I Spin and magnetic field of a photon

[PeterDonis]

It's not an eigenstate of $\vec{B}$. Why should it be? It's an eigenstate of $\hat{N}$!

That was my intuition, yes, but I wanted to check how the math worked out.

 

[DrDu]

There's another caution here as well, in addition to the one I gave in my last post. If we are going to say that the magnetic field is orthogonal to the photon's momentum, we have to be able to measure each one without disturbing the other, i.e., their measurements must commute. Do they?

(Similar remarks would apply to statements about the relative directions of the magnetic field and the spin.)

No, definitely not. Let's take another example, namely the angular momentum of an electron. It is defined as $L=r \times p$.
L is only non-zero if r isn't parallel to p.
Would you also insist that for the angular momentum to be well defined, you must be able to measure simultaneously r and p, which is impossible because they are non-commuting observables.

 

[vanhees71]

The total angular momentum of an electron (in the non-relativistic approximation) is represented by self-adjoint operators,
$$\hat{\vec{J}}=\hat{\vec{L}}+\hat{\vec{S}}=\hat{\vec{x}} \times \hat{\vec{p}} + \hat{\vec{s}}.$$
Since it's impossible to determine (not measuring!) position and momentum of the electron simultaneously, your above description of determining orbital angular momentum of course doesn't make any sense. Nevertheless the electron can be in a definite orbital-angular momenum eigenstate, i.e., a simultaneous eigenstate of $\hat{\vec{L}}^2$ and $\hat{L}_z$, which are commuting self-adjoint operators, thus representing compatible observables.

For photon states, of course the electric and magnetic components are not compatible. Nevertheless you can have definite photon-number eigenstates (as demonstrated above for a single-photon state).

 

[PeterDonis]

Would you also insist that for the angular momentum to be well defined, you must be able to measure simultaneously r and p, which is impossible because they are non-commuting observables.

Well, if $L = r \times p$, and we can't measure $r$ and $p$ simultaneously, then how do we measure the observable $L$? I understand that we can compute its expectation value, but we're not talking about expectation values (at least I thought we weren't).

 

[vanhees71]

You can simultaneously determine $\vec{L}^2$ and one component in an arbitrary direction (usually one takes the $z$ direction) but never all three components since $\hat{\vec{L}}^2$ and $\hat{L}_z$, because these two operators commute, but
$$[\hat{L}_j,\hat{L}_k]=\mathrm{i} \hbar \epsilon_{ijk} \hat{L}_i.$$

 

[LeandroMdO]

But if we are going to talk about the "direction of propagation" of a photon (for example if we want to say that the magnetic field is perpendicular to the direction of propagation), and the state is not a plane wave, then the only way of making sense of that direction is as an expectation value.

Even in classical mechanics the idea that E and B are orthogonal to the direction of propagation is not realized exactly. For example, if you have a circularly polarized wave packet of finite size, the Poynting vector on the edges of the packet points in an azimuthal direction, and carries the http://people.westminstercollege.edu/faculty/ccline/courses/phys425/AJP_54(6)_p500.pdf angular momentum of the wave. Something similar happens in a waveguide (see fig 2).

So clearly the idea that electric and magnetic fields are orthogonal to the direction of propagation must be handled with care in realistic cases. That is fine, but I prefer handling conceptual issues in easy ones.

(Similar remarks apply to the "direction" of the magnetic field as well, if I'm correct that the state you wrote down is not an eigenstate of the B operator.)

Let me try a different argument. We've been considering circularly polarized states thus far because they're the "natural" ones, but we don't have to. We can just as well quantize photons in a linearly polarized basis. You can read out easily from the expressions for the quantized electromagnetic field what this would look like, simply by replacing the set of e^\mu vectors by e^x and e^y. It is then clear that while the value of the electric field does not commute with photon number, the direction of it does (it is the definition of polarization after all).

I still maintain that this is inessential, because even for circularly polarized light the electric and magnetic fields must lie on a plane perpendicular to the momentum vector (near the center of a wavepacket of sharply peaked momentum), but it might help.

 

[DrDu]

Well, if $L = r \times p$, and we can't measure $r$ and $p$ simultaneously, then how do we measure the observable $L$? I understand that we can compute its expectation value, but we're not talking about expectation values (at least I thought we weren't).

You could use e.g. some Stern Gerlach type experiment, i.e. measure the deflection when passing an atom through a magnetic field.
But let's look at even a simpler property, velocity: You can define it as $v=(x(t)-x(0))/t$ (I consider it to be constant for simplicity of argument). Given this expression I think you would demand that we measure x at two times, so that the particle has to be in a position eigenstate. However, it is impossible for a particle to have both a precisely known position and velocity. What comes to our rescue is that v is invariant under a shift of x, i.e., $v=(x(t)-x(0))/t =x(t)+\Delta X -(x(0)+\Delta X))/t$, that is, we don't need absolute information about the position of the particle. In fact, velocity can be measured with high accuracy e.g. from Doppler effect measurements, e.g. on planets orbiting distant stars whose distance is only very vaguely known.
The same holds in QM. Usually, we can measure any (hermitian) operator of r and p, although it does not commute with neither r nor p. The point is that we are looking only at some specific correlation between r and p and don't need absolute information on these quantities. Especially the relative direction of B and p or S is such a correlation.

 

[vanhees71]

I don't know about a concrete procedure to prepare a precise angular-momentum state. The SG apparatus of course comes immediately to mind. I also guess you can nowadays prepare atoms of ultracold gases in traps in all kinds of quantum states using various techniques, particularly laser excitations with taylored pulsed em. fields and the like. Maybe, we have an expert in atomic physics around, who can tell more.

 

[DrClaude]

I also guess you can nowadays prepare atoms of ultracold gases in traps in all kinds of quantum states using various techniques, particularly laser excitations with taylored pulsed em. fields and the like. Maybe, we have an expert in atomic physics around, who can tell more.

Polarized atomic samples can be obtained by optical pumping. This, in combination with magnetic trapping, is commonly used to get clouds of atoms in a single hyperfine state $|F, M_F \rangle$.