2
$\begingroup$

We know that coherent states $\vert\alpha\rangle$ are eigenvectors of the annihilation operator $\hat{a}$, i.e. $$ \hat{a} \vert\alpha\rangle = \alpha \vert\alpha\rangle $$ while the creation operator $\hat{a}^\dagger$ has no eigenvector.

Now, I have few questions:

  1. Would it be correct to say that $\langle\alpha\vert$ is (left) eigenvector of $\hat{a}^\dagger$ ? Can we use this formalism, and does it have some (physical) meaning?
  2. In view of the above, I would say that the naive argument that is often found: "a coherent state resembles a classical state because if you annihilate excitations it does not change", is rather wrong. In fact the converse should also be true, which is only the case if $\vert\alpha\rangle$ is eigenstate of $\hat{a}^\dagger$ as well.
$\endgroup$
1
$\begingroup$

Would it be correct to say that $\langle\alpha\vert$ is (left) eigenvector of $\hat{a}^\dagger$ ? Can we use this formalism, and does it have some (physical) meaning?

Hmmm... I am not familiar with a specific physical interpretation of that. However, note that if $u$ is a left eigenvector of $A$, i.e. $uA=\alpha u$, then taking the transpose + Hermitian conjugate gives

$$A^\dagger u^\dagger=\left(uA\right)^\dagger=\left(\alpha u\right)^\dagger=\alpha^\ast u^\dagger$$

Now denote $B\equiv A^\dagger$, $v\equiv u^\dagger$ and $\lambda=\alpha^\ast$ so

$$Bv=\lambda v$$

This means that studying left eigenvectors is equivalent to studying right eigenvectors. There is nothing special with the side choice.

In view of the above, I would say that the naive argument that is often found: "a coherent state resembles a classical state because if you annihilate excitations it does not change", is rather wrong. In fact the converse should also be true, which is only the case if $\vert\alpha\rangle$ is eigenstate of $\hat{a}^\dagger$ as well.

I also don't like this argument. The resemblances of coherent states to classical states is a consequence of their dynamics, which is the result of the Hamiltonian.

$\endgroup$
  • $\begingroup$ Note that this only holds for arbitrary cases if the hilbert space has finite dimension. For infinite dimensions you need to work with functionals of the dual space for the left states. Since $\alpha$ is a continous complex variable I'm not sure this simple argument is rigorous. $\endgroup$ – Halbeard Dec 4 '18 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.