Initial wave function for an electron in a magnetic field

[Syrius]

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

 

[ck00]

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.

 

[andrien]

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)}

 

[Syrius]

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 $\hat{p}_{\text{mech}} = \hat{p} - \frac{q}{c} \mathbf{A}$, where $\hat{p}$ is the usual momentum operator and $\mathbf{A}$ 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 $\mathbf{A} = 0$. As an initial wave function I will take the eigenfunction $|p>$ to the operator $\hat{p}$.

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

But now I can do everything in the backward way. I start with $\mathbf{A} = (1,0,0)$ and the initial wave function $|p>$ that is an eigenfunction to $\hat{p}$ but it is not an eigenfunction to $\hat{p}_{\text{mech}} = \hat{p} - \frac{q}{c} (1,0,0)$. That means, depending on the gauge, $|p>$ 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