I am going to make an important distinction between the concepts of
As a macroscopic illustration, consider two cars. I choose to call them intrinsically identical whenever they have the same make and model, same color, et cetera. Nevertheless you might be able to distinguish them on other grounds. For example you might be able to distinguish them on the basis of position: a car ’over here’ is distinguishable from a car ’over there’. As a fancier example, if the cars are both weaving their way around a racetrack, so that you cannot easily distinguish them on the basis of position, it would be easy to distinguish them if one were driving normally while the other were turned around and driving in reverse.
Let us now apply this terminology to quantum mechanics. We will say that all electrons are intrinsically identical ... but this does not make all electrons indistinguishable. For example:
Some people take a different approach, saying that the electrons per se are always indistinguishable, but they may be put into states that are distinguishable. I disapprove of this approach. For starters, we don’t need to distingush the states, because they are always distinguishable; we’re asking whether we can distinguish the electrons. The quantum states themselves are always 100% distinct and distinguishable. The lowest state of a particle-in-a-box system is 100% distinct from the first excited state. So we are back where we started: the electrons may be distinguishable because of the state that they’re in.
Here is my acid test for what "distinguishable" and "indistinguishable" mean:
Suppose we have an emitter labeled "1" and another emitter labeled "2", each of which emits a particle. (Note that the labels apply to the emitter machinery. We can always label the machinery, even if QM doesn’t allow us to label the emitted particles.) The particles interact and are detected, perhaps as shown in figure 1 or figure 2.
detector B ___ [ ] | | | | | 1 >-----------\ \-----------< 2 | | | | [___] detector AFigure 1: One way a collision can occur.There are three possibilities:
- If the interaction proceeds according to bosonic indistinguishable-particle rules ("add before you square"), then the particles are considered bosons and are considered indistinguishable.
- If the interaction proceeds according to fermionic indistinguishable-particle rules ("subtract before you square"), the particles are considered fermions and are considered indistinguishable.
- If the interaction proceeds according to distinguishable-particle rules ("square before you add"), then the particles are considered distinguishable. This doesn’t tell us whether the particles are fermions, bosons, or one of each.
The foregoing "acid test" is not the only possible test, and is not always practical, but it serves to clarify the notions.
| Anti-symmetry is a mathematical property. It is absolute. | Exchange is a physical process. It is not absolute; i.e. it can happen to a greater or lesser degree ... depending on things like time, distance, energy, and temperature. |
It turns out that in the case of ordinary chemistry-lab conditions, the time, temperature, atomic energies, and atomic distances are such that when dealing with the electrons inside atoms, you need to worry about exchange quite a lot. This is an important case, but not the only case that needs to be considered. By way of contrast, on a larger scale, the atoms within molecule do not ordinarily exchange with each other to any appreciable degree (although this can sometimes happen, for instance in the ammonia maser, as discussed in reference 1).
Here’s another misconception, i.e. another example of how not to think about identical particles. Consider again the collision experiment shown in figure 1.
We have an electron (supposedly labeled 1) that comes in from the left and another (supposedly labeled 2) that comes in from the right. The collide, and electron 1 winds up in detector A, while electron 2 winds up in detector B. Now the labels are completely meaningless. They are dummy indices. Reassigning dummy indices is a trivial mathematical operation that never has any effect on the math or on the physics. You could relabel things so that electron 2 comes in from the left and winds up in detector A, while electron 1 comes in from the right and winds up in detector B. This would not be an example of exchange. Both versions involve electrons following the same paths. You should not imagine that the relabeled version interferes in any way with the original version. (Contrast this with figure 2, in which there are two physically-different contributions that do interfere.)
To say the same things in more technical terms, the physics of exchange only shows up in the interference terms. If you have one electron ’over here’ and other electrons ’over there’ you are free to choose the phase of each electron without regard to the phase of the others, without regard to whether this makes the resulting wavefunction anti-symmetric, symmetric, or neither. The relative phase affects only the interference terms, and is irrelevant if the two particles are not interfering. For details on this, see equation 6.a
(Tangential remark: It is super-important that the previous paragraph be true; otherwise it would be impossible to do quantum mechanics at all. You would need to anti-symmetrize your wavefunctions with respect to every electron in the universe, which would clearly be an impossible task.)
Here’s yet another warning: You should never think of exchange as changing "the" sign of "the" overall wavefunction. There’s a lemma that says changing the overall sign of the wavefunction |W⟩ cannot possibly have any observable consequences, since all observables are of the form ⟨W|O|W⟩. More generally, applying any phase factor of the form exp(i theta), for any real theta, to the overall wavefunction is completely inconsequential. You can easily prove this lemma for yourself; it’s a one-line proof, starting from ⟨W|O|W⟩.
We shall see that what matters is the interference between various contributions to the wavefunction. As discussed in section 5, we never care about the overall sign of the wavefunction ... rather we care when the exchanged part of the wavefunction interferes with the unexchanged part.
To learn how to handle these issues properly, an excellent place to start is reference 1, especially volume III section 3-4 and continuing into chapter 4. In particular, there is much discussion of collisions where there are two contributions that differ by an exchange. 1
An example of this is shown in figure 2:
detector detector
B B
___ ___
[ ] [ ]
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1 >-----------\ \-----------< 2 1 >-----------/ /-----------< 2
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[___] [___]
detector detector
A A
To see why exchange must be considered a physical (not mathematical) process, look closely at the figure. You can see that the two contributions are not exactly equal, because the particles depart the collision region from slightly different places (differing by roughly the effective diameter or interaction-length). This may be significant if the detectors can detect the position and/or angle at which the particle arrives. Whether this is actually significant depends on the extent to which the two possibilities interfere, which in turn depends on things like the “size” of the particles and the resolving power of the detectors, et cetera.
For atomic scattering, there are actually two different sizes to consider: The classical “particle size” of the atom, and also the size of the wavepacket that describes the atom. At high temperature, the atom behaves more-or-less like a classical particle. At low temperature, the wavepacket properties dominate. Some peculiar things happen in the intermediate regime.
Suppose you have a scattering experiment that normally exhibits identical-particle behavior ("add before you square"). If you re-run the experiment with the particles at a sufficiently higher temperature, you will see distinguishable-particle behavior ("square before you add"). And at intermediate temperatures you will see intermediate behavior.
Consider the contrast:
| It is a mistake to oversimplify the problem by focusing on what would happen, mathematically, if we had two contributions to the wavefunction that differed only by an exchange of identical particles – other things being equal. | The physical reality is that the other things are never equal. There is always something else going on. So we are left with the question of how much the two possibilities interfere, even though the contributions differ slightly as to position, momentum, et cetera. |
However, if we are careful, we can break things down into a part that involves only exchange, and a part that covers everything else. For example, consider a gas of fermions that are only partially spin-polarized. Some collisions will involve particles with unlike spins, and these will surely follow the distinguishable-particle rules. Other collisions will involve particles with the same spin, and these may or may not follow the identical-particle rules ... depending on whether the thermal de Broglie length is long or short compared to the hard-sphere diameter of the particles.
For a discrete variable such as spin, it is obvious that spin-up is distinguishable from spin-down.
In contrast, we have to work harder when considering continuous-valued variables such as position and momentum. It is not obvious whether two particles should be considered indistinguishable when they have only a slightly different position or a slightly different momentum. However, in many cases of interest, position can be decomposed in terms of a discrete basis. In particular, for electrons near an atom, if you say that an electron is in the |2px⟩ orbital, there is nothing more to be said about its position or momentum. If you have another electron that is aligned with some funny direction in the XY plane (17.37 degrees or some such), you can decompose it into its projection on the |2px⟩ orbital and the |2py⟩ orbital and then just turn the crank. The component will be position-wise indistinguishable from any other |2px⟩ electron, while the |2py⟩ component will be distinguishable.
Similarly: The states for a particle-in-a-box are similarly discrete, so again it is easy to decide what’s identical and what’s not.
If you really, really have a continuous-valued variable and no natural discretization, then we enter the realm of quantum field theory. In this case once again it is best to focus attention on the physical processes as expressed by the creation and annihilation operators, rather than on the symmetry of the initial state or final state. In its simplest form, the rule is simple indeed: bosonic creation operators obey a special commutation relation, while fermionic creation operators obey a corresponding anticommutation relation. This time we are faced with a much more serious version of the question about what happens when states are only slightly different. There is no way to answer this question without introducing a notion of temperature. The details are beyond the scope of this note, but you can do pretty well using the following guideline: the QM object that most closely corresponds to a classical particle is a fuzzy cloud (i.e. a wavepacket) having a size on the order of a thermal deBroglie length Λ. Roughly speaking, if two such clouds are less than one Λ apart, you cannot distinguish them on the basis of position. More precisely, you should work out the geometrical angles and the wavefunction phase shifts.
Note that Λ doesn’t play the role of a wavelength. As explained in reference 2, it has more to do with the envelope size of the wavepacket.
If you really want to get a good feel for this topic, including
then you should read reference 3. It’s a masterpiece, elegant and full of important ideas. It’s not easy to read, but it’s worth the trouble. Beware of huge numbers of typos, especially in the formulas.
QM teaches us to think about the states. A lot of people get themselves tangled trying identify the particles instead of the states.
Specifically: It makes no sense whatsoever to talk about two states that differ by exchange of identical particles.
To illustrate the right way to think about things, consider the following states, for particles in two boxes:
_________ | | | | X | | | X | | (a) |____|____|
_________ | | | | X | | | | X | (b) |____|____|
_________ | | | | | X | | | X | (c) |____|____|
Returning to the case of indistinguishable photons (which are bosons), it is begging for trouble to think of state (b) as having photon #1 on the left and photon #2 on the right (or vise versa). That would be OK for distinguishable photons (distinguishable by their color, polarization, or whatever), and it would be OK for classical particles, but if the photons are indistinguishable then there is no such thing as photon #1 and photon #2; state (b) is just a state with two photons in it. Asking "which photon is which" makes no sense; it is like asking what is 5 miles north of the north pole.
Thinking about this the right way is an acquired skill. It requires an investment. Acquiring the skill is made harder by the fact that people learn about classical behavior before they learn about quantum behavior, and old habits die hard. For classical objects like socks, it is OK to talk about sock #1 and sock #2, but extrapolating that procedure into the quantum realm just doesn’t work.
If you make a slightly larger investment, you can express things using the "second quantized" formalism, i.e. in terms of creation operators and annihilation operators. This gives you a systematic way of constructing the states and counting the states. For instance, the formalism will automatically tell you that states such as (a) and (c) do not exist for identical fermions. Similarly the formalism will automatically assign equal weight to states (a), (b), and (c) for identical bosons. And perhaps most importantly, the formalism does not tempt you – or even permit you – to talk about electron #1 and electron #2.
Let’s apply the ideas we’ve been discussing. We can derive the Pauli exclusion principle using a backhanded "proof by contradiction" technique.
Suppose (hypothetically and temporarily!) that we have created a state containing two identical fermions – for example, two electrons with parallel spins sharing a |2px⟩ orbital. Now we try to observe this state. We hit it with a photon or some such, trying to knock loose one or both of the electrons so we can observe the current. The amplitude for this process will contain a contribution from electron #1, plus a contribution from electron #2. Since the two electrons are in exactly the same state, these two contributions differ by only the exchange of identical fermions. Therefore the two contributions will be exactly equal and opposite, and they will add to zero. We conclude that there is no way to observe this hypothetical state.
Also, since all physically-realizable operators are unitary, our argument is just as valid the other way: If you can’t disassemble such a state, you can’t assemble it either. Bottom line: this hypothetical state cannot be created, and it were created it wouldn’t be observable.
Suppose the single-particle wavefunction is |W⟩ = |x⟩. Then we have
| (1) |
The two-particle wavefunction is guaranteed to be interesting whenever the two particles try to do the same thing, because interference produces strongly non-classical results. We can see this when we evaluate some observable (call it “O”).
That means evaluating some expression like ⟨W|O|W⟩. We can see what’s going to happen by looking at equation 1. For bosons, we find that the two-particle probability is |⟨W|O|W⟩|2 is four times as large as the single-particle probability |⟨x|O|x⟩|2, which is a rather significant departure from the classical rule that "one and one makes two". For fermions, we find that the two-particle probability is zero, again a rather significant departure from the classical result.
Suppose the wavefunction has two pieces:
| (2) |
and suppose analysis of the situation tells us that |q⟩ should be very nearly the same as |p⟩ except for the exchange of identical particles. So we write
| (3) |
and we assume b is either +1 or −1. Plugging in, we get
| (4) |
and we saw in the previous section that the factor b is very significant.
(Remark: It is not entirely trivial to prove that the exchange operator is either +1 or −1 times the identity operator, so let’s just assume it for now. This assumption is certainly consistent with innumerable observations, and with the theoretical idea that two exchanges gets you back where you started. That idea is not, alas, sufficient, because many other operators – notably reflection operators and certain rotations – have the same property. Don’t get fooled by glib, bogus proofs.)
Once again, if we have an overall phase a, as in
| (5) |
then we don’t care about a – it is completely inconsequential. It drops out when we measure something, as in expressions such as ⟨W|O|W⟩. The only thing that we can observe is b, because that tells us the relative phase, i.e. the phase of |q⟩ relative to |p⟩.
To be explicit, multiply out the probability:
| (6) |
since b2=1 and |a|2=1.
We see that it is the cross terms, i.e. the terms that express the interference between |p⟩ and |q⟩ , that carry all the sensitivity to b. If |W⟩ had consisted of two pieces that didn’t overlap, i.e. that didn’t interfere, then the value of b would have been inconsequential.
In the macroscopic world, it is semi-unusual to see phase-sensitivity, but it can be arranged.
In any case, this reinforces the point that exchange is a physical process: The exchange of a neutron doesn’t happen by magic or by mathematics: