Answer to Question #116098 in Quantum Mechanics for Tolulope

Question

we construct a latter operator in the form of a = p − iα tanh(x), with α ∈ R, how do I get
a+a =

Answer 1

$\hat{a}=\hat{p}-i\alpha\tanh(\hat x),\,\,\hat{a}^+=\hat{p}+i\alpha\tanh(\hat x)$

$\hat a^+ \hat a=(\hat{p}+i\alpha\tanh(\hat x))(\hat{p}-i\alpha\tanh(\hat x))=\\ \hat{p}^2-i\alpha \hat p \tanh(\hat x)+i\alpha \tanh(\hat x)\hat{p}+\alpha^2\tanh^2(\hat x)=\\ \hat{p}^2+\alpha^2\tanh^2(\hat x)+i\alpha ( \tanh(\hat x)\hat p-\hat p \tanh(\hat x))$

Let $\hat x \psi(x)=x\psi(x),\,\,\hat p \psi(x)=-i\hbar \partial_x \psi(x)$. Hence,

$( \tanh(\hat x)\hat p-\hat p \tanh(\hat x))\psi(x)=\\ i\hbar(-\tanh x \,\partial_x \psi(x)+\partial_x(\tanh x\, \psi(x))=\\ i\hbar(-\tanh x \,\partial_x \psi(x)+\partial_x(\tanh x)\, \psi(x)+\tanh x\,\partial_x\psi(x))\\ i\hbar \operatorname{sech}^2 x\, \psi(x)$

Finally, $a^+a=\hat{p}^2+\alpha^2\tanh^2(\hat x)-\alpha\hbar \operatorname{sech}^2 (\hat x)$