# Schrödinger equation for 2 particles – separation of variables

Required math: calculus

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 5.1.

The Schrödinger equation that we’ve looked at so far involves the wave function for a single particle moving in a potential. To extend this to multi-particle systems, we need to make the wave function and the potential functions of the positions of all the particles and the time. Thus the Schrödinger equation for a system of ${n}$ particles becomes

$\displaystyle -\frac{\hbar^{2}}{2}\sum_{i=1}^{n}\frac{1}{m_{i}}\nabla_{i}^{2}\Psi\left(\mathbf{r}_{1},\mathbf{r}_{2},\ldots,\mathbf{r}_{n},t\right)+V\left(\mathbf{r}_{1},\mathbf{r}_{2},\ldots,\mathbf{r}_{n},t\right)\Psi\left(\mathbf{r}_{1},\mathbf{r}_{2},\ldots,\mathbf{r}_{n},t\right)=i\hbar\frac{\partial}{\partial t}\Psi\left(\mathbf{r}_{1},\mathbf{r}_{2},\ldots,\mathbf{r}_{n},t\right) \ \ \ \ \ (1)$

Needless to say, finding solutions of this equation for even as few as 2 particles is extremely difficult. In one case, however, we can make some progress. In a 2-particle system, if the potential ${V}$ is not time-dependent and depends only on the separation ${\mathbf{r}\equiv\mathbf{r}_{1}-\mathbf{r}_{2}}$ of the two particles, we can fiddle with it a bit and produce a simpler form.

First, we define the centre of mass

$\displaystyle \mathbf{R}\equiv\frac{m_{1}\mathbf{r}_{1}+m_{2}\mathbf{r}_{2}}{m_{1}+m_{2}} \ \ \ \ \ (2)$

If we also introduce the reduced mass

$\displaystyle \mu\equiv\frac{m_{1}m_{2}}{m_{1}+m_{2}} \ \ \ \ \ (3)$

then we get

 $\displaystyle \mathbf{r}_{1}$ $\displaystyle =$ $\displaystyle \mathbf{R}+\frac{\mu}{m_{1}}\mathbf{r}\ \ \ \ \ (4)$ $\displaystyle \mathbf{r}_{2}$ $\displaystyle =$ $\displaystyle \mathbf{R}-\frac{\mu}{m_{2}}\mathbf{r} \ \ \ \ \ (5)$

In the coordinates ${\mathbf{r}}$ and ${\mathbf{R}}$, we can find the gradient operators. We use a dummy function ${f}$ to give the gradient something to operate on. We’ll consider the ${x}$ component and use the chain rule (remember that ${\mathbf{r}_{1}}$ and ${\mathbf{r}_{2}}$ are independent vectors, each with 3 components, so there is a total of 6 independent position variables):

 $\displaystyle \frac{\partial f}{\partial r_{1_{x}}}$ $\displaystyle =$ $\displaystyle \frac{\partial f}{\partial r_{x}}\frac{\partial r_{x}}{\partial r_{1_{x}}}+\frac{\partial f}{\partial R_{x}}\frac{\partial R_{x}}{\partial r_{1_{x}}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial f}{\partial r_{x}}(1)+\frac{\partial f}{\partial R_{x}}\frac{m_{1}}{m_{1}+m_{2}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial f}{\partial r_{x}}+\frac{\partial f}{\partial R_{x}}\frac{\mu}{m_{2}}\ \ \ \ \ (8)$ $\displaystyle \frac{\partial f}{\partial r_{2_{x}}}$ $\displaystyle =$ $\displaystyle \frac{\partial f}{\partial r_{x}}(-1)+\frac{\partial f}{\partial R_{x}}\frac{m_{2}}{m_{1}+m_{2}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial f}{\partial r_{x}}+\frac{\partial f}{\partial R_{x}}\frac{\mu}{m_{1}} \ \ \ \ \ (10)$

The relations for the other two components are similar, so dropping the test function ${f}$ , we get for the gradients:

 $\displaystyle \nabla_{1}$ $\displaystyle =$ $\displaystyle \nabla_{r}+\frac{\mu}{m_{2}}\nabla_{R}\ \ \ \ \ (11)$ $\displaystyle \nabla_{2}$ $\displaystyle =$ $\displaystyle -\nabla_{r}+\frac{\mu}{m_{1}}\nabla_{R} \ \ \ \ \ (12)$

To get the Laplacian operators, we differentiate the ${x}$ component expressions above.

 $\displaystyle \frac{\partial^{2}f}{\partial r_{1_{x}}^{2}}$ $\displaystyle =$ $\displaystyle \left(\frac{\partial}{\partial r_{x}}+\frac{\mu}{m_{2}}\frac{\partial}{\partial R_{x}}\right)\left(\frac{\partial f}{\partial r_{x}}+\frac{\partial f}{\partial R_{x}}\frac{\mu}{m_{2}}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}f}{\partial r_{x}^{2}}+\frac{\partial f}{\partial r_{x}\partial R_{x}}\frac{2m_{1}}{m_{1}+m_{2}}+\frac{\partial^{2}f}{\partial R_{x}^{2}}\left(\frac{\mu}{m_{2}}\right)^{2}\ \ \ \ \ (14)$ $\displaystyle \frac{\partial^{2}f}{\partial r_{2_{x}}^{2}}$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}f}{\partial r_{x}^{2}}-\frac{\partial f}{\partial r_{x}\partial R_{x}}\frac{2m_{2}}{m_{1}+m_{2}}+\frac{\partial^{2}f}{\partial R_{x}^{2}}\left(\frac{\mu}{m_{1}}\right)^{2} \ \ \ \ \ (15)$

The combination of these two expressions that appears in the Schrödinger equation is, after cancelling terms and putting the remaining terms over common denominators:

 $\displaystyle -\frac{\hbar^{2}}{2m_{1}}\frac{\partial^{2}f}{\partial r_{1_{x}}^{2}}-\frac{\hbar^{2}}{2m_{2}}\frac{\partial^{2}f}{\partial r_{2_{x}}^{2}}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2(m_{1}+m_{2})}\frac{\partial^{2}f}{\partial R_{x}^{2}}-\frac{\hbar^{2}}{2\mu}\frac{\partial^{2}f}{\partial r_{x}^{2}} \ \ \ \ \ (16)$

The calculations for the other two components are similar, so the final result is:

$\displaystyle -\frac{\hbar^{2}}{2(m_{1}+m_{2})}\nabla_{R}^{2}\psi-\frac{\hbar^{2}}{2\mu}\nabla_{r}^{2}\psi+V(\mathbf{r})\psi=E\psi \ \ \ \ \ (17)$

We can now try the usual technique of separation of variables, so we try

$\displaystyle \psi=A\left(r\right)B\left(R\right) \ \ \ \ \ (18)$

We get, after substituting and dividing through by ${AB}$:

$\displaystyle -\frac{\hbar^{2}}{2(m_{1}+m_{2})B}\nabla_{R}^{2}B-\frac{\hbar^{2}}{2\mu A}\nabla_{r}^{2}A+V(\mathbf{r})=E \ \ \ \ \ (19)$

As usual, the terms involving each of the variables ${r}$ and ${R}$ must separately be equal to constants, so we get

 $\displaystyle -\frac{\hbar^{2}}{2(m_{1}+m_{2})}\nabla_{R}^{2}B$ $\displaystyle =$ $\displaystyle E_{R}B\ \ \ \ \ (20)$ $\displaystyle -\frac{\hbar^{2}}{2\mu}\nabla_{r}^{2}A+AV(\mathbf{r})$ $\displaystyle =$ $\displaystyle E_{r}A \ \ \ \ \ (21)$

The first equation is that of a free particle with mass ${m_{1}+m_{2}}$, while the second is that of a particle with mass ${\mu}$ moving in a potential ${V}$. Thus the system separates into one equation for a free particle with the total mass and a position at the centre of mass and another for a single particle with the reduced mass ${\mu}$. The total energy is ${E=E_{R}+E_{r}}$.

## 3 thoughts on “Schrödinger equation for 2 particles – separation of variables”

1. mosaic wolf

Now that makes it perfectly clear. I got the free particle confused with the other one moving within the potential. Thax for the simple explanation.