# Decoupling the two-particle Hamiltonian

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.1.3.

Shankar shows that, for a two-particle system, the state vector ${\left|\psi\right\rangle }$ is an element of the direct product space ${\mathbb{V}_{1\otimes2}}$. Its evolution in time is determined by the Schrödinger equation, as usual, so that

$\displaystyle i\hbar\left|\dot{\psi}\right\rangle =H\left|\psi\right\rangle =\left[\frac{P_{1}^{2}}{2m_{1}}+\frac{P_{2}^{2}}{2m_{2}}+V\left(X_{1},X_{2}\right)\right]\left|\psi\right\rangle \ \ \ \ \ (1)$

The method by which this equation can be solved (if it can be solved, that is) depends on the form of the potential ${V}$. If the two particles interact only with some external potential, and not with each other, then ${V}$ is composed of a sum of terms, each of which depends only on ${X_{1}}$ or ${X_{2}}$, but not on both. In such cases, we can split ${H}$ into two parts, one of which (${H_{1}}$) depends only on operators pertaining to particle 1 and the other (${H_{2}}$) on operators pertaining to particle 2. If the eigenvalues (allowed energies) of particle ${i}$ are given by ${E_{i}}$, then the stationary states are direct products of the corresponding single particle eigenstates. That is, in general

$\displaystyle H\left|E\right\rangle =\left(H_{1}+H_{2}\right)\left|E_{1}\right\rangle \otimes\left|E_{2}\right\rangle =\left(E_{1}+E_{2}\right)\left|E_{1}\right\rangle \otimes\left|E_{2}\right\rangle =E\left|E\right\rangle \ \ \ \ \ (2)$

Thus the two-particle state ${\left|E\right\rangle =\left|E_{1}\right\rangle \otimes\left|E_{2}\right\rangle }$. Since a stationary state ${\left|E_{i}\right\rangle }$ evolves in time according to

$\displaystyle \left|\psi_{i}\left(t\right)\right\rangle =\left|E_{i}\right\rangle e^{-iE_{i}t/\hbar} \ \ \ \ \ (3)$

the compound two-particle state evolves according to

 $\displaystyle \left|\psi\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle e^{-iE_{1}t/\hbar}\left|E_{1}\right\rangle \otimes e^{-iE_{2}t/\hbar}\left|E_{2}\right\rangle \ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\left(E_{1}+E_{2}\right)t/\hbar}\left|E\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-iEt/\hbar}\left|E\right\rangle \ \ \ \ \ (6)$

In this case, the two particles are essentially independent of each other, and the compound state is just the product of the two separate one-particle states.

If ${H}$ is not separable, which will occur if ${V}$ contains terms involving both ${X_{1}}$ and ${X_{2}}$ in the same term, we cannot, in general, reduce the system to the product of two one-particle systems. There are a couple of instances, however, where such a reduction can be done.

The first instance is if the potential is a function of ${x_{2}-x_{1}}$ only, in other words, that the interaction between the particles depends only on the distance between them. Shankar shows that in this case we can transform the system to that of a reduced mass ${\mu=m_{1}m_{2}/\left(m_{1}+m_{2}\right)}$ and a centre of mass ${M=m_{1}+m_{2}}$. We’ve already seen this problem solved by means of separation of variables. The result is that the state vector is the product of a vector for a free particle of mass ${M}$ and of a vector of a particle with reduced mass ${\mu}$ moving in the potential ${V}$.

Another case where we can decouple the Hamiltonian is in a system of harmonic oscillators. We’ve already seen this system solved for two masses in classical mechanics using diagonalization of the matrix describing the equations of motion. The classical Hamiltonian is

$\displaystyle H=\frac{p_{1}^{2}}{2m}+\frac{p_{2}^{2}}{2m}+\frac{m\omega^{2}}{2}\left[x_{1}^{2}+x_{2}^{2}+\left(x_{1}-x_{2}\right)^{2}\right] \ \ \ \ \ (7)$

The earlier solution involved introducing normal coordinates

 $\displaystyle x_{I}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{1}+x_{2}\right)\ \ \ \ \ (8)$ $\displaystyle x_{II}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{1}-x_{2}\right) \ \ \ \ \ (9)$

and corresponding momenta

 $\displaystyle p_{I}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{1}+p_{2}\right)\ \ \ \ \ (10)$ $\displaystyle p_{II}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{1}-p_{2}\right) \ \ \ \ \ (11)$

These normal coordinates are canonical as we can verify by calculating the Poisson brackets. For example

 $\displaystyle \left\{ x_{I},p_{I}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial x_{I}}{\partial x_{i}}\frac{\partial p_{I}}{\partial p_{i}}-\frac{\partial x_{I}}{\partial p_{i}}\frac{\partial p_{I}}{\partial x_{i}}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (13)$ $\displaystyle \left\{ x_{I},x_{II}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial x_{I}}{\partial x_{i}}\frac{\partial x_{II}}{\partial p_{i}}-\frac{\partial x_{I}}{\partial p_{i}}\frac{\partial x_{II}}{\partial x_{i}}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (15)$

and so on, with the general result

 $\displaystyle \left\{ x_{i},p_{j}\right\}$ $\displaystyle =$ $\displaystyle \delta_{ij}\ \ \ \ \ (16)$ $\displaystyle \left\{ x_{i},x_{j}\right\}$ $\displaystyle =$ $\displaystyle \left\{ p_{i},p_{j}\right\} =0 \ \ \ \ \ (17)$

We can invert the transformation to get

 $\displaystyle x_{1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{I}+x_{II}\right)\ \ \ \ \ (18)$ $\displaystyle x_{2}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{I}-x_{II}\right) \ \ \ \ \ (19)$

and

 $\displaystyle p_{1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{I}+p_{II}\right)\ \ \ \ \ (20)$ $\displaystyle p_{2}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{I}-p_{II}\right) \ \ \ \ \ (21)$

Inserting these into 7 we get

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \frac{1}{4m}\left[\left(p_{I}+p_{II}\right)^{2}+\left(p_{I}-p_{II}\right)^{2}\right]+\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{m\omega^{2}}{4}\left[\left(x_{I}+x_{II}\right)^{2}+\left(x_{I}-x_{II}\right)^{2}+x_{II}^{2}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p_{I}^{2}}{2m}+\frac{p_{II}^{2}}{2m}+\frac{m\omega^{2}}{2}\left(x_{I}^{2}+2x_{II}^{2}\right) \ \ \ \ \ (24)$

We can now subsitute the usual quantum mechanical operators to get the quantum Hamiltonian:

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(P_{I}^{2}+P_{II}^{2}\right)+\frac{m\omega^{2}}{2}\left(X_{I}^{2}+2X_{II}^{2}\right) \ \ \ \ \ (25)$

In the coordinate basis, this is

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x_{I}^{2}}+\frac{\partial^{2}}{\partial x_{II}^{2}}\right)+\frac{m\omega^{2}}{2}\left(x_{I}^{2}+2x_{II}^{2}\right) \ \ \ \ \ (26)$

The Hamiltonian is now decoupled and can be solved by separation of variables.

We could have arrived at this result by starting with 7 and promoting ${x_{i}}$ and ${p_{i}}$ to quantum operators directly, then made the substitution to normal coordinates. We would then start with

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x_{1}^{2}}+\frac{\partial^{2}}{\partial x_{2}^{2}}\right)+\frac{m\omega^{2}}{2}\left[x_{1}^{2}+x_{2}^{2}+\left(x_{1}-x_{2}\right)^{2}\right] \ \ \ \ \ (27)$

The potential term on the right transforms the same way as before, so we get

$\displaystyle \frac{m\omega^{2}}{2}\left[x_{1}^{2}+x_{2}^{2}+\left(x_{1}-x_{2}\right)^{2}\right]\rightarrow\frac{m\omega^{2}}{2}\left(x_{I}^{2}+2x_{II}^{2}\right) \ \ \ \ \ (28)$

To transform the two derivatives, we need to use the chain rule a couple of times. To get the first derivatives:

 $\displaystyle \frac{\partial\psi}{\partial x_{1}}$ $\displaystyle =$ $\displaystyle \frac{\partial\psi}{\partial x_{I}}\frac{\partial x_{I}}{\partial x_{1}}+\frac{\partial\psi}{\partial x_{II}}\frac{\partial x_{II}}{\partial x_{1}}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\frac{\partial\psi}{\partial x_{I}}+\frac{\partial\psi}{\partial x_{II}}\right)\ \ \ \ \ (30)$ $\displaystyle \frac{\partial\psi}{\partial x_{2}}$ $\displaystyle =$ $\displaystyle \frac{\partial\psi}{\partial x_{I}}\frac{\partial x_{I}}{\partial x_{2}}+\frac{\partial\psi}{\partial x_{II}}\frac{\partial x_{II}}{\partial x_{2}}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\frac{\partial\psi}{\partial x_{I}}-\frac{\partial\psi}{\partial x_{II}}\right) \ \ \ \ \ (32)$

Now the second derivatives:

 $\displaystyle \frac{\partial^{2}\psi}{\partial x_{1}^{2}}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial x_{I}}\left(\frac{\partial\psi}{\partial x_{1}}\right)\frac{\partial x_{I}}{\partial x_{1}}+\frac{\partial}{\partial x_{II}}\left(\frac{\partial\psi}{\partial x_{1}}\right)\frac{\partial x_{II}}{\partial x_{1}}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\frac{\partial}{\partial x_{I}}\left(\frac{\partial\psi}{\partial x_{I}}+\frac{\partial\psi}{\partial x_{II}}\right)+\frac{\partial}{\partial x_{II}}\left(\frac{\partial\psi}{\partial x_{I}}+\frac{\partial\psi}{\partial x_{II}}\right)\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\frac{\partial^{2}\psi}{\partial x_{I}^{2}}+2\frac{\partial^{2}\psi}{\partial x_{I}\partial x_{II}}+\frac{\partial^{2}\psi}{\partial x_{II}^{2}}\right]\ \ \ \ \ (35)$ $\displaystyle \frac{\partial^{2}\psi}{\partial x_{2}^{2}}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial x_{I}}\left(\frac{\partial\psi}{\partial x_{2}}\right)\frac{\partial x_{I}}{\partial x_{1}}+\frac{\partial}{\partial x_{II}}\left(\frac{\partial\psi}{\partial x_{2}}\right)\frac{\partial x_{II}}{\partial x_{1}}\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\frac{\partial}{\partial x_{I}}\left(\frac{\partial\psi}{\partial x_{I}}-\frac{\partial\psi}{\partial x_{II}}\right)-\frac{\partial}{\partial x_{II}}\left(\frac{\partial\psi}{\partial x_{I}}-\frac{\partial\psi}{\partial x_{II}}\right)\right]\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\frac{\partial^{2}\psi}{\partial x_{I}^{2}}-2\frac{\partial^{2}\psi}{\partial x_{I}\partial x_{II}}+\frac{\partial^{2}\psi}{\partial x_{II}^{2}}\right] \ \ \ \ \ (38)$

Combining the two derivatives, we get

$\displaystyle \frac{\partial^{2}\psi}{\partial x_{1}^{2}}+\frac{\partial^{2}\psi}{\partial x_{2}^{2}}=\frac{\partial^{2}\psi}{\partial x_{I}^{2}}+\frac{\partial^{2}\psi}{\partial x_{II}^{2}} \ \ \ \ \ (39)$

Inserting this, together with 28, into 27 we get 26 again.