Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauge Freedom Quantum Electrodynamics

  1. Apr 17, 2015 #1
    It's well known when if we are working on problems related to particles in presence of an electromanetic field, the way we state the problem can be done using the next Hamiltonian:
    [itex]H=\dfrac{(p-\frac{e}{c}A)^2}{2m} +e \phi[/itex] where the only condition for A is: [itex]\vec{\nabla } \times \vec{A} =\vec{B}[/itex]
    So we have this "gauge freedom" under (of course) the above condition .
    What happens when are studying two or more partcles in presence of the electromagnetic field.

    Naturally:
    [itex]H=\dfrac{(p_{1}-\frac{e}{c}A)^2}{2m} +\dfrac{(p_{2}-\frac{e}{c}A)^2}{2m} +e \phi[/itex] where the only condition for A is: [itex]\vec{\nabla } \times \vec{A} =\vec{B}[/itex]

    The question is: can we relax the condition of one unique gauge for all the particles?

    Just saying:
    can we select two different gauges? obviously with the condition : [itex]\vec{\nabla } \times \vec{A} =\vec{B}[/itex]

    I've been working on this, I satetd two different gauges and I can say that there is no difference in terms between the equations of motion having the same gauge. (Hamilton)

    In particular I've done the same for the correspoding quantum problem obtainig the heisemberg equations, having the same result: no difference.

    I'm interested on the quantum problem because having two different gauges could provoque loosing the symmetry of a Hamiltonian given.
     
    Last edited: Apr 17, 2015
  2. jcsd
  3. Apr 22, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Apr 22, 2015 #3
    Not sure what you mean by choosing two different gauges/vector potentials for the two particles. If you mean, whether you can use to different vector potentials at the two points which are the locations of the two particles, then the answer is negative. Once you apply a gauge transformation to fix gauge or change to a new gauge, you applying that transformation over all space points including the two locations of the two point charges.
     
  5. Apr 23, 2015 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Gauge invariance means that the scalar and vector potentials for the electromagnetic field (or better said the four-vector potential of the electromagnetic field) are only defined up to a four-gradient, i.e., if you have a four-vector potential ##A_{\mu}## then also
    $$A_{\mu}'=A_{\mu} + \partial_{\mu} \chi$$
    describes the same physical situation, i.e., you get the same field-strength tensor,
    $$F_{\mu \nu}=\partial_{\mu} A_{\nu}' - \partial_{\nu} A_{\mu}'=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} + \partial_{\mu} \partial_{\nu} \chi - \partial_{\nu} \partial_{\mu} \chi = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}.$$
    So you can impose one gauge condition. Depending on the situation you look at, the right choice of gauge can simplify the problem considerably. Usual gauge conditions are the Lorenz gauge
    $$\partial_{\mu} A^{\mu}=0$$
    and the Coulomb gauge
    $$\vec{\nabla} \cdot \vec{A}=0.$$

    The equations of motion for your point-particle problem are gauge independent, because only the field components enter (in terms of the Lorentz force, ##q (\vec{E}+\vec{v} \times \vec{B})## (with ##c=1##).
     
  6. Apr 24, 2015 #5

    ChrisVer

    User Avatar
    Gold Member

    Are you sure about your Hamiltonian for the two (or more) particles???
     
  7. Apr 27, 2015 #6
    It's the standard construction.
    It's the standard construction.
     
  8. Apr 27, 2015 #7
    For Example: Lets suppose we have the magnetic field in the z direction, for example two Landau's gauges:
    [tex]\vec{A_{1}}=B(-y,0,0)[/tex] and [tex]\vec{A_{2}}=B(0,x,0)[/tex] where both satisfy [tex] \vec{B}=B\vec{k} [/tex].

    So in particular I could say that my Hamiltonian for the two particle system is:

    [itex]H=\dfrac{(\vec{p_{1}}-\frac{e}{c}B(-y_{1},0,0))^2}{2m}+\dfrac{(\vec{p_{1}}-\frac{e}{c}B((0,x_2,0))^2}{2m} +e \phi[/itex]

    So I'm using two different gauges for my problem. I'm using it in that sense. What I'd like to emphasise is that the equations of motion are the same in comparison to those obtained when I use only one.

    Thank you.
     
    Last edited: Apr 27, 2015
  9. Apr 27, 2015 #8
    I get your point, is the usual treatment for one particle. But what I'm saying is related to the two particle system, where the question is: Is it correct use two different gauges, one for each particle?

    Thank you
     
  10. Apr 28, 2015 #9

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    No, you should use one gauge for the whole calculation!
     
  11. Apr 28, 2015 #10
    and the reason is? I repeat, equations of motion don't change.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gauge Freedom Quantum Electrodynamics
Loading...