- #1
spaghetti3451
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Homework Statement
Given an operator ##D(\alpha)=\exp\ (\alpha a^{\dagger}-\alpha^{*}a)## and a function ##g(a,a^{\dagger})##, where ##a## and ##a^{\dagger}## are operators and ##\alpha## and ##\alpha^{*}## are complex numbers, show that
##D^{-1}(\alpha)g(a,a^{\dagger})D(\alpha)=g(a+\alpha, a^{\dagger}+\alpha^{*})##
Homework Equations
The Attempt at a Solution
##D^{-1}(\alpha)g(a,a^{\dagger})D(\alpha)##
##=\exp\ (-\alpha a^{\dagger}+\alpha^{*}a)\ g(a,a^{\dagger})\ \exp\ (\alpha a^{\dagger}-\alpha^{*}a)##
##=(1-\alpha a^{\dagger}+\alpha^{*}a)\ g(a,a^{\dagger})\ (1+\alpha a^{\dagger}-\alpha^{*}a)##
##=g(a,a^{\dagger})+\alpha^{*}[a,g(a,a^{\dagger})]-\alpha[a^{\dagger},g(a,a^{\dagger})]##.
On the other hand,
##g(a+\alpha, a^{\dagger}+\alpha^{*})##
##=g(a, a^{\dagger}) + \alpha \frac{\partial g(a,a^{\dagger})}{\partial \alpha} + \alpha^{*} \frac{\partial g(a,a^{\dagger})}{\partial \alpha^{*}}##.
This seems to suggest that
##\frac{\partial g(a,a^{\dagger})}{\partial \alpha^{*}}=[a,g(a,a^{\dagger})]##
and
##\frac{\partial g(a,a^{\dagger})}{\partial \alpha}=-[a^{\dagger},g(a,a^{\dagger})]##.
Am I missing something here?