Wednesday, February 22, 2012

The Semi-Classical Calculation

Yesterday, I set up a sample of hydrogen atoms in an excited state and calculated the radiation output according to the standard Copenhagen picture. Or, at least, I walked you through Richard Fitzpatrick's lecture on the University of Texas website. I took the sample to be 1,000,000 atoms, with 1% of them in the 2p state. (I ignored the possibility that some of the excited atoms should have been in the 2s state, but I don't think that matters. Except it would have screwed up my temperature calculation.) Using these numbers, I calculated a power output of 10 microwatts.

Then I said I was going to redo the same calculation, this time treating the atoms as tiny classical antennas. Since everybody knows that Maxwell's Equations don't apply to atomic systems, we should get nonsense. Let's see what actually happens.

I'm going to admit that I sweated bullets doing this calculation. I probably did it at least twenty times and got a different answer each time. The antenna calculation just wouldn't come out right. The best I could do was still out by a factor of ten. In desperation, I looked up the formula for the radiation from an accelerating charge. It's called the Larmor Formula, and I found it on the Wolfram Alpha website:


This gives us a formula to work with: but what numbers do we put in the formula? Well, there's not too much to worry about; we've got a couple of constants, which are provided in the table; we've got the charge on the electron, which everyone knows is 1.6x10^-19. The only other thing we need is the acceleration.

That's the hard part. What is the acceleration of the electron, according to the Schroedinger picture? Fortunately, Professor Fitzpatrick has already done all the hard work for us. I said there were two alternative pictures of the physics. In the Copenhagen picture, one percent of the hydrogen atoms are in the excited state. In the Schroedinger picture, each of the atoms is in a mixed state to the extent of one percent excited. To calculate the dipole moment of such a mixed state, you do the bra-ket thing with x in the middle. Actually, the order of operations doesn't really matter in this calculation (as it sometimes does in quantum mechanics). You can just think of this as the square of the amplitude (that's the product of the bra and ket states) integrated against x to get the dipole moment. The calculation looks like this:

What exactly are we seeing here? I said that the atoms were one percent excited, but here I'm showing the P state with a ten percent amplitude. Is that a mistake? No, because don't forget you square the amplitude to get the density. And why don't I have any s^2 or p^2 terms in my result? Because there is no dipole moment for a pure state: because the s^2 and p^2 states are symmetric (even) about the origin, the integral against x (an odd function) gives you zero. No, the calcuation reduces to the dipole coupling of s versus p.

But this is the calculation which the Professor has already done for us. Recall from yesterdeay:

He's actually given us the dipole moment for all three p states in terms of the Bohr Radius of 5.3*10^-11 m. (It's the third value which pertains to our case, the one with the extra factor of root 2.) Working out all those powers, it comes to about 75%, or 0.4 Angstroms. We also mustn't forget the factor of 20% which came out of the 1% excitation: so the dipole moment in our case actually comes to 8 x 10^-12 meters. (Multiplied by the charge of the electron, of course.) It's funny how much dipole moment you actually get by mixing in just 1% of the p state, but there it is.

Most importantly, let's notice that this dipole moment is an oscillating dipole moment. Because the s and p states precess through time at different exponential frequencies. So based on the difference in those two frequencies, what is initially a positive dipole moment becomes, after one-half cycle, a negative moment. It oscillates.

We've now got almost all the numbers we need to do the Larmor calculation. We still need the acceleration of the charge. We've got the oscillation amplitude - that's the 8 picometers - and we've got the frequency from yesterday - that was 2.5 x 10^15 Hz. Now I'm going to do something pretty slick. Instead of working out the acceleration due to sinusoidal motion, I'm going to pretend I'm working with uniform cirular motion instead. That's not what the atom is doing: in fact, it will throw me off by a factor of two, because the energy output of the circular motion is just the superposition of two orthogonal harmonic oscillators. No problem...I'll just divide by two at the end. The nice thing is I get to use the omega-squared-r formula for circular motion that I still remember from high school physics. Remembering the factor of 2-pi to change from hertz to radian frequency, I get an acceleration of 2*10^21 m/sec^2.

Putting all the numbers together, I get a power output of...20 picowatts. (You can check this if you like by plugging the numbers into the applet on the Wolfram site.) Divide by two to convert circular motion into simple harmonic motion, multiply by one million for the number of atoms in the sample, and I get, incredibly, exactly the same power output that I calculated yesterday. The Copenhagen picture with its quantum leaps gives exactly the same result as the Schroedinger picture with its tiny oscillating dipoles.

Not convinced? There's one more calculation that really needs to be done to bring this argument full circle. I said originally that I was going to use antenna theory to do the calculation, and when I tried, the numbers wouldn't come out. I kept getting two microwatts instead of ten. How do you get a factor of five for an error? It didn't make sense, and it was driving me crazy. But now I've worked it out so the antenna calculation comes out right. That's a story for my next blogpost.



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