I’ve had a chance now to read through the new papers mentioned in the Wolfgang Ketterle post last week, and there’s some interesting stuff there. The second item on the list from the AIP news article, “First observation of Mott insulator shells,” is particularly interesting, as I did some early work in that area when I was a post-doc. We’re coming down to the end of the term here, and I have a ton of things to do, but I feel weirdly inspired to try to explain what that item is about. I really ought to do something about my blog addiction.

Anyway, explaining the new Ketterle experiment will require explaining a few things about Bose-Einstein Condensation. Which, in turn, will require a bit of explanation about bosons and fermions, so that’s where we’ll start.

When you start talking about quantum systems containing multiple particles, you quickly run into the idea of indistinguishable particles. The basic statement of the idea is very simple: in dealing with quantum particles, you can’t tell two identical particles apart. It sounds more or less like common sense, but the consequences are surprisingly profound, and not at all intuitive. (Continued below the fold.)

Say, for example, that you have a situation like the picture at left, in which two identical particles– electrons, for example– come together and collide with each other. One electron comes in from the left, the other from the right, and they repel each other so that one leaves the collision region headed up and to the right, and the other leaves headed down and to the left.

Perfectly straightforward, right? The thing is, when you’re dealing with quantum particles, you can’t tell the difference between the picture above and the picture to the right. The end result is the same– one electron leaves headed up and to the right, the other down and to the left, but the two paths are very different.

In quantum mechanics, you can’t tell the difference between those two cases. All electrons are identical, and there’s no way to stick a label on one or the other, so you can’t say for sure whether the electron coming in from the left headed out in an upward or downward direction. The quantum situation is sort of like this:

Two particles come in, and two particles go out, but a sort of quantum fog hides the region in which they interact, and there’s no way to say which is which when they leave. Formally, when you calculate the probability for finding the two particles leaving in a given direction, you need to include both of these collision processes, and that leads to some remarkable consequences, like the fact that it’s impossible for certain types of particles to bounce off each other at certain angles.

But that’s getting a little ahead of the story.

The key idea here is indistinguishability: that is, that there’s no way to tell the difference between two quantum particles of the same type. It doesn’t matter what they are– electrons, protons, hydrogen atoms, whatever– there’s no way to label them, so you can’t tell the difference between them. (If the two particles are different types, all bets are off– you can certainly tell the difference between an electron and a rubidium atom.)

A slightly more formal statement of this requirement is that while you can label quantum particles to keep them straight while you write down the wavefunction, nothing that you measure can possibly depend on the labels you put on those particles. Another way of putting this is to say that whatever the wavefunction may be, the states must have symmetry: if you switch the labels on any two of the particles, you need to get the same final answer.

This would seem to be ridiculously restrictive, but there’s a little more wiggle room than you think. In quantum mechanics, the things that we get to measure are probabilities, and the probability is the **square** of the wavefunction. Which means there are two ways to meet the symmetry requirement: switching two labels can give you exactly the same wavefunction, or it can give you the same wavefunction with a negative sign out front. (The negative sign goes away when you square the wavefunction to get the probability, so it doesn’t affect anything measurable.)

This might be clearer if I use a concrete example, and some really simple equations. Let’s think about a system that has to states– we’ll make like computer scientists and call them **0** and **1**. And we’ll imagine that we have two particles, A and B, each of which can be in **0** or **1**, and we can label them with a little subscript.

There would seem to be four possible states of our two-particle system, then:

- 1)
**0**_{A}**0**_{B} - 2)
**0**_{A}**1**_{B} - 3)
**1**_{A}**0**_{B} - 4)
**1**_{A}**1**_{B}

But remember, we need to meet the symmetry requirement: when we swap the labels “A” and “B”, we can’t change the final answer. States 1) and 4) are already there, because switching the labels on **0 _{A}**

**0**gives you

_{B}**0**

_{B}**0**, which is the same as what we started with. (Whether we write the state of particle A to the left of particle B or vice versa doesn’t matter.)

_{A}States 2) and 3), on the other hand, don’t appear to meet our requirement. If we switch the labels on 2), we go from **0 _{A}**

**1**to

_{B}**0**

_{B}**1**. Switching the labels on state 2) gets us state 3), which is not at all the same thing.

_{A}So, that means we’re down to two possible states? No, because we can add 2) and 3) together, to get a new state, call it 5):

- 5)
**0**_{A}**1**+_{B}**1**_{A}**0**_{B}

It’s kind of a strange state, from a classical perspective, because it involves particle A and B each being in a superposition of **0** and **1**, but everybody has heard of Schrödinger’s Cat, and are resigned to superposition states. Switching the labels on **0 _{A}**

**1**+

_{B}**1**

_{A}**0**gets you

_{B}**0**

_{B}**1**+

_{A}**1**

_{B}**0**, which is the same thing you started with.

_{A}So, we’ve got three states that meet the symmetry requirement, right? No, we’re not done, because we can also **subtract** 3) from 2), to get a new state:

- 6)
**0**_{A}**1**–_{B}**1**_{A}**0**_{B}

State 6) behaves slightly differently when you switch the labels. It becomes:

0_{B}1–_{A}1_{B}0_{A}

But we can re-arrange the terms to be:

-1*(

0_{A}1–_{B}1_{A}0)_{B}

Which is just state 6) with a negative sign out front. And, remember, we’re going to need to square this in order to obtain a probability, so the negative sign will go away, and give exactly the same measurement results as 6).

So there are still four states of our two-particle system, but two of those states are these funny quantum superposition states:

- 1)
**0**_{A}**0**_{B} - 4)
**1**_{A}**1**_{B} - 5)
**0**_{A}**1**+_{B}**1**_{A}**0**_{B} - 6)
**0**_{A}**1**–_{B}**1**_{A}**0**_{B}

Notice that you can divide these states into two groups, based on what they do when you switch the labels. States 1), 4) and 5) remain the same when you swap labels, and are called **symmetric** states. State 6) picks up a negative sign when you do the swap, and is called an **anti-symmetric** state. You’ll also note that there are three symmetric states and only one anti-symmetric state; for this reason, you’ll often hear the symmetric states of a two-particle system referred to as a “triplet” state, and the anti-symmetric state referred to as a “singlet” state.

Whenever you find two different ways of categorizing behavior in physics, you seem to find two different types of objects corresponding to those two different behaviors, and the same is true here. We can divide all the particles in the universe up into two types. **Fermions** are particles whose wavefunctions are anti-symmetric, and **Bosons** are particles whose wavefunctions are symmetric. Electrons, protons, and neutrons are examples of fermions, while photons are bosons.

Snobbish particle physics types will sometimes say that material particles are all fermions, and the only bosons in nature are force carriers, but this is true only in a particle physics context (if even there). At the sort of energies that normal people deal with, you can get composite bosons– it turns out that if you stick two fermions together tightly enough, the resulting particle behaves like a boson, provided you don’t do anything to it that will break the two fermions apart.

Any composite particle with an even number of fermions, then, can be thought of as a boson. If you consider an atom with many protons and neutrons, it can be either a boson or a fermion, depending on exactly how many particles there are making it up. Krypton-84 is a boson, because it has 36 protons, 36 electrons, and 48 neutrons. Krypton-85 is a fermion, because it has 36 protons, 36 electrons, and 49 neutrons. That one extra neutron changes the character of the atom, and has dramatic consequences for how it behaves in collisions, among other things.

(You don’t need the individual components to come in even numbers, either. Rubidium-87 is a boson, because it has 37 protons, 37 electrons, and 50 neutrons, for a total of 124 particles. Rubidium doesn’t have any stable isotopes with even mass number, but if it did, they would be fermions. Rubidium turns out to be a very nice atom, and very important for Bose-Einstein Condensation experiments.)

That’s the basic idea of identical particles in quantum mechanics, and the basic symmetry operations for systems of particles. You can carry the same basic procedure out for more than two particles, and you get the same results– a bunch of symmetric wavefunctions, and one anti-symmetric wavefunction (it gets to be a real pain in the ass to write down the anti-symmetric wavefunctions after a while– I’ll spare you the horror of Slater determinants). No matter how many particles you have, if they’re fermions, they can only be in the anti-symmetric state, and if they’re bosons, they can only be in one of the symmetric states.

What’s the effect of all this, other than giving first-year grad students blinding headaches from writing down all these wavefunctions. The consequences are actually really dramatic, but I’ll save them for another post.