Harmonic oscillator expectation value

Question

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?

Answer 1

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}