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Homework Statement
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?
Homework Equations
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?
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