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fluidistic
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Homework Statement
Assume that inside an infinite potential well there are 2 identical particles that doesn't interactuate between themselves and that have spin 1/2 (for instance electrons).
1)Write down the Schrödinger's equation associated with such a system. Write the eigenfunctions in terms of the wavefunction of a single particle.
2)Make a diagram of the seven first energy levels, indicating the degeneracy and the total spin for each state.
3)Calculate the expectation value of the energy for a particle in the ground state and the other in the first excited state. Compare with the case of distinguishable particles.
Homework Equations
Coming in attempt part.
The Attempt at a Solution
1)Non interacting particles implies that [itex]V(x_1,x_2)=V(x_1)+V(x_2)=0+0=0[/itex] in this exercise.
I consider electrons, they are fermions so the total wavefunction [itex]\Psi[/itex] must be antisymmetric. Its spatial part can be either antisymmetric ([itex]\frac{1}{\sqrt 2 } \left [ \psi _1 (x_1)\psi _2 (x_2) - \psi _2 (x_1)\psi _1 (x_2) \right ][/itex] ) in which case the spin part must be symmetric ([itex]\chi ^S[/itex]). Or the spatial part can be symmetric ([itex]\frac{1}{\sqrt 2 } \left [ \psi _1 (x_1)\psi _2 (x_2) + \psi _2 (x_1)\psi _1 (x_2) \right ][/itex] ) and then the spin part must be antisymmetric ([itex]\chi ^A[/itex]).
Where [itex]\chi ^A (1,2)=\frac{1}{\sqrt 2} (\uparrow _1 \downarrow _2- \downarrow _1\uparrow _2 )[/itex].
[itex]\chi _1 ^S (1,2)=\uparrow _1\uparrow _2[/itex]
[itex]\chi _0 ^S (1,2)=\frac{1}{\sqrt 2 } (\uparrow _1 \downarrow _2+ \downarrow _1\uparrow _2 )[/itex]
[itex]\chi _{-1}^S (1,2)=\downarrow _1\downarrow _2[/itex].
Also the energy of a single particle is [itex]E_n=\alpha n^2[/itex] where [itex]\alpha = \frac{h^2}{8L^2m}[/itex] where L is the length of the potential well. The energy of the system in this exercise is the sum of the energy of each particle.
So my answer to part 1 is:
[itex]-\frac{h^2}{2m} \frac{d^2}{dx^2} \Psi (x_1,x_2)=[E(1)+E(2)] \Psi (x_1, x_2)[/itex].
Where [itex]E(1)+E(2)=\alpha (n_1 ^2+n_2^2)[/itex].
And [itex]\Psi (x_1,x_2)= \begin{cases} \frac{1}{\sqrt 2 } \left [ \psi _1 (x_1)\psi _2 (x_2) - \psi _2 (x_1)\psi _1 (x_2) \right ] \chi ^S \\ \frac{1}{\sqrt 2 } \left [ \psi _1 (x_1)\psi _2 (x_2) + \psi _2 (x_1)\psi _1 (x_2) \right ] \chi ^A \end{cases}[/itex].
2)Ground state: [itex]n_1+n_2=2 \Rightarrow E=2\alpha[/itex].
First excited state: [itex]n_1+n_2=3 \Rightarrow E= 3 \alpha[/itex]. This can be done with 2 different ways (without considering the spin): [itex]n_1=1[/itex] and [itex]n_2=2[/itex] or [itex]n_1=2[/itex] and [itex]n_2=1[/itex].
I've noticed that without considering the spin, for the nth excited state, [itex]E_n= n \alpha[/itex] with [itex]n-1[/itex] degeneracies.
I am not sure what to do with the spin... By intuition I know that if both particles are in the ground state, they can only have opposite spin. I am not sure I can distinguish between "the spin of particle 1 is up and the spin of particle 2 is down" and "the spin of particle 1 is down and the spin of particle 2 is up". So I am not sure what is the degeneracy of the ground state. Maybe I should use the wavefunction of part 1), but I don't know how.
3)Rather than calculating [itex]\int _{0}^L \Psi (x_1,x_2) ^ * \hat E \Psi (x_1, x_2)dx[/itex], I guess I can just say that [itex]\langle E \rangle = |c_1|^2E(1)+|c2|^2E(2)=\frac{E(1)+E(2)}{2}[/itex].
If those particles were distinguishable like in classical mechanics, [itex]\langle E \rangle = (E(1)+E(2))[/itex].
I'd appreciate if someone could correct me for any error(s) and for part 2) especially. Thank you!