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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?

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If you start with the wavefunction $\psi\gamma$, and evolve it according to a Schrodinger equation, the associated probability distribution will develop differently to that obtained by just starting with $\psi$, and evolving it according to the Schrodinger equation. So $\psi$ and $\psi\gamma$ are physically different states. (The only exception is if $\gamma(x)$ is the same for all $x$.) As for why physics works this way, no-one knows. –  Mitchell Porter Oct 4 '12 at 6:30
    
If these two wavefunctions evolve differently, they must be meaningful in different ways--despite the fact that they both generate the same distribution $|\psi|^2$. I guess I'm trying to figure out the significance of this extra information and to see if there is a purely mathematical reason why it should be there rather than resorting to experiment or a particular physical example or setup. –  Justin Solomon Oct 4 '12 at 13:19
    
With your background, you might be familiar with quaternions. In essence, these are just the wave functions for a single spin or 2-level system. One needs all complex degrees of freedom to describe the state. See en.wikipedia.org/wiki/Bloch_sphere and en.wikipedia.org/wiki/Hopf_fibration –  Arnold Neumaier Oct 4 '12 at 15:11

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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.

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So the function $\psi$ somehow encodes position and velocity/momentum? I understand that they are different states, but am unclear on how to "read" the different states. What do we know about the particle at $x$ relative to the particle at $y$ if $\psi(x)=1$ and $\psi(y)=\frac{\sqrt{2}}{2}(1+i)$? Is there any mathematical explanation for what's going on? –  Justin Solomon Oct 4 '12 at 14:15
    
@JustinSolomon: Nothing. The wave function is $L^2$, which means it is only definied up to arbitrary changes on a set of measure zero. You need to have $\psi(x)$ is a whole neighborhood of a point in order to deduce minimal information. –  Arnold Neumaier Oct 4 '12 at 14:30
    
I added the mathematical explanation to my answer. –  Arnold Neumaier Oct 4 '12 at 14:34
    
Thanks for posting the link to your book -- it looks closer to a language I might be able to speak, so I have downloaded it and will be taking a look on a flight later today! –  Justin Solomon Oct 4 '12 at 15:16

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.

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This is a very clear explanation of some of the phenomena I was having trouble understanding -- thanks! As you mention, the second and third equations for $p(D|E)$ yield different values, the third being the result of not trying to detect the slit. Is there a probabilistic explanation that's hiding here? E.g. that the second equation actually is somehow conditioned on having made an observation, which adds or removes some degree of probabilistic independence? –  Justin Solomon Oct 4 '12 at 13:26
    
Note that relative phases have meaning, but the absolute phases do not. This is not the only place in physics where one gets to choose an arbitrary starting point, however, and people don't generally bother themselves with the "Why?" of it. –  dmckee Oct 4 '12 at 13:42
    
But hopefully there is at least an explanation for relative phase? –  Justin Solomon Oct 4 '12 at 13:48
    
@JustinSolomon yes, I guess really the first and second expressions should be for $p(D|E,M)$ and the third for $p(D|E,\neg M)$, where $M$ is a Boolean that's true if you made the measurement of which slit the photon passed through, and false if not. Then the third expression is actually $$p(D|E,\neg M)= |a(S_1|E,M)a(D|S_1,M) +a(S_2|E,M)a(D|S_2, M)|^2,$$ with the r.h.s. conditioned on $M$ rather than $\neg M$ to indicate that the $a$'s themselves don't depend on whether the measurement was made. –  Nathaniel Oct 4 '12 at 19:23
    
@JustinSolomon in terms of an explanation, the problem is that the historical development of QM mostly took the form of people making wild guesses about how to perform the calculations, which somehow rapidly converged to a very successful formalism that nobody knew how to correctly interpret. The attitude of most physicists today is either "shut up and calculate" (i.e. it's not a physicist's job to worry about what the equations actually mean) or words to the effect that no interpretation is needed and the formalism of QM already contains everything you need to know about the physics. –  Nathaniel Oct 5 '12 at 12:16

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