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Classical and quantum electromagnetic field

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karlzr
#1
Jan29-14, 04:59 PM
P: 86
In classical physics, charged particles induce electric field ##\vec{E}_c## around them. How do we interpret this classical electric field ##\vec{E}## in quantum mechanics. Is this just the vacuum expectation value ##\vec{E}_c=<0|\vec{E}|0>##? if so, it means ##<A>\neq 0##. This would lead to Lorentz violation. So I don't know what's going on.

Another thing, we always try to define quantum modes in vacuum (the state with minimum energy). For instance, Higgs field is defined in some arbitrary vacuum of the potential: ##\phi=v+h## where ##\phi##, ##v## and ##h## are the original scalar field, vev ##<\phi>## and the higgs field respectively. Do we always have to do in this way(define excitation in vacuum ? I know this is essential for fermions, and W/Z bosons in SM to get mass. But does this really make sense? this makes me wonder what is field on earth.
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ChrisVer
#2
Jan30-14, 12:09 AM
P: 1,010
For the first question I am not quiet sure for the answer... In general it's better to work with the 4-vector potentials rather than fields.
Also the expression that you give for the expectation value of the vacuum does not feel right. I mean it takes no account of the existence of particle you are talking about.

As for the Higgs field..... You have the vev of the field, that minimizes the potential. Now something moving in the minimum potential is not interesting - it corresponds to a Goldstone boson, and as such could not have masses. At least that you get from perturbing the vev.
In general the perturbated field, will not get only the
[itex]Φ= v+h[/itex]
in fact the higgs field would be:
[itex] Φ= e^{iθ ξ} \left( \begin{array}{c}
0 \\
v+h \\
\end{array} \right)[/itex]

for which ξ can be a goldstone boson field, that is being dropped out once you work in the unitary gauge.... It also can be seen as performing a perturbation along the potential minimum (the slightest push along the minimum will start circulating you around it). The [itex]h[/itex] is the perturbation that happens along the field's values and that's the higg's field (by the last term I mean that it's the higgs that when coupled to itself will give the mass of the higg's boson) since it's the field that excites the vacuum.
karlzr
#3
Jan30-14, 11:36 PM
P: 86
Quote Quote by ChrisVer View Post
Also the expression that you give for the expectation value of the vacuum does not feel right. I mean it takes no account of the existence of particle you are talking about.
Do you mean the virtual electron-position pairs or photons?

Quote Quote by ChrisVer View Post
In general the perturbated field, will not get only the
[itex]Φ= v+h[/itex]
The [itex]h[/itex] is the perturbation that happens along the field's values and that's the higg's field (by the last term I mean that it's the higgs that when coupled to itself will give the mass of the higg's boson) since it's the field that excites the vacuum.
My doubt is whether it is legitimate to define a new field ##h## by a constant? At high energies before symmetry breaking, maybe we would quantize ##\Phi## canonically with some creation and annihilation operators. But after a constant translation, I suppose the creation and annihilation operators in ##h## would correspond to totally distinct particles. That's why I have this concern.


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