 #1
 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!
I wondered if anyone could point out my mistake? Thanks!