- #1

etotheipi

- 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!