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## Homework Statement

For a quantum harmonic oscillator in an electric field, using ##\hat{V}=q\epsilon\hat{x}##, with the following trial state: $$|\psi\rangle=|0\rangle+b|1\rangle$$

Show that the energy can be written as $$E=\frac{\frac{\hbar \omega}{2}+\frac{3b^2\hbar\omega}{2}+\frac{\sqrt{2}bq\epsilon}{\alpha}}{1+b^2}$$

## Homework Equations

##\hat{x}=\frac{1}{\sqrt{2}\alpha}(a^{\dagger}+a)##

## The Attempt at a Solution

There are two things that are confusing me:

(1) how to distribute the brakets in the following:

##\langle \psi | ~~= \langle 0 | +b^*\langle 1 |##

##E^{(1)}_{\psi}=\frac{q\epsilon}{\sqrt{2}\alpha}\langle \psi|a^{\dagger}+a|\psi\rangle##

##~~~~~~~\stackrel{?}{=}\frac{q\epsilon}{\sqrt{2}\alpha}[(\langle 0|+b^*\langle 1|)a^{\dagger}(|0\rangle+b|1\rangle)+(\langle 0|+b^*\langle 1|)a(|0\rangle+b|1\rangle)]##

##~~~~~~~=\frac{q\epsilon}{\sqrt{2}\alpha}[\langle 0|1 \rangle+\sqrt{2}b\langle 0|2 \rangle+b^*\langle 1|1 \rangle+\sqrt{2}b^*b\langle 1|2 \rangle+b\langle 0|0 \rangle+b^*b\langle 1 |0 \rangle]##

##~~~~~~~=\frac{q\epsilon(b^*+b)}{\sqrt{2}\alpha}##

Is this how you would distribute it? I get the feeling it is completely wrong..

(2) What is the unperturbed energy of the state ##|\psi \rangle##

Would it be ##E_{\psi}^{(0)}=\frac{1}{2}\hbar\omega+\frac{3}{2}b\hbar\omega## ?

Any guidance is appreciated!