- #1
AStaunton
- 105
- 1
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?
(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?