Daily Archives: Thu, 2 March 2017

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.