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Fermions and the Fixed Node Approximation

Exact Fermions

Example: Two fermions in a one-dimensional simple harmonic oscillator

For two distinguishable particles in a haromonic confinining potential in three dimensions, the imaginary time propagator is

(1)K(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_1', \mathbf{r}_2';\tau)
= \left(\frac{m\omega}{2\pi\hbar\sinh\omega\tau}\right)^3
\exp\left(-\frac{m\omega((r_1^2 + r_2^2 + r_1'^2 + r_2'^2)
\cosh\omega\tau
-2\mathbf{r}_1\cdot\mathbf{r}_1'
-2\mathbf{r}_2\cdot\mathbf{r}_2')}
{2\hbar\sinh\omega\tau}\right)

The partition function is the trace of the propagator for \tau = \beta\hbar,

(2)Z &= \operatorname{tr} K \\
&= \int d\mathbf{r}_1^3 \int d\mathbf{r}_1^3 \,
K(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_1, \mathbf{r}_2; \beta\hbar) \\
&= \left(\frac{m\omega}{2\pi\hbar\sinh\beta\hbar\omega}\right)^3
\int d\mathbf{r}_1^3 \int d\mathbf{r}_1^3
\exp\left(-\frac{2m\omega(\cosh\beta\hbar\omega -1)(r_1^2 +r_2^2)}
{2\hbar\sinh\omega\tau}\right)\\
&= \left(2(\cosh\beta\hbar\omega-1)\right)^{-3} \\
&= \left[2\sinh\left(\frac{\hbar\omega}{2k_BT}\right)\right]^{-6}

For identical particles, we need to symmetrize the states for fermions, or antisymmetrize the states for bosons. The trace of the permuted propagator is,

Z_P &= \operatorname{tr} PK \\
&= \int d\mathbf{r}_1^3 \int d\mathbf{r}_1^3 \,
K(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_2, \mathbf{r}_1; \beta\hbar) \\
&= \left(\frac{m\omega}{2\pi\hbar\sinh\beta\hbar\omega}\right)^3
\int d\mathbf{r}_1^3 \int d\mathbf{r}_1^3
\exp\left(-\frac{2m\omega(\cosh\beta\hbar\omega(r_1^2 +r_2^2)
-\mathbf{r}_1\cdot\mathbf{r}_2)}
{2\hbar\sinh\omega\tau}\right)

Next we change coordinates to \mathbf{R}=\frac{1}{2}(\mathbf{r}_1+\mathbf{r}_2) and \mathbf{r}=\mathbf{r}_1-\mathbf{r}_2. Then r_1^2+r_2^2 = R^2 + r^2/2 and \mathbf{r}_1\cdot\mathbf{r}_2 = R^2 - r^2/2, and we find,

(3)Z_P = \left[2\cosh\left(\frac{\hbar\omega}{2k_BT}\right)
\sinh\left(\frac{\hbar\omega}{2k_BT}\right)\right]^{-3}