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,