energy levels for two particles in infinite potential well


by AStaunton
Tags: energy, infinite, levels, particles, potential
AStaunton
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#1
Apr7-11, 10:11 AM
P: 105
problem is:

(a)write down the spatial or orbital for two-non interacting particles, with the same mass, in a one dimensional well, where the potential energy is zero for 0<x<2a and infinite anywhere else.

(b)What are the energies of the four lowest energy levels for the system in units of [tex]E_{0}=\frac{\pi^{2}\hbar^{2}}{8ma^{2}}[/tex]


My trouble is with part (b) of the question:

as the particles are identical, I can solve for particle 1, say and then clearly particle 2 will have same EVs:

energy levels for particle 1:

[tex]E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{8ma^{2}}[/tex]

and so can immediately infer that energy levels for particle 2 are:

[tex]E_{\bar{n}}=\frac{\bar{n}^{2}\pi^{2}\hbar^{2}}{8ma^{2}}[/tex]

and now I think to find enerfy levels of the system, simply use superposition of energy levels for particle 1 and energy levels for particles 2, so for example the lowest energy level of the system is when n=1 and nbar=1:

[tex]E_{1system}=\frac{\pi^{2}\hbar^{2}}{4ma^{2}}[/tex]

however, I am not sure if degenerate cases like when n=nbar are acceptable.

Also as I think the particles are identical, is it considered a different energy level when nbar=2 and n=1 to when n=2 and nbar=1? if these are two different energy levels and also degenerate levels such as when n=nbar are acceptable I think the four lowest energy levels in terms of E_0 are:

[tex]E_{1system}=2E_{0}[/tex]

[tex]E_{2system}=5E_{0}[/tex]

[tex]E_{3system}=5E_{0}[/tex]

[tex]E_{4system}=8E_{0}[/tex]

and to clarify again - I am assuming that degenerate cases are OK and also that it is a different energy state when n=1 and nbar=2 than when nbar=2 and n=1 and so on..
so for E_1system,n=1,nbar=1 E_2system,n=1,nbar=2 E_3system,n=2,nbar=1 E_4system,n=2,nbar=2.

are these assumptions valid?
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ideasrule
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#2
Apr8-11, 01:32 PM
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No, they're not valid. An energy level is just that: a definite energy. It doesn't matter how that energy was achieved or what the components of the system look like; the only thing that matters is the value of the energy.

In this case:

[tex]
E_{N}=\frac{N^{2}\pi^{2}\hbar^{2}}{8ma^{2}}
[/tex]

where N^2=n^2+n_bar^2, and the lowest energy levels correspond to the lowest values of N.
AStaunton
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#3
Apr9-11, 02:34 PM
P: 105
OK using this: N^2=n^2+n_bar^2 and it is not allowed that n is equal to nbar?

ideasrule
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#4
Apr9-11, 02:46 PM
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energy levels for two particles in infinite potential well


Quote Quote by AStaunton View Post
OK using this: N^2=n^2+n_bar^2 and it is not allowed that n is equal to nbar?
It is allowed. Every possible value of N^2 corresponds to one energy level. It doesn't matter what values n and n_bar take, or whether they're the same.
AStaunton
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#5
Apr9-11, 02:59 PM
P: 105
OK, so in that case weren't my assumptions that I stated at the bottom of my first post valid?
it seems that you are saying the same thing... unless I'm not understanding you properly...

ie. would you agree that the four energy levels I stated E_1system to E_4system are indeed the four lowest energies?
ideasrule
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#6
Apr9-11, 04:39 PM
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I wouldn't agree because you posted:

[tex]
E_{3system}=5E_{0}
[/tex]
[tex]
E_{4system}=5E_{0}
[/tex]

Those represent the same energy level, because they have the same energy.
kloptok
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#7
Apr9-11, 05:16 PM
P: 188
Quote Quote by ideasrule View Post
It is allowed. Every possible value of N^2 corresponds to one energy level. It doesn't matter what values n and n_bar take, or whether they're the same.
However, this is only true if the particles are identical bosons, right? If they're fermions we have to take the Pauli principle into account.

(Although I see now that it was stated in the OP that degenerate levels are allowed, implying that we're dealing with bosons.)


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