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Aim: Show that $-{\hbar^2\over 2m} {d^2\over dx^2}\psi(x)-{\hbar^2\over m}\mathrm{sech}^2(x)\psi(x)=E\psi(x)$ is equivalent to $\hat{O}^\dagger\hat{O}\psi(x)=({2mE\over \hbar^2}+1)\psi(x)$ where $\hat{O}={d\over dx}+\tanh(x)$ and $\hat{O}^\dagger=-{d\over dx}+\tanh(x)$

What I did:

$\hat{O}^\dagger\hat{O}\psi(x)=({2mE\over \hbar^2}+1)\psi(x) \implies (-{d^2\over dx^2}+\tanh^2(x))\psi(x)=({2mE\over \hbar^2}+1)\psi(x)$

$\implies (-{d^2\over dx^2}-\mathrm{sech}^2(x))\psi(x)={2mE\over \hbar^2}\psi(x)$

$\implies -{\hbar^2\over 2m} {d^2\over dx^2}\psi(x)-{\hbar^2\over 2m}\mathrm{sech}^2(x)\psi(x)=E\psi(x)$

There is an extra "${1\over 2}$" in the 2nd term of LHS! Also, is there a better way of showing this. What about showing it in the other direction -- i.e. starting with $-{\hbar^2\over 2m} {d^2\over dx^2}\psi(x)-{\hbar^2\over m}\mathrm{sech}^2(x)\psi(x)=E\psi(x)$?


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up vote 1 down vote accepted

Wait a minute, $(-A+B)(A+B)=-A^2+B^2$ holds only when $A$ and $B$ commute, that is, when we have $BA-AB=0$ ($\color{Red}{\text{foil out the expression and check this}}$), which does not apply here with the operators $A=d/dx$ and $B=\tanh x \cdot \mathrm{Id}$ ($\color{Red}{\text{check this too}}$). Instead, we have (note: write $D=d/x$): $$\hat{O}^\dagger\hat{O}\;\psi=\left(-D+\tanh x\right)\left(D+\tanh x\right)\psi$$ $$=-D(D+\tanh x)\psi +(\tanh x)(D+\tanh x)\psi$$ $$=-D^2\psi-D((\tanh x)\cdot\psi)+\tanh x\cdot D\psi+\tanh^2 x \cdot \psi$$ $$=-D^2\psi-(\mathrm{sech}^2x\cdot\psi+\tanh x\cdot D\psi)+\tanh x\cdot D\psi+(1-\mathrm{sech}^2x)\cdot\psi$$ $$=(-D^2-2\mathrm{sech}^2x+1)\psi.$$ Now do you see how to reason towards the desired conclusion?

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thanks loads, anon! – bugged Oct 17 '11 at 8:12

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