Why do wave functions have to be single-valued?

Question

In some situations, like the wave function of an electron in a hydrogen atom, we must require that the wave function $\psi(r,\theta,\phi)$ to be single-valued, i.e. $\psi\propto e^{i2\pi m\phi}(m\in\mathbb{Z})$. But the result strongly relies on the symmetry of the system. Does there exist multi valued wave functions which do not violate the basic principles of quantum mechanics?

Answer 1

As far as I know, whenever you have a wavefunction defined in the "real world", it has to be single-valued, namely it has to come back to the same expression after a $2\pi$ rotation (in the real world). We see in real life that $2\pi$ rotations are irrelevant.

However this is not the case for wavefunctions defined in spaces which are not our real space, like spin space for example. Spinors change sign after a $2\pi$ rotation so in this sense they are not single-valued. I also encountered an interesting case when I was reading about geometric phases in molecular systems. Using Born-Oppenheimer approximation the total wavefunction is split into an electronic wavefunction and a nuclear one. Since only the total one is "in real space", it has to be single-valued, but the electronic and nuclear one separately do not have this restriction. They can both change sign after a cyclic evolution, with the condition that their product remains the same.

I hope it helps.

edit: by the way, I think the single-valuedness of real stuff is the reason why orbital angular momentum always takes integer values, even though the general theory of angular momentum allows half integer values also.

Answer 2

The probabilities are based on the modulus of the wave function, that is upon $\psi^\star\psi$. In principle this allows for example double valued angular momentum wave functions with

$$\psi(\phi+2\pi)=-\psi(\phi)$$

In practice the double-valued spin functions aren't very nice, but have been used (Pavšič, M (2007). "Rigid Particle and its Spin Revisited". Foundations of Phys. 37 (1): 40–79). If your question is really about integer values of angular momentum, see Ballentine, L. E. (1998). Quantum Mechanics: A Modern Development. pp. 169-172 for an argument that doesn't rely on the single-valued assumption, nor on the more sophisticated Lie algebra maths above by Silvia G.

Answer 3

Hopefully it is not off-topic, but I think I can also provide a (sketch of) mathematical explanation for the difference between rotations in real space and in spin space. In real space we have a 3-dimensional matrix representation of the rotation group $SO(3)$, where the rotation with $2\pi$ corresponds to the identity matrix. The wavefunction is this case is single-valued.

When it comes to 1/2 spin, we start with a 2-dimensional representation of the Lie algebra of $SO(3)$, the Pauli matrices. However, it cannot be exponentiated to a representation of the group because the group is not simply-connected. It will be exponentiated to a representation of the universal covering group of $SO(3)$, namely the simply-connected group $SU(2)$. And indeed this leads to a projective representation of $SO(3)$, which is actually what we want (we deal with the space of states, not with the Hilbert space itself). Now, $SU(2)$ is a double –covering of $SO(3)$, roughly speaking $SU(2)$ is made out of two copies of $SO(3)$. Rotation with $2\pi$ is sent to minus the identity matrix, while rotation with $4\pi$ is sent to the identity matrix. Both are projected to the same element in $SO(3)$. The wavefunction in this case is not single-valued, but double-valued.

It’s a similar discussion with the Lorentz group and $SL(2,\mathbb{C})$. The Dirac spinors also change sign at $2\pi$ rotations. The book of B.Thaller "The Dirac equation" provides a rigorous discussion of what I tried to say here with covering groups and sign changes.

Answer 4

It is so because in case it is multi-valued then it will lead to different probabilities of finding the Particle at given point of space at a given time which is not possible