The location of an object is gauge dependent. Therefore, it's not measurable?

The location of an object $$x$$ depends on how we choose our coordinate system. If we move the zero point, $$x$$ also changes. However, since we have translational invariance, we can always do such shifts without changing anything.

Now, in quantum theories, there is usually a lot of emphasis on quantities that are gauge independent. For example, the phase of a wave function in quantum mechanics can be changed using global $$U(1)$$ transformations and is therefore gauge dependent. So completely analogous to how we can shift the position of an object using translations, we can here shift here the phase of the wave function.

How are these two situations different? Since both, the location and the phase of the wave function, depend on how we choose our coordinate systems, they both shouldn't be measurable?!

2 Answers

The location of an object x depends on how we choose our coordinate system.

This is not true. The position of an object is invariant; it is the label which we assign to that position which depends on our coordinate system.

Take out a blank sheet of paper and draw a dot on it. Where is the dot? It seems like a bit of a non-answer, but a reasonable response would be "it is where it is."

This isn't particularly useful for computing things, so we typically decide to assign the point a numerical label. We can do this in a number of ways - we could choose a rectilinear grid, or we could choose a polar grid, or something more exotic; we could choose the coordinate origin to be in the center of the page, or we could choose the bottom left instead; and given any valid coordinate system, we can stretch or compress it to get a different one.

However, the dot that you drew on the paper hasn't moved - swapping labels around, which physicists do in our imaginations, has no effect on the world, or on any measurements we could perform in it. A coordinate system is simply a choice of labels for the observable quantity position.

How are these two situations different?

At this level, they aren't. Saying that the coordinates of a particular point are $$(2,3)$$ means absolutely nothing unless I specify a coordinate system, which essentially amounts to a choice of gauge. Similarly, saying that the global phase of a wavefunction is $$\pi/4$$ is meaningless unless I establish a reference point of some kind. This wouldn't be impossible - I could demand that the wave function evaluated at $$x=0$$ be purely real at $$t=0$$, and then calculate the global phase of the wave function at any other time based on this reference point.

The difference lies in the fact that position is an observable quantity while the wave function is not. If it were somehow possible to ascertain the precise value of $$\psi(x)$$, then we could set a reference point as mentioned above and define a meaningful notion of observable global phase. However, as we can only actually measure $$|\psi|^2$$, we can't go backward to unambiguously determine $$\psi$$, and so the global phase of a wave function does not have measurable physical content.

• Thanks! What confuses me is that it’s usually argued that the phase of the wave function (or the wave function itself) are not observable BECAUSE they can be changed arbitrarily by U(1) transformations. However this does not seem to be true, given the above analogy with the location of an object.
– jak
Oct 9 '18 at 6:00
• In some sense, I would even argue that the situation is even more analogous. We can measure the phase of a wave function if we provide a point of reference. In other words, we can measure the phase relative to the phase of another wave function. Analogously, we can only measure the location relative to the location of other objects.
– jak
Oct 9 '18 at 6:03
• @JakobH From a top-down theory perspective, the (pure) states of a quantum system are elements of a projective Hilbert space, which means that changing the global phase corresponds to precisely the same physical state - this is why the phase is not measurable. Oct 9 '18 at 6:15
• The U(1) invariance refers to global phase - multiplying the entire state by some $e^{i\phi}$. Relative phases, e.g. between different points in space, are indeed measurable. Oct 9 '18 at 6:17
• @J.Murray If we solve the Schrodinger equation in a coordinate system we will get the p orbitals. If we solve it again with slightly shifting the coordinate system (a small rotation around z axis) then we will get again p orbitals but now the wavefunctions $p_x$ and $p_y$ will be different. It seems that the solution depends on the choice of the coordinate system but I know that this can't happen. Where I get it wrong? Thanks in advance. Mar 1 '20 at 22:27
1. Translation symmetry is not a gauge symmetry. Nor is the choice of coordinates/origin, since we may use coordinate-independent notations from differential geometry. A physical system has a gauge symmetry if and only if its Hamiltonian formulation is constrained or if its Lagrangian formulation has solutions which contain arbitrary functions of (space)time that cannot be determined from the initial conditions. Equivalently the Hessian of the Lagrangian with respect to the generalized velocities does not have full rank, i.e. is not invertible. For more on the different conditions on gauge theories in the Hamiltonian and Lagrangian formulations, see this answer of mine. Possessing translation symmetry does not imply that the system has a gauge symmetry.

2. Quantum mechanics does not stress the principle of gauge invariance more or less than classical mechanics, but the global phase in quantum mechanics is not an example of a gauge symmetry. Nor is the global phase a "choice of coordinate system": The true "states" in quantum mechanics are rays in Hilbert space, corresponding to points in the projective Hilbert space. When we work with "states" in the Hilbert space instead, we choose - consistently - any one element of the ray as its representant. This is not analogous to a choice of coordinate system - the choice of coordinates on the (projective) space is entirely unrelated to this choice of representants. The choice of representants is a priori arbitrary and does not influence any of the predictions of the theory. In this sense it is similar to a choice of coordinate system, but it really is not - the Hilbert space is simply larger than the actual space of states, and we need to account for that. This is not nitpicking, but at the heart of why the representation of symmetry groups is more subtle in quantum mechanics than in classical mechanics, see this Q&A of mine.