# Eigenstates of the creation operator

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.
• $a^\dagger$ does not have right eigenvectors, but it has a left eigenvector, $\langle\alpha|$, precisely, as suggested in the question. This fact is routinely used in derivation of the path integral formulation in terms of coherent states. This is really a homework question. – Roger Vadim Jan 20 at 12:49
• @Vadim Yes, (1) is just a trivial dual representation of coherent states, as indicated. The $64 question is whether you can come close to exploring notional ket eigenvectors of the creation operator. – Cosmas Zachos Jan 20 at 14:33 • @CosmasZachos it is an interesting question to ponder, and I enjoyed reading your answer. However, I think that this is not what OP asks. – Roger Vadim Jan 20 at 15:09 • @Vadim ...not in soo many words. Indeed, I stretched his question to cover the contrast between eigenstates of$a$to eigenstates of$a^\dagger$. The evident answer dispatches (2). – Cosmas Zachos Jan 20 at 15:13 ## 3 Answers 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 ― mysteriously, 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 (unnormalizable) state $$|\psi\rangle=\delta(a^\dagger - \beta) |0\rangle =\frac{1}{2\pi} \int \!\! dk ~e^{ik(a^\dagger -\beta)} |0\rangle \\ =\frac{1}{2\pi}\int\!\! dk~\exp(-ik\beta) \left (|0\rangle+ ik |1\rangle + ... +(ik)^n/\sqrt{n!}|n\rangle +...\right ) \\ = \delta(\beta)|0\rangle -\partial_\beta \delta (\beta)|1\rangle +...+{(-\partial_\beta )^n \delta(\beta)\over \sqrt{n!}}|n\rangle+...$$ formally 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. NB Add-on addressing the no-go proof Through the orgy of distributions involved, one may see the strictly formal loophole of the no-go proof, $$a^\dagger \left (\delta(\beta)|0\rangle -\partial_\beta \delta (\beta)|1\rangle +...+{(-\partial_\beta )^n \over \sqrt{n!}}\delta(\beta)|n\rangle+... \right ) \\ = \beta \left ( \delta(\beta)|0\rangle -\partial_\beta \delta (\beta)|1\rangle +...+{(-\partial_\beta )^n \over \sqrt{n!}}\delta(\beta)|n\rangle+...\right ).$$ Note the "snag" $$|0\rangle$$ term on the right vanishes by virtue of $$\beta \delta(\beta)=0$$, the second term $$-\beta \partial_\beta \delta(\beta) = \delta (\beta)$$, etc. Hyperformal to be sure, but operator-valued distributions are the lifeblood of QFT, and the task of "cleaning up" the landscape undertaken by axiomatic QFT is still not complete... In QM people have come to expect some rigging stunts will settle everything, but I am not too good at that. • @ Cosmas given this follow-up question, maybe we could ask you to add some additional information about the state$\delta(a^\dagger-\beta)|0⟩$? In particular, is it normalizable? If not, then how "pathological" is it? Say, is it as "bad" as$|x⟩$, or does it require even further subtleties to handle properly? – Emilio Pisanty Jan 19 at 17:19 • @Emilio I did something, to dramatize its pathological operator-valued distribution nature. It is, of course, a freak formal state, aggressively not normalizable... unless there is a slick trick collecting the δs... which I have not found. It is strictly an improper, "formal" loophole, that is ignored at the reader's risk. Dirac's book is not riven by such qualms and scruples... You can tell he had an engineering (and math) degree, and you can see why Feynman was so impressed/obsessed by him... – Cosmas Zachos Jan 19 at 20:32 • @ Cosmas Thanks for the added commentary. I for one will sign my name under "expecting rigging stunts to settle everything" -- if they don't then I would definitely be at a loss. – Emilio Pisanty Jan 20 at 15:13 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. • 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. – Halbeard Dec 4 '18 at 19:51
The first question has been already covered in other answers, but I recap it for completeness: $$a^\dagger$$ does not have right eigenvectors, but it has left eigenvectors, $$\langle\alpha | = |\alpha\rangle^\dagger = e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^{+\infty}\frac{(\alpha^*)^n}{\sqrt{n!}}\langle n|,$$ precisely, as suggested in the question. This fact is routinely used in derivation of the path integral formulation in terms of coherent states. This is really a homework question.
Regarding the second question: the claim about the coherent states similarity to the classical states is usually made in the context of quantizing electromagnetic fields, with the field operators represented by, e.g., $$\hat{E}(x,t) =\lambda e^{-ikx}a^\dagger(t) + \lambda e^{ikx}a(t).$$ With the evolution of the creation and annihilation operators given by unperturbed Hamiltonian, $$a(t)=ae^{_i\omega t}, a^\dagger(t)=a^\dagger e^{i\omega t}$$, averaging over the coherent states results in $$\langle \alpha | \hat{E}(x,t)|\alpha\rangle = \lambda e^{-ikx+i\omega t}\alpha^* + \lambda e^{ikx-i\omega t}\alpha = E_0\cos(kx-\omega t+\phi),$$ i.e., the classical expression for the electric field intensity. On the other hand, the number states, $$|n\rangle$$ do not have a direct classical correspondence - the values of the electromagnetic fields in these states are undefined. This correspondence to the coherent electromagnetic field is the reason for naming these states coherent.