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What is the time-structure (i.e., how does the magnitude and direction change with time) of the expectation value of the electric field $\langle\textbf{E}\rangle(t)$ of a radiation field described by a quantum coherent state $|\alpha\rangle$?

Actually, I heard that laser can be described by coherent states and from the time-structure, I want to see whether the expected electric field is polarized or not. I know little to no quantum optics and not sure how to extract the information about the nature of the observable electric field due to a coherent state such as a laser.

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  • $\begingroup$ What do you mean by "the time-structure", and what precisely do you mean by "average"? There is one unambiguous reading of your question but it is very narrow and answering that runs the danger of being very flat on several adjacent nontrivial conceptual issues. $\endgroup$ – Emilio Pisanty Jun 30 '18 at 23:45
  • $\begingroup$ Then don't say 'average', say 'expected' or (better) 'expectation value'. Those are protected terms with unambiguous meaning - that's why we have them. $\endgroup$ – Emilio Pisanty Jul 1 '18 at 11:50
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Your question is ill-posed.

Basically, when you say that you want to consider

a radiation field described by a quantum coherent state $|\alpha\rangle$

you are already implicitly considering the polarization of the electric field. This is because to speak of a quantum coherent state $|\alpha\rangle$ over a single mode you are already implicitly basing your treatment on a separation of the electromagnetic field observables over a basis of classical modes, i.e. on an expansion of the form $$ {\mathbf E}(\mathbf r,t) = \sum_n\bigg[a_n(t)\mathbf f_n(\mathbf r)+a_n(t)^*\mathbf f_n^*(\mathbf r)\bigg], $$ where $\mathbf f_n(\mathbf r)$ are vector-valued functions of position and the (complex) mode amplitudes $a_n(t)$ carry all of the dynamical information about the classical state of the field, and which you then translate via canonical quantization to an expansion of the form $$ \hat{\mathbf E}(\mathbf r) = \sum_n\bigg[\hat{a}_n\mathbf f_n(\mathbf r)+\hat {a}_n^\dagger\mathbf f_n^*(\mathbf r)\bigg], $$ where the classical mode amplitudes $a_n(t)$ get replaced by bosonic annihilation operators $\hat a_n$. (A previous appearance of this description on answer to a question of yours is here.)

Generically speaking, moreover, saying that the radiation is described by a coherent state $|\alpha\rangle$ implicitly implies that it is one of those modes that carries the coherent state in question, and that all the other modes are in the QED vacuum state.

Thus,

  • is the electric field polarized? yes, by implicit assumption.
  • what polarization does it have? it depends ─ whatever polarization is defined by the mode function $\mathbf f_n(\mathbf r)$.

Moreover, that polarization can be anything: it can be linearly polarized, it can be circularly polarized, and it can also be a space-dependent polarization like a radial or azimuthal polarization or more complicated combinations like, say, ellipse-point polarization fields.

On the other hand all of this can, of course, be quantified: in the Schrödinger picture we normally have the time dependence $t\mapsto |\alpha e^{-i\omega t}\rangle$, where the frequency $\omega$ comes from demanding that $\mathbf f_n(\mathbf r)$ be a divergenceless Helmholtz eigenfunction, and this will then give the time dependence $$ \langle \hat{\mathbf E}(\mathbf r) \rangle = \alpha e^{-i\omega t}\mathbf f_n(\mathbf r)+\alpha^* e^{+i\omega t}\mathbf f_n^*(\mathbf r). $$ If your mode was linearly polarized to begin with then you can set e.g. $\mathbf f_n(\mathbf r) = \mathbf e_x e^{ikx}$, or if you wanted a circular polarization you could swap the $\mathbf e_x$ for $\mathbf e_x+ i \mathbf e_y$, or you could have a space-dependent polarization. In the 'worst'-case scenario, $\mathbf f_n(\mathbf r)$ could be a space-dependent polarization field that varies on length scales much shorter than what your detector can resolve, in which case the field will look de-polarized to your detector.

However:

it is important to note that the assignment of the coherent state $|\alpha\rangle$ to a laser field is a good model in most conditions but not in all conditions. The clearest division is that this description is only valid for CW lasers, but there is also a huge cross-section of lasers that are pulsed instead of CW, and those require a multi-mode description to even begin to make sense.

This is part of a wider point, in that lasers are a varied bunch, and there are few properties that hold universally. In the particular case of polarization, most lasers do indeed produce a polarized output, but there are multiple cases that produce unpolarized or partially-polarized light, through a variety of physical mechanisms (like the space-dependent polarization mentioned above, or a time-varying output due to temperature or other fluctuations).

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Its hard to answer your question 'cause its a little ambiguous. Can you give an example of the coherent state you're talking about?

Meanwhile, all light is polarized no matter what. I think you are asking if the radiation field is linearly polarized, or maybe circularly polarized. In general, you can have a coherent quantum state with any polarization you like. So the answer is, "no, there is no special polarization associated with a generic coherent state." But depending on the coherent state, it might be linearly polarized.

When we talk about "unpolarized" light, what we really mean is that the polarization is fluctuating randomly with time. At any instant, there is a well-defined polarization its just that on average over some period of time, the polarization cancels out. If this is what you were getting at, then your question "Is it polarized" might be more aptly expressed as "is the polarization constant in time?" Or at least slowly varying in time.

I think when we talk about coherent states, we usually are not talking about a state where the polarization fluctuates randomly in time. So probably, the light is polarized in that sense.

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  • $\begingroup$ I've edited the question a bit! Let me know whether it is clearer now. :-) @GerryHarp $\endgroup$ – SRS Jul 1 '18 at 11:57
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When we just talk about a coherent state $|\alpha\rangle$, we usually only represent the particle-number degrees of freedom. If we want the coherent state to represent the spin degrees of freedom (polarization) as well, we need to expand the formalism a little. Perhaps one can add a little index: $|\alpha_s\rangle$. This index runs over the two orthogonal polarization state in terms of which one can specify the polarization. However, such a state is a pure state and it can only represent fully polarized light. If you want to represent more general states of polarization, including unpolarized light, you would need to turn it into a mixed state. One can do this by ensemble averaging; something like $$ \hat{\rho} = \sum_n |\alpha_{s,n}\rangle P_n \langle\alpha_{s,n}| , $$ where $P_n$ represents the probablilities for the different states of polarization.

Not sure if this answers your question.

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You have already got two good answers -- one from a classical and one from a quantum point of view. Here’s a third, which tries to bridge the gap.

Consider well-collimated light. If you compute the 2x2 covariance matrix of transverse electric field components as $\left\langle {{E}_{m}}{{E}_{n}} \right\rangle $, you will find that completely polarized light gives you a rank-1 matrix, partially polarized light a rank-2 matrix with unequal eigenvalues, and completely polarized light a multiple of the identity matrix. Translated into QM notation, $\left| \alpha \right\rangle \left\langle \alpha \right|$ is your rank-1 matrix, and the sum over polarization states your rank-2 matrix.

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