Built on Facts

Bose-Einstein Condensates, pt. 2

Today we’re going to have to do some groundwork to set up for a step we’re going to need a post or two from now. It has to do with density in a slightly more abstract context than usual.

Imagine you want to know how many people are living in a particular region. You can multiply the population density by the land area and find out. But things get complicated if the density isn’t uniform. You’d have to separately multiply the density of each small uniform sub-region by the area of each sub-region and add them all together. Mathematically we’d call this integrating the density over the area. To do it we need a function describing how the density changes over the landscape of our region.

Yesterday we talked about how if you dump a bunch of particles in a box they’ll be in various quantum states. There are bazillions of possible quantum states each corresponding to a particular energy, and since we can’t count them all by hand we need some way to say how the states are distributed over something we can count, like energy. Maybe there’s 2 bazillion states with energies between 1 and 2 eV, and 4 bazillion states between 2 and 3 eV. We need a way to figure out just how many there are so we can integrate the states over the energy later on.

So say you’ve got a particle in a 3-d box. It’s a macroscopic box, so quantum mechanically speaking a particle in that box can be described by six effectively continuous quantum numbers. One each for its position on the x, y, and z axes. One each for its momentum in the x, y, and z, directions. In that sense each particle inhabits a six-dimensional phase space, which is an alarmingly jargonized way of saying that those six numbers completely characterize your particle. We want to integrate over that phase space to see how much phase space (or number of states) we accumulate for each bit of energy we add. The total phase space for particles of total momentum between 0 and P will be:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

The factor of 1/h^3 is the quantum mechanical size of a unit of phase space. We can prove it without all that much difficulty, but it’s kind of outside the scope of this post. A dimensional argument that length times momentum has the same units as h will be suggestive enough for now.

On to evaluating the integral. The integral over the volume is just the volume V. The integration over momentum space is constrained by the fact that p must be less than or equal to P, which is the same as saying that the momentum for a particle must lie inside a sphere (in momentum space) of radius P. That turns our integral into this:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

Now that’s the total number of states less than P. We want to turn this into the number of states between p and p + dp. Which is going to be the actual density of states over each tiny region in phase space. This is just a differentiation:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

This is all in terms of momentum. We want it in terms of energy, so we can just substitute the classical formula relating the two:

i-8d7070ce28f0af28047c7a5fb8246d53-4.png

Which also implies with a little calculus that:

i-9ef27a331268eac0961c3fd4d1a55446-7.png

Substitute this into g(p) above and you get:

i-bb5c2c6b0452df43a61e3974bd9b473f-5.png

And we’re done! Integrate that and you can get the number of states between whatever two energies you’d like. I used “bazillions” to describe the number above, and for fun we can actually plug in some numbers and be more precise. Using 1 cubic meter for V and the mass of a helium 4 atom for m, I find that there’s about 1.4×10^33 states between 0 and 1 electron volts.

A little tricky at first, but not so bad. We’ll put this to use tomorrow.

Comments

  1. #1 Uncle Al
    April 23, 2009

    Lovely exposition! Within social context, how can we base a taxation scheme upon it? Momentum is “extrapolated potential income”. It is not enough to take what they have. National invigoration requires confiscating what they might have.

  2. #2 rob
    April 23, 2009

    very cogent explantation. a good derivation to practice for your oral exam.

    howerver, i think the proper term is “gazillions” and not “bazillions.” i need to check my copy of Kitel to find out. :)

  3. #3 Blake Stacey
    April 23, 2009

    Well, we have it on good authority (Calvin and Hobbes) that 300 billion gajillion is a 3 followed by 85 zeroes. . . .

  4. #4 Jurgis
    March 2, 2010

    It’s g(p) times dE not a(E) times dE in the end.

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