Built on Facts

I always tell my students that they should never just write down an equation blindly and start plugging things in. If you don’t understand what you’re doing, you’re much more likely to make a mistake.

Sometimes I don’t take my own advice. On a science forum I read occasionally, a person who was just taking their first quantum mechanics class asked just how exactly you would go about preparing a system such as a hydrogen atom in a particular quantum state. It was correctly suggested that at room temperature, pretty much all of the atoms would automatically be in the ground state by virtue of the first excited energy level of the hydrogen atom is quite a bit bigger than kT. From there you could do things like shine a laser tuned to the transition frequency to put the atoms in your desired state. This is all true as far as it goes.

I decided to run the numbers and see just exactly what the percentage of hydrogen atoms are in the ground state at room temperature of roughly 300 K. To describe what I did (and how I did it wrong), I need to give you just a little background in statistical mechanics.

Imagine you’re playing poker. You draw 5 cards, and if you’re very lucky you might get a royal flush. There are four ways this might happen, one for each suit. Therefore, the odds of you getting a royal flush are four out of… how many times? To find out, you need to calculate how many possible hands there are. Then the probability of that royal flush will be 4 divided by however many possible hands there are.

Statistical mechanics works along the same lines. You figure out the unweighted probability of the state you’re interested in – this is just the number of combinations that form the state (4 for a royal flush) times a factor called the Boltzmann factor which tells you how the probability of the state varies with the temperature of the system. Then you divide by the total number of states, with each one weighted by the Boltzmann factor appropriate to its energy. This latter sum is called the partition function.

All this is to say that the probability of finding a particle in a particular state n is given by the equation below. Don’t worry about trying to derive it from the rather vague paragraph above, if you’re interested in the details and somewhat masochistic you can find them in any good statistical mechanics textbook. The equation:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Where Z is the partition function:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

The g_n is the degeneracy, which happens to be n^2 for the hydrogen atom (neglecting small effects which are not crucial here). The energy levels E_n for the hydrogen atom are well-known. They are (defining the ground state as E_1 = 0):

i-d18063683dcc0d42b9be45451a84d1e3-3.png

Where R is the Rydberg constant, approximately 13.6 eV. We might as well calculate kT as well – at T = 300K, kT is about 0.02585 eV.

All right! Knowing all that, we can evaluate the sum and calculate Z. The first term involves n = 1, so E_0 = 0 and you end up with e^0 which is 1, multiplied by 1^2 which is also 1. Fine. Then move on to n = 2 and repeat the process. Keep going until you get tired and n will get very big, which means E_n will tend to R, so the factor in the exponent will tend to R/kT. And R/kT is about 526, and I assure you that e^-526 is a tiny, tiny, insanely tiny number.

But it’s not zero, and we’re trying to add an infinite number of them. The partition function is infinite, and the probability of finding a hydrogen atom in any finite n state is therefore 0.

Yikes. What in the world did we do wrong? Well, for starters I’ll admit that I didn’t even look closely enough at the equation to even notice the divergence until Mathematica promptly spat out an overflow error. I was at a loss – surely the equations were right, but equally surely this was such an obvious calculation that if there were anything weird a textbook would have used it as a problem that I’d surely have been assigned. But no, to my knowledge I’d never encountered this issue before. I thought about it for a while, and decided to take a look through the literature. Sure enough, this calculation really does have some subtle features. To be completely honest I’m not sure I’m 100% comfortable with this resolution, but S.J. Strickler in the Journal of Chemical Education (Volume 43, Number 7, July 1966) has published what seems to be a pretty decent resolution of the seeming paradox.

He notes that the quantum mechanical solution for the hydrogen atom has energy levels which scale as 1/n^2, but the spatial extent of the energy levels – the radius of the electron “orbits” – scales as n^2. For the ground state n = 1, the orbit is just about half an angstrom. But by the time you get to n = 10^5 or so, the radius is of macroscopic size, say, 50cm or so. This is not possible in the laboratory. The electron will begin interacting with the vacuum tube or whatever it happens to be in, and the original solution for a hydrogen atom in otherwise free space doesn’t apply. Instead, the potential will begin to look something more like the square well potential, for which the partition function does not diverge. Effectively, you can cut off the sum at n = 10^5 and you’ll have a perfectly acceptable partition function.

This seems a little artificial. Plenty of hydrogen atoms aren’t in laboratory flasks, they’re floating freely in interstellar space. Does this resolution work there? Sure it does. At n = 10^18, the radius of the electron orbit is about the size of the known universe. It seems reasonable to cut off the sum there, and even for such high n the partition function different only infinitesimally from what it would be if you cut it off after just the first few terms.

Perfectly satisfied? Neither am I. Still, it’s an interesting and subtle problem involving some pretty basic concepts, and one I’m happy to have stumbled over.

Comments

  1. #1 Eric Lund
    January 25, 2011

    I’ve encountered divergence problems in my own field of plasma physics. Often it is convenient to approximate some function by its asymptotic expansion for large argument. It works rather well as long as you cut off the expansion at or before the smallest term-it turns out that the asymptotic expansion is formally divergent. Keeping too many terms will cause the approximation to blow up, as your calculation did here. In practice, keeping only the leading one or two terms is usually good enough given the other approximations you need to make the problem tractable.

    And if you are dealing with real computers, you will find that many Taylor series expansions, even the ones that are formally convergent over the entire complex plane, will blow up for sufficiently large argument. This is particularly true of alternating-sign series like trigonometric functions or Bessel functions. Certain special function recursion relations also have a similar instability.

  2. #2 Grant Goodyear
    January 25, 2011

    I’ve always preferred a different interpretation of that result. It is, in fact, correct that ionized H is the thermodynamic preferred state (entropy FTW!) at any finite temperature. So why do we see so much seemingly stable H in the universe? The quick answer is that atoms (and isolated molecules, for that matter) are metastable states. A chemist would look at this problem as an exercise in kinetics. The rate of electron escape is a product of the number of times per second that the electron attempts to escape, multiplied by the probability that the electron will have the requisite energy to do so. The first term is going to be somewhere on the attosecond [10^(-18) s] scale, or so, while the probability of having the right energy is exp(-Ea/kT) = exp(-13.6eV/0.025eV) ~ 10^(-237). Waiting on thermal energy to ionize an H atom is going to take many, many ages of the universe (roughly 10^10 years, or (pi*10^7 s/yr)(10^10 yr)) to see such an event.

  3. #3 Matt Springer
    January 25, 2011

    Grant, I have to admit I like that interpretation. My first thought was that in fact we’re not looking at just the |n> states of the H atom, we’re also looking at the number states of the photon field. This suggests moving to the grand canonical ensemble formalism to account for varying numbers of photons due to absorption and spontaneous emission. I haven’t tried to work it out yet – I suspect it will be a pain – but it might be another way to resolve the issue.

  4. #4 maxwell
    January 25, 2011

    I think Grant is on to it, but there is a more physical explanation.

    If we take the n=10^5 eigenstate, as Matt points out, its energy is 100 peV from the ionization energy for a H atom. At room temperature, there are 26 meV available for ionization. Given these two numbers, there are very, very, very few H atoms that will reach the n=10^5 energy eigenstate for any appreciate amount of time. They will ionization very, very quickly, probably less than attoseconds.

    Now, to me, the questions becomes, where do we truncate the series to calculate the partition function? I’d say we’d find the state whose energy difference from the ionization energy is equal to the energy at rm. T. That is which state has n such that

    26 meV=R-R(1-1/n^2)

    I get n=23. So I’d say that given a H atom at rm. T, all states that have n>23 are going to ionize so fast they won’t contribute to our calculation because H ions don’t have electrons. That means they don’t have electronic energy levels anymore!

  5. #5 BlackGriffen
    January 25, 2011

    You’re ignoring the unbound states, anyway. If you don’t you get the simple solution that the atom is always ionized.

    Where’s the flaw? Well, my gut’s telling me that what’s causing the calculation to fail is that you’re not taking into account the photon bath the atom is by necessity coupled to. When you do that, the infinite phase space of the photons should counterbalance the infinite phase space of the hydrogen atom.

    BlackGriffen

  6. #6 maxwell
    January 26, 2011

    BlackGriffen,

    I think the essential physical confusion here stems from the fact that some very large energy bound states create a divergence in the partition function. In practice at room temperature those states don’t exist because they ionize, therefore they shouldn’t contribute.

    So it’s not a question of whether we ignore the unbound states. It’s rather a physical point that high energy bound states ARE unbound states at room temperature. High energy states being those with n>23.

    Moreover, I’m not seeing how adding the quantized photon states really does anything special.

    ‘When you do that, the infinite phase space of the photons should counterbalance the infinite phase space of the hydrogen atom.’

    Could you explain what that means physically?

    If you’re going to do that, why not just take into account the relativistic effects like spin-orbit coupling and throw away the states where those factors are negligible? I think you’d get the same basic answer I get above.

  7. #7 Paul Murray
    January 27, 2011

    If there’s a “cutoff” point at the radius of the known universe, then these atoms would have behaved differently back when the universe was smaller.

  8. #8 Carl Mungan
    January 28, 2011

    Schroeder’s text “An Introduction to Thermal Physics” discusses this exact issue in Problem 6.9 (page 227). His explanation is that the Boltzmann factor is actually not simply exp(-E/kt) but instead is exp[-(E+PV)/kT] where P and V are the pressure (of the bath that the system is in) and volume (of the system). He argues that dS = (dE + PdV)/T and that we neglected the PdV term in the usual derivation of the Boltzmann factor (because it usually is negligible – but NOT when the system volume change by a lot when it goes from the initial to the final state of interest, say here from the ground state to the n-th state where n is big). Putting in P=constant (say 1 atm for a H atom in the atmosphere, or even P= for interstellar space) and V=a0*n^2 (where a0 is the Bohr radius), the partition function is now finite as you can check in Mathematica. -Carl

  9. #9 Anonymous Coward
    January 29, 2011

    Grant Goodyear and BlackGriffen have the right idea, but you don’t need kinetics (or a finite bath) to fix the problem. I think Carl Mungan has a fix that works even in equilibrium. I’ll try to say a similar thing in words.

    What the original post has (re-)discovered is a universal thing:

    Held at a finite temperature and surrounded by perfect vacuum, everything evaporates. A plate of metal surrounded by vacuum will always evaporate, as long as it’s not exactly at T=0. The evaporation process will be slow (as Goodyear points out, depending on the temperature, it’s not unusual to find that it’s longer-than-the-age-of-the-universe slow), but in the end everything will evaporate if it’s surrounded by vacuum. Similarly, a single atom held at a finite temperature in perfect vacuum will always eventually evaporate its electron (we usually call this ionization).

    The fix is what Carl’s comment suggests: you need to fix the volume that your single atom is in. Equivalently, you can fix the density of hydrogen atoms in your sample. This is not an artificial condition for simulating what’s going on in our universe, that’s how matter is typically found: at finite density.

    The original post’s “cutoff from the radius of the universe” would give the correct answer for the equilibrium state of a hydrogen atom IF it were the only hydrogen atom in our universe (actually, it wouldn’t: it would also need to include unbound states).

    But a lone hydrogen atom isn’t typical for our universe: there are lots of hydrogen atoms. And they’re usually found in clouds that are gravitationally bound and at some finite (nonzero) density.

  10. #10 capsiplex
    January 29, 2011

    Grant Goodyear and BlackGriffen have the right idea, but you don’t need kinetics (or a finite bath) to fix the problem. I think Carl Mungan has a fix that works even in equilibrium. I’ll try to say a similar thing in words.

  11. #11 Chat
    February 8, 2011

    The fix is what Carl’s comment suggests: you need to fix the volume that your single atom is in. Equivalently, you can fix the density of hydrogen atoms in your sample. This is not an artificial condition for simulating what’s going on in our universe, that’s how matter is typically found: at finite density.

  12. #12 sohbet arkadaslik
    February 13, 2011

    The fix is what Carl’s comment suggests: you need to fix the volume that your single atom is in. Equivalently, you can fix the density of hydrogen atoms in your sample. This is not an artificial condition for simulating what’s going on in our universe, that’s how matter is typically found: at finite density.

  13. #13 film izle
    February 15, 2011

    I mean, really?? I’m a scientist, and just reading that even made *my* eyes glaze over. If one thing they’re trying to convey is the importance and relevance of the scientist’s research to GQ readers, what percentage of the readers are really going to walk away with a deeper understanding of what Dr. Jamieson does by reading that description? It would have been a small thing to ask each participant to submit a layman-friendly version of their research (their “elevator talk” description, for example) for GQ to include.

    Finally–one of the “scientists” is Dr. Oz. What is he doing in there? One, I would think he’s already well-known enough; why not save that spot for another scientist? Two, yes, I know he’s actually done research and published, and is on the faculty at Columbia. Fantastic. He’s also a serious woo peddler, who has even featured everyone’s favorite “alternative” doc, Joseph Mercola, on his talk show, and discussed how vaccines may be playing a role in autism and allergies (despite mounds of evidence to the contrary). This seems to completely contradict their goal of “research funding as a national priority,” since Oz is often (and Mercola is always) highly critical of “mainstream medicine.” I really don’t understand his inclusion, and think it’s to the detriment of the rest of the campaign.

  14. #14 tütüne son
    February 21, 2011

    Grant Goodyear and BlackGriffen have the right idea, but you don’t need kinetics (or a finite bath) to fix the problem. I think Carl Mungan has a fix that works even in equilibrium. I’ll try to say a similar thing in words.

  15. #15 moliva
    February 22, 2011

    The fix is what Carl’s comment suggests: you need to fix the volume that your single atom is in. Equivalently, you can fix the density of hydrogen atoms in your sample. This is not an artificial condition for simulating what’s going on in our universe, that’s how matter is typically found: at finite density.

  16. #16 altın çilek
    February 22, 2011

    I think Carl Mungan has a fix that works even in equilibrium. I’ll try to say a similar thing in words.

  17. #17 dmabius
    February 25, 2011

    SH*THEADS

    atheistthinktank.net/thinktank/index.php?topic=9870

  18. #18 Sohbet Arkadaslik
    March 1, 2011

    Bu yüzden rm bir H atomunun verilen söylemişti. T> 23 n tüm devletler H iyonları elektronlar yok, çünkü onlar bizim hızlı hesaplama katkıda bulunmamaktadırlar çok iyonize olacak. Ki onlar artık elektronik enerji düzeyleri yok demektir!

  19. #19 Anonymous
    March 6, 2011

    A) You are including states that are far beyond the realm of the theory. Everything you’ve ever learned about physics involves lots of approximations which allow us to do the mathematics that we have. I truly doubt that there is any good reason to believe that the n=83 state even exists.

    B) Your expression for the energy of a state is actually the inverse wavelength of the photon emitted in the fall to the ground state from state n, thus the positive quantity.

    C) The energy of state n, n=1,2,… is given by E_n=-E/n^2 where E=13.6ev.

    D) Calculating a quantity such as Z directly will almost always diverge as Eric Lund said above. One trick is to calculate relative probabilities. The energy of state 1 is -E/4 so the relative probability of finding the first excited state vs. the ground state is e^(E/4kT)/e^(E/kT)=e^(-3E/4kT)~10^(-172) at 300K. Similarly, the relative probability of the 2nd excited state is e^(-8E/9kT) and the relative probability of any excited state (vs ground) tends very quickly to e^(-E/kT)~10^(-229) at 300K.

    In a nutshell, I think you are trusting the physics to represent things it doesn’t, and trusting a computer to do something they really don’t do very well.

  20. I agree,an equation isn’t just an equation.It needs analization and getting the exact point is a consideration to avoid mistake.Sometimes,even the formula is there,we still get wrong due to lack of analysis.

  21. #21 altın çilek
    March 21, 2011

    he quick answer is that atoms (and isolated molecules, for that matter) are metastable states. A chemist would look at this problem as an exercise in kinetics. The rate of electron escape is a product of the number of times per second that the electron attempts to escape, multiplied by the probability that the electron will have the requisite energy to do so.

  22. #22 tütüne son
    March 21, 2011

    it’s not a question of whether we ignore the unbound states. It’s rather a physical point that high energy bound states ARE unbound states at room temperature. High energy states being those with n>23… Moreover, I’m not seeing how adding the quantized photon states really does anything special..

  23. #23 altın çilek
    March 23, 2011

    A chemist would look at this problem as an exercise in kinetics… The rate of electron escape is a product of the number of times per second that the electron attempts to escape, multiplied by the probability that the electron will have the requisite energy to do so..

  24. #24 supratall
    March 23, 2011

    The electron will begin interacting with the vacuum tu be or whatever it happens to be in; and the original solution for a hydrogen atom in otherwise free space doesn’t apply. Instead, the potential will begin to look something more like the square well potential, for which the partition function does not diverge…

  25. #25 altın çilek
    March 27, 2011

    Calculating a quantity such as Z directly will almost always diverge as Eric Lund said above. One trick is to calculate relative probabilities… The energy of state 1 is -E/4 so the relative probability of finding the first excited state vs. the ground state is e^(E/4kT)/e^(E/kT)=e^(-3E/4kT)~10^(-172) at 300K…

  26. #26 altın çilek
    April 6, 2011

    The fix is what Carl’s comment suggests: you need to fix the volume that your single atom is in. Equivalently, you can fix the density of hydrogen atoms in your sample. This is not an artificial condition for simulating what’s going on in our universe, that’s how matter is typically found: at finite density.

  27. #27 tütüne son
    April 8, 2011

    The fix is what Carl’s comment suggests: you need to fix the volume that your single atom is in. Equivalently, you can fix the density of hydrogen atoms in your sample. This is not an artificial condition for simulating what’s going on in our universe, that’s how matter is typically found: at finite density.

  28. #28 sohbet arkadaslik
    April 8, 2011

    Eric Lund yukarıda söylediğim gibi Z doğrudan neredeyse her zaman farklılık olacak gibi bir miktar hesaplanması. Bir hüner göreceli olasılıklar hesaplamaktır. zemin devlet vs ilk uyarılmış durum bulma göreceli olasılık / e ^ (E / kT) = e ^ (-3E/4kT) e ^ (E/4kT) böylece devletin 1 enerji-E / 4 ~ 300K 10 ^ (-172). Benzer şekilde, 2 heyecanlı devletin göreli olasılık e ^ (-8E/9kT) ve herhangi bir uyarılmış durum (vs toprak) göreceli olasılık e ^ (-E/kT) ~ 10 ^ (-229) çok hızlı eğilimindedir 300K de.

  29. #29 Spaceman Spiff
    August 20, 2011

    To follow up on #27, is the fact is that there is no such thing as an isolated hydrogen atom. Interactions with the rest of the universe result in a cut-off in n.

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