j(x) = -\frac{\hslash^2a^2}{2m}\frac{j(j+1)}{\cosh^2(ax)}$

for any positive value of j (I think I'm supposed to show that j has to be integer at some point, but I don't know). By a change of variable, $ax\rightarrow y$, I have rewritten the Schrödinger equation for this potential to:

$-u''(y)-\frac{j(j+1)}{\cosh^2(y)}u(y) = Eu(y).$

I will denote the corresponding Hamiltonian $H_j$. I have shown that the ground state for $V_j(x)$ is $\psi_0(x) = A\cosh(ax)^{-j}$ (this is all correct, has been verified). Now here is where serious problems begin. I am given these ladder operators:

$a_{j+} = p + ij\tanh(y)$

$a_{j-} = p - ij\tanh(y)$

where $p = -i\frac{d}{dy}$ (no h-bar). I have shown that $H_j = a_{j+}a_{j-} - j^2$ and $H_{j-1} = a_{j-}a_{j+} - j^2$ and I have used this to show that if $|\psi>$ is an eigenstate for $H_j$ then $a_{j-}|\psi>$ is an eigenstate for $H_(j-1)$ and $a_{(j+1)+}|\psi>$ is an eigenstate for $H_(j+1)$, both with the same eigenvalue (energy). Thus the purpose of the a-operators is to carry an eigenfunction for the potential characterized by the value$ j (H_j)$ over to an eigenfunction for the potential characterized by the value j-1 or $j+1 (H_(j-1), H_(j+1)).$

Now I should be able to find all bound states and their energies (I have been told that there is only one bound state for every (integer?) value of j, so this amounts to showing that the ground state I found earlier is the only bound state there is). I'm clueless on this last step.

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j is not an integer, but it reduces by one unit each ladder step. –  Ron Maimon Jun 26 '12 at 21:42

1 Answer 1

Assume that you have an arbitrary eigenstate, and apply the lowering operator often enough to be able to say something definite about the resulting vector.

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