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.

  • $\begingroup$ Are you sure that's $a$ instead of $\alpha$ on your last equation? $\endgroup$ – Emilio Pisanty Aug 18 '18 at 23:35
  • $\begingroup$ My bad, you're right it's actually $-\alpha$ $\endgroup$ – Charlie Aug 20 '18 at 20:47
  • $\begingroup$ 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. $\endgroup$ – 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


To show this, let's compute the operator


We can use BCH expansion and it reduces to


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


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 $\alpha$.


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