Quantum Mechanics for Two Particles

We can know the state of two particles at the same time. The positions and momenta of particle 2 commute with the positions and momenta of particle 1.

\begin{displaymath}\bgroup\color{black} [x_1,x_2]=[p_1,p_2]=[x_1,p_2]=[x_2,p_1]=0 \egroup\end{displaymath}

The kinetic energy terms in the Hamiltonian are independent. There may be an interaction between the two particles in the potential. The Hamiltonian for two particles can be easily written.

\begin{displaymath}\bgroup\color{black}H={p_1^2\over 2m_1}+{p_2^2\over 2m_2}+V(x_1,x_2) \egroup\end{displaymath}

Often, the potential will only depend on the difference in the positions of the two particles.

\begin{displaymath}\bgroup\color{black}V(x_1,x_2)=V(x_1-x_2) \egroup\end{displaymath}

This means that the overall Hamiltonian has a translational symmetry. Lets examine an infinitesimal translation in \bgroup\color{black}$x$\egroup. The original Schrödinger equation

\begin{displaymath}\bgroup\color{black}H\psi(x_1,x_2)=E\psi(x_1,x_2) \egroup\end{displaymath}

transforms to

\begin{displaymath}\bgroup\color{black}H\psi(x_1+dx,x_2+dx)=E\psi(x_1+dx,x_2+dx) \egroup\end{displaymath}

which can be Taylor expanded

\begin{displaymath}\bgroup\color{black}H\left( \psi(x_1,x_2)+{\partial \psi\over...
...rtial x_1}dx+{\partial \psi\over\partial x_2}dx\right) .\egroup\end{displaymath}

We can write the derivatives in terms of the total momentum operator.

\begin{displaymath}\bgroup\color{black}p=p_1+p_2={\hbar\over i}\left({\partial\over\partial x_1}+{\partial\over\partial x_2}\right) \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}H\psi(x_1,x_2)+{i\over\hbar}Hp\;\psi(x_1,x_2)dx=E\psi(x_1,x_2)+{i\over\hbar}Ep\;\psi(x_1,x_2)dx \egroup\end{displaymath}

Subtract of the initial Schrödinger equation and commute \bgroup\color{black}$E$\egroup through \bgroup\color{black}$p$\egroup.

\begin{displaymath}\bgroup\color{black}Hp\;\psi(x_1,x_2)=Ep\;\psi(x_1,x_2)=pH\psi(x_1,x_2) \egroup\end{displaymath}

We have proven that

\begin{displaymath}\bgroup\color{black}[H,p]=0 \egroup\end{displaymath}

if the Hamiltonian has translational symmetry. The momentum is a constant of the motion. Momentum is conserved. We can have simultaneous eigenfunctions of the total momentum and of energy.

Jim Branson 2013-04-22