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This is a question on gauge invariance in quantum mechanics. I do some simple math on a 1D wave-function with periodic boundary conditions, and get that gauge invariance is violated. What am I doing wrong?

Consider one coordinate dimension configured as a ring. The gauge dependent momentum operator can be written:

$p_{op}=-i \frac{\partial}{\partial x} - k$

Units have been chosen so that $\hbar = 1$, $k$ is an arbitrary real constant different for each gauge and $x$ represents the coordinate.

The gauge dependent eigenfunction can be written

$\psi(x)= Ae^{i(n+k)x}$

where A is a constant determined by normalization. As is well known in quantum mechanics, an operator applied to one of its eigenfunctions should yield a real constant eigenvalue multiplying the same eigenfunction: Thus

$[-i \frac{\partial}{\partial x} - k] Ae^{i(n+k)x}= nAe^{i(n+k)x}$

so that the real number n is the eigenvalue, which must be determined by the boundary conditions.

The boundary condition for this periodic system must be that the wave function should join onto itself smoothly everywhere. Thus, if the coordinate is chosen such that x extends from –$\pi$ around the ring to $\pi$ then the eigenfunction in equation 3 must have (n + k) = m, where m is an integer.

Under these conditions, the eigenvalue n in equation 3 will be n = m – k. This eigenvalue depends explicitly on k, and so is not gauge invariant.

I'm assuming this simple situation should be gauge invariant, but I don't see where I goofed. I'd appreciate any help.

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Thanks to kmm for upgrading the format of my equations.--AD –  Arthur Davidson Jun 19 '11 at 1:53
@Arthur Davidson: You're welcome. –  Kasper Meerts Jun 19 '11 at 10:09
You didn't goof up. You just discovered that von Neumann theorem doesn't need to hold for non-trivial systems. Recall than for systems on ${\mathbb R}^N$ (described by Heisenberg group) von Neumann states that all strongly continous unitary irreducible representations of the Heisenberg group are equivalent and you can always work in the standard representation given by multiplication and derivation operators for $x$ and $p$. This doesn't hold for topologically non-trivial manifolds (such as circle) and there is continuum of inequivalent irreducible representations (index by $k$ here)! –  Marek Jun 19 '11 at 11:31
The same thing (although more complex) happens in QFT where there is a problem with a choice of vacuum. Again, this is because QFT can be realized as a system with infinite number of degrees of freedom, so $N \to \infty$ and von Neumann doesn't hold again. –  Marek Jun 19 '11 at 11:32
@Arthur: I don't really have an answer. I just wanted to bring into your attention the fact that this stuff is actually pretty deep. Well, at least from the mathematical point of view, physicists don't really care usually and proceed in the standard fashion. And usually they get the correct results. I always wondered how they do that... :) –  Marek Jun 20 '11 at 15:11
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5 Answers

2nd Update

If you just want to think about a 1D universe with periodic boundary conditions, there is simply the point that there is no gauge transformation that takes from one you value of $A$ constant to another. Recall a gauge transformation takes $A_\mu\rightarrow A_\mu +\partial_\mu f$, where $f$ is an arbitrary smooth function. To go from $A = A_1$ to $A=A_2$ you would want to gauge transform with $f(x) = (A_1-A_2)x$. But since your system is periodic in $x$ this is not a smooth function. That would be okay if $\exp(i f(x)$ was a single valued function but that only happens when $A_1 -A_2$ is an integer. It is an interesting fact about some topologically nontrivial spacetimes that there are gauge fields which give the same electromagnetic fields but are not equivalent by gauge transformation. And Quantum Mechanics actually cares about the gauge fields. So it does not break gauge invariance for different $A$ to give different physical answers, since they are not connected by a gauge transformation.

The update below is mainly concerned with thinking about your 1D quantum ring as embedded in actual space. Still worth reading


Arthur, I applaud the fact you're still attempting to understand this. So I'm trying to re-explain what I wrote here. QMechanic and Marek have both provided good explanations, but they don't hit exactly the source of what seems to be your confusion. Let me try again to explain.

-Your question is: I have this simple system with a gauge field, and my answer seems to be depend on my choice of gauge. How can this be? I thought nothing physical can depend on your choice of gauge.

  • Let's step back and remember where we first met gauge invariance. We had these $E$ and $B$, the electromagnetic fields, and these are observable physical quantities. But these are a pain to work with so we introduce the gauge fields $A$. But these are massively redundant - I can gauge transform my fields $A$ to something completely different and still have the same $E$ and $B$. Since $E$ and $B$ are what we care about in the first place we demand our theories be "gauge invariant" - nothing physical can change when I do a gauge transformation.

  • Gauge invariance is created by the fact that different gauge fields lead to physically equivalent situations. There are lots of things that look like gauge fields - but only in those situations where gauge transformation connect physically equivalent situations must gauge invariance be respected.

  • In the situations you mention yourself, the rigid rotator, there is something that looks a lot like a gauge field. We have a hamiltonian $H = \frac{1}{2m}(p - m\omega R)^2$, where $\omega$ is the angular velocity. This looks just like a gauge field $A= m\omega R$. But different values of $A$ and $\omega$ do not correspond to the same physical situation. The rotator is spinning at a different speed. This is visibly different. Because the different $A$ are physically inequivalent, there is no reason to expect the answer to be independent of the value of $A$. And thats what you find. Correctly.

  • In the case where $A$ actually is the electromagnetic vector potential the situation is slightly more subtle, since thats really supposed to gauge invariant. But I show below in the original post that, here as well, different choices of $A$ correspond to different physical situations.

Bottom Line: Just because something looks like a gauge field doesn't mean we must have gauge invariance. Gauge invariance is only a must when gauge transformations connect physically equivalent situations. In the systems that realize a 1D quantum ring different choices of $A$ are not physically equivalent.

Original Post

Your original calculation is correct. The energy levels do depend on the magnitude of the gauge vector $A$.

How does this not break gauge invariance? Calculate $\oint A(r)\cdot dr$, that is the integral of your gauge field around the loop. This is $\int dS\cdot \nabla \times A$ by Stokes theorem. This is $\int dA \cdot B$ by the definition of the gauge field. So its equal to the magnetic flux $\Phi$ threading through your ring.

So if you take $A$ to be constant like you did, then we have $A = \frac{\Phi}{2\pi R}$, where $R$ is the radius of your ring. But $\Phi$ is a physical quantity - it can't change under gauge transformations. There is no gauge transformation that keeps $A$ constant and changes its magnitude. So its okay that different $A$ give different energy levels - they correspond to different physical situations.

This is the basis for many physical phenomena: the Aharanov-Bohm effect, flux quantization, Little-Parks effect, weak localization etc.. It's all on wikipedia.

By the way, the weird part here is not so much that your answer depends on the guage, since the different gauges are not equivalent. The weird part is that if I take a ring a light year in radius and put a magnetic field through the a meter-squared patch in the center of the ring, all the particles know about this magnetic field even though they are a light year away. But if I cut a meter long chunk out of my ring they suddenly forget all about that magnetic field. (Only in the limit where everything is impossibly clean, but still, its wacky).

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I appreciate your intuitive approach, but I think it is misleading. The problem I posed has no potential. I am considering a rotator without torque or applied field of any kind. The question is, where in my 3 equations and continuous boundary condition is the error that produces a violation of gauge invariance? –  Arthur Davidson Jun 20 '11 at 18:03
@Arthur Davidson: I apologize for leaping to the magnetic field language. You didn't actually specify what the gauge field was, so I assumed. But the point remains: If two gauge fields give the same physical configuration than no physical quantity can distiguish them. But the point is although normally adding a constant to the gauge field gives the same physical, with non-trivial boundary conditions this gives a different physical configuration. In the rotator case adding a constant to the gauge corresponds to spinning faster, which is a physically different situation. So there is no reason –  BebopButUnsteady Jun 20 '11 at 22:32
-cont-- for the energy levels to be the same and indeed they aren't. If you want the more general and more mathematical statement, than see Marek's comment. –  BebopButUnsteady Jun 20 '11 at 22:34
Comment to the answer (v3): In the periodic 1D space, the Lie-algebra-valued gauge-parameter called $f(x)\in u(1)$ does not have to be single-valued. Only the Lie-group-valued gauge-element $e^{if(x)}\in U(1)$ and the derivative $\partial_x f(x)$ must be single-valued. This implies that $A_x$ is actually allowed to change with a discrete amount under a so-called 'large' gauge transformation, i.e. a gauge transformation not continuously connected to the identity. –  Qmechanic Jul 6 '11 at 11:12
Okay now come back to 1D. It is a fact of 1D with trivial boundary conditions, that all functions $f(x)$ are equivalent to each other by gauge transformation. But now when I put on periodic boundary conditions this is no longer true! Thats what QMechanic and I have shown in our answers. There is no gauge transformation that takes you from, say, $A=0$ to $A=1/2$. Because there is no gauge transformation that connects the two, gauge invariance doesn't require that they give the same answers, and they don't. –  BebopButUnsteady Jul 6 '11 at 20:14
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1) Let us for simplicity put various constants to one: Speed of light $c=1$; Planck constant $\hbar=1$; Mass $m=1$ of non-relativistic scalar (Bosonic) particle in $1+1$ dimensions; charge of particle $q=1$; Circumference of spatial circle $\ell=2\pi$.

2) The mechanical momentum (sometimes called the kinetic momentum)


(or equivalently, the covariant derivative) commutes with the Hamiltonian $\hat{H}=\frac{1}{2}\hat{v}^2+\Phi$ in the temporal gauge $\Phi=A^t=0$, which we will assume from now on. (Recall that $\Phi=A^t$ is the temporal component of the gauge potential.) So we can find common eigenstates. Suppressing the time dependence in what follows, we want to solve the mechanical momentum eigenvalue equation

$$\hat{v}\psi_v(x)=v \psi_v(x),$$

where the mechanical momentum eigenvalue $v\in\mathbb{R}$ is related to the energy $E=\frac{1}{2}v^2$. The solution is $e^{ivx}$ times a Wilson line:

$$\psi_v(x)~=~ \psi_v(0)\exp\left[ivx+i\int_0^x A_x(x')dx'\right] ,\qquad v\in\mathbb{R}.$$

3) Under a local gauge transformation

$$\psi(x)\longrightarrow \tilde{\psi}(x):=e^{i\alpha(x)}\psi(x), \qquad A_x(x)\longrightarrow \tilde{A}_x(x):=A_x(x)+\partial_x\alpha(x), $$

it is well-known that the covariant derivative (or mechanical momentum) transforms covariantly,

$$D_x\psi(x)\longrightarrow e^{i\alpha(x)}D_x\psi(x), \qquad D_x \longrightarrow \tilde{D}_x ~=~ e^{i\alpha(x)}D_x e^{-i\alpha(x)}, \qquad \hat{v}\longrightarrow e^{i\alpha(x)}\hat{v} e^{-i\alpha(x)}.$$

It is true that they are not gauge invariant. They are only gauge covariant. However, we may easily construct manifestly gauge invariant quantities, for instance, $|\psi(x)|^2$; $\psi^*(x) \hat{v}\psi(x)$. In particular, the mechanical momentum eigenvalue


is a gauge invariant (still assuming temporal gauge $A^t=0$).

4) Finally, we assume for simplicity that $A_x(x)=A_x\in\mathbb{R}$ and $\tilde{A}_x(x)=\tilde{A}_x\in\mathbb{R}$ are independent of $x$. This corresponds to a partial gauge fixing. We still have residual gauge transformations left where $\alpha(x)$ is an affine function of $x$. The eigenfunction becomes

$$\psi_v(x)~=~ \psi_v(0)e^{i(v+A_x)x},\qquad \tilde{\psi}_v(x)~=~ \tilde{\psi}_v(0)e^{i(v+\tilde{A}_x)x},\qquad v\in\mathbb{R}. $$

We now recall that the $x$-coordinate is periodic $x\sim x +2\pi$. The wave function should be single-valued




as OP observes. In other words, $A_x$ and $\tilde{A}_x$ belong to the same shifted lattice $\mathbb{Z}-v$. The residual affine function $\alpha(x)$ must have $x$-independent integer-valued derivative


5) So what have we learned?

  1. On one hand, we may write $v\in\mathbb{Z}-A_x$, so we see that the energy $E=\frac{1}{2}v^2$ and the mechanical momentum $v$ depend on the gauge potential $A_x\in\mathbb{R}$.

  2. On the other hand, we saw in section 3 that $E$ and $v$ are gauge invariants, i.e., they are invariant under gauge transformations (still assuming temporal gauge $A^t=0$).

The two statements (1.) and (2.) do not clash in terms of physics, only in terms of semantics. In particular, if we perform a gauge transformation on the formula $v\in\mathbb{Z}-A_x$, we cannot claim that $v$ changes since a gauge transformation changes $A_x$ by an integer. ($A_x$ itself is not necessarily an integer.)

6) Another example is the canonical momentum operator $\hat{p}=\frac{1}{i}\partial_x$.

  1. On one hand, it is independent of the gauge potential $A_x$.

  2. On the other hand, it is not a gauge covariant quantity, i.e., it does not transform covariantly under a gauge transformation

$$\hat{p}\longrightarrow e^{i\alpha(x)}\hat{p} e^{-i\alpha(x)}~=~\hat{p}-\partial_x\alpha(x).$$

Again, the two statements (1.) and (2.) do not clash in terms of physics, only in terms of semantics.

7) Finally, let us connect to Marek's comment. We have a Hilbert space $L^2(\mathbb{R}/\mathbb{Z})$ of wave functions on a circle $\mathbb{R}/\mathbb{Z}$, and a Heisenberg algebra $[\hat{x},\hat{v}]=i$. (More precisely, the Stone-von Neumann theorem is a statement about the corresponding Heisenberg group to avoid issues with unbounded operators.) I interpret Marek's comment as roughly saying that

$$\hat{x}=x, \qquad \hat{v}=\frac{1}{i}\partial_x-A_x,$$

yields inequivalent irreducible representations of the Heisenberg algebra labeled by a continuous label $A_x\in[0,1[$. Changing $A_x \to A_x+1$ yields an equivalent representation due to gauge symmetry.

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@Qmechanic: I think I follow most of you math, except for "residuial affine function," but I think in the end you get the same non gauge invariant eigenvalue that I do. Do you not have the eigenvalue v, which is real, and v+Ax must be an integer? –  Arthur Davidson Jun 20 '11 at 4:04
@Arthur Davidson: I added some more explanation to the answer (v2). –  Qmechanic Jun 20 '11 at 9:01
@Qmechanic: Thanks for adding more info. Your bottom line (literally) is "This implies that the wave function ψ˜v(x)=ψv(x) is actually gauge invariant under the remaining affine gauge transformations." Doesn't this mean that gauge invariance of the momentum eigenvalue is valid only if we restrict Ax to integers? –  Arthur Davidson Jun 20 '11 at 14:51
@Arthur Davidson: I removed the incorrect bottom line in v3. –  Qmechanic Jun 20 '11 at 18:04
@Qmechanic: So would you say that your analysis leads to the same flawed conclusion as mine? –  Arthur Davidson Jun 20 '11 at 21:28
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I believe you're seeing a problem where there isn't one. Indeed we have $n + k \in \mathbb{Z},$ so if you shift $k$, say $$k \mapsto k + \delta,$$ you need to adjust $$n \mapsto n - \delta.$$ I wouldn't say that $n$ explicitly depends on $k$, but shifting $k$ by an arbitrary real (= non-integer) constant does force an adjustment for $n$ - not very surprising.

However, I think you can set up the problem in a more natural way. Forget about the gauge part, just consider the wave function $\phi(x) = e^{i n x}$; to satisfy the p.b.c., you find that $n \in \mathbb{Z}.$ Now add a gauge part, $U_k(x) = e^{ikx};$ you then find that the derivative changes to $\partial_x \mapsto D_x = \partial_x - ik$ (modulo a sign error) etc. The total wave function changes to $\phi(x) \mapsto U_k(x) \phi(x) = \Psi(x)$; again, you find that $k \in \mathbb{Z}$ etc.

In this new set-up, you find that you can change $k$ by an integer without punishment, and I think this is also meant in your problem; once you restrict yourself to the integers, you have integer freedom in your choice of gauge.

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I don't understand "you find that $k \in {\mathbb Z}$". How did you find that? I am curious because this statement doesn't hold ;) –  Marek Jun 19 '11 at 11:38
If n is integer, and n + k is integer, than k is also integer - right? Please explain what's going on! –  Gerben Jun 19 '11 at 13:00
In that case I have no idea what's this answer supposed to mean because in the original problem there was the freedom to choose arbitrary real $k$. –  Marek Jun 19 '11 at 13:28
I think I see the gist of your answer. But I don't have a theoretical physics or advanced math background: can you go step by step, and use English instead of logic symbols to show how your solution gives gauge invariance? –  Arthur Davidson Jun 20 '11 at 3:57
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It's possible to do quantum mechanics on other domains than $\mathbb R^3$, but you have to make a few adjustments, which you forgot to do. Namely, the gauge transformation

$$ \psi(x) \mapsto e^{i\chi(x)} \psi(x) $$

must fulfill $e^{i\chi(0)}=e^{i\chi(2\pi)}$ to be single-valued on the circle, i.e.

$$\chi(2\pi)-\chi(0)=2\pi m \quad\text{ with } m\in\mathbb Z$$

In other words, your choice $\chi(x)=kx$ is only allowed for particular values of $k$, namely $k\in\mathbb Z$.

Moreover, the eigenvalue does not depend on the gauge, despite your claim to the contrary. That's because a gauge transformation will change the velocity operator

$$ v \mapsto v - \partial_x \chi(x) $$

and its eigenvalues stay the same. No physically measurable quantity will change after a gauge transformation, ever.

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I agree that gauge invariance works with periodic boundary conditions on the ring for integer values of k. My point is that gauge invariance will work for real values of k if you're willing to change the boundary conditions. Moreover, making the probability density and probability current density periodic instead of the raw wave function seem like perfectly reasonable things to do. My arxiv.org paper goes into the details: arxiv.org/abs/1106.4510 . By the way, I expanded this paper and the revision should show up 8/1/2011. –  Arthur Davidson Jul 31 '11 at 17:56
@Arthur Davidson: It's ok to change the boundary conditions of the wave function (to $\psi(2\pi)=e^{i\theta}\psi(0)$). But the gauge transformation must still fulfill periodic boundary conditions. –  Greg Graviton Aug 1 '11 at 7:42
_@Greg Graviton:I don't believe there is anything in gauge theory that says the ring domain can only have integer gauges. Rather, that is the only thing that works if periodicity is required of ψ. Putting the boundary condition on the periodicity of ψ*ψ relieves the necessity of using only integer gauge variables, so ultimately, the question can only be settled by observation. –  Arthur Davidson Aug 1 '11 at 15:50
@Arthur Davidson: Nope. The gauge transformation shouldn't change the vector bundle, i.e. the boundary condition. This means that it has to be periodic. (Unlike $\psi(x)$, which doesn't need to be periodic.) –  Greg Graviton Aug 1 '11 at 16:47
_@Greg Graviton: I need an explanation or reference about your assertion about "the vector bundle." It would be good if there were an explanation aimed at an advanced undergrad physics major. –  Arthur Davidson Aug 1 '11 at 19:26
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up vote 0 down vote accepted

I would like to try an answer for my own question.

The straightforward interpretation of the 3 equations in the original post is that they do present a legitimate conflict. This conflict disappears when the spatial domain is extended to infinity, and the role of the boundary conditions goes away. So, the conflict, the violation of gauge invariance for the finite ring, suggests that either the ring is simply not covered by quantum mechanics, or there is something about the boundary conditions implied by the ring geometry that doesn't work.

So are there other boundary conditions that could be tried? The obvious ones are to demand periodicity not of the wave function, but of the probability density and probability current density. These are both real (not complex) quantities, would let the phase of the wave function have a discontinuity at the boundary, so long as the gradient of the phase was smooth.

Such a choice for the boundaries would allow the continuous eigenvalue spectrum of the infinite line to apply as well to the ring, which would restore gauge invariance. But this comes at the expense of inhomogeneous and nonlinear boundary conditions.

The nonlinear boundary conditions could be accepted if they are consistent with the standard type of Hilbert space of Hermitean operators, and the principle of superposition.

In a paper I recently posted on arxiv.org I go into some detail about how the nonlinear boundaries are indeed consistent with both gauge invariance and the standard structure of Hilbert space. The bottom line is that the nonlinearity creates a continuous spectrum of eigenvalues, not all of which are superposable in Hilbert space. The nonlinearity in the boundary allows a subset to be superposable. The combination of continuous eigenvalues and superposable discreet eigenvalues makes a band structure for the Hilbert space, rather than simple discreet levels.

PLEASE CLICK HERE for the larger explanation.

So I think the correct solution of this problem is quite significant for quantum systems coupled to their environments, such as Josephson junctions used as qubits.

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