Measurements corresponding to Exchange Operator

[Jolb]

The exchange operator P is hermitian--therefore it corresponds to some observable of the system. I've read on Wikipedia that this means "... we can, in principle, perform a measurement to find out if a state is symmetric or antisymmetric," but I have some doubts whether that is absolutely accurate.

My question is: what kind of measurement can be done that corresponds to the exchange operator?

 

[Bill_K]

Assuming you mean the full particle exchange operator, the eigenvalues of the exchange operator are +1 for bosons and -1 for fermions. So if you know which type of particle you have, you know the eigenvalue.

 

[Jolb]

Bill_K,
That doesn't quite answer my question. I don't exactly know what you mean by "full particle exchange operator," but in Griffiths (blue griffiths, second edition, page 205) it's denoted P and it's action on a two-particle wavefunction F(r1, r2) is

P F(r1, r2) = F(r2, r1).

Intuitively, it makes sense that the eigenvalues are +1 or -1 since a two-particle wavefunction for identical bosons is symmetrical whereas a two-particle wavefunction for identical fermions is antisymmetrical. If you know what kind of identical particles you have (bosons or fermions), you do automatically know which eigenvalue you'll measure if you perform a measurement that corresponds to the operator P.

My question is, what experiment corresponds to this P operator? According to your answer, Bill, would it be simply measuring what kind of particle you have? But (as far as I know) figuring out what kind of particle you have is not just one particular measurement, and it doesn't correspond to any hermitian operator. (Normally, to find out which particle you're observing, you would do momentum and position measurements in various fields to determine mass and charge--these correspond to the momentum and position operators.)

However P is hermitian, and thus it should correspond to some observable, and there should be some measurement to determine this observable. So what's this measurement?

 

[cgk]

This is not an operator in the sense of an observable. It does not correspond to any measurement or physical quantity. Note that for a given kind of physical particle (boson or fermion) the action of the operator is trivial: it transforms any physical wave function (i.e., a symmetric or antisymmetric wave function) into an equivalent wave function (in the bosonic case, the identical wave function, in the fermionic case, in one with a possible phase factor of -1). So it is not possible to gain any information about physical particles by applying it. Note also the action "permute coordinate 1 and coordinate 2 of the wave function" cannot be expressed in terms of second quantization---that is the case because it distinguishes between identical particles, which a physical observable is not supposed to distinguish.

 

[Jolb]

cgk,
That's a very reasonable answer. The one problem I still have is that I was under the impression that absolutely every hermitian operator must correspond to some measurable observable... is that wrong?

Also, your answer would totally refute the Wikipedia quotation in my first post. I'd appreciate a scholarly reference to support that reasoning so that I could update wikipedia (to set the record straight).

 

[Physics Monkey]

There is a way to potentially measure a related phase.

First an aside. The multiparticle wavefunction is a cumbersome object so the modern way to think about particle statistics is via Berry's phase. This is simply the phase encountered when you physically adiabatically exchange the two identical particles i.e. by literally dragging them around. This phase is 1 for bosons and -1 for fermions.

Call U the operation which adiabatically exchanges the two particles. Now consider the controlled U operation cU. This operator has an extra input, a qubit, whose value determines whether the exchange is carried out or not. In other words, \mbox{cU} (| \mbox{2 particle state} \rangle | 0 \rangle ) \rightarrow | \mbox{2 particle state} \rangle | 0 \rangle and \mbox{cU}( | \mbox{2 particle state} \rangle | 1 \rangle \rightarrow | \mbox{exchanged 2 particle state} \rangle | 1 \rangle.

Now the procedure. Apply cU to the state | \mbox{2 particle state} \rangle( | 0 \rangle + | 1 \rangle ) to get ( | \mbox{2 particle state} \rangle | 0 \rangle +| \mbox{exchanged 2 particle state} \rangle | 1 \rangle ). If the particles are identical then the exchanged state is equal to the original up to a phase. Measure in the | 0 \rangle \pm | 1 \rangle basis to determine the phase.

Note also that while I haven't specified exactly how, it is certainly feasible to coherently exchange two particles. You may find it fun to think of some possible procedures.

Hope this helps.

 

[Jolb]

Physics Monkey,
That's very interesting! It seems like using the magic of mixed quantum states you can, "in principle", design such an experiment. So wikipedia is right, as usual. It's just that the "in principle" part relies on being able to physically drag two particles into exactly their partners' position without at all disturbing their quantum state--a mixed one. Have I got that right?

In that case P would be physically measurable, but it's just a particularly hard one to measure (hence why there's no literature about measurements of it).

 

[cgk]

There is a way to potentially measure a related phase.

[...]

Note also that while I haven't specified exactly how, it is certainly feasible to coherently exchange two particles. You may find it fun to think of some possible procedures.

Is that possible? My impression was that we were talking about the exchange operator that is not only talking about exchanging (any) two particles, but explicitly exchanging particle 1 and particle 2 (or some other specified coordinate slots in a N-body wave function). I'd be very interested in how one would possibly do that while leaving all the other coordinates intact in an antisymmetric wave function.

Or am getting this wrong?

 

[Amok]

cgk,
The one problem I still have is that I was under the impression that absolutely every hermitian operator must correspond to some measurable observable... is that wrong?

What gave you that impression? Every observable corresponds to a hermitian operator, but I've never heard that every hermitian operator corresponds to an observable in physics.

 

[Physics Monkey]

Is that possible? My impression was that we were talking about the exchange operator that is not only talking about exchanging (any) two particles, but explicitly exchanging particle 1 and particle 2 (or some other specified coordinate slots in a N-body wave function). I'd be very interested in how one would possibly do that while leaving all the other coordinates intact in an antisymmetric wave function.

Or am getting this wrong?

I think the wavefunction is a confusing way to think about the matter. This is in part because it already contains redundant information by explicitly labeling the particles.

The better way to think is in terms of an adiabatic phase. In the simplest case the state of n identical particles can be specified by giving the n positions of the particles. Note that we are not labeling the particles, we are simply saying there is a particle at position x1, a particle at position x2, etc. Now any motion of the particles that ends with the same configuration returns the system to the same initial state. However, it is meaningful to ask what the overall phase accumulated on this round trip through Hilbert space is. The little sketch I gave above is a simple interference experiment to detect this phase. I suspect you will agree that this phase I described is in principle measurable even if it is not yet clear that it has something to with particle statistics as more traditionally defined.

If you like you can make your own fermion in 2d using charges and fluxes. Then you can easily calculate the phase in this case via the AB effect. You can also make more exotic particles in 2d.

Actually, this kind of business of making fermions from bosons and gauge fields illustrates quite well the importance of the adiabatic phase concept. In general, to get the right statistics in complicated situations one must not simply formally exchange labels in a wavefunction but instead carefully parallel transport charges through electromagnetic fields. An example: a bosonic unit charge orbiting a bosonic unit magnetic monopole is actually a fermionic object (it has spin 1/2), but you can only see this using the adiabatic phase approach.

Hope this helps.