Example: Nodes for Two Particles

The following example describes the restricted path integral method and illustrates why it leads to the exact solution in the case that the nodes are exactly known. It represents a simplified version of the illustration by Ceperley (1996) that uses the example of the hydrogen molecule. Here, we talk about the simplest possible problem that has a node: two free identical particles. For those, the exact nodes are given by,

$$\rho({\bf r}_1,{\bf r}_2,{\bf r}_1^*,{\bf r}_2^*;\beta) \equ... ...({\bf r}_2,{\bf r}_2^*;\beta) \end{array}\right\vert = 0\quad,$$ (98)

where $\rho_1$ is given by the free particle density matrix Eq. 2.12. The solution of this equation is given by,

$$({\bf r}_1 - {\bf r}_2) \cdot ({\bf r}_1^* - {\bf r}_2^*) = 0 \quad.$$ (99)

Introducing the relative coordinate ${\bf r}={\bf r}_1-{\bf r}_2$, the node is a plane at the origin ${\bf r}=0$ perpendicular to the vector given by reference point separation ${\bf r}^*={\bf r}^*_1-{\bf r}^*_2$.

In this example, we discuss closed paths that end in the reference point ${\bf r}(\beta )={\bf r}^*$. In the case of a permutation, the path must start at $-{\bf r}^*$, otherwise at ${\bf r}^*$. One can distinguish three types of paths as shown in Fig. 2.1,

A

The path does not cross the node, therefore it starts and ends at ${\bf r}^*$.

B

It crosses the node an even number of times $>0$, therefore also starts and ends at ${\bf r}^*$.

C

It goes an odd number of times across the node and therefore must start at $-{\bf r}^*$.

Figure 2.1: Illustration of the nodal constraint for paths going to ${\bf r}(\beta)={\bf r}^*$. Type A paths are node-avoiding, Type B cross the nodes an even number of times, and Type C permutes, therefore must start at ${\bf r}(0)=-{\bf r}^*$ and cross the node an odd number of times.

The various kinds of paths have different physical interpretations. In a system of distinguishable particles, no permutations can occur and no nodes exist. Therefore only paths of type A and B contribute. For bosons, one sums over all permutations, nodes are irrelevant and no negative signs come in. For fermions, one has the choice of the direct fermion method or the restricted path method. Both methods are equivalent if the exact nodes are used. In case of the direct method, one considers all permutations with positive and negative signs and does not restrict the path. One sums up contributions from all types including those from type C with a negative sign. For restricted path integrals, one enforces the node, which rules out any permutations in two particle systems. Therefore, there are no negative contributions. Furthermore, the nodal constraint also prevents paths of type B from occurring and one is left with paths of type A, that stay within the half space given by the nodal plane. Essentially, one employs the cancellation of B and C in this method. Both cancel exactly because the flux of the paths is given by the gradient of the density matrix, which is the same on both sides of the node since the derivative is a continuous function. Summarizing, it reads,

$~\bullet~$Distinguishable particles:    A+B

$~\bullet~$		Bosons: 		A+B+C

$~\bullet~$		Fermions without nodes:   		A+B-C

$~\bullet~$		Fermions with nodes: 		Only A.

The magnitude of the different contributions can also be understood studying the exact solution to the problem stated in Eq. 2.46. For this system of two fermions, it reads,

$$\rho({\bf r},{\bf r}^*;\beta) = \rho_{\rm D}({\bf r},{\bf r}^*;\beta) - \rho_{\rm D}(-{\bf r},{\bf r}^*;\beta) \quad.$$ (100)

Setting ${\bf r}={\bf r}^*$, the positive first term is greater in magnitude than the second because it is a diagonal density matrix element. It represents paths of type A and B, while the second term is smaller in magnitude and describes paths of type C. This equation also shows how a node emerges at ${\bf r}=0$. The second term can be regarded a result from an image charge with opposite sign on the other side of the node similarly to the method of image charges in electrostatics. Both contributions are solutions to the Bloch equation that added together lead to a zero at the node. The determining parameter for the importance of the image charge is the distance to the node. This will be used in the derivation of the nodal action in the following section.