# Harmonic oscillator expectation value

In calculating the expectation value of the quantum harmonic oscillator, I've come across a problem for finding $$\left \langle x \right \rangle$$ for the coherent state $$\left| \alpha \right \rangle$$

$$x = \sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_- )$$

$$\sqrt{\frac{\hbar}{2m\omega}}\left \langle \alpha \right| (a_+ + a_- ) \left| \alpha \right \rangle =\sqrt{\frac{\hbar}{2m\omega}} \alpha ^* (a_+ \alpha + a_- \alpha )$$

This doesn't agree with the result my book has given, which is $$\sqrt{\frac{\hbar}{2m\omega}} (\alpha + \alpha ^* )$$.

I'm not used to working in bra-ket notation, should it's extremely likely that I'm missing something small here. The general principle is that $$\left \langle \beta \right| A \left| \beta \right \rangle = \beta ^* (A\beta)$$, right?

Each coherent state $$|\alpha\rangle$$ is, by definition, an eigenvector of the lowering operator $$a_-$$ with eigenvalue $$\alpha$$, namely \begin{align} a_-|\alpha\rangle = \alpha|\alpha\rangle. \end{align} If we assume that $$|\alpha\rangle$$ is normalized to one (i.e $$\langle \alpha|\alpha\rangle=1$$), as one usually does, then we find that the expectation value of $$a_-$$ in a coherent state is \begin{align} \langle \alpha|a_-|\alpha\rangle = \alpha\langle \alpha|\alpha\rangle = \alpha \end{align} On the other hand, the raising operator is the adjoint (aka hermitian-conjugate), of the lowering operator and vice versa \begin{align} (a_+)^\dagger = a_- \end{align} It follows that the expectation value of the raising operator in a coherent state is \begin{align} \langle \alpha|a_+|\alpha\rangle = \langle\alpha| (a_+)^\dagger|\alpha\rangle^* =\langle\alpha|a_-|\alpha\rangle^* = \alpha^* \end{align} Putting these together, we get the result you are looking for; \begin{align} \langle \alpha| x|\alpha\rangle = \sqrt{\frac{\hbar}{2m\omega}}(\langle\alpha|a_+|\alpha\rangle + \langle\alpha|a_-|\alpha\rangle) = \sqrt{\frac{\hbar}{2m\omega}}(\alpha^* + \alpha). \end{align}