How do electrons couple to gauge field?

[a dull boy]

I have three questions which I have to put into context, much of which is paraphrased from a book by Kerson Huang.

In QED, the source of the gauge field (is the gauge field different from the vector potential?) is the current and charge density j, ρ. When a particle (electron) couples to the gauge field (photons), a gauge transformation occurs. A gauge transformation in QM involves both the vector potential and the charged particle. It consists of the joint operation A -> A+ ∂χ (where A is the vector potential) and ψ -> Uψ where U is the phase factor = e(iq/hcχ) for the charged particle. The quantum phase factor is e(-iq/hc) which is a compact representation of the group U1, all rotations about a fixed axis.

1) Does gauge symmetry refer to the term χ acting both on the vector potential and the phase of the charged particle?

When the vector potential climbs or falls a gauge fiber, the phase makes corresponding changes around the ring.

2) Is this is what Feynman does with his clocks…changing the vector potential changes the phase, causing the clock to spin..although Feynman’s clocks seemed associated with photons, not particle phases…).

3) The vector potential seems to only affect the phase of the particle, which doesn’t affect its energy. How is the energy of the particle, (for example when an electron accepts a photon it can jump to a higher orbital/energy level), modeled through interactions with the gauge field?

I'm a biologist trying to get a handle on qft for a project I'm doing.
Thanks very much, I've enjoyed this forum.

 

[tom.stoer]

1) Does gauge symmetry refer to the term χ acting both on the vector potential and the phase of the charged particle?

Yes, b/c only this combination is a symmetry.

2) ... although Feynman’s clocks seemed associated with photons, not particle phases…).

I am not sure to which book you are referring to, but I guess that Feynman is talking about the polarization vector of photons.

3) ... How is the energy of the particle, (for example when an electron accepts a photon it can jump to a higher orbital/energy level), modeled through interactions with the gauge field?

In quantum mechanics which is used to describe atoms there is no "photon"; instead the electromagnetic field is treated classically using electromagnetic fields. In QED it's the interaction between the electric current (carried by the quantized electron field) and the quantized photon field which is responsible for the energy. There is a formulation which looks nearly classical.

Look at the Coulomb potential q₁q₂ / |r₁ - r₂| between two static charges located at r₁ and r₂. This replaced by an energy density (that means in the end one has to integrate over three-space) ρ(r₁)ρ(r₂) / |r₁ - r₂|

The last expression contains the charge densities which are operators, not just functions.

 

[Polyrhythmic]

is the gauge field different from the vector potential?

No, it isn't.

When a particle (electron) couples to the gauge field (photons), a gauge transformation occurs.

I would rephrase that:
The symmetry under U(1) gauge transformations is a fundamental property of the theory. Its Lagrangian (the defining quantity of the theory) is invariant under those transformations.

1) Does gauge symmetry refer to the term χ acting both on the vector potential and the phase of the charged particle?

The U(1) symmetry of QED acts on both the gauge field (photon) and the spinor field (electron).

The transformation of the photon field represents nothing physical. That's the whole point of the story: a gauge transformation connects states that are physically equivalent. You can pick any of those fields out of a vast amount of canditates and still get the same physical configuration. This fact allows one to impose further restrictions on the field, that's what one usually calls "gauge-fixing". One usually choses those restrictions (gauges) such that one gets a mathematical advantage out of it. An important gauge in electrodynamics would be for example the Coulomb gauge.

 

[Galron]

This is actually a better question than you know the math is tight but the philosophy is...

But whs.

It works in the field variance in field theory. Etc..

 

[Polyrhythmic]

This is actually a better question than you know the math is tight but the philosophy is...

But whs.

How is the topic at hand connected to philosophy?

 

[Galron]

How is the topic at hand connected to philosophy?

Interpretation matters.

How is it not connected to philosophy would of been a better question, ie as soon as you use the term particle (especially in a wave theory although that is trivial) you enter a world of pain.

Well you did ask?

In QED, the source of the gauge field (is the gauge field different from the vector potential?) is the current and charge density j, ρ. When a particle (electron) couples to the gauge field (photons), a gauge transformation occurs. A gauge transformation in QM involves both the vector potential and the charged particle. It consists of the joint operation A -> A+ ∂χ (where A is the vector potential) and ψ -> Uψ where U is the phase factor = e(iq/hcχ) for the charged particle. The quantum phase factor is e(-iq/hc) which is a compact representation of the group U1, all rotations about a fixed axis.

This assumes that any particular maths actually pictorially represents any particular quantum state. That is of course an interpretation matter and is not per se evidential.

The equation simply takes the results applies a canonical value related to i, it does not in any way relate to anything from experiment except the results, ie it is a deduction not an induction.

But never mind. I suppose it was the equivalent of a physics in joke, one you are not party to.

 

[Polyrhythmic]

Interpretation matters.
How is it not connected to philosophy would of been a better question, ie as soon as you use the term particle (especially in a wave theory although that is trivial) you enter a world of pain.

The original poster wanted to know some technical details about the theory of quantum electrodynamics, this really isn't related to interpretational matters.

This assumes that any particular maths actually pictorially represents any particular quantum state. That is of course an interpretation matter and is not per se evidential.

The equation simply takes the results applies a canonical value related to i, it does not in any way relate to anything from experiment except the results, ie it is a deduction not an induction.

Quantum electrodynamics, like any other theory of physics, is a model of reality (which is itself just an idealization). It agrees with experiment quite well, therefore it is a good one.

 

[a dull boy]

Thanks everyone, your responses have helped me inch along on this (as has been my experience in my reading). Polyrhythmic is right, at this point I'm just trying to understand the material, though I will say I've enjoyed finding that QED addresses (inadvertantly or not) some fundamental philosophical questions (subject/object and scale, for example) in a very explicit way.

I'd like to follow up a bit more:

I am not sure to which book you are referring to, but I guess that Feynman is talking about the polarization vector of photons.

The book I’m referring to is called QED: The Strange Theory of Light and Matter. He talks about how the classical light path results from interference effects from non-classical paths cancelling each other out. He uses a clock hand spinning (which is what I thought referred to the phase of the photon) to describe this. My gap in understanding the phase factor U = e(-iq/hc) in part comes from thinking that this shifts the particle’s wavelength forward or back from an origin, and this would result in interference. But I’m pretty sure I’m wrong about this now, hearing your responses. This phase factor seems a way to state a coupling of the fermion to the gauge field, rather than referring to the phase of photons and interference.

In quantum mechanics which is used to describe atoms there is no "photon"; instead the electromagnetic field is treated classically using electromagnetic fields. In QED it's the interaction between the electric current (carried by the quantized electron field) and the quantized photon field which is responsible for the energy. There is a formulation which looks nearly classical.

Look at the Coulomb potential q₁q₂ / |r₁ - r₂| between two static charges located at r₁ and r₂. This replaced by an energy density (that means in the end one has to integrate over three-space) ρ(r₁)ρ(r₂) / |r₁ - r₂|

The last expression contains the charge densities which are operators, not just functions.

Thanks very much, that helps greatly. It seems there are two types of interaction between electrons and photons, one described by U1 gauge symmetry which does not involve energy but defines the way electrons have charge, and a second type where a photon is absorbed by an electron, changing its energy. Is this correct? Would this operator ρ(r1)ρ(r2) / |r1 - r2| be included as part of a creation/annihilation matrix in QED? Is this matrix a form of the vector potential?

 

[a dull boy]

I didn't use the quote button quite right...I'll just ask my follow-up questions here:

The Feynman book I’m referring to is called QED: The Strange Theory of Light and Matter. He talks about how the classical light path results from interference effects from non-classical paths cancelling each other out. He uses a clock hand spinning (which is what I thought referred to the phase of the photon) to describe this. My gap in understanding the phase factor U = e(-iq/hc) in part comes from thinking that this shifts the particle’s wavelength forward or back from an origin, and this would result in interference. But I’m pretty sure I’m wrong about this now, hearing your responses. This phase factor seems a way to state a coupling of the fermion to the gauge field, rather than referring to the phase of photons and interference.

It seems there are two types of interaction between electrons and photons, one described by U1 gauge symmetry which does not involve energy but defines the way electrons have charge, and a second type where a photon is absorbed by an electron, changing its energy. Is this correct? Would this operator ρ(r1)ρ(r2) / |r1 - r2| be included as part of a creation/annihilation matrix in QED? Is this matrix a form of the vector potential?

 

[Polyrhythmic]

It seems there are two types of interaction between electrons and photons, one described by U1 gauge symmetry which does not involve energy but defines the way electrons have charge, and a second type where a photon is absorbed by an electron, changing its energy.

I don't think you understood this correctly. The interaction is given by some function ( the Lagrangian), which shows a symmetry, U(1). With the help that function, you can calculate quantities that describe processes like scattering or absorption. They are just different aspects of one and the same thing.

 

[a dull boy]

Yes, b/c only this combination is a symmetry.


I am not sure to which book you are referring to, but I guess that Feynman is talking about the polarization vector of photons.


In quantum mechanics which is used to describe atoms there is no "photon"; instead the electromagnetic field is treated classically using electromagnetic fields. In QED it's the interaction between the electric current (carried by the quantized electron field) and the quantized photon field which is responsible for the energy. There is a formulation which looks nearly classical.

Look at the Coulomb potential q₁q₂ / |r₁ - r₂| between two static charges located at r₁ and r₂. This replaced by an energy density (that means in the end one has to integrate over three-space) ρ(r₁)ρ(r₂) / |r₁ - r₂|

The last expression contains the charge densities which are operators, not just functions.

I don't think you understood this correctly. The interaction is given by some function ( the Lagrangian), which shows a symmetry, U(1). With the help that function, you can calculate quantities that describe processes like scattering or absorption. They are just different aspects of one and the same thing.

I think I am conflating Schrodinger's approach with QED, but I'll go ahead anyhow. I think that absorption of a photon pushes E up into the next highest energy state in the equation
Hpsi = Epsi , which is an interaction between a photon and an electron. But in QED, I also see an interaction between an electron and a photon described in terms of 2pi rotations of the phase of the electron, which does not involve any change in energy. Can you tell me,
maybe with an example or an equation that involves both phase and energy eigenstate, of how these are different aspects of the same thing? Or maybe suggest a good book...I might simply lack the knowledge to understand your answer.

Thanks, Mark

 

[a dull boy]

I found this notation to describe the amplitude to propagate from point q1 to q2 as
<q2 l e(exp-iHT l q1> where e(exp-iHT) is the unitary operator, and contains the hamiltonian where I would think a PE term for absorbing a photon could be located. This "format" seems to hold both "aspects of the same thing" - is that true?

 

[tom.stoer]

It seems there are two types of interaction between electrons and photons, one described by U1 gauge symmetry which does not involve energy but defines the way electrons have charge, and a second type where a photon is absorbed by an electron, changing its energy. Is this correct? Would this operator ρ(r1)ρ(r2) / |r1 - r2| be included as part of a creation/annihilation matrix in QED? Is this matrix a form of the vector potential?

It will be hard for to understand the details w/o looking at the math.

First of all you must not confuse quantum mechanics and QED. I tried to explain what happens in QED and I would like to continue.

In QED you start with a Lagrangian with U(1) gauge symmetry from which everything can be derived but which itself cannot be interpreted easily. Using the Lagrangian one can derive scattering quite easily, but in order to talk about energy (interaction energy, total energy) you need the so-called Hamiltonian,which is the energy-operator. In order to derive this Hamiltonian you have to get rid of unphysical degrees of freedom!

In the Lagrangian you have four "photons", namely four components of the 4-vector potential. But you know that in reality there are only two photons, i.e. two polarizations. It is interesting that reducing the number of photons from four to two is identical to gauge-fixing. That means that the gauge-symmetry itself is unphysical (or artificial; it's not like rotational symmetry; you cannot "see" it). Doing this you eventually arrive at the Hamiltonian which looks completely different; you can't directly see that it has been derived from a Lagrangian. In addition it is not unique b/c have you different choices how to fix the gauge.

You asked regarding the interaction and the "energy". That's why I posted this interaction term (which is not the only interaction term) but which is rather familiar as it contains the Coulomb potential. That means that there is a way to derive the specific form of the Coulomb potential via gauge fixing from a Lagrangian that does not contain something like this potential. On top of this "static" Coulomb potental where no dynamical photons are present there are other interaction terms coupling electrons to photons; but when you want to calculate the energy of something, the Coulomb potential is the starting point, whereas the other terms result in corrections only.

An approximation could be to "freeze" the photon degrees of freedom, i.e.to keep the Coulomb potential and to throw away the other interaction terms. That's a way to derive equations valid in quantum mechanics as an approximation to the full QED.

 

[Galron]

The original poster wanted to know some technical details about the theory of quantum electrodynamics, this really isn't related to interpretational matters.



Quantum electrodynamics, like any other theory of physics, is a model of reality (which is itself just an idealization). It agrees with experiment quite well, therefore it is a good one.

Obviously you just don't get my point.

 

[Dickfore]

@Title:

Through the Noether current corresponding to Dirac fermions.

 

[Polyrhythmic]

Obviously you just don't get my point.

Apparently not. Mind elaborating?

 

[Galron]

Apparently not. Mind elaborating?

I already did, you didn't get it.

 

[jjustinn]

Is this is what Feynman does with his clocks…changing the vector potential changes the phase, causing the clock to spin..although Feynman’s clocks seemed associated with photons, not particle phases…).

I am not sure to which book you are referring to, but I guess that Feynman is talking about the polarization vector of photons.

Assuming this is referring to the "clocks" Feynman uses in the book QED, the "clocks" were simply used to represent the complex phase of the wavefunction, which has nothing to do with classical / spatial polarization.

However, this phase is indeed what gets changed in a Schrodinger gauge transformation (and also what gets changed when coupling with an electromagnetic field) -- it's the phase of the wavefunction / probability amplitude. If you change the phase everywhere equally, it makes no difference in your final answer, because when you square it you get the same answer.

To your broader question, I'll leave that to those with a better understanding than I do, but I do have one comment: it seems that you're thinking of a gauge transformation as something that actively "happens" / is "caused" -- my understanding is that a gauge transformation is more akin to a passive coordinate transformation...the underlying physical reality isn't affected. A gauge transformation is just a transformation that can be applied to certain functions without changing their physical attributes (just like translating / rotating a free particle doesn't change its behavior)...the "magic" of gauge theory comes about by taking this global symmetry (e.g. adding the same function to each point in space) and forcing it to be local (e.g. adding different functions at each point).

 

[Polyrhythmic]

I already did, you didn't get it.

How nice of you, I appreciate that.

 

[Galron]

How nice of you, I appreciate that.

I see little point in iterating what I already said, simply go back and read the post and see if it sinks in, if not then we are done.

 

[Polyrhythmic]

I see little point in iterating what I already said, simply go back and read the post and see if it sinks in, if not then we are done.

I understand quite clearly what you wrote, and I agree that interpretational matters regarding particles are something worth discussing! Nevertheless, there is only a vague connection to the issues the original poster has with the theory.

 

[a dull boy]

Thanks everyone for your help. I'll keep reading, I'll try to find treatments that follow the lagrangian to derive the hamiltonian. I don't get that the 4-vector, which I thought I was familar with, referred to 4 "photons"...I thought it was the x,y,z,ct that described a single, relativistic particle's position in space. Or is it the Fuv 4x4 matrix you are referring to? -Mark

 

[The_Duck]

I haven't read every post here, OP, but I think you do not know, and may be interested to know, the fundamental relationship between phase and energy in quantum mechanics. Energy is the rate of change of phase. For instance, when the "phase clock" of a photon turns at a given rate, that is because the photon has a certain energy that is proportional to this rate. (The proportionality factor is h-bar. You can verify that multiplying h-bar by a measure of rotation rate like Hz gives you units of energy). Electrons, too, and indeed all particles, have a phase clock. Just as with a photon, an electron's energy is how fast its phase clock is turning.

The gauge symmetry of electrodynamics complicates this. When an electron interacts with the electromagnetic field, the phase of the electron's wave function doesn't tell you anything unless you also know the vector potential A. This is because we can do a "gauge transformation" that modifies two mathematical objects, the electron wave function phase and A, while doing nothing physically measurable to the actual electron state or the actual electromagnetic field. When we say there is a "gauge symmetry" we mean that there exists a certain http://en.wikipedia.org/wiki/Group_(mathematics)" of these physically meaningless transformations of the mathematical objects in the theory.

So you can make the phase vary arbitrarily with time as long as you vary A in a corresponding way (and you can do this *without* changing what electromagnetic field A specifies). Put the other way, it's possible to make "frivolous" changes in A that don't change the actual electromagnetic field, and which result in "frivolous" changes in the electron's wave function's phase that aren't actually related to the electron's energy. In the end, to find the electron's energy you have to first subtract out the part of the wave function's phase that comes from these irrelevant parts of A and take the rate of change of the remaining phase.

So some parts of A and their corresponding effects on the wave function phase don't matter. But the rest of A, the physically significant part of A that specifies the actual electric and magnetic fields, also produces effects on the wave function phase. And these effects matter because they can't be gotten rid of by varying A. This is the part of the interaction between the electromagnetic field and the electron that affects the electron's energy.

 

[jjustinn]

Thanks everyone for your help. I'll keep reading, I'll try to find treatments that follow the lagrangian to derive the hamiltonian. I don't get that the 4-vector, which I thought I was familar with, referred to 4 "photons"...I thought it was the x,y,z,ct that described a single, relativistic particle's position in space. Or is it the Fuv 4x4 matrix you are referring to? -Mark

I'm not sure what you mean by the 4-vector "referring to four 'photons'" -- the field tensor ($F_{uv}$) is the 'curl' of the 4-vector-potential ($\partial_{u}A_{v} - \partial_{v}A_{u}$), and the 4-vector-potential is just $(V, A_{x}, A_{y}, A_{z})$, where V is the classical electrostatic potential (e.g. 'voltage') and A is the classical vector/magnetic potential. This 4-potential varies across space/time, but its components aren't a space-time position (just like the components of a classical 3-dimensional force vector don't tell you where that force is located).

At the level you're looking at, it probably doesn't make sense (or at least, isn't particularly helpful / clarifying) to even be talking about individual photons / electrons -- gauge freedom and electron-field coupling don't require quantization at all...once you clearly understand how an electron (Dirac) field couples to a "photon" (Maxwell) field, then it may make sense to take the next step of seeing how second quantization affects this.

But again, IANAP (I am not a physicist), so if anyone contradicts me, they're probably right.

 

[a dull boy]

I had a breakthrough of sorts the past day, spurred by your help. I was reading David Griffith's text Intro. to Elem. Part. and finally understand how moving from global
to local gauge invariance has the effect of linking a dirac field to a proca field and thus coupling the em force to the electron. Whoever first did that math must have had a great day that day! I had yet read the duck and jjustin's posts, now reading them I see I have the chance of understanding them! I'll respond to those soon, after having a chance to digest them.
Great forum! -Mark

 

[Dickfore]

a proca field

I think a mass term for the electromagnetic field breaks the gauge invariance of the theory. You are following a route which is technically simpler, but obscures the essential fact that the 4-potential is a gauge field. Namely, you can always set the mass of the photons to zero in the end of all the calculations and obtain the same result without ever discussing gauge invariance. But, then, your coupling term, the very topic of this thread, is inserted 'by hand' and does not have an obvious meaning. Also, by the Ward identities, I think the photon cannot acquire mass in any order of perturbation theory.

 

[Polyrhythmic]

I think a mass term for the electromagnetic field breaks the gauge invariance of the theory. You are following a route which is technically simpler, but obscures the essential fact that the 4-potential is a gauge field. Namely, you can always set the mass of the photons to zero in the end of all the calculations and obtain the same result without ever discussing gauge invariance. But, then, your coupling term, the very topic of this thread, is inserted 'by hand' and does not have an obvious meaning. Also, by the Ward identities, I think the photon cannot acquire mass in any order of perturbation theory.

The mass of the photon field is zero by definition, not because you set it to zero due to some gauge symmetry. The gauge transformation of the photon field is the addition of the gradient of some local function, which doesn't change Maxwell's equations. Furthermore, requiring invariance of full QED Lagrangian (with coupling to spinor fields) under that transformation leads to an imposed local gauge symmetry of the spinor. The mass of the photon doesn't play a role in this, simply because it is absent due to the relativistic origins of QED.

 

[Dickfore]

The mass of the photon field is zero by definition, not because you set it to zero due to some gauge symmetry.

ORLY? What definition is that? The vacuum polarization is nothing more but a self-energy of the photon. Ward-Takahashi identities, which are a consequence of gauge invariance protect the photon propagator from acquiring non-zero mass.

The gauge transformation of the photon field is the addition of the gradient of some local function, which doesn't change Maxwell's equations.

This is true in Classical Electrodynamics. However, In Quantum Theory, it has far more reaching consequences, because it also changes the phase of the quantum fields.

Furthermore, requiring invariance of full QED Lagrangian (with coupling to spinor fields) under that transformation leads to an imposed local gauge symmetry of the spinor.

I guess this is in accordance with my previous remark.

The mass of the photon doesn't play a role in this, simply because it is absent due to the relativistic origins of QED.

Please show how a term
$$-\frac{1}{2} m^{2}_{\mathrm{photon}} \, A^{\mu} \, A_{\mu}$$
is gauge invariant?

 

[Polyrhythmic]

$$-\frac{1}{2} m^{2}_{\mathrm{photon}} \, A^{\mu} \, A_{\mu}$$
is gauge invariant?

I never claimed that it was gauge invariant. I claimed that it wasn't there, which is a true statement for QED.

 

[Dickfore]

Cool, I was talking about something else.

 

[Polyrhythmic]

Furthermore, I don't see how the Ward-Takahashi identities have any impact on the form of the original QED Lagrangian.

 

[Dickfore]

Furthermore, I don't see how the Ward-Takahashi identities have any impact on the form of the original QED Lagrangian.

You should familiarize yourself with renormalization.

 

[Polyrhythmic]

You should familiarize yourself with renormalization.

I'm familiar with renormalization, that's why I wrote "original" instead of "renormalized" Lagrangian.

 

[Dickfore]

I'm familiar with renormalization, that's why I wrote "original" instead of "renormalized" Lagrangian.

If you really understood the gist of Renormalization, you would know that the "original" Lagrangian is not connected to reality.

 

[Polyrhythmic]

I understand that. Apparently we were both confused with what the other one meant! ;)