# Coherent states of the form $|{-\alpha}\rangle$

I've a brief question about coherent states in quantum mechanics.

As everyone knows, a coherent state is just the proper state of the anhilitation operator $a$, thus they're defined with the eigenvalue equation $a|\alpha\rangle=\alpha|\alpha\rangle$ (or inverting, $\langle\alpha|a^\dagger=\langle\alpha|a^{*}$).

However, sometimes I've seen coherent states of the form $|{-\alpha}\rangle$, and I was wondering about their physical meaning and if they satisfy the same eigenvalue equation as before, i.e., $a|{-\alpha}\rangle=-\alpha|{-\alpha}\rangle$ (and thus, $\langle{-\alpha}|a^\dagger=\langle{-\alpha}|{-\alpha}^{*}$).

Precisely, we can find these states, for example, in the case of the Schrödinger cat state.

• Are you sure that's $a$ instead of $\alpha$ on your last equation? – Emilio Pisanty Aug 18 '18 at 23:35
• My bad, you're right it's actually $-\alpha$ – Charlie Aug 20 '18 at 20:47
• Pro LaTeX tip: notice the difference in spacing in $|-\alpha\rangle$ ($|-\alpha\rangle$) and $|{-\alpha}\rangle$ ($|{-\alpha}\rangle$). This is because, in the former, LaTeX (or, here, MathJax) interprets the - sign as a binary operation acting on | and \alpha, which is not the case. That's why using braces to enforce the correct grouping produces a better spacing. – Emilio Pisanty Aug 21 '18 at 13:11

Yes, coherent states can be found for any value of $\alpha$. To see this, it's nice to use the displacement operator and construct the states:

$$D(\alpha)=e^{\alpha a^\dagger-\alpha^\star a}.$$

It is the exponential of an anti-Hermitian operator and so is unitary by construction. What is interesting is it's effect on the vacuum state, namely we can write

$$|\alpha\rangle=D(\alpha)|0\rangle.$$

To show this, let's compute the operator

$$D^\dagger(\alpha)aD(\alpha).$$

We can use BCH expansion and it reduces to

$$a+\alpha.$$

As such, when we act with this operator on the vacuum state,

$$D^\dagger(\alpha)aD(\alpha)|0\rangle=(a+\alpha)|0\rangle=\alpha|0\rangle.$$

Since $D(\alpha)$ is unitary, acting with it on the left on both sides, we get

$$aD(\alpha)|0\rangle=\alpha D(\alpha)|0\rangle,$$

i.e. $D(\alpha)|0\rangle=|\alpha\rangle$ for any complex number $\alpha$.