Mathematical probabilistic interepretation of probability amplitude

Question

As a warning, I come from an "applied math" background with next to no knowledge of physics. That said, here's my question:

I'm looking at the possibility of using probability amplitude functions to represent probability distributions on surfaces. From my perspective, a probability amplitude function is a function $\psi:\Sigma\rightarrow\mathbb{C}$ satisfying $\int_\Sigma |\psi|^2=1$ for some domain $\Sigma$ (e.g. a surface or part of $\mathbb{R}^n$)-- obviously these are some of the main objects manipulated in quantum physics! In other words, $\psi$ is a complex function such that $|\psi|^2$ is a probability density function on $\Sigma$.

From this purely probabilistic standpoint, is it possible to understand why multiple $\psi$'s can represent the same probability density $|\psi|^2$? What is the most generic physical interpretation?

That is, if I write down any function $\gamma:\Sigma\rightarrow\mathbb{C}$ with $|\gamma(x)|=1\ \forall x\in\Sigma$, then $|\psi\gamma|^2=|\psi|^2|\gamma|^2=|\psi|^2$, and thus $\psi$ and $\psi\gamma$ represent the same probability distribution on $\Sigma$. So why is this redundancy useful mathematically?

Answer 1

Different wave functions with the same $|\psi(x)|^2$ represent different physical states (unless they are proportional). Different states means that one gets different measurable results on at least one kind of measurements.

The same $|\psi(x)|^2$ gives the same probability density for position measurements (only), but generally not for measurements of other observables such as momentum. For the momentum probability density, the absolute squares of the Fourier transform counts, and this is usually different if only the $|\psi(x)|^2$ are the same.

The mathematical content of the wave function is the following (from which the above follows): The inner product of $\psi$ with $A\psi$ gives the expectation value of the operator $A$ for a system in state $\psi$. For example, if you take $A$ to be multiplication by the characteristic function of a region in $R^3$ you get the probability for being in that region. The position operator is simply multiplication by $x$, while the momentum operator is a multiple of differentiation.

For going deeper, try my online book http://lanl.arxiv.org/abs/0810.1019, written for mathematicians without any background knowledge in physics.

Answer 2

The redundancy is useful because, apparently, the phases have a physical meaning, and relative phases do actually make a difference to the probabilities in some situations. For example, consider a simplified two-slit experiment. We have a photon emitter, which fires a photon toward two slits. Behind the two slits is a detector, which will either fire or not fire. (If it doesn't fire, we think of the photon as having "missed" the detector and been absorbed by something else.) We also have the option to try and detect which of the slits the photon passed through, or not to try and do this.

Let $E$ stand for "a photon is emitted", $D$ stand for "the detector fires" $S_i$ stand for "the photon was detected passing through slit $i$." If we do try to detect which slit the photon passed through, the probability of the detector firing is $$ p(D|E) = p(S_1|E)p(D|S_1) + p(S_2|E)p(D|S_2),$$ as you would expect from elementary probability theory. If we want, we can formally define a complex number $a(X|Y)$ for each pair of events, such that $p(X|Y) = |a(X|Y)|^2.$ There is some redundancy in this definition because any choice of phase gives the same probability. Now we have $$ p(D|E) = |a(S_1|E)a(D|S_1)|^2 + |a(S_2|E)a(D|S_2)|^2.$$ Note that this is completely non-standard notation that you won't find anywhere, but it's a perfectly reasonable way to express the path integral formalism for this type of simplified system.

If we don't try to detect which slit the photon passed through, so that it remains isolated throughout its journey, then it's a bit different. Now it turns out that instead of the above expression we have $$ p(D|E) = |a(S_1|E)a(D|S_1) + a(S_2|E)a(D|S_2)|^2,$$ for some particular choice of the numbers $a(S_i|E)$ and $a(D|S_i)$ defined above. Note that this can be greater or less than the "classical" $p(D|E)$, depending on the relative phases of $a(S_1|E)a(D|S_1)$ and $a(S_2|E)a(D|S_2)$. Therefore the different phases lead to different physical predictions, and part of the power of quantum theory is that it does actually tell you these relative phases.

This argument shows that there must be some physical interpretation of the phases, but it doesn't tell you what that physical interpretation actually is. I'm afraid I don't know the answer to that question.