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Identical particles in a 2D potential well

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    So, I'm asking for a bit of help before I confuse myself completely.

    The question statement is:

    Consider a two-dimensional potentialbox

    [itex]V(x,y) = 0[/itex] if [itex]0 \leq x \leq a, 0 \leq y \leq 2a[/itex]
    and infinity otherwise.

    a) Determine the energy eigenstates and energy eigenvalues of a particle in this box. The solutions of the 1D potential well can be considered as known.

    b) If we place 3 identical bosons in the box, what will the ground state energy be if we disregard interaction between the bosons.

    c) Same as in b), but for 3 identical spin 1/2 fermions.

    d) Write down the complete wavefunction (with both spatial and spin parts) for the ground state if two identical fermions with spin 1/2 and without interaction are put in the box.

    e) Same as d) but for 3 identical fermions with spin 1/2.

    2. Relevant equations

    1D potential well equations:

    [itex]\psi_n (x) = \sqrt{\frac{2}{a}} {sin(\frac{n \pi x}{a})}[/itex]

    [itex]E_n = \frac{n^2 \pi^2 \hbar^2}{2ma^2} [/itex]


    3. The attempt at a solution

    Okay, for a) I did a variable separation and ended up with

    [itex]\psi_{n_{x}n_{y}} (x) = \frac{\sqrt{2}}{a} sin(\frac{n_x \pi x}{a})sin(\frac{n_y \pi y}{2a})[/itex]

    [itex]E_{n_{x}n_{y}} = \frac{\pi^2 \hbar^2}{2ma^2}(n_{x}^2 + \frac{n_{y}^2}{4}) [/itex]

    Then in b)

    Since there are three bosons they can all be in the same state, and the lowest state would be for [itex]n_x=n_y=1[/itex], so the total energy would be

    [itex]E_{tot} = 3E_{1,1} = \frac{15 \pi^2 \hbar^2}{8ma^2}[/itex]

    and for c)

    Again, the lowest energy will be for [itex]n_x=n_y=1[/itex], but since only two spin 1/2 fermions can be in that energy at the same time, I'll also have a third particle, which I'm thinking will be in [itex]n_x= 1, n_y=2[/itex], since this will give me a lower energy than [itex]n_x= 2, n_y=1[/itex].

    So then, the total energy would be:

    [itex]E_{tot} = 2E_{1,1}+E_{1,2} = \frac{10 \pi^2 \hbar^2}{8ma^2} + \frac{8 \pi^2 \hbar^2}{8ma^2} = \frac{9 \pi^2 \hbar^2}{4ma^2}[/itex]

    And that's where I'm not completely sure if my reasoning is completely correct, and where I want to confirm. I haven't started d) and e) yet, but I want to confirm this first, and I'm going to continue doing the rest with the assumptions I have above until I get a reply or figure something else out.

    Thank you for any help that you may provide!
     
    Last edited: Oct 15, 2011
  2. jcsd
  3. Oct 15, 2011 #2

    vela

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    Looks fine!
     
  4. Oct 16, 2011 #3
    Thank you :D

    Well then, I actually need some help with d) as well, it turns out.

    Because I know, from c), that the two fermions will be in [itex] n_x=n_y=1 [/itex], so that the wavefunctions will be

    [itex]\psi_{1,1}^{(1)}(x_1,y_1) = \frac{\sqrt{2}}{a}sin(\frac{\pi x_1}{a})sin(\frac{\pi y_1}{2a})[/itex]

    [itex]\psi_{1,1}^{(2)}(x_2,y_2) = \frac{\sqrt{2}}{a}sin(\frac{\pi x_2}{a})sin(\frac{\pi y_2}{2a})[/itex]

    Where the exponent on the psi refers to the particle. And this is because if I solve the Schroedinger equation for two particles I can do a variabel (particle) separation as well, and see that

    [itex]\psi (x_1,y_1,x_2,y_2) = \psi^{(1)} (x_1,y_1) \psi^{(2)} (x_2,y_2)[/itex]

    So then

    [itex]\psi (x_1,y_1,x_2,y_2) = \psi_{1,1}^{(1)}(x_1,y_1) \psi_{1,1}^{(2)}(x_2,y_2) = \frac{2}{a^2}sin(\frac{\pi x_1}{a})sin(\frac{\pi y_1}{2a})sin(\frac{\pi x_2}{a})sin(\frac{\pi y_2}{2a})[/itex]

    I also see that this is a symmetric function when exchanging the particles (if this is indeed correct, which I can't honestly say I'm 100% sure about), so I know that they have to be in the singlet spin state.

    How on earth do I write this? Can I just write it as

    [itex]\psi (x_1,y_1,x_2,y_2)\left| singlet \right\rangle[/itex]

    And can I put in what the singlet state is? I mean, I know it's

    [itex]\left| singlet \right\rangle = \frac{1}{\sqrt{2}} (\left| \uparrow \downarrow \right\rangle - \left| \downarrow \uparrow \right\rangle)[/itex]

    so can I put this into the equation?
     
  5. Oct 16, 2011 #4

    vela

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    Yes, the complete state is a Cartesian product of the spatial state and the spin state. You typically just write them next to each other as you did or perhaps stick a symbol between them indicating it's a Cartesian product. Do you have any examples in your textbook?
     
  6. Oct 16, 2011 #5
    I honestly can't find any, the closest I came to it was from my lecture notes, where my teacher wrote

    |spatial>|spin>

    in an example.
     
  7. Oct 16, 2011 #6

    vela

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    It's funny. I just checked two of my books, and I can't find an example either. I'd just do what your professor did and write the two pieces next to each other. It's pretty clear what it means.
     
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