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Really stupid question about bound electron

  1. Aug 23, 2012 #1
    I have a question, and I'm positive it has a really simple answer, but I can't think of it right now.

    In the infinite square well (the simplest bound problem), the wave functions have discrete energy values. We can have a wave function that's a linear superposition of any number of these so that the average value of the energy we measure is any value (above the ground state energy, of course), but at the end of the day each measurement will yield one of the discrete energy values allowed.

    So, let's say we have an electron moving in 0 potential with a well defined momentum p. This means it has a well defined energy [itex]E = p^2/2m[/itex], and let's say this E isn't one of the discrete energy levels of the infinite square well we're looking at. Now let's say and it gets captured in this infinite square well. It could have the same average energy that it had before capture, but necessarily some measurements would yield a higher energy than its initial energy, if you constructed some wave function that had the average energy equal to its initial (pre-capture) energy.

    So what happens? Can it not be captured? If it is, does it form some new wave function with the average energy equal to the initial one? This would seem sketchy to me, because there are obviously an infinite number of combos that could give that average energy.

    Thanks!
     
  2. jcsd
  3. Aug 23, 2012 #2

    bcrowell

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    Energy is conserved not just on the average but exactly.

    Right, it can't be captured. This is sort of analogous to the fact that a gas has a discrete absorption spectrum (although some of the details of the story are different, e.g., the electron doesn't get annihilated). A closer but possibly less familiar analogy would be neutron absorption resonances of nuclei.
     
  4. Aug 23, 2012 #3
    In QFT, where you can express the notion of particle creation, I believe that what would happen is the electron would have some probability of getting captured as an exact eigenstate of the potential, and simultaneously emitting a photon to carry away the excess energy. Does that sound right?
     
    Last edited: Aug 23, 2012
  5. Aug 23, 2012 #4
    This doesn't quite make sense as you've posed it. You get an infinite square well when you make it impossible for a particle to exist outside a bounded region (because the potential is infinite outside that region). Therefore it doesn't make sense for an electron to be outside an infinite square well and then fall in--it couldn't have been outside the well in the first place.

    Of course, the infinite square well is not a physically realizable situation. A finite square well is more realistic, and I suggest considering it instead. In a finite square well, the potential is zero everywhere except for a finite region, in which the potential takes on a finite negative value -V0.

    A finite square well has bound states much like the infinite square well, except that there are a finite number of them. An important fact about these bound states is that their energies are all negative (relative to the potential outside the well, which we set at zero).

    Free particles can exist outside the well. All the free particle states have positive energy. So you can see that energy conservation prevents a free particle, with positive energy, from entering a bound state of the well, which has negative energy. So a particle cannot be captured by the well. The classical analog is: if you have a ball rolling around along the ground without friction, it cannot be captured by a hole in the ground--it will just roll back out the other side because of energy conservation.

    Of course, the ball can be captured if it gives up some of its energy somehow. Similarly a particle can be captured by a potential well if it gives off energy in the process. This is how electrons are captured by ions to form atoms: a photon is emitted in the process, after which the electron has negative energy and is bound to the ion.
     
  6. Aug 30, 2012 #5
    I mean...this is kind of nitpicky. Here, consider this if it bothers you so much: a NEARLY infinite square well. from [itex]a < |x| < b[/itex], V(x) is crazy large, [itex]V(x) = \hbar^{-\hbar^{-1}}[/itex], and V(x) = 0 everywhere else. Also, say a is a normal infinite square well width, and [itex]b = a/\hbar^{100}[/itex].

    So, our particle has E>0, and, as unlikely as it is, tunnels into our nearly infinite well. Sure, we don't have an absolutely complete set of states in the new well, but it's still pretty robust.
     
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