Fiber bundle understanding of the wavefunction

Question

Usually people say that given a wavefunction $\Psi$ although $|\Psi(\cdot, t)|^2$ is the probability density for the position random variable at time $t$, the wavefunction $\Psi$ itself has no physical meaning. It occurs, however, that if $M$ is spacetime, then we know that $\Psi : M\to \mathbb{C}$. In that setting, if one considers $E = M\times \mathbb{C}$ and $\pi : E\to M$ given by $\pi(a, z)=a$ then $(E,M,\pi)$ is the trivial bundle with typical fibre $\mathbb{C}$

In that case if we let $s\in \Gamma(E)$ be a section of the bundle, we have $s(a)=(a,\xi(a))$. In that case the wavefunction can be thought of as a section of the bundle $\Psi(a)=(a,\xi(a))$ where $\xi(a)$ is what we called wavefunction earlier.

In that case, it seems we are, at each point of spacetime, plugging one copy of the plane and laying down a vector on the plane. This vector is the wavefunction at the point.

It is tempting, therefore, to generalize this a little further and consider a general vector bundle $(E,M,\pi)$ where $M$ is spacetime and $\mathbb{C}$ is the typical fibre, only this time we allow twists. Then locally, given an event $a\in M$ there is one open set $U\subset M$ with a local trivialization $\varphi : \pi^{-1}(U)\to U\times \mathbb{C}$ where the arguments work as before.

So my question is: this way to view the wavefunction has any advantage? The generalization to allow non-trivial bundles give something useful? This point of view allows one to understand the meaning of the wavefunction itself?

Answer 1

Your introductory claims are not correct. The usual wavefunction is not a function of spacetime, for two reasons: It is not Lorentz invariant (time is singled out, so it is not a function of a point on a spacetime manifold, and it is not even time dependent at all in the Heisenberg picture) and it can depend on more than one "set" of positions - for a system of $n$ particles in $d$ spatial dimensions, the wavefunction will depend on $n\cdot d$ position variables. Thus, viewing it as a section of something over spacetime is inappropriate.

If you want to understand a quantum state as the section of a complex line bundle, that is possible through the procedure of geometric quantization.

Let $X$ be the classical phase space with symplectic form $\omega$.

The actual line bundle you have to consider is the prequantum line bundle, which is a line bundle over the phase space equipped with a $\mathrm{U}(1)$-connection $\nabla$ such that the curvature of the connection is the symplectic form $\omega$. If you have now a choice of polarization on the phase space, i.e. a split of the coordinates into positions $q$ and momenta $p$, the actual wavefunctions of quantum mechanics are the sections of the prequantum line bundle which only depend on position, i.e. are constant on surfaces of constant $q$.¹

If you're wondering where time evolution is here: We have only said what the space of states is. The time evolution is of course given by the unitary operator generated by the Hamiltonian (which is encoded in $\omega$) acting on this space of states. Generically, the action of the quantum version of a phase space observable $f$ is given by the covariant derivative (defined by the connection $\nabla$) of the section along the vector field $X_f$ associated to $f$ by $\mathrm{d}f(\dot{}) = \omega(X_f,\dot{})$. More precisely, $f$ acts on a section $\psi$ as $$ \hat{f}(\psi) = \mathrm{i}\nabla_{X_f}\psi + f\cdot\psi$$

where $\hat{f}$ is now the (pre-)quantum operator associated to $f$.

Now, lastly, for the (non-)triviality of the bundle:

Nothing forbids the prequantum line bundle from being non-trivial, but the classical phase space needs to have non-trivial cohomology for that - complex line bundles are completely characterized by their first Chern class, which is an element in $H^2(X,\mathbb{Z})$. Thus, in most situations you will encounter in typical quantum mechanics applications (where the phase space is just a symplectic vector space, and hence in particular contractible, so most cohomologies vanish), the prequantum line bundle will indeed be trivial.

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^{1Also, they should be square integrable in an appropriate sense. This requires discussing how we equip the space of sections with a Hilbert space structure and what measure we integrate against - these choices are done by a metaplectic correction.}