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Energy of the electron in a random hydrogen atom

  1. May 31, 2015 #1

    ftr

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    Does the energy of the electron in a random hydrogen atom is in superposition of all eigenvalues(some value upon measurement) or you will find it most likely in the ground state.

    Additional clarification:

    From my reading the textbooks said the electron energy is in superposition, yet the hydrogen will go into ground state. That seems to be different from the position operator which in the same atom the electron position is unknown until measured. Why this difference if any.
     
    Last edited: May 31, 2015
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  3. May 31, 2015 #2

    mfb

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    Yes (both).
    Usually the ground state is the most likely result of a measurement, but other states are possible as well, even if their probability is small in most setups.
     
  4. May 31, 2015 #3

    Jano L.

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    What measurement do you mean? It is not as if one could take an "Hamiltonian-meter" and measure one of the eigenvalues of the Hamiltonian.
     
  5. May 31, 2015 #4

    ftr

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    Thanks. Was the H atom in one of those eigenvalues before measurement or in all eigenvalues like the position operator. That is my main question.
     
  6. Jun 1, 2015 #5

    mfb

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    You can do spectroscopy, at least in principle - see which photon gets absorbed for specific transitions.
    It was in a superposition of the eigenvalues.
     
  7. Jun 1, 2015 #6

    ftr

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    My understanding is that the H atom will go quickly to the ground state and hence no energy superposition. Yet the electron position operator stays in the superposition indefinitely. My question is, are these two superposition different in principle, how do you characterize such a difference in superposition behavior.

    Moreover, if that H atom was in the superposition of ALL energy states how do we distinguish it from an excited H atom upon measurement.
     
    Last edited: Jun 1, 2015
  8. Jun 1, 2015 #7

    fzero

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    The proper way to answer these questions is to ask about the experimental setup. First, where did the H atom come from? If I have a box of H atoms, then they will interact with one another. Let's ignore the fact that they like to form H2 molecules for simplicity and mainly consider that they will have kinetic energy and bounce off of each other. With enough atoms we can use temperature as a measurement of the average kinetic energy of the gas. At low temperatures, most of the atoms are in the ground state, but once in a while, two atoms with above average energy will collide and some of the energy of the collision can cause an atom to go to an excited state. This is the reason that a random H-atom from the sample will be in a superposition of energy states.

    If I open the box and let a single atom into my measurement device, the atom will remain in the superposition state until I make a measurement on it. The atom can spontaneously decay eventually and if I measure the photon energy I might be able to compare to the difference between energy levels, but unless I had an infinite precision measurement, I could not conclude for certain that the atom decayed to the ground state and not some lower excited state. Generally you have to interact directly with the atom to put it in a definite energy state. The most that you can say for certain is that the probability of finding the atom in the ground state will increase with time and will approach 1 at the time goes to infinity.

    If I made my measurement and found the atom in the ground state, then the electron position would still be unknown because the electron position operator does not commute with the Hamiltonian. Energy and electron position are incompatible observables.

    One way to measure the energy of the atom would be to study the hyperfine structure. If I hit the atom with a laser tuned to the correct frequency to flip the spin of the electron without changing the radial or orbital energy, I could deduce the electron state without the risk of ionizing (and so destroying) the atom.
     
  9. Jun 1, 2015 #8

    ftr

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    Thanks fzero, that is clear . However, my question is a bit more subtle, I am asking why the position operator does not go to some eigenvalue like the energy, is it because it takes on continuous values or what.

    EDIT: and asking a reverse question, if the H atom was in superposition like the position operator all the time, what will happen to the chemistry of the universe.
     
    Last edited: Jun 1, 2015
  10. Jun 1, 2015 #9

    fzero

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    The fact I mentioned that energy and electron position are not compatible observables means that a electron position eigenstate is not an eigenstate of the Hamiltonian. If we were to measure the position of the electron, the system would be in the eigenstate corresponding to the value that we found. However, since this does not match an energy eigenstate, the state is not stationary, it will evolve in time. It has to, even at the classical level, because of the attraction between the electron and nucleus.
     
  11. Jun 1, 2015 #10

    fzero

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    I'm having trouble parsing this, but atoms and molecules are generally in superpositions of eigenstates of whatever observable we might choose to label them, precisely because of the thermal effects that I mentioned in my earlier posts. In fact, with molecules, the energy required to excite rotational and vibrational states is much smaller than that required to excite electrons in atoms, so the probability to be in a rotational or vibrational ground state is relatively smaller than for the electronic ground state. In fact we often exploit the fact that heat weakens chemical bonds in order to make chemical reactions more likely. Cooking is the most obvious example.
     
  12. Jun 1, 2015 #11

    ftr

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    Thanks for your effort. I must be missing something. When we excite an atom do we consider it as going into superposition of unknown spectrum of energy or a particular energy. I guess learning QM without experiment is hard.
     
  13. Jun 1, 2015 #12

    fzero

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    Again, it generally helps to answer a question if we try to fill out all of the details of what we're trying to do. On the one hand, we can have a thought experiment where we take a single atom that we have prepared in the ground state, perfectly shield it from all external effects, and then hit it with a single photon of precisely the correct energy to excite it to a particular excited state. If we can be sure that the photon has been absorbed, we would say that this atom is now in the definite excited state.

    Even this simple thought experiment is a bit oversimplified, since we don't take into account what we call translational degrees of freedom. That is, it is technically possible for some of the energy of the photon to go into the kinetic energy of the entire atom, rather than just exciting the electron. This is actually one of the reasons that spectral lines actually look like rounded peaks instead of sharp lines.

    However, more directly important is that a real experiment would not tend to match the perfect setup we described. Experimentalists are exceedingly clever, but individual atoms are hard to work with. So we usually have a gas of atoms and, instead of a single photon, a laser pulse that consists of many thousands of photons at least. Not all of the atoms will be in the ground state (that thermal superposition I have been harping on) and not all of the atoms will be excited by the laser pulse. Nevertheless, the clever experimentalists can combine the probabilities that QM gives us with the parameters like the density of the gas and the intensity of the laser in order to give a statistical description of the system. We can estimate, given the thermal distribution of states, after the laser pulse, what the new distribution of states will be. We might then measure the decays of the system over time, measuring the photon energies and detection times in order to elucidate the spectrum of excitations and their lifetimes.

    What I am really getting at is whenever our measurement process involves uncertainties, then we will have to deal with the fact that we can't completely specify the state of the entire system that we are using. So it is far more likely in practice to have to deal with superpositions of states rather than definite states like we would in a perfect thought experiment. This isn't to say that such perfect setups don't have a use in learning the subject. It's quite like learning the laws of motion in examples where we would neglect friction to keep the math simpler. But you should also take away the lesson that the more precise you can make your question, the more precise an answer you will find.
     
  14. Jun 2, 2015 #13

    ftr

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    Thanks fzero, I might open a thread on mass spectrum later, probably with a sloppy question.:nb)
     
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