5
$\begingroup$

In elementary treatments of quantum mechanics, we are taught that the wavefunction of a single particle is complex valued ($\Psi : \mathbb{R}^3 \to \mathbb{C}$). In particular, the wavefunction has a definite phase at each point (up to a global phase redefinition).

However, I have also seen references (see the quote from the Dirac paper in the footnote below) to the idea that it is only path-dependent phase differences between points that are well-defined. That is, $\Psi$ is actually to be viewed as a section of a bundle of $\mathbb{C}$ over $\mathbb{R}^3$. The wikipedia article on the Aharonov-Bohm effect refers to this idea (http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect#Mathematical_interpretation), but I can't tell if it's something physically new.

$\bf{Question:}$ Are there physical consequences to $\Psi$ being a section instead of a regular function, or is it just a mathematical convenience? In other words, is there any measured phenomena (so, exclude magnetic monopoles) that forces us to view $\Psi$ as a section?

$\bf{Footnote:}$ "We may assume that $\gamma$ [the phase of the wavefunction] has no definite value at a particular point, but only a definite difference in values for any two points. [...] For two distant points there will be then be a definite phase difference only relative to some curve joining them and different curves will in general give different phase differences." -Dirac's 1931 paper (http://rspa.royalsocietypublishing.org/content/133/821/60)

EDIT: Another paper that refers to this idea is R. Jackiw's "(Constrained) Quantization Without Tears" (http://arxiv.org/abs/hep-th/9306075). The magnetic vector potential $a_i$ in nonrelativistic quantum mechanics (see equation 16(b)) is referred to as a connection one-form in the appendix.

$\endgroup$
  • 1
    $\begingroup$ If I understand it correctly you are asking if the non-relativistic single particle wave function is physically invariant under a scalar gauge transformation multiplying it with a constant of modulus of 1? I believe the answer is "yes". $\endgroup$ – CuriousOne Apr 27 '15 at 3:06
  • $\begingroup$ I think it is worthwhile to quote part of the wiki you cited, "If we want to ignore the physics inside the superconductor and only describe the physics in the outside region, it becomes natural and mathematically convenient to describe the quantum electron by a section in a complex line bundle." So it's really about ignoring the magnetic flux and replacing it with new math for the outside region alone. Math designed to replicate what the magnetic flux does. I think it is misleading. $\endgroup$ – Timaeus Apr 27 '15 at 5:14
  • $\begingroup$ What are the base and total space of the supposed bundle? Is it just a bundle or a fiber bundle (i.e. it has a topological structure)? $\endgroup$ – yuggib Apr 27 '15 at 7:07
  • $\begingroup$ @yuggib: I believe the idea here is to view the A-B effect as a $\mathrm{U}(1)$-gauge theory effect and hence have the wavefunction (or scalar field, rather) be the section of an associated $\mathrm{U}(1)$ vector bundle, where the path dependent phase then appears as the holonomy of paths. $\endgroup$ – ACuriousMind Apr 27 '15 at 16:08
  • $\begingroup$ marlow, I'm not sure what you are asking here - there's one description, and there's another. To say that "there is a phase up to global phase redefinition" and keeping in mind the gauge freedom of electromagnetism is the same as saying "it's the section of a certain $\mathrm{U}(1)$-associated bundle". If you're asking whether these descriptions are physically inequivalent, the answer is no. $\endgroup$ – ACuriousMind Apr 27 '15 at 16:10
2
$\begingroup$

The reason that Dirac is saying this is: every prediction that QM makes is some sort of expectation value computed in the modern notation as $\langle \Psi|\hat A|\Psi\rangle$; a phase rotation of $|\Psi\rangle \mapsto e^{i\theta} |\Psi\rangle$ therefore maps all of these predictions to $e^{-i\theta} e^{i\theta} \langle \Psi|\hat A|\Psi\rangle = \langle \Psi|\hat A|\Psi\rangle.$

His argument is essentially, "well, since only phase differences matter in the normal theory, you can imagine that we're looking at objective-phases $\gamma$ when we're supposed to be looking at phase-differences $\nabla \gamma$, so let's imagine that we look at those directly, path-integrating them from a starting point to get a total phase difference. Assuming that $\gamma$ exists, therefore, is assuming that our new phase-field is conservative, and every difference is path-independent. But is there a weaker form of this assumption that we can use? Yes: we can insist that all of the wave functions pick up the exact same "extra" phase from one path over another path, so that every closed loop has the same phase difference for every wavefunction. We need this because $\hat A$ can be viewed as being made of a sum of scaled eigenfunctions $\int da ~a~|a\rangle\langle a|,$ so we would then always get $\langle a|$ getting some extra phase $e^{-i~\Delta\gamma}$ matching exactly $|a\rangle$ getting some extra phase $e^{+i~\Delta\gamma}$, and then our observables $\hat A$ do not change and $\langle \Psi|\hat A|\Psi\rangle$ doesn't either, so everything is unvaried."

$\endgroup$
1
$\begingroup$

Let me attempt an answer, based on the comments from Timaeus and ACuriousMind..

Answer: No, there is nothing that forces us to view non-relativistic quantum mechanics as a $U(1)$ gauge theory. It's an elegant, equivalent way of looking at some situations like the Aharanov-Bohm effect.

In more detail: We could treat the Aharanov-Bohm effect in the usual, elementary way, with $\Psi$ as a function defined on the whole space, including the magnetic flux. Alternatively, if we are only interested in the region outside the magnetic flux, we can view $\Psi$ as a section, with the magnetic flux manifesting itself as curvature. This seems to parallel the distinction made in Olaf's answer to a question about constrained quantization (Quantum mechanics on a manifold); namely, we have an "extrinsic" point of view that can describe all of space, including the magnetic flux region, and an "intrinsic" point of view that can only describe the region outside the flux.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.