Two-Particle Systems

Consider a system consisting of two particles, mass $m_1$ and $m_2$, interacting via the potential $V(x_1-x_2)$ which only depends on the relative positions of the particles. According to Eqs. (419) and (426), the Hamiltonian of the system is written

$$H(x_1,x_2) = -\frac{\hbar^2}{2 m_1}\frac{\partial^2}{\part... ...^2}{2 m_2}\frac{\partial^2}{\partial x_2^{ 2}}+ V(x_1-x_2).$$ (437)

Let

$$x' = x_1-x_2$$ (438)

be the particles' relative position, and

$$X = \frac{m_1 x_1+m_2 x_2}{m_1+m_2}$$ (439)

the position of the center of mass. It is easily demonstrated that

$$\frac{\partial}{\partial x_1} = \frac{m_1}{m_1+m_2}\frac{\partial}{\partial X} + \frac{\partial}{\partial x'},$$ (440)

$$\frac{\partial}{\partial x_2} = \frac{m_2}{m_1+m_2}\frac{\partial}{\partial X} - \frac{\partial}{\partial x'}.$$ (441)

Hence, when expressed in terms of the new variables, $x'$ and $X$, the Hamiltonian becomes

$$H(x',X) = -\frac{\hbar^2}{2 M} \frac{\partial^2}{\partial ... ...\hbar^2}{2 \mu}\frac{\partial^2}{\partial x'^{ 2}} + V(x'),$$ (442)

where

$$M = m_1+ m_2$$ (443)

is the total mass of the system, and

$$\mu = \frac{m_1 m_2}{m_1+m_2}$$ (444)

the so-called reduced mass. Note that the total momentum of the system can be written

$$P= -{\rm i} \hbar\left(\frac{\partial}{\partial x_1} + \fra... ...l x_2}\right) = -{\rm i} \hbar \frac{\partial}{\partial X}.$$ (445)

The fact that the Hamiltonian (442) is separable when expressed in terms of the new coordinates [i.e., $H(x',X) = H_{x'}(x') + H_X(X)]$ suggests, by analogy with the analysis in the previous section, that the wavefunction can be factorized: i.e.,

$$\psi(x_1,x_2,t) = \psi_{x'}(x',t) \psi_X(X,t).$$ (446)

Hence, the time-dependent Schrödinger equation (423) also factorizes to give

$${\rm i} \hbar \frac{\partial\psi_{x'}}{\partial t} = -\fr... ...c{\partial^2\psi_{x'}}{\partial x'^{ 2}} + V(x') \psi_{x'},$$ (447)

and

$${\rm i} \hbar \frac{\partial\psi_X}{\partial t} = -\frac{\hbar^2}{2 M}\frac{\partial^2\psi_X}{\partial X^2}.$$ (448)

The above equation can be solved to give

$$\psi_X(X,t) = \psi_{0} {\rm e}^{ {\rm i} (P' X/\hbar - E' t/\hbar)},$$ (449)

where $\psi_0$, $P'$, and $E' = P'^{ 2}/2 M$ are constants. It is clear, from Eqs. (445), (446), and (449), that the total momentum of the system takes the constant value $P'$: i.e., momentum is conserved.

Suppose that we work in the centre of mass frame of the system, which is characterized by $P'=0$. It follows that $\psi_X=\psi_0$. In this case, we can write the wavefunction of the system in the form $\psi(x_1,x_2,t) = \psi_{x'}(x',t) \psi_0\equiv \psi(x_1-x_2,t)$, where

$${\rm i} \hbar \frac{\partial\psi}{\partial t} = -\frac{\h... ... \mu} \frac{\partial^2\psi}{\partial x^{ 2}} + V(x) \psi.$$ (450)

In other words, in the center of mass frame, two particles of mass $m_1$ and $m_2$, moving in the potential $V(x_1-x_2)$, are equivalent to a single particle of mass $\mu$, moving in the potential $V(x)$, where $x=x_1-x_2$. This is a familiar result from classical dynamics.