Particle in a potential

Homework Statement:

A particle moves in the potential $V(x) = -\frac{\hbar^2}{m} \text{sech}^2{x}$. Show that $A^{\dagger}A \psi = (\mathcal{E} + 1)\psi$ where $\mathcal{E} = \frac{2mE}{\hbar^2}$ and $A = \frac{d}{dx} + \tanh{x}$, $A^{\dagger} = -\frac{d}{dx} + \tanh{x}$.

Hence show that the ground state has $\mathcal{E} \geq -1$ and that a wavefunction $\psi_0(x)$ has an energy eigenstate with $\mathcal{E} = -1$ iff $\frac{d\psi_0}{dx} + \tanh{(x)} \psi_0 = 0$ and find $\psi_0(x)$.

Relevant Equations:

N/A

I'm falling at the first hurdle here; the time independent Schrodinger becomes $$-\frac{\hbar^2}{2m} \psi''(x) - \frac{\hbar^2}{m}\text{sech}^2(x) \psi(x) = E\psi(x)$$ $$\left(-\frac{d^2}{dx^2} - 2\text{sech}^2(x) \right)\psi(x) = \frac{2mE}{\hbar^2}\psi(x) = \mathcal{E} \psi(x)$$ $$\left(-\frac{d^2}{dx^2} + 2\tanh^2(x) \right)\psi(x) = (\mathcal{E}+2) \psi(x)$$ But this can't be right since $A^{\dagger}A = \tanh^2(x) - \frac{d^2}{dx^2}$ and I've also got a 2 on the other side.

I wondered if anyone could point out my mistake? Thanks!

 

[anuttarasammyak]

Applying coefficient 1/2 on V(x) seems to satisfy the given relation.

 

Applying coefficient 1/2 on V(x) seems to satisfy the given relation.

I noticed that but the source of these questions is usually very reliable so I thought there's no way it could be a mistake! But if you say so also, then I am happy to correct it and work from that. Thanks!

 

A further question; with an integrating factor and then with some normalisation I can find $$\psi(x) = \frac{1}{\sqrt{2}\cosh{x}}$$ for the equation $\frac{d\psi}{dx} + \tanh{x} \psi(x) = 0$. But how was it possible to obtain this differential equation from the second order one above? We know that $A^{\dagger}A$ has eigenvalues greater than or equal to zero.

 

[anuttarasammyak]

I found my post #2 was wrong. From OP
$$(-\frac{d^2}{dx^2}+2\tanh^2 x -1)\psi(x)=(\epsilon+1) \psi(x)$$
$$(-\frac{d}{dx}+\tanh x)(\frac{d}{dx}+\tanh x)\psi(x)=(\epsilon+1) \psi(x)$$
using
$$-\frac{d}{dx}(\tanh x\ \psi(x))= (-1 + tanh^2x)\psi(x) - \tanh x \frac{d}{dx}\psi(x)$$

 

I found my post #2 was wrong.

How so? It gives the correct result: $$ \begin{align*}-\frac{\hbar^2}{2m} \psi''(x) - \frac{\hbar^2}{2m}\text{sech}^2(x) \psi(x) &= E\psi(x) \\ \left(-\frac{d^2}{dx^2} - \text{sech}^2(x) \right)\psi(x) &= \frac{2mE}{\hbar^2}\psi(x) = \mathcal{E} \psi(x) \\ A^{\dagger}A \psi(x) = \left(-\frac{d^2}{dx^2} + \tanh^2(x) \right)\psi(x) &= (\mathcal{E}+1) \psi(x) \end{align*}$$

 

[anuttarasammyak]

We have forgotten the term
$$-\frac{d \tanh x }{dx} \psi(x)$$

 

We have forgotten about the term
$$-\frac{d \tanh x }{dx} \psi(x)$$

I'm sorry, I don't follow. Isn't it just a difference of two squares? That along with $\tanh^2(x) + \text{sech}^2(x) = 1$.

 

[anuttarasammyak]

$$(-\frac{d}{dx}+\tanh x)(\frac{d}{dx}+\tanh x)\psi(x)=-\frac{d^2 \psi(x)}{dx^2}+\tanh x\frac{d \psi(x)}{dx}+\tanh^2 x\ \psi(x)-\frac{d}{dx}[\tanh x\ \psi(x)]$$

 

Ah okay, yes you're right it's not commutative.