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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.

2 Answers 2

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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.

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    On the answer of my 2nd question: So if I "map" my original problem to another one with different boundary conditions via a gauge transformation, I should not expect the eigenvalues to remain the same? But the whole theory of gauge invariance(as we see it in physics) does not mention boundary conditions, only the gauges. – 2017-02-14
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    Also, why do you think that the problem I describe is not a proper gauge transformation? – 2017-02-14
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    @TheQuantumMan - I already told you why I don't think it is a gauge transformation. I believe you are missing something in what qualifies as gauge invariance. If you change the boundary conditions, then you change the problem, and you change the solution. That is not "invariant". Gauge invariance (as I recall it - again, it has been many years) is about things that do not change the solution itself, but only how we represent it. This may be hard to spot, because we are used to seeing the representation, so if it changes, it looks like the solution does too. – 2017-02-14
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    But the actual solution is independent of some of the elements we use to represent it. Two different representations look vastly different, but those differences "mod out" when you consider the abstract version of the solution. They only involve things we introduced to allow us to represent the solution in a manner we can do calculations from. But the boundary conditions are part of the problem definition, and so are part of the solution itself. If two transformations both solve the equation for different conditions, they at least differ on the boundary and so cannot be the same. – 2017-02-14
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    In your example, you changed from a problem about a free particle to a problem about a particle in a magnetic field. The problem changed and solution changed. This is to be expected. A particle in a magnetic field behaves differently from a free particle. You may have solved the problem of the bound particle, but this by itself does not solve the free particle problem. The only place where I see that is helpful is when you are really interested in a greater system where the differences in behavior between free and bound particles do not matter. – 2017-02-14
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    On your last comment, I would like you to clarify something last for me please. When we have a magnetic field inside the ring, then there is not a magnetic field passing through the curve that the ring defines. This means that we only have a magnetic vector potential A. Thus, in the area that the wavefunction exists(on the ring) the physical quantity, which is the magnetic field, is the same in the two problems(without and with the magnetic field respectively). So, physically the two problems are equivalent. What would you answer to the argument that since we are only interested – 2017-02-14
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    in the area that the wavefunction exists and the two problems are physically equivalent there, then the gauge transformation should not change the eigenvalues of the original problem? And, if you work it out, we can solve the whole problem of the ring with magnetic field in the middle by gauge transforming the problem to the free particle on a ring problem. The only difference is in the energies which are mathematically associated with the eigenvalues of the problem. This is, at the end of the day, the heart of my question. I can't argue against the above. – 2017-02-14
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    If I follow this correctly, you are interested only in the solution for the particle in the ring, but are looking at a system that entails all space. This larger system is not easily solvable, but if you add a magnetic field outside the region of interest, it becomes so, and then you can restrict the solutions to the region. What you've changed with the gauge is the solution outside the ring, not in. The boundary conditions for that outside solution may change, but if you change those for the ring, then it is not gauge invariant. – 2017-02-15
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    Well, I just want the solution for the wavefunction that lives on the ring and also the eigenvalues of the problem(the energies). So, I have a ring on which the particle lives and I also have a magnetic field in the middle of the ring. I can gauge transform to a problem with no magnetic field since the magnetic field of the two problems in the region of the solution is in both cases equal to zero. But, it turns out that the eigenvalues of the problem are not the same. So , the boundary conditions in the region of interest is in both cases the periodic ones Psi(x)=Psi(x+2pi) with x the azimuth – 2017-02-16
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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 $$\phi_{x, y}=\frac{e\hbar}{w}\Phi_{x, y}$$