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Fermions in infinite square well in compact geometry

  1. Sep 17, 2016 #1
    1. The problem statement, all variables and given/known data

    The global topology of a ##2+1##-dimensional universe is of the form ##T^{2}\times R_{+}##, where ##T^{2}## is a two-dimensional torus and ##R_{+}## is the non-compact temporal direction. What is the Fermi energy for a system of spin-##\frac{1}{2}## particles in this universe?

    What happens in the limit of large ##N##? Can you find a closed-form expression for the energy?

    2. Relevant equations

    3. The attempt at a solution

    Assuming that the Schrodinger equation can be applied in this geometry and the potential is zero, we obtain ##\psi(x,y)=X(x)Y(y)##, where ##x## and ##y## label the coordinates along the torus. So,

    ##X(x)=A_{x}\text{sin}(k_{x}x)+B_{x}\text{cos}(k_{x}x)## so that ##E_{x}=\frac{\hbar^{2}k_{x}^{2}}{2m}##
    ##Y(y)=A_{y}\text{sin}(k_{y}y)+B_{y}\text{cos}(k_{y}y)## so that ##E_{y}=\frac{\hbar^{2}k_{y}^{2}}{2m}##

    Using the boundary condition ##X(0)=X(L)##, where ##L## is the circumference of the torus in the ##x##-direction,

    ##B_{x}=A_{x}\text{sin}(k_{x}L)+B_{x}\text{cos}(k_{x}L)##
    ##\implies B_{x}(1-\text{cos}(k_{x}L))=A_{x}\text{sin}(k_{x}L)##
    ##\implies B_{x}(2\text{sin}^{2}(k_{x}L/2))=A_{x}2\text{sin}(k_{x}L/2)\text{cos}(k_{x}L/2)##
    ##\implies A_{x}=\text{tan}(k_{x}L/2)B_{x}##

    so that

    ##X(x) = B_{x}[\text{sin}(k_{x}x)\text{tan}(k_{x}L/2)+\text{cos}(k_{x}x)]##

    Now, I cannot an expression for ##k_{x}## and hence I cannot calculate the minimum energy of the system. Any hints on how to procees?
     
    Last edited: Sep 17, 2016
  2. jcsd
  3. Sep 18, 2016 #2

    mfb

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    You also get a boundary condition for the first derivative.

    I would express the wave function via a complex exponential, that should be easier than the sin/cos stuff.
     
  4. Sep 19, 2016 #3
    I see!

    In that case, ##k_{x}L=2\pi n_{x}##, or ##k_{x}l=n_{x}##, where ##l## is the radius of the torus in the ##x##-direction.

    So, ##\displaystyle{k_{x}=\frac{n_{x}}{l}}##, where ##n_{x} = 0, \pm 1, \pm 2 , \ldots##

    I can see that the state with ##n_{x}=0## is not normalisable, so this state is not physical.

    However, I do not quite follow as to why we can ignore negative values of ##n_{x}##. It's true that the energy values are the same as for the corresponding ##|n_{x}|##, but the wavefunctions are different, by a phase factor at least. I know that phase factors are not observable, but we still have different wavefunctions or states for negative values of ##n_{x}##, don't we?
     
  5. Sep 19, 2016 #4

    mfb

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    Who said that you can ignore them? They have different momentum, for example, so they are certainly relevant.
     
  6. Sep 20, 2016 #5
    But aren't negative quantum numbers ignored in the infinite square well potential?
     
  7. Sep 20, 2016 #6

    mfb

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    There the walls are not identified with each other, which leads to different wave functions. In particular, they are purely real in the box, while your eigenstates have complex numbers.
     
  8. Sep 24, 2016 #7
    It would be helpful if you could explain this point in a bit more detail.
     
  9. Sep 24, 2016 #8

    mfb

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    You have two different situations, why do you expect similar results?

    Your particles can have net momentum, similar to classical particles. Particles in a box cannot have that, they would hit the wall.
     
  10. Sep 26, 2016 #9
    It's true that the energy values are the same as for the corresponding |nx|
     
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