Fermions in infinite square well in compact geometry

[rock_pepper_scissors]

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 Schrödinger 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?

 

[mfb]

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.

 

[rock_pepper_scissors]

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?

 

[mfb]

Who said that you can ignore them? They have different momentum, for example, so they are certainly relevant.

 

[rock_pepper_scissors]

But aren't negative quantum numbers ignored in the infinite square well potential?

 

[mfb]

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.

 

[rock_pepper_scissors]

It would be helpful if you could explain this point in a bit more detail.

 

[mfb]

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.

 

[phamtuanktdt]

It's true that the energy values are the same as for the corresponding |nx|