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Cecilie Glittum
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Is the spin of a photon pointing in the same direction as the magnetic field of the photon?
Cecilie Glittum said:Is the spin of a photon pointing in the same direction as the magnetic field of the photon?
DrDu said: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.
Cecilie Glittum said:I have understood that the sigma photons have m=+-1 and pi photons have m=0.
Cecilie Glittum said:I'm wondering how this is connected to the polarization of the light.
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 said: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.
DrDu said:Magnetic field is an observable
https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_fieldPeterDonis said:What quantum operator does it correspond to?
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 said:The ##\pi## states, with average angular momentum ##0##, correspond to linear polarization.
DrDu said:
DrDu said: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.
"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.
PeterDonis said:When we measure magnetic fields at the classical level we are measuring expectation values, not eigenvalues.
PeterDonis said: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.
PeterDonis said: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.
Why do you all insist on taking expectation values?LeandroMdO said:A "magnetic field" can be defined as simply the expectation value of (say) F_{12} in a given state.
DrDu said:Why do you all insist on taking expectation values?
DrDu said:when talking about electric and magnetic fields, I always had operators in mind, not classical expectation values
DrDu said:Nevertheless ##\langle (p\cdot B)^2 \rangle=0##
, i.e. B is purely transversal
DrDu said:Why do you all insist on taking expectation values?
DrDu said:As helicity (the "spin" of the photon) refers to the direktion of p
DrDu said:Note that this time I tried to base my argument only on expectation values
LeandroMdO said:A "magnetic field" can be defined as simply the expectation value of (say) F_{12} in a given state.
LeandroMdO said:F_{12} is a linear function of A so this operator only has one of either a or a^\dagger
DrDu said:Magnetic field is an observable, which means that you can measure it whenever you like
Why doesn't it have eigenstates?PeterDonis said:Since the quantum operator corresponding to this observable has no eigenstates, what value will you get when you make this measurement you describe?
DrDu said:Why doesn't it have eigenstates?
DrDu said: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
PeterDonis said: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.
PeterDonis said: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.
LeandroMdO said: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
LeandroMdO said: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.
True, but this is a technicality already solved by von Neumann:PeterDonis said: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.
Again spectral theorem. Additionally, you have to use smeared out operators in QFT.PeterDonis said: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.
DrDu said:spectral theorem
DrDu said:you have to use smeared out operators in QFT
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 said: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.
DrDu said:It is defined on a dense subset of the hilbert space and its adjoint is defined on the same subset.
PeterDonis said: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.
PeterDonis said: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.
LeandroMdO said:these difficulties can be sidestepped by talking about results of measurements rather than ontological properties
LeandroMdO said:some plane-wave one-photon state
LeandroMdO said: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.
LeandroMdO said:the direction of the photon's momentum (known from the preparation procedure)