Quantum sophistry

A few days ago, we got to know the members of the St. Hypotheticus drinking and nerdery club. Today we listen in on their inaugural meeting. They have just finished the administrative part and are now about to enter on the drinking and nerdery.

Carl: (Knocking at a beer keg) All right, I already prepared the drinking …

[applause]

Carl: …now for the nerdery. Any ideas?

Bob: I have a question about quantum mechanics and philosophy.

Albert: (groan) You know, most I heard about this is bunk. It has far less philosophical consequences than you might think.

Bob: I know. The thing is, shortly I was accosted by some young folks at the college who where looking for an analytical philosopher…

Matthew: …Uhu…

Bob: An analytical Thomist does totally count, and besides if it was so important they could have asked, couldn’t they? Anyway, they asked if one could empirically prove two particles indistinguishable even in principle and I said no. Then they walked away laughing and talking about configurations and flows and distinguishable wave functions. Most of it went over my head, but I googled it and it turns out they were working from a script. I brought some copies, so I wondered what y’all have to say about that.

Matthew: Argh, it’s the…

Albert: … many-worlders …

Clara: … Bayesians …

Matthew: …Gnostics.

Bob: Um yes, there is a lot of strange stuff on that site, but I don’t even know how much of that the nice folks who followed the script actually believe. So can we look at the point? That script claims:

If you have a particle P1 and a particle P2, and it’s possible in the experiment for both P1 and P2 to end up in either of two possible locations L1 or L2, then the observed distribution of results will depend on whether “P1 at L1, P2 at L2” and “P1 at L2, P2 at L1” is the same configuration, or two distinct configurations. If they’re the same configuration, we add up the amplitudes flowing in, then take the squared modulus. If they’re different configurations, we keep the amplitudes separate, take the squared moduli separately, then add the resulting probabilities. As (1 + 1)² != (1² + 1²), it’s not hard to distinguish the experimental results after a few trials.

Clara: Well first of all someone should tell them Bayesian geniuses that neither \(\left(1+1\right)^2=4\) nor \(\left(1^2+1^2\right)=2\) are acceptable values for a probability. You know, those are supposed to be between 0 and 1.

Carl: Yeah, but the point seems to be that addition and modulus squaring don’t commute, and that much seems true.

Albert: Is there some spin \(\frac{1}{2}\) stuff in there?

Bob: It explicitly says half-integer spins don’t matter.

Albert: That sure is a relief, otherwise our author might have wounded his fingers typing up all the math. Anyway, other than in the detail Clara just noted, that quote is quite true. But the claim it is supposed to support just doesn’t follow from it, except with an implicit wrong assumption that it couldn’t be just one configuration even if the particles are distinguishable in principle.

Bob: I’m not quite sure I get that. Can we look at this in more detail?

Albert: OK, the most famous mathematical objects we have in quantum mechanics are wave functions.  In the simplest case they describe  single spinless particles and then they are just  functions that gives you a complex number for every point in  space.

Jenny:  Why complex?

Albert: So that they can interfere. But don’t worry about that now, it isn’t that important for what we’re talking about today.

Jenny: OK.

Kate: Fine, so what do we do with these complex numbers?

Albert: Take their modulus (don’t worry about that step for now) and square it. And that gives us a probability density.

Matthew: Um, density?

Albert: Yeah, there are infinitely many points, so we can’t give probability to every one, or else we would have trouble with them adding up to more than one, and we can’t be more than sure the particle is somewhere. But the math trick isn’t that important, if there where only finitely many points this would simply be a probability. Now there are some neat things we can do with wave functions. One thing we can do with them is multiply them with (complex) numbers. So for example if \(\psi\) is a valid wave function, then so is \(5\psi\).

Clara: Whoa! That can’t work, you’re bringing in those probabilities larger than one again!

Albert: OK, we do some normalization. We said for a wave function \(\psi\left(\vec x\right)\) the probability density for finding the particle at \(\vec x\) is \(\left|\psi\left(\vec x\right)\right|^2\).  Now if those probabilities would add up to more than one, whatever they add up to is called \(\left\langle\psi\middle|\psi\right\rangle\). There’s a meaning to that notation, but today we’ll just treat it as a name. Then instead of \(\left|\psi\left(\vec x\right)\right|^2\) our probability density is actually \(\frac{\left|\psi\left(\vec x\right)\right|^2}{\left\langle\psi\middle|\psi\right\rangle}\).

Clara: OK, now you just canceled out the factor you had allowed in. What is this good for again?

Albert: Quite true, alone it would be useless. But we can also add wave functions. So if \(\psi_1\) and \(\psi_2\) are valid wave functions, then so is \(\psi\) with \(\psi\left(\vec x\right):=\psi_1\left(\vec x\right)+\psi_2\left(\vec x\right)\).

Clara: Aha! So you’re telling me they form a vector space.

Albert: Yes, though I didn’t want to confuse everyone else with that terminology. So now we can write some wave functions in terms of others. What we do next is select some set of wave functions as a base^[1] and then write all the others by multiplying and adding those. To put the same in more pompous words, every wave function can be written as a linear combination of base functions.

Now, for many-particle systems it works in basically the same way.

Bob: Care to elaborate on that?

Albert: I think discussing it for two particles will do.  The principles generalize to an arbitrary number of particles, but then the math gets a bit more complicated.

Bob: Fine by me.

Albert: The way it works for two particles is is basically the same as for one. We just get a wave function depending on two locations and then the probability density for particle 1 being at \({\vec x}_1\) and particle 2 being at \({\vec x}_2\) is \(\left|\psi\left({\vec x}_1,{\vec x}_2\right)\right|^2\).

Bob: Well mathematically it sounds very similar, but only because we haven’t talked about any particular wave functions and operators in either case. If we don’t have any connection between the particular objects for different numbers of particles, then the one-particle case just isn’t a simplification of the two-particle case.  And then by extension it would never make sense to talk of a part of a whole because the whole wouldn’t just be more than but actually unrelated to the sum of its parts. Since we can’t conceive of the whole universe at once that would make all physics impossible.

Albert: Right, but luckily^[2] there is such a connection. It turns out, we can express the base functions of the two-particle system by those of two one-particle systems. Now how we do that depends on whether the particles are distinguishable and that is what those folks you met where getting at.

Bob: Now it gets interesting.

Albert: Yup. If the particles are distinguishable it’s really easy. In that case if we have a base function \(\psi_1\left({\vec x}_1\right)\) for particle one and a base function \(\psi_2\left({\vec x}_2\right)\) for particle two, then \(\psi\left({\vec x}_1,{\vec x}_2\right)=\psi_1\left({\vec x}_1\right)\psi_2\left({\vec x}_2\right)\) is a base function for the two-particle system and all two-particle base functions can be expressed in that way. Since we don’t always want to mention the arguments I’ll introduce some notation here: Instead of \(\psi\left({\vec x}_1,{\vec x}_2\right)=\psi_1\left({\vec x}_1\right)\psi_2\left({\vec x}_2\right)\) we’ll write \(\psi=\psi_1\otimes\psi_2\).

Bob: I psi^[3]. But of course this can’t work if they are indistinguishable, because if \(\psi_1\neq\psi_2\) we would have the two particles distinguished right there.

Albert: Right. In the two-particle case it turns out that if \(\psi_1\) and \(\psi_2\) are one-particle base functions, then  \(\psi=\frac{1}{\sqrt{2}}\left(\psi_1\otimes\psi_2\pm\psi_2\otimes\psi_1\right)\) is a two-particle base function. It’s only one base function, it’s either \(+\) or \(-\) depending on what kind of particle we’re talking about.

Bob: Slow now, I don’t quite understand this form. First, what is \(\frac{1}{\sqrt{2}}\) doing in there?

Albert: You remember how a few minutes ago we talked about dividing out \(\langle\psi|\psi\rangle\) to make sure all the probabilities added up to 1? Well, most of the time it is easier to multiply the whole function  by some constant that makes  \(\langle\psi|\psi\rangle=1\), so that we don’t have to bother with that step. In this case \(\frac{1}{\sqrt{2}}\) is that constant.

Bob: Fine. And the part in the parentheses?

Albert: That part takes care of the indistinguishability. See, if even in principle nobody knows which particle is where, then the probabilities really shouldn’t change if we swap them. So we want \(\left|\psi\left({\vec x}_1,{\vec x}_2\right)\right|=\left|\psi\left({\vec x}_2,{\vec x}_1\right)\right|\) and this form guarantees precisely that.

Bob: Great.

Albert: There is one more feature I would like to draw your attention to. Please note \(\psi_1\otimes\psi_2\) and \(\psi_2\otimes\psi_1\) are base functions for the system of two distinguishable particles. So the base functions for the system of indistinguishable particles are linear combinations of those for the system of distinguishable particles.  That means all wave functions for indistinguishable particles are also valid wave functions for distinguishable particles.

Bob: But not the other way around?

Albert: Nope, not the other way around. In fact if \(\psi_1\neq\psi_2\) then \(\psi_1\otimes\psi_2\) and \(\psi_2\otimes\psi_1\) themselves are wave functions for the system of distinguishable particles that don’t work for indistinguishable particles.

Bob: You know, the wave functions for indistinguishable particles being a subset of those for distinguishable ones makes sense.  After all we can always not distinguish distinguishable particles, but distinguishing indistinguishable ones is, by definition, impossible.

Albert: Exactly. But now look at this: Even if they would be distinguishable in principle, we can prepare two particles we can’t practically distinguish into an indistinguishable configuration. But if we could prepare them into any of the additional states available to distinguishable particles that would amount to distinguishing them. Since we have assumed we know of no method of distinguishing them, we also know of no method of preparing them to any of the additional states they might have available if they are distinguishable in principle. So whether our inability to distinguish them is an intrinsic feature or our ignorance, the set of configurations we can prepare is identical. Both cases correspond to only one alternative of the experiment they were proposing and therefore it can’t be used to tell the difference between those two cases.

Bob: So you can’t be sure the particles we now hold indistinguishable will never turn out to be distinguishable after all?

Albert: I can think that very unlikely chiefly because of Friar Ockham’s razor, but no, I can’t be absolutely sure of it.

Bob: It’s nice to be right.

Albert: Ahem. The argument you gave for your position was actually invalid. It’s only by luck that you turned out to be technically correct.

Bob: That is, of course, always the best kind of correct.