I Spin and magnetic field of a photon

[Cecilie Glittum]

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

 

[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.

 

[PeterDonis]

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

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".

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.

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.

 

[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.

 

[PeterDonis]

I have understood that the sigma photons have m=+-1 and pi photons have m=0.

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.

I'm wondering how this is connected to the polarization of the light.

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.

 

[DrDu]

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.

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.)

 

[PeterDonis]

Magnetic field is an observable

What quantum operator does it correspond to?

 

[DrDu]

What quantum operator does it correspond to?

https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

 

[DrDu]

The $\pi$ states, with average angular momentum $0$, correspond to linear polarization.

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.

 

[PeterDonis]

https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

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.

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.

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.

 

[DrDu]

↑

"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."
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.

When we measure magnetic fields at the classical level we are measuring expectation values, not eigenvalues.

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.

 

[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."

 

[LeandroMdO]

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.

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.

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.

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.

 

[DrDu]

A "magnetic field" can be defined as simply the expectation value of (say) F_{12} in a given state.

Why do you all insist on taking expectation values?

 

[LeandroMdO]

Why do you all insist on taking expectation values?

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).

 

[PeterDonis]

when talking about electric and magnetic fields, I always had operators in mind, not classical expectation values

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.

Nevertheless $\langle (p\cdot B)^2 \rangle=0$

, i.e. B is purely transversal

But...

Why do you all insist on taking expectation values?

I thought you objected to taking expectation values?

As helicity (the "spin" of the photon) refers to the direktion of p

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.

Note that this time I tried to base my argument only on expectation values

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.

 

[PeterDonis]

A "magnetic field" can be defined as simply the expectation value of (say) F_{12} in a given state.

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.

F_{12} is a linear function of A so this operator only has one of either a or a^\dagger

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.

 

[PeterDonis]

Magnetic field is an observable, which means that you can measure it whenever you like

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

 

[DrDu]

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

Why doesn't it have eigenstates?

 

[PeterDonis]

Why doesn't it have eigenstates?

Because the creation operator has no eigenstates.

 

[DrDu]

But the B operator is made up of a linear combination of the creation and anihilation operator. Therefore it is hermitian and obviously has eigenstates.
compare to the position operator of the harmonic oscillator $x \sim (a+a^+)$.

 

[PeterDonis]

the B operator is made up of a linear combination of the creation and anihilation operator. Therefore it is hermitian and obviously has eigenstates.

compare to the position operator

I see what you mean, but the "eigenstates" of the position operator are not normalizable (nor are those of the momentum operator), which means they're not actually in the Hilbert space of the system. I would expect a similar objection to apply to the "eigenstates" of the E and B operators.

 

[LeandroMdO]

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.

What he stated to you is correct. In a quantum field theory fields are promoted to operators. Saying that a photon has no magnetic field is analogous to saying that a particle in the ground state of a harmonic oscillator has no position. Its position is simply zero on average. A measurement of the magnetic field of a photon would reveal something both finite and perpendicular to the direction of motion. The x operator also doesn't commute with the Hamiltonian in the harmonic oscillator, but we still talk about the position of a particle in the ground state.

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.

They are. I misspoke: what I meant is that there's no _term_ with both a and a^\dagger, so nothing that would yield something nonzero when taking the expectation value in an eigenstate of photon number.

 

[PeterDonis]

Saying that a photon has no magnetic field is analogous to saying that a particle in the ground state of a harmonic oscillator has no position

I see your point as it relates to the use of ordinary language. But consider: when we measure the "position" of a quantum object, do we actually measure using the operator $\hat{x}$? No; we don't. Since the eigenstates of this operator are not normalizable, there is no way to physically realize a measurement of it. What we actually measure is some different operator whose eigenstates only confine the position to some finite range of values and are therefore normalizable. So when we describe this object as "having a position", we don't actually mean that it has a definite value corresponding to the operator $\hat{x}$.

In other words, our usual use of language is somewhat loose, because the operator we think of when we use the word "position" is not actually the one we physically realize. And I would expect that to be the case also for the E and B field operators, for similar reasons. I would be interested to know if this issue is explicitly treated for the EM field case in any available sources; I have not seen such a treatment, but I am not very familiar with the relevant literature in this area.

what I meant is that there's no _term_ with both a and a^\dagger, so nothing that would yield something nonzero when taking the expectation value in an eigenstate of photon number.

Ok, good, I agree with this.

 

[DrDu]

I see what you mean, but the "eigenstates" of the position operator are not normalizable (nor are those of the momentum operator), which means they're not actually in the Hilbert space of the system. I would expect a similar objection to apply to the "eigenstates" of the E and B operators.

True, but this is a technicality already solved by von Neumann:
https://en.wikipedia.org/wiki/Spectral_theorem

 

[DrDu]

In other words, our usual use of language is somewhat loose, because the operator we think of when we use the word "position" is not actually the one we physically realize. And I would expect that to be the case also for the E and B field operators, for similar reasons. I would be interested to know if this issue is explicitly treated for the EM field case in any available sources; I have not seen such a treatment, but I am not very familiar with the relevant literature in this area.

Again spectral theorem. Additionally, you have to use smeared out operators in QFT.
The main problem is that nobody knows how to give sense to products of these ill defined operators (that's what renormalisation is all about!) when describing interacting fields. However, for free particles the operators can be made waterproof. See axiom W1 in https://en.wikipedia.org/wiki/Wightman_axioms

 

[PeterDonis]

spectral theorem

This requires that the operators in question are self-adjoint, which means that they must be linear maps from some vector space $V$ into itself that are equal to their own adjoint. But, as I pointed out, the operator $\hat{x}$ is not normalizable, so it is not a linear map from $L^2(R)$, which is the relevant vector space, into itself. So it is not self-adjoint by the strict mathematical definition that is required, as I understand it, to prove the spectral theorem. Similar remarks would apply to the E and B operators.

you have to use smeared out operators in QFT

This is a valid point, since we started out talking about photons, which are inherently relativistic so a QFT treatment would be required. The question would be, what are these operators for E and B, and what eigenstates do they have?

 

[DrDu]

This requires that the operators in question are self-adjoint, which means that they must be linear maps from some vector space $V$ into itself that are equal to their own adjoint. But, as I pointed out, the operator $\hat{x}$ is not normalizable, so it is not a linear map from $L^2(R)$, which is the relevant vector space, into itself.

No, x is is water-proof self-adjoint. It is defined on a dense subset of the hilbert space and its adjoint is defined on the same subset. Furthermore, it coincides with its inverse on the domain of definition. This is sufficient for the operator to be self-adjoint.

 

[PeterDonis]

It is defined on a dense subset of the hilbert space and its adjoint is defined on the same subset.

Ah, got it.

 

[LeandroMdO]

In other words, our usual use of language is somewhat loose, because the operator we think of when we use the word "position" is not actually the one we physically realize. And I would expect that to be the case also for the E and B field operators, for similar reasons.

That's right. What is physically measured is more like a test function that approaches a position observable in the limit of infinite precision. That also explains the usefulness of the position operator, which should be understood in the distribution sense. There's a clear message here: the more precise you make your "position" operator, the closer your measurements approach that of the idealized position operator. People don't normally prevaricate on this point because there's very little here of fundamental importance: we all know that perfectly precise measurements are impossible, even in classical physics. There are interpretational matters concerning e.g. counterfactual definiteness, but that doesn't depend on whether the operator in question has normalizable eigenvectors, and it doesn't even depend on whether you're measuring observables that commute.

More concretely, these difficulties can be sidestepped by talking about results of measurements rather than ontological properties. If I measure an operator that approaches the idealized magnetic field operator in some plane-wave one-photon state, what do I get? Well, I most likely get some nonzero vector, guaranteed to be approximately orthogonal to the photon's momentum vector. If I make repeated measurements, I know they will average to zero. This seems like a reasonable answer modulo the difficulties mentioned above.

I would be interested to know if this issue is explicitly treated for the EM field case in any available sources; I have not seen such a treatment, but I am not very familiar with the relevant literature in this area.

I don't know of any sources that deal with this in a more practical context either. Maybe something quantum optics?

 

[PeterDonis]

these difficulties can be sidestepped by talking about results of measurements rather than ontological properties

Yes, but the same caution should then apply to states; we should be talking about preparation procedures instead. For example:

some plane-wave one-photon state

This should be replaced with a description of the preparation procedure, i.e., how did we make this thing we are now measuring?

 

[PeterDonis]

If I measure an operator that approaches the idealized magnetic field operator in some plane-wave one-photon state, what do I get? Well, I most likely get some nonzero vector, guaranteed to be approximately orthogonal to the photon's momentum vector.

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.)

 

[LeandroMdO]

The same type of reasoning applies to a measurement of position in the ground state of the harmonic oscillator. You suitably prepare the state first, and measure position second, after which the particle is no longer in any energy eigenstate. Nevertheless, something can be said about measurements of position in the ground state. Or, for a perhaps better example, in the infinite square well the probability of measuring the position to be outside the box is zero, which justifies a statement such as "the particle is inside the box". The probability of measuring a magnetic field to be anything but perpendicular to the direction of the photon's momentum (known from the preparation procedure) is similarly zero, which justifies the statement that the magnetic field and momentum are orthogonal.

 

[PeterDonis]

the direction of the photon's momentum (known from the preparation procedure)

And what is the preparation procedure? How are we preparing a "plane-wave one-photon state"?

 

[LeandroMdO]

As before, the statement must be understood in a limiting sense. It is a technicality.

 

[vanhees71]

Now everything is a big mess again! Photons are not as simple as other quanta since they are massless and have spin 1. The first subtlety is that they have only 2 not three spin states. That's because there's no restframe for massless quanta. The full group theoretical foundation can be found in my lecture notes on QFT in appendix B:

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

The 2nd subtlety is that there's no position operator for photons. You can't treat photons in a first-quantization way to begin with. It's the very point where you really need QFT, if you want to get things right (although there are claims in the literature to the contrary).

The 3rd subtlety is that the naive canonical quantization doesn't work, because the classical Maxwell theory is a gauge theory. As is also shown in my QFT notes, that must be so from the group-theoretical point of view, if you don't want to introduce continuous spin-like degrees of freedom, which are not observed.

A quick hand-wavy way out (for the free field) is to completely fix the gauge. For the free em. field the most convenient choice is the radiation gauge, which consists of two gauge conditions for the vector field
$$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
The equation of motion is
$$\Box \vec{A}=(\partial_t^2-\Delta)\vec{A}=0.$$
Now we look for plane-wave modes and get
$$\vec{A}(t,\vec{x})=a(\vec{k},\lambda) \vec{\epsilon}_{\lambda} \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x})+c.c.$$
$\lambda$ labels two polarization vectors, which can be chosen arbitrarily, but they must fulfill $\vec{\epsilon}_{\lambda} \cdot \vec{k}=0$ (i.e., you have two linearly independent choices). The most natural choice are the two circular polarized modes, which corresponding to helicity eigenmodes to the eigenvectors $h \in \{-1,1 \}$. Then there's the dispersion relation $\omega=|\vec{k}|$.

Now we skip all the subtleties of canonical quantization by observing that we have a set of harmonic oscillators there and thus we can make the ansatz
$$\hat{\vec{A}}(t,\vec{x}) = \sum_{\vec{k},\lambda} \frac{1}{\sqrt{2 \omega}} \hat{a}(\vec{k},\lambda) \vec{\epsilon}(\vec{k},\lambda) \exp(-mathrm{i} k \cdot x)|_{k^0=\omega}+\text{h.c.}$$
Here, I've assumed a large quantization volume as a cube of length $L$ with periodic boundary conditions for the field operator, so that the momenta are discrete, $\vec{k} \in \frac{2 \pi}{L} \mathbb{Z}^3$.

The creation and annhiliation operators fulfill the Bose commutator relations
$$[\hat{a}(\vec{k},\lambda) ,\hat{a}(\vec{k}',\lambda')]=0, \quad [\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=\delta_{\vec{k},\vec{k}'} \delta_{\lambda,\lambda'}.$$
The total four-momentum is given by
$$\hat{P}^{\mu} = \sum_{\vec{k},\lambda} \hat{N}(\vec{k},\lambda) k^{\mu}, \quad \hat{N}(\vec{k},\lambda)=\hat{a}^{\dagger}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda).$$
The ground state is the vacuum $|\Omega \rangle$ defined by
$$\hat{a}(\vec{k},\lambda) |\Omega \rangle=0.$$
A single-mode single-photon state is given by
$$|\vec{k},\lambda \rangle=\hat{a}^{\dagger}(\vec{k},\lambda) |\Omega \rangle.$$
A single-mode coherent state that is most closely related to a classical em. field is given by the eigenvector of $\hat{a}(\vec{k},\lambda)$,
$$\hat{a}(\vec{k},\lambda) |\alpha \rangle = \alpha |\alpha \rangle.$$
Since $\hat{a}$ is NOT self-adjoint $\alpha \in \mathbb{C}$. It's not a state of definite photon number. The probability to find $n$ photons (of momentum $\vec{k}$ and with polarization $\lambda$ of course) is given by a Poisson distribution
$$P(n)=\exp(-n|\alpha|^2) \frac{|\alpha|^{2n}}{n!}$$
The mean photon number and four-momentum is
$$\langle N(\vec{k},\lambda) \rangle=|\alpha|^2, \quad \langle \vec{P} \rangle=k^{\mu} |\alpha|^2.$$

The operators for the electromagnetic field components are given by
$$\hat{\vec{E}}(t,\vec{x})=-\partial_t \hat{\vec{A}}(t,\vec{x}), \quad \hat{\vec{B}}(t,\vec{x})=\vec{\nabla} \times \hat{\vec{A}}(t,\vec{x}).$$
For a given state you can calculate expectation values as for any observable according to Born's rule.

 

[vanhees71]

And what is the preparation procedure? How are we preparing a "plane-wave one-photon state"?

There is no plane-wave one-photon state (in the limit $L \rightarrow \infty$ in the formalism of my previous posting) since in this limit you have the usual $\delta$-distribution as in ordinary QT,
$$\langle \vec{k},\lambda|\langle \vec{k}',\lambda' \rangle=\delta^{(3)}(\vec{k}-\vec{k}') \delta_{\lambda \lambda '}.$$
So for a true one-photon state, you must use a wave packet.

The preparation procedure for a single photon was very complicated not too long ago. Nowadays it's quite simple, because of the discovery of parametric down-conversion. There a laser shines into certain types of crystals (BBO=beta barium borat) and due to non-linear optics effects there's a process where one photon out of the laser field (emitting light close to a coherent state as described above) is absorbed in the crystal and two phase-matched polarization entangled photons emitted, obeying energy-momentum conservation. You use one photon (the "signal") to be sure to have created a photon pair (only about 1 out of 10^{12} laser photons are converted according to Wikipedia), using a polarizer. Then you are sure to have the other photon in a given (quite well-determined) momentum and definite polarization state (the "idler"). This is also known as a "heralded photon". This prepares true single-photon states!

https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion

Sometimes you read that you simply might dim down a laser very much so that the average number of photons $|\alpha|^2 \simeq 1$, but that's far from being a single-photon source. The corresponding coherent state is a superposition of mostly the vacuum state and the single-photon state but also contains small mixtures of all multi-photon states.

 

[PeterDonis]

There is no plane-wave one-photon state

That's what I thought.

 

[LeandroMdO]

That's what I thought.

In the real world there are no plane waves, period. Statements about plane waves are still useful, however, both in the limiting sense and in the Fourier decomposition sense.

 

[PeterDonis]

Statements about plane waves are still useful

For some purposes, yes. But I'm still not sure that justifying claims like "the magnetic field of a photon is perpendicular to its direction of propagation" (or, for that matter, "the spin of a photon is parallel with its direction of propagation") is one of them. If there is no "one photon state" that we can actually prepare and measure that has this property, then I don't think the statement is justified. If there is, it isn't a plane wave state, so what state is it?

 

[PeterDonis]

Then you are sure to have the other photon in a given (quite well-determined) momentum and definite polarization state

How would this state be described mathematically?

 

[vanhees71]

It's something like
$$|\Psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{k} \frac{1}{\sqrt{(2 \pi)^3 2 \omega(\vec{k})}} A(\vec{k}) \hat{a}^{\dagger}(\vec{k},\lambda)|\Omega \rangle,$$
where $A(\vec{k}6)$ fullfills
$$\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3 2 \omega(\vec{k})} |A(\vec{k})|^2=1.$$

 

[PeterDonis]

It's something like

I see how this state has a definite polarization, since there is no integration over $\lambda$; but since there is integration over $k$, isn't this state a superposition of different momentum states, and therefore not a state with a definite value of momentum?

 

[DrDu]

I thought we were talking about single photon states not single photon states being exact momentum eigenstates? Why are you constantly and artificially complicating things?

 

[PeterDonis]

I thought we were talking about single photon states not single photon states being exact momentum eigenstates?

If they're not exact momentum eigenstates, then they don't have a definite direction of the momentum, so it makes no sense to say that the magnetic field is transverse to the direction of the momentum--at least not unless you allow that statement to be made using expectation values, which you said you objected to (and which I don't think it's reasonable to use for one photon states anyway, since expectation values imply averaging over many observations).

Similar remarks apply to the direction of the magnetic field itself--is the "single photon state" in question an eigenstate of the magnetic field operator? I haven't worked that out for the state that @vanhees71 wrote down. But if it isn't, then the magnetic field doesn't have a definite direction either.

 

[vanhees71]

I see how this state has a definite polarization, since there is no integration over $\lambda$; but since there is integration over $k$, isn't this state a superposition of different momentum states, and therefore not a state with a definite value of momentum?

You can make $A(\vec{k})$ very sharply peakded around an arbitrary momentum $\vec{k}=\vec{k}_0$. Then you get a true one-photon state that has nearly definite momentum and a precise polarization state.

 

[PeterDonis]

a true one-photon state that has nearly definite momentum and a precise polarization state

Yes, "nearly definite". But that still means that when you talk about the "direction" of the momentum you're talking about an expectation value.

Also, what about the magnetic field operator? How would it act on this state?

 

[vanhees71]

You can derive the action of $\hat{\vec{B}}(x)$ on my example state by inserting the mode decomposition given above, using the commutation relations for creation and annihilation operators.

I'm also not understanding, why you insist on plane-wave modes. They are generalized eigenfunctions of momentum. Also in classical electrodynamics plane waves are not realizable in nature exactly. They have diverging total energy and momentum.

 

[PeterDonis]

You can derive the action of $\hat{\vec{B}}(x)$ on my example state by inserting the mode decomposition given above, using the commutation relations for creation and annihilation operators.

Ok, that's what I thought, and when I do that it does not look like your example state is an eigenstate of $\hat{\vec{B}}(x)$.

I'm also not understanding, why you insist on plane-wave modes

I don't "insist" on them as a description of the actual states involved; I understand perfectly well that those states are not plane waves. 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. And as has been discussed earlier in this thread, for a "single photon state" I don't think an expectation value can be given that interpretation, since taking an expectation value implies averaging over many measurements. (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.)

 

[vanhees71]

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