Eigenvalues and Gauge Transformations and the role of Boundary Conditions in a gauge transformation

Question

I would like to ask a question that has to do with gauge transformations both in general and in the context of the Schroedinger equation found in the theory of Quantum Mechanics. I think that my question is related to functional analysis.

1) In Linear Algebra, I know that changing the basis changes the matrix that represents a transformation but keeps its eigenvalues the same but I don't know if in infinite dimensional spaces there could be exceptions to this(that's basically my question).
Now, trying to solve an eigenvalue problem in the context of differential equations(sorry if the terminology is a bit different than you are used to) we can perform a gauge transformation to help us solve the equation more easily. I have also read that the gauge transformation(at least in the context of Quantum Mechanics) is equivalent to changing the basis.

So, the question is whether the gauge transformation could(in rare occasions that could be regarded as exceptions) change the eigenvalues.

2) Also, if the boundary conditions of the original problem are different from the boundary conditions of the (gauge) transformed problem, does it change anything regarding the eigenvalues of the original problem?

(Note: The above can be seen in the case of the quantum mechanical problem of the free particle confined on a ring. We can gauge transform to a problem with a magnetic field in the middle of the ring which gives rise to a magnetic vector potential. This changes the eigenvalues of the original(free particle) problem.)

Please, use simple terms and explanations in order for a non-mathematician to understand the answers.

Thank you.

Answer 1

If $L$ is a linear transformation on a vector space $V$, then the eigenvalues $\lambda$ and eigenvectors $v$ of $L$ are defined to be the non-zero solutions of the equation $Lv = \lambda v$. The important thing to note here is that $\lambda$ and $v$ are defined in terms of the transformation $L$, not the components of some matrix representation of of $L$. Thus eigenvalues and eigenvectors are properties of the transformation itself, not its representation under any basis.

1. So the answer to question 1 is no. Note that there is also no mention of the dimension of $V$ in that definition. It doesn't matter if $V$ is finite dimensional or infinite.
2. The boundary conditions determine what linear transformation solves the Schroedinger equation. If you change the boundary conditions, you change the solution, and there is no a priori reason to expect the solution to the new problem to have the same eigenvalues as the solution to the old one.

It has been a very long time since I studied QM, but I recall gauge transformations as being transformations that do not change the problem. The transformation you describe would not be called a "gauge transformation" in my (rather vague) memory of it.

Answer 2

Here is the case study of your question, hope it helps. Say you start with the Hamiltonian $$H=\frac{1}{2m}(p-\frac{e}{c}\bar{A}(\bar{x}))^{2}$$ And, say, for the sake of argument, we are in 2d and the wave function satisfies the periodic boundary conditions $$\Psi(x, 0)=\Psi(x, l_{y})$$ $$\partial_{y}\Psi(x, 0)=\partial_{y}\Psi(x, l_{y})$$ $$\Psi(0, y)=\Psi(l_{x}, y)$$ $$\partial_{x}\Psi(0, y)=\partial_{x}\Psi(l_{x}, y)$$ It is possible to nullify the vector potential by the gauge transformation $$\bar{A}'=\bar{A}-\bar{\nabla}{f}$$ Thus the Hamiltonian $H$ is simplified. But the wave function would undergo certain changes, it is clear that the probability $|\Psi|^{2}$ of particle location should not be changed by gauge (gradient) transformation, i.e. the new wave function $\psi$ differs from the old one $\Psi$ only in phase factor $$\psi(\bar{x})=\Psi(\bar{x})\exp\Big\{-\frac{ie}{\hbar{c}}f(\bar{x})\Big\}$$ where $f$ is the function of gauge transformation, $\psi$ is the eigenfunction of the new modified Hamiltonian $H'$, which is a free particle operator. However, the boundary conditions change, they will be now lagging $$\psi(x, 0)=e^{i\phi_{y}}\psi(x, l_{y})$$ $$\partial_{y}\psi(x, 0)=e^{i\phi_{y}}\partial_{y}\psi(x, l_{y})$$ $$\psi(0, y)=e^{i\phi_{x}}\psi(l_{x}, y)$$ $$\partial_{x}\psi(0, y)=e^{i\phi_{x}}\partial_{x}\psi(l_{x}, y)$$ Solving the free particle equation, and using the boundary conditions, one has $$\psi=\exp(ik_{x}x+ik_{y}y)$$ with $$k_{x}=\frac{\phi_{x}}{l_{x}}+\frac{2\pi{n_{x}}}{l_{x}}$$ and $$k_{y}=\frac{\phi_{y}}{l_{y}}+\frac{2\pi{n_{y}}}{l_{y}}$$ with $n_{1}, n_{2}\in\mathbb{Z}$. In fact, $\phi_{x, y}$ are related to the magnetic flux. $$\Phi=\oint_{S}\bar{H}\cdot\bar{dS}=\oint_{S}\nabla\times\bar{A}\cdot\bar{dS}=\oint_{\partial{S}}\bar{A}\cdot\bar{dl}$$ Using this we get $$\Phi_{1}=\int_{0}^{l_{y}}A_{y}(x, y)dy=f(x, l_{y})-f(x, 0)$$ and $$\Phi_{1}=\int_{0}^{l_{x}}A_{x}(x, y)dx=f(l_{x}, y-f(0, y))$$ Again using the boundary conditions, we get $$\phi_{x, y}=\frac{e\hbar}{w}\Phi_{x, y}$$