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Specifically, my question is: Should one expect energy quantization for a particle in the following potential well? potential

More generally, how can one tell whether or not energy should be discrete/quantized just by looking at the potential function?

Qmechanic
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zeta
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  • Related: http://physics.stackexchange.com/q/39208/2451 , http://physics.stackexchange.com/q/65636/2451 and links therein. – Qmechanic Nov 16 '13 at 22:56

5 Answers5

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The rule of thumb for me is how would the solutions of the Schrodinger equation look for this potential. When the energy of the electron is above the potential well the solutions can be continuous ( although there can be discrete resonances with an energy width). Below the lip of the potential, as for example in the hydrogen atom, the solutions will be discrete, quantized.

If the walls of the potential go to infinity, as in your example, then all the solutions will be quantized as in this example of the particle in a box .

In the limited in energy potential ( not going to infinity) the energy levels with large energy quantum numbers are very dense and resemble a continuum, although they can be labeled with a discrete number, as for example in the hydrogen atom.

hydrogenatom

Above the energy of the well are the continuum solutions, not discrete.

anna v
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A summarised derivation for the 1 D triangular potential well can be found here. Interesting graphs of the wave functions $\psi_n(x)$ are provided.

Also treated at [this *.pdf] (p.5, section 1.2.8.2)2.

Gert
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Well, if we start from Bohr's explanation to the stability of the atom, we remember that he said that the obit of the electron should make a standaing wave around the nucleus. Quantum mechanics came up, then, and showed that the situation of the standing wave is true always when you have a potential that isn't time dependent.

So, in any potential you have, you should have standing waves, and standing waves create your quantization condition. That's the answer in the simplest form.

  • Isn't it true that a standing wave does not necessarily lead to quantization/discreteness? For example, a potential that is completely flat does not lead to quantization. True? – zeta Nov 16 '13 at 23:06
  • Right, we can send two waves from two infinities from the opposite sides and get a standing wave with any energy $E$. But in a well it is the well walls who send waves back and they interfere. – Vladimir Kalitvianski Nov 16 '13 at 23:09
  • @Jonathan, that wouldn't be a standing wave, merely a wave. What TheQuantumPhysicist could have mentioned was that in your potential, you are solving a linear ODE (Time independent Schrodinger equation) subject to the two boundary conditions $\psi(0) = 0$ and $\psi(\infty)=0$. I am not sure of the exact theory but I believe you can show that this is only possible for certain values of $E$. – lionelbrits Nov 16 '13 at 23:10
  • @Jonathan Let me put it this way: standing wave is the default configuration, and that leads to quantization. The special case is when you have your potential so huge that the quantization condition becomes so smooth, and you won't notice any quantization anymore. So in other words: everything is quantized, but the quantization cannot be always observed. I actually don't understand your question further. Please elaborate if you have something not clear from that. – The Quantum Physicist Nov 17 '13 at 11:20
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The regions with $U > E$ are classically forbidden - a particle cannot get there, it is reflected (stopped and sent back). In QM it is the same except the wave function can penetrate slightly into forbidden regions. As far as the wave is reflected, it exists within the well. Such a wave can be represented as a sum of standing waves with some time-dependent coefficients. A standing wave has a certain energy $E_n$ because the wave with $E_n <E<E_{n+1}$ interferes differently (it is not standing) and its energy is in fact uncertain.

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One way of getting to the answer, which is probably not going to cut it for Jonathan, is to use the WKB approximation to get at the wave-function. This involves matching coefficients inside and outside the forbidden regions of the potential, and this matching can only be possible for certain values of $E$. This leads to the Born-Sommerfield quantization rules. Roughly, the action of the particle moving from one end of the classically allowed region to the other is a multiple of $2\pi n \hbar$.

lionelbrits
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