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Energy Quantisation

  1. Dec 9, 2005 #1
    Ok, im reading this part in my book that tries to explain the reason for energy quantisation. It first shows that the time independent S-eq is a an eigenvalue equation, and writes it as: d^2/dx^2 Y(x) = 2m/h^2 [V(x) - E(x)] Y(x) where i use Y instead of "Psi". It gives a graph of an arbitrary potential approaching V- as x-> -inf, V+ as x-> +inf and has a min of Vmin. It considers 4 cases: E < Vmin, Vmin< E< V-, V-< E < V+, E > V+. I understand the first case is physically unacceptable because Y(x) has to be concave up if at some x=x* Y(x*)>0, concave down if Y(x*)<0, and "escapes" from the x-axis if Y(x*)=0 and due to the probabilistic nature of the wavefunction, Y(x) must be finite. Now for the second case, Vmin < E < V-, there are two points of intersection between E and V, call them x1 for the left intersection and x2 for the right. I understand the Y(x) must oscillate when x is between x1 and x2. Also if x<x1, then it can converge either to the left or to the right and if x>x2 then also it can converge either to the left or to the right. Now since we're assuming V is finite and continuous everywhere it follows that d^2Y(x)/dx^2 and dY(x) are also continuous. The book tries to explain that because the d^2Y(x)/dx^2 depends on E, then the curvature of Y(x) depends on E, then there may be certain discrete values of E for which the solution in the internal region will join smoothly with to the solutions in the external regions, and they say that such solutions which are finite, continuous and having a continuous derivative everywhere are eigenfunctions of the time independent S-eq.
    I dont really understand what they mean...why cant you have any value of E creating this smooth connection between the curves?
     
  2. jcsd
  3. Dec 10, 2005 #2

    Physics Monkey

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    The idea is that the exponential decay outside the classically allowed region is controlled by a different parameter than the parameter that controls the oscillations inside the classically allowed region. Both these parameters depend on E, but the matching conditions at the boundaries are going to involve these two different parameters. In other words, you will obtain a self consistent equation for E. Now, the equation is rarely algebraic, but you can still convince yourself that it isn't going to allow for a continuum of solutions. So in some sense, quantization is generated from the fact that you have to match two qualitatively different solutions with matching conditions that involve the energy. However, the hidden key is normalization. Without this condition you have enough freedom (from the extra solution) to choose your E in any way you like. This is why scattering states have a continuum of energies, when you change the energy you only change the relative weights of the two linearly independent solutions (which are both oscillatory). In the language of scattering theory, all this is saying is that reflection and transmission coeffecients are energy dependent. Interestingly however, one can still see the vestiges of quantized energy in certain resonance phenomenon in scattering theory.
     
    Last edited: Dec 10, 2005
  4. Dec 10, 2005 #3
    I kind of understand, but is there a more precise way to show this?
     
  5. Dec 10, 2005 #4

    Physics Monkey

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    Sure, just count! Here is what I mean: with normalization, you have 1 solution in the right classically forbidden region, 2 solutions in the classically allowed region, and 1 solution in the left classically forbidden region. Including the energy that's five constants you have to nail down. Now, one of these will be determined by normalization, so you are really left with only four constants. How many matching conditions do you have? Well, you have exactly four since you require continuity of the wavefunction and its first derivative at two different points. Four equations and four unknowns, and while you will generally have multiple solutions (several bound states) its clear that you aren't going to get a continuum of solutions.

    I encourage you to try a similar kind of counting for scattering states to convince yourself that the system has a solution for each E.
     
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