# I Spin and magnetic field of a photon

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1. Jun 1, 2017

### Cecilie Glittum

Is the spin of a photon pointing in the same direction as the magnetic field of the photon?

2. Jun 1, 2017

### DrDu

No, it doesn't. The electromagnetic field is a transversal field, i.e., both the magnetic and the electric fields are perpendicular to the direction of the propagation, while the spin is either parallel or anti-parallel to the propagation direction.

3. Jun 1, 2017

### Staff: Mentor

Strictly speaking, a single photon does not have a magnetic field. A "magnetic field", such as the field produced by a magnet, is a state involving a very large number of photons (and they're not really photons that you detect as photons, it's more complicated than that), and such states won't, in general, have a single "spin" that points in a single direction.

If, alternately, you're thinking of the magnetic field of a light beam, where you're thinking of the beam as a traveling wave of electric and magnetic fields, then you're not using a photon model of light at all (you're using a classical wave model), and it doesn't really make sense to ask what direction the "spin" of a photon is in. The light wave will have a polarization, but that's somewhat more complicated in this model than just a "spin".

I think this is mixing models--the classical wave model of light and the quantum photon model--so I don't know that it is really a valid answer.

4. Jun 1, 2017

### Cecilie Glittum

I understand. What I'm actually thinking about is that in the Zeeman effect, both sigma and pi photons are emitted, and their polarization is orthogonal. I have understood that the sigma photons have m=+-1 and pi photons have m=0. I'm wondering how this is connected to the polarization of the light.

5. Jun 1, 2017

### Staff: Mentor

These values of $m$ actually refer to the change in the state of the atom when the photon is emitted. Heuristically, photons carry angular momentum $\pm 1$, so emitting a photon in a state of definite angular momentum can change the angular momentum of the atom (more precisely, the orbital angular momentum of the electron in the atom that goes from a higher to a lower energy level by emitting the photon) by $\Delta m_l = \pm 1$ (in units of $\hbar$). These transitions are the $\sigma$ transitions. However, this being quantum mechanics, it is also possible for the electron to emit a photon in a superposition of angular momentum states which allows for a transition with $\Delta m_l = 0$. These transitions are the $\pi$ transitions.

By "angular momentum" above we really mean "angular momentum measured about an axis parallel to the direction of the magnetic field that is producing the Zeeman effect". So the photons emitted by the $\sigma$ transitions will have a definite angular momentum about that axis, which we can call the $z$ axis--in other words, if we measured the angular momentum of these photons about the $z$ axis, we would always get a definite value of $\pm 1$, corresponding to the $\Delta m_l$ of the electron in the atom.

The photons emitted by the $\pi$ transitions, however, will be in a superposition of those two angular momentum states with equal amplitudes, so if we measured their angular momentum about the $z$ axis, there would be an equal probability of getting $+1$ or $-1$ as the result. The average of these is $0$, corresponding to a change $\Delta m_l = 0$ in the electron in the atom.

Per my previous post, we have to be careful about mixing models, but heuristically, if we think of the $z$ axis as the direction of propagation of the photon (corresponding to looking at the light along a direction parallel to the magnetic field), then the $\sigma$ states, with definite angular momentum $\pm 1$ about the $z$ axis, correspond to circular polarization, with $+1$ being right-handed and $-1$ being left-handed (at least, I think I've got that sign convention right). The $\pi$ states, with average angular momentum $0$, correspond to linear polarization.

6. Jun 1, 2017

### DrDu

Magnetic field is an observable, which means that you can measure it whenever you like and you will get a sensible answer also for a one photon state. Specifically, you will never find a non-zero value in propagation direction, but well so for components perpendicular to it. (I am assuming an appropriate gauge here, e.g. Coulomb, where only transversal photons exist.)

7. Jun 1, 2017

### Staff: Mentor

What quantum operator does it correspond to?

8. Jun 1, 2017

9. Jun 1, 2017

### DrDu

I am not sure about this. As I tried to explain, for photons the direction of propagation and spin are tethered. So the sigma photons are emitted preferentially along the z axis while the pi photons in the plane perpendicular to it. Linear polarisation can be obtained also from superpossition of sigma photons.

10. Jun 1, 2017

### Staff: Mentor

Ok. Then the E and B field operators given here, since they are linear combinations of the creation and annihilation operators, do not commute with any spin operator. That means there is no such thing as a "one photon state" that has a definite value for both magnetic field and spin. In fact, since the creation operator has no possible eigenstates, there is no such thing as a quantum state of the EM field that has a definite value for the magnetic field, period. When we measure magnetic fields at the classical level we are measuring expectation values, not eigenvalues.

Yes, this is a good point that I glossed over. Strictly speaking, if you are looking exactly along the direction of the magnetic field, you won't see any $\pi$ photons (at least to a very good approximation). So any results you get from linear polarization measurements are really due to, as you say, measuring superpositions of multiple $\sigma$ photons. But if we consider an idealized measurement where we can always measure the spin of just one photon at a time, we will never see linear polarization if we measure spin along this axis.

11. Jun 2, 2017

### DrDu

Nobody ever talked about magnetic fields at the classical level besides you. With the question being about single photon states, which are as non-classical as QM can get, when talking about electric and magnetic fields, I always had operators in mind, not classical expectation values.

With $\langle B \rangle=0$ and $\langle B^2 \rangle >0$ in an n-photon state, it is clear that the B field is non-zero but wildly fluctuating. Nevertheless $\langle (p\cdot B)^2 \rangle=0$, i.e. B is purely transversal. As helicity (the "spin" of the photon) refers to the direktion of p, it should be clear that for a photon B cannot be parallel to spin. Note that this time I tried to base my argument only on expectation values.

12. Jun 2, 2017

### DrDu

From
http://www.arm.ac.uk/lectures/landstreet/text/landstreet-larochelle-1.pdf
"The various Zeeman components into which a spectral line splits when the atom is in a magnetic field are polarised. This polarisation depends on the orientation of the field with respect to the direction of the emitted light. Two simple cases occur. If the field is perpendicular to the line of sight, then transitions with ∆m = 0 (pi components) are linearly polarised with their electric field vectors parallel to the field lines. Transitions with ∆m = ±1 (sigma components) are linearly polarised perpendicular to the field. If the field is aligned parallel to the line of sight, the line components produced by ∆m = 0 transitions vanish (they have zero intensity), while the transitions for which ∆m = 1 and −1 have opposite senses of circular polarisation. Intermediate field orientations lead to elliptical polarisation of the light from the various Zeeman components."

13. Jun 2, 2017

### LeandroMdO

A "magnetic field" can be defined as simply the expectation value of (say) F_{12} in a given state. Write down F_{12} in terms of A and you'll have a suitable canonical form for the magnetic field operator. Of course, F_{12} is a linear function of A so this operator only has one of either a or a^\dagger. So in a photon number eigenstate the expectation value of B vanishes. The expectation value of B², however, does not vanish, so it's incorrect to say that a photon doesn't have a magnetic field. It does, but it averages to zero, just like the position of a particle in the ground state of a harmonic oscillator averages to zero.

There's no mixing, really. This statement is valid in either classical or quantum field theory. The transversality can be seen straight from the operator representation I mentioned above.

14. Jun 2, 2017

### DrDu

Why do you all insist on taking expectation values?

15. Jun 2, 2017

### LeandroMdO

I did that solely to indulge the desire for a "single" answer to the question "what is the magnetic field of a photon". But I agree that your posts already contain all the relevant physics (I hit reply before reading most of the thread -- my bad).

Last edited: Jun 2, 2017
16. Jun 2, 2017

### Staff: Mentor

Then your statement that "both the magnetic and the electric fields are perpendicular to the direction of the propagation" makes no sense, because it would require the fields to have definite values, and they don't--they can't, since their operators have no eigenstates.

But...

I thought you objected to taking expectation values?

No, it doesn't. The direction of p is the direction of the wave vector, not the spin. The spin only "points" in that direction if it is in an eigenstate of the spin operator along that direction. But only two of the possible "one photon states" have that property. The rest don't.

Why? You objected to my talking about the "classical level", which was the only reason I brought up expectation values at all. If we're going to talk about a "one photon state", then I don't think expectation values can justify your statements; only eigenvalues would, and they don't, as I've already pointed out.

Also, if you're going to allow expectation values, then your statements about spin no longer make sense, because the expectation value of spin does not have to be $+1$ or $-1$, yet you are talking as if those are the only two possible spin values.

17. Jun 2, 2017

### Staff: Mentor

That's one possible definition, but not the only one. I've stated my objections to it for this discussion in my responses to DrDu just now.

I don't see how. The operators described in the Wikipedia article that DrDu linked to have both $a$ and $a^\dagger$. They look correct to me.

18. Jun 2, 2017

### Staff: Mentor

Since the quantum operator corresponding to this observable has no eigenstates, what value will you get when you make this measurement you describe?

19. Jun 2, 2017

### DrDu

Why doesn't it have eigenstates?

20. Jun 2, 2017

### Staff: Mentor

Because the creation operator has no eigenstates.