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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.
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Well, here is a seat-of-the pants "lark" formal answer, going to "rigged" Fock spaces and places you (or anybody else) shouldn't really be at; except you may already have been there, when you learned about bras and kets, if you looked in Dirac's QM book ― actually, the section virtually all modern texts skip!

I will cover your Fock space question, but mostly as an introduction to segue into Dirac's sublime picture that merits exposure. In this sense, this informal formal answer is exceptionally indulgent.

Indeed, for coherent states, $$ |\alpha\rangle= e^{-|\alpha|^2/2 } e^{\alpha a^\dagger}|0\rangle , $$ where $a|0\rangle =0$ and $[a,a^\dagger]=1$, you find that $a|\alpha\rangle=\alpha | \alpha\rangle$. The adjoint/dual relation in your (1) is nothing but these, so they really do not tell you something new, and they are not the "left" eigenstates of $a^\dagger$ in any meaningful sense, except the trivial one. You are just looking at the very same states in dual space.

The strict proofs of linked questions excluding eigenstates of $a^\dagger$ have a flakey formal loophole, however. The freak state $$ |\psi\rangle=\delta(a^\dagger - \beta) |0\rangle =\frac{1}{2\pi} \int \!\! dk ~e^{ik(a^\dagger -\beta)} |0\rangle $$ satisfies $$ a^\dagger | \psi\rangle=\beta| \psi\rangle, $$ and is the sought-after right-eigenstate of $a^\dagger$, so if you create/add an excitation it indeed does not change, as per your (2).

It is essentially a formal continuous superposition of an infinity of tilted coherent states, analogous to Dirac's $|x\rangle$, which is where it came from, of course; see below.

I do not know if these states have actually entered in some twisted technical fashion into quantum optics as a tool, but, frankly, I am terminating this part of the discussion, regarding it as the warmup handle to the beautiful theory of Dirac's kets, which he details in his classic QM book (4th ed.), Ch III  §20 & Ch IV  §22, §23.


Most of the formal maneuvers we've seen so far algebraically map in a popular loose analogy into the standard Dirac bra-ket entities and operators, $$ a^\dagger \rightsquigarrow \hat x , \qquad a \rightsquigarrow \hat p , \qquad [a^\dagger,a]=1 \rightsquigarrow [\hat x , \hat p]= i\hbar. $$

Let's forget about creators and annihilators, if they confuse you, and start with Dirac's standard ket, $$ \rangle\equiv \sqrt{2\pi\hbar } |\varpi\rangle. $$ I introduce my own notation on the r.h.s. to mitigate the culture shock which has alienated generations: What I mean by it is the standard translationally invariant vacuum, an infinite x-vector with the same constant component in every entry, so that a translation leaves it invariant. (Since you know the tail end of this notation already, it will turn out to be $\lim_{p\to 0} |p\rangle\equiv |\varpi\rangle$. I will not use 0 instead of $\varpi$, as I want to be reminding you it is a p-eigenstate and not an x-eigenstate. So, in this picture, $\hat p$s are annihilators and $\hat x$s creators.) It is thus the analog of the coherent state vacuum, the null state of of $\hat p$, $$ \bbox[yellow,5px]{ \hat p |\varpi\rangle = 0}. $$

Dirac then defines $$ \bbox[yellow,5px]{ |x\rangle = \delta(\hat x -x)|\varpi \rangle \sqrt{2\pi \hbar}}, $$ and $$ \bbox[yellow,5px]{ |p\rangle = e^{ip\hat x /\hbar} | \varpi \rangle}, $$ a momentum-translated zero momentum state, $e^{p \partial_{\varpi}} | \varpi \rangle$, the analog of the coherent state!

Check that, therefore, $$ \langle x | p\rangle = \sqrt{2\pi \hbar} \langle \varpi | \delta (\hat x - x ) e^{ip\hat x /\hbar} |\varpi\rangle =e^{ip x /\hbar} \langle x |\varpi\rangle = e^{ip x /\hbar} /\sqrt{2\pi \hbar}, $$ since the projection of all x-eigenstates on the standard ket is the same, and $$ \hat p |p\rangle = \hat p e^{ip\hat x /\hbar}|\varpi\rangle = p|p\rangle , \qquad \hat x |x\rangle = \hat x \delta(\hat x -x)|\varpi \rangle \sqrt{2\pi \hbar}=x |x\rangle, $$ the properties that most QM textbooks start with. (But, to my surprise when I first learned this, Dirac evidently understood the structure of coherent states a long time before their official inception...)

If you must have a mental picture of these infinite-dimensional vectors in the x-basis, the $|x\rangle$ vector only has a nonzero entry in the x=n th slot, so is terminally sharp; but the $|\varpi \rangle$ vector is terminally broad, and has, say, 1 in every single entry, suitably normalized by the continuum normalization of these monsters; and the $|p\rangle $ vector is as broad, and has entries around the 0th slot which go like $...,e^{-2ip}, e^{-ip}, 1,e^{ip},e^{2ip},e^{3ip},...$. Afficionados of Finite Fourier Transform will recognize these as akin to beloved vectors.

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  • $\begingroup$ The grammar in "You just turned upside down and imagine them different" feels a bit wonky to me. $\endgroup$ – Emilio Pisanty Feb 20 at 10:52
  • $\begingroup$ is there a difference? ;-) $\endgroup$ – Emilio Pisanty Feb 20 at 12:09
  • $\begingroup$ I supplanted the sentence. If I were to paraphrase the above one, I'd say, instead: "You just flipped and stood on your head, and imagine these states to be different". No Victorian but Heaviside would have put it the original way... $\endgroup$ – Cosmas Zachos Feb 21 at 23:01
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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.

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  • $\begingroup$ 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

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