The homogeneous strength of the Higgs field

[mfb]

Thank you very much! Could you please explain to me what you mean with "emitting a Higgs"? It sounds to me like a virtual particle associated with the Higgs background field, but (as I have been informed on this forum) this is not the Higgs boson.

The real Higgs boson is an excitation of that field, the virtual ones are a way to visualize perturbation theory.

No, that graph would be part of the electron self-energy, not the same thing.

Still a part of its mass, and don't forget the other part of my post.

 

[kurros]

Still a part of its mass, and don't forget the other part of my post.

But before symmetry breaking there are still higgs loop contributions to the fermion propagators, yet they don't get mass from these. This is actually a question: is the reason for this that the gauge symmetry protects the fermion masses from these corrections? Until we break the gauge symmetry of course?

Perhaps related: photons have no direct interaction with the Higgs, but they do at loop level, and so the photon propagator has Higgs bosons floating around in it. Nevertheless it remains massless.

 

[dauto]

But before symmetry breaking there are still higgs loop contributions to the fermion propagators, yet they don't get mass from these. This is actually a question: is the reason for this that the gauge symmetry protects the fermion masses from these corrections? Until we break the gauge symmetry of course?

Perhaps related: photons have no direct interaction with the Higgs, but they do at loop level, and so the photon propagator has Higgs bosons floating around in it. Nevertheless it remains massless.

Well, the problem is that the mass term connects the left- and right- handed spinors and those spinors belong to two different representations of the gauge group symmetry so there is no way to create a mass term that is a singlet (invariant) under the symmetry. So a explicit mass term in the Lagrangean would indeed break the gauge symmetry. The way around that is to create an interaction of those two fields with a third field. That way it is possible to create a term in the lagrangean that involves both the left- and the right-handed spinors which is also a gauge invariant. That will work if that third field is a complex doublet under the weak isospin gauge group. That third field is, of course, the Higgs field. When the Higgs field acquires a non-vanishing vacuum expectation value, its interaction term with the fermions turn into mass terms.

A mass for the photon would break electromagnetic gauge invariance which survives the symmetry breaking, so it must remain massless.

 

[kurros]

Hmm, yeah I get all that, but I guess my question is more about why such mass terms are not generated during renormalisation of the fields. The fields that have tree-level masses receive corrections to their masses as mfb describes, but it seems that if no tree-level mass exists then one doesn't get generated by such corrections. My postulate was that this is because the original gauge symmetry somehow arranges for all such corrections to cancel each other out or some such; I guess this is related to anomalies or something right? The gauge symmetry is preserved at the quantum level.

 

[dauto]

Hmm, yeah I get all that, but I guess my question is more about why such mass terms are not generated during renormalisation of the fields. The fields that have tree-level masses receive corrections to their masses as mfb describes, but it seems that if no tree-level mass exists then one doesn't get generated by such corrections. My postulate was that this is because the original gauge symmetry somehow arranges for all such corrections to cancel each other out or some such; I guess this is related to anomalies or something right? The gauge symmetry is preserved at the quantum level.

Yes to everything you said, including the part where anomalies must cancel out otherwise the symmetry is not preserved by quantum corrections. Assuming that the anomalies do cancel, radiative corrections to the photon mass mass vanish by gauge symmetry. Note though that in general it is possible to build theories where a particle mass is created by radiative effects even if that mass was absent from the tree-level (bare particles) Lagrangian.

 

[Mandragonia]

There is no "rate of produced virtual particles". That does not exist.

I am surprised. Physical processes and interactions tend to occur at typical time-scales. This is also the case for continuous forces. Take for example the pressure that gas molecules exert on the walls of a gas cylinder. The pressure may appear constant, but is actually the result of many gas molecules bouncing against the walls. This occurs at a rate which is well-described by Boltzmann statistics. In quantum physics radioactive decay points to internal processes in the nucleus that occur spontaneously due to vacuum fluctuations. The strength and frequency of the fluctuations determine the transition probability of the nucleus and the decay time.

 

[Bill_K]

In quantum physics radioactive decay points to internal processes in the nucleus that occur spontaneously due to vacuum fluctuations. The strength and frequency of the fluctuations determine the transition probability of the nucleus and the decay time.

No. This is a misunderstanding common among newcomers to QM, and unfortunately it gets perpetuated in popular accounts. They attribute things to "fluctuations" where the blame should fall to "steady superpositions". Telling this fib makes things sound more classical than they really are, and therefore easier for QM novices to understand.

"The electron zips around in an orbit." - No, it's position has a steady state probability distribution.

"The electron spends part of the time within the nucleus." - No, it has a certain constant probability of being found there.

"Alpha decay happens because the alpha particle repeatedly bounces against the Coulomb barrier, and eventually penetrates it." - Again no, its wavefunction has a certain constant amplitude at the barrier.

These are all examples of an important difference between Classical and Quantum Mechanics.

 

[Mandragonia]

You make it sound as if the only purpose of QM is to solve the time-independent Schroedinger equation (in my opinion this is not quite true). But even if you focus on the resulting steady-state probability functions, you will see that they often contain time-like parameters, for example an angular frequency or a velocity. So the static solution already hints at underlying dynamics.

"The electron zips around in an orbit." - No, it's position has a steady state probability distribution.

Of course that is true. But it is also true that the orbiting electron has a well-defined non-zero kinetic energy and velocity. Therefore it (or something) is moving! This is one of the amusing paradoxes of QM. In fact, if the electron where non-moving, one would run into serious problems. For example, for the outer orbits (higher quantum numbers) of the atom there would be no comparison possible between the quantum orbits and their classical counterparts, where the electron moves around the nucleus in a planet-like elliptical orbit.

 

[mfb]

But it is also true that the orbiting electron has a well-defined non-zero kinetic energy and velocity.

The expectation value for the velocity is zero in all time-independent orbits.

Therefore it (or something) is moving!

I disagree.

For example, for the outer orbits (higher quantum numbers) of the atom there would be no comparison possible between the quantum orbits and their classical counterparts, where the electron moves around the nucleus in a planet-like elliptical orbit.

Every classical part can be written as superposition of orbitals - and in those superpositions the electron can be moving, as the wavefunction is not static any more.

 

[Mandragonia]

The expectation value for the velocity is zero in all time-independent orbits.

Velocity regarded as a vector quantity has zero expectation value, due to the symmetry of the system.
However its absolute value (normally referred to as SPEED) is certainly non-zero.

 

[mfb]

That's why I said "velocity" and not "speed".

 

[Mandragonia]

Why does the electron have a non-zero speed? Because in the sub-atomical realm things move! This has been recognized by countless leading physicists. It is also the insight that led Mr. Schroedinger to formulate his famous result: the time-dependent Schroedinger equation. Its key-aspect is that it is a DYNAMICAL equation. It describes the evolution (in space and time) of the wave function of a particle, in relation to its initial state (t=0). Necessarily the equation contains parameters that are associated with time, such as Planck' s constant and the inertial mass of the particle.

Technically it is very useful to consider first the solutions to the time-independent Schroedinger equation. This is a convenient simplification. This way one obtains the energy eigenfunctions. They form (mathematically speaking) a basis, and so they can be superimposed to create time-dependent functions. But my point is, that the solutions of the time-independent Schroedinger equation contain exactly the same parameters as the time-dependent version. So no wonder that inspection of the steady-state solutions reveals certain dynamical properties of the particle, such as its average speed in orbit. In my view this property is no less "physically real" than the probability density.

 

[mfb]

No one ever questioned that particles can move in quantum mechanics in general. The main point was that they do not move around in time-independent states, which are solutions to the time-independent SE.

 

[Mandragonia]

No one disputes the existence of time-independent states, which are solutions of the time-independent SE.
The discussion is how the electron can have a non-zero speed while being in a time-independent state.

 

[mfb]

A non-zero expectation value for the speed.
It does not have a well-defined, single "speed value".

I don't see the problem.

 

[Bill_K]

No one disputes the existence of time-independent states, which are solutions of the time-independent SE.
The discussion is how the electron can have a non-zero speed while being in a time-independent state.

This discussion has long ago departed from the OP, which was about the Higgs field. For answers to these questions you should start a new thread. Furthermore, the questions you are now asking are basic QM and should be asked in the QM forum.

 

[Mandragonia]

Wikipedia statement: "The mean speed of the electron in hydrogen is 1/137th of the speed of light."
Therefore the electron is moving. Yet its probability distribution is time-independent.

I don't have a problem reconciling these two (seemingly contradictory) facts. For me it obvious that the electron is moving. Due to the impossibility to have information on the position of the electron, the best one can do is to assume that it is simultaneously present at the different positions allowed in the orbital. Of course with proper weighting. This leads to the time-independent solution, in which the effects of motion becomes hidden.

 

[mfb]

Wikipedia statement: "The mean speed of the electron in hydrogen is 1/137th of the speed of light."
Therefore the electron is moving.

Wikipedia as source? That does not work.

Anyway, Bill_K is right. Please start a new thread if you want to discuss interpretations of the wave-function as "moving" or "not moving".

 

[Mandragonia]

Please start a new thread if you want to discuss interpretations of the wave-function as "moving" or "not moving".

Yet there is no guarantee that if you start a new thread in the QM forum on this subject, you will get any meaningful answers.

The purpose of Physics Forums is to promote interesting and helpful discussions, but in reality these are scarce and occur only within the inner circle of experts.