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Initial wave function for an electron in a magnetic field

  1. May 14, 2013 #1
    Hello everybody,

    I am currently struggling with a problem that I came across while spending some free time on non-relativistic quantum mechanic problems.

    Suppose we have an electron that is describe at time [itex] t_0 = 0 [/itex] by a wave function in position space [itex]\psi(x,y,z)[/itex]. Furthermore, assume that we have a Hamiltonian
    [itex] \hat{H}_1 = \frac{1}{2m}\left [ \mathbf{p}+\frac{e}{c} \mathbf{A}_1(\mathbf{r}) \right ]^2[/itex]
    with charge e, mass m and speed of light c. Then we can construct another Hamiltonian
    [itex] \hat{H}_2 = \frac{1}{2m} \left [\mathbf{p}+\frac{e}{c} \mathbf{A}_2(\mathbf{r}) \right ]^2[/itex]
    that can be obtained from [itex] \hat{H}_1[/itex] via a gauge transformation of the vector potential
    [itex] \mathbf{A}_2(\mathbf{r}) = \mathbf{A}_1(\mathbf{r}) + ∇ f(\mathbf{r}) [/itex]
    with a scalar function f.

    The question is now which of these two Hamiltonians do I use to evolve my initial wave function in time? I mean, if I fix the initial wave function and the Hamiltonian it is like fixing a gauge for the initial wave function.

    In other words, are there some conditions that the initial wave function has to fulfill that depend on the chosen gauge?

    Greetings, Syrius
     
  2. jcsd
  3. May 14, 2013 #2
    Maybe the form of wavefunctions will depend on the gauge, probably just an addition of phase factor in wavefunction. But all the physical measurements should be independent of gauge.
     
  4. May 15, 2013 #3
    If we change A by a gauge transformation then
    A'=A+∇λ(x),then the wave function will change accordingly as ψ'=ψe[ie/h-c∫∇'λ(x').ds']=ψe(ieλ(x)/h-c)
     
  5. May 15, 2013 #4
    Hello Adrien and ck00,

    thank you for your answers. I do know the transformation law that Adrien mentioned. However, my question is a little bit different and I will illustrate it with an example.

    Let [itex] \hat{p}_{\text{mech}} = \hat{p} - \frac{q}{c} \mathbf{A} [/itex], where [itex] \hat{p} [/itex] is the usual momentum operator and [itex] \mathbf{A} [/itex] the vector potential. This operator is gauge covariant, i.e. it has the same form for every gauge.

    Let us now assume that we are in a field free region and we choose our vector potential [itex] \mathbf{A} = 0 [/itex]. As an initial wave function I will take the eigenfunction [itex] |p> [/itex] to the operator [itex]\hat{p}[/itex].

    If we now change the gauge to [itex] \mathbf{A} = (1,0,0) [/itex] and transform [itex] |p> [/itex] with [itex] |p>' = e^{ieλ(x)/\hbar c} [/itex] |p>, then we get that [itex] |p>' [/itex] is indeed an eigenfunction to [itex] \hat{p}_{\text{mech}} = \hat{p} - \frac{q}{c} (1,0,0) [/itex]. So everything is correct here.

    But now I can do everything in the backward way. I start with [itex] \mathbf{A} = (1,0,0) [/itex] and the initial wave function [itex]|p>[/itex] that is an eigenfunction to [itex]\hat{p}[/itex] but it is not an eigenfunction to [itex] \hat{p}_{\text{mech}} = \hat{p} - \frac{q}{c} (1,0,0) [/itex]. That means, depending on the gauge, [itex] |p> [/itex] can represent two different wave functions when one chooses it as the initial wave function. So my question is rather, how can I be sure that I choose the right initial wave function with the desired properties, when this properties may depend on the gauge.

    Greetings, Syrius
     
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