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Homework Help: Ground state wavefunction & energy for 2 electrons

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    2 electrons are in a box of length L. Ignoring Coulomb force, 1 and 2 are labels for the electrons and m is the mass of an electron. What is the ground state and 1st excited state for the energy and wavefunction for the two electrons? Is there more than one wavefunction for the ground state and the 1st excited state? Your answer will include the spatial and spin components of hte wavefunction. Use |+-z> to describe spin.

    2. Relevant equations

    H=p12/2m+p22/2m

    3. The attempt at a solution

    Since they are electrons, they are spin 1/2 particles and are fermions. Thus, their wavefunction must be antisymmetric with repsect to the exchange of particle labels. This means the two particles have a total spin of 0 or 1, so there are two wavefunctions for the ground state?

    S=0: |0,0,s=0,m=0> and S=1: |0,1,s=0,m=0>. Am I on the right track here?
     
    Last edited: Apr 6, 2010
  2. jcsd
  3. Apr 6, 2010 #2
    Explain to us your notation for the wavevectors. I agree that if you only look at spin, you will have a singlet (S=0) state and a triplet state (S=1). Which of those spin states are antisymmetric and which is symmetric?

    Also, you can separate the spatial wavefunction from the spin wavefunction. So be careful that you keep the total product of those wavefunctions antisymmetric since these are fermions.
     
  4. Apr 6, 2010 #3
    it would be |n,l,s,ms>, is that right? I'm so confused on how to do this...
     
  5. Apr 6, 2010 #4
    This isn't a hydrogen atom. It is just a 1D box with 2 particles, so you will have [tex]\psi_{n_1,n_2}(x_1,x_2)[/tex] or you can write it as [tex]|n_1,n_2>[/tex] for the spatial wavefunction and wavevector. The 2 particle wavefunction is just a product of the single particle wavefunctions in a box. The total energy is just the sum of the single particle energies.

    The spin vector can be written as [tex]|\uparrow \uparrow>,|\uparrow \downarrow>,|\downarrow \uparrow>, |\downarrow \downarrow>[/tex] and any combination of those. Preferably combinations that form symmetric and antisymmetric spin vectors.
     
  6. Apr 6, 2010 #5
    so would the ground state wavefunction be |+z,+z>|-z,+z>|+z,-z>|-z,-z>?
     
  7. Apr 6, 2010 #6
    Alright, I will start you off since you have me confused again :)

    Let's assume these are bosons. So the total wavefunction must be symmetric. Remember, the total wavefunction is the product of the spatial wavefunction and the spin wavevectors. If the total is symmetric, that means both the spatial and spin wavevectors are symmetric or they both are antisymmetric (similar to +1*+1 = +1 and -1*-1 = +1).

    We want the ground state. The lowest energy state for a particle in a box is n=1. Since we have two particles in a box, the energy will be:

    [tex]E_{1,1} = E_1 + E_1[/tex]

    where [tex]E_1[/tex] is the ground state energy for a single particle in a box. This corresponds to the wavevector [tex]|1,1>[/tex]. This is symmetric, because if I interchange those two numbers in the vector, I get back the same vector.

    Next we need a symmetric spin vector, because we have a symmetric spatial vector. That leads to the spin triplet (which is symmetric):

    [tex]|\uparrow \uparrow>[/tex]
    [tex]|\downarrow \downarrow>[/tex]
    [tex]\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow>+|\downarrow\uparrow>\right)[/tex]

    Therefore we will have 3 possible ground states that are all degenerate for the boson pair:
    [tex]|11>|\uparrow \uparrow>[/tex]
    [tex]|11>|\downarrow \downarrow>[/tex]
    [tex]\frac{1}{\sqrt{2}}|11>\left(|\uparrow\downarrow>+|\downarrow\uparrow>\right)[/tex]

    Now you will do the same for a pair of fermions, but take care to keep the total wavefunction antisymmetric and obey the Pauli exclusion principle.
     
  8. Apr 6, 2010 #7
    So since these are fermions, we have an antisymmetric state. It is a singlet state?

    The states would then be |+z, -z> and |-z, +z>, but the electrons are identical so the only state is |+z, -z>, right? (+z is up arrow, -z is down arrow)

    Then we also have to include the spatial component. Is that just a position label and an assignment of n? Would the wavevector then be |0>1|0>2(|-z,+z>)???
     
  9. Apr 6, 2010 #8
    |+z,-z> is not an antisymmetric or symmetric state. For example, if I swap them I get:

    [tex]|\uparrow \downarrow> \neq |\downarrow \uparrow>[/tex]

    So you will need to find some combination of the spin states that are symmetric or antisymmetric. But before you do that, you should find the spatial ground state for a pair of fermions. That way you will know which spin state to choose to make the total wavefunction antisymmetric.
     
  10. Apr 6, 2010 #9
    Is my spatial ground state |n>label? Would it be |0>1|0>2?

    Would the combination of states be 1/sqrt(2)*(|+z,-z>-|-z,+z>)?
     
  11. Apr 6, 2010 #10
    Not sure why you use n=0. But yes, that would be essentially correct. The spatial part is symmetric, and the spin part is antisymmetric. So the whole thing will be antisymmetric. You would just need to multiply those two pieces together.
     
  12. Apr 6, 2010 #11
    okay, so I was looking at an example with a harmonic oscillator and it used n=0. Should it be n=1 then since this is a ground state? My wavefunction is;

    (1/sqrt(2))|1>1|1>2*(|+z,-z>-|-z,+z>)

    Would my energy then be 1/2 hbar w?
     
  13. Apr 6, 2010 #12
    There is no harmonic oscillator in the problem. This is just 2 particles in a well. And so you should be looking at the single particle in a well solutions.

    Also the energy is incorrect. You would use the energy for particles in wells and not particles in a harmonic oscillator. You seem to be confusing things between problems.
     
  14. Apr 6, 2010 #13
    Energy for a single particle in a square well is En=hbar2pi2n2/2mL2 Do I square the whole thing since I intend 2 particles now instead of just 1?

    Also- The Psi function for a single particle is sqrt(2/L)sin(npix/L) Would that be more representative of the wavefunction I need than

    (1/sqrt(2))|1>1|1>2*(|+z,-z>-|-z,+z>)?
     
  15. Apr 7, 2010 #14
    Total energy is just the sum of the individual energies. You can write out the whole spatial wavefunction but it isn't necessary. You can just write out what [tex]|n_1 n_2\rangle[/tex] at the top, then write out your wavefunctions just using kets to keep it cleaner looking.
     
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