Question 4.P.16: Find the energy levels and the wave functions of two harmoni...

This problem reduces to finding the eigenvalues for the Hamiltonian

$=\frac{1}{2m_{1}}\hat{P}^{2}_{1} +\frac{1}{2}m_{1}\omega ^{2} \hat{X}^{2}_{1}+\frac{1}{2m_{2}}\hat{P}^{2}_{2}+\frac{1}{2}m_{2}\omega ^{2} \hat{X}^{2}_{2}+\frac{1}{2}K(\hat{X}_{1}-\hat{X}_{2} )^{2}$.            (4.325)

This is a two-particle problem. As in classical mechanics, it is more convenient to describe the dynamics of a two-particle system in terms of the center of mass (CM) and relative motions. For this, let us introduce the following operators:

$\hat{P}=\hat{p}_{1}+\hat{p}_{2}$,        $\hat{X} =\frac{m_{1}\hat{x}_{1}+ m_{2}\hat{x}_{2}}{M}$,                 (4.326)

$\hat{p}=\frac{m_{2}\hat{p}_{1}- m_{1}\hat{p}_{2}}{M}$,        $\hat{x}=\hat{x}_{1}-\hat{x}_{2}$,              (4.327)

where $M=m_{1}+m_{2}$ and $\mu =m_{1}m_{2} /(m_{1}+m_{2})$ is the reduced mass; $\hat{P}$ and $\hat{X}$ pertain to the CM; $\hat{p}$ and $\hat{x}$ pertain to the relative motion. These relations lead to

$\hat{p}_{1} =\frac{m_{1}}{M}\hat{P}+\hat{p}$,     $\hat{p}_{2}=\frac{m_{2}}{M}\hat{P}+\hat{p}$,           (3.328)

$\hat{x}_{1} =\frac{m_{2}}{M}\hat{x}+\hat{X}$,     $\hat{x}_{2} =-\frac{m_{1}}{M}\hat{x}+\hat{X}$.        (4.329)

Note that the sets (X, P) and (x, p) are conjugate variables separately: $[\hat{X},\hat{P}]=i\hbar ,[\hat{x},\hat{p}]=i\hbar =[\hat{X} ,\hat{p}]=[\hat{x},\hat{P}]=0$. Taking $\hat{p}_{1}, \hat{p}_{2},\hat{x}_{1},$ and $\hat{x}_{2}$ of (4.328) and (4.329) and inserting them into (4.325), we obtain

$=\hat{H}_{CM}+\hat{H}_{rel}$,                         (4.330)

where

$\hat{H}_{CM}=\frac{1}{2M}\hat{P}^{2} +\frac{1}{2}M\omega ^{2}\hat{X}^{2}$,     $\hat{H}_{rel}=\frac{1}{2\mu } \hat {p}^{2} +\frac{1}{2}\mu \Omega ^{2}\hat{x}^{2}$,        (4.331)

with $\Omega ^{2}=\omega ^{2}+k/\mu$. We have thus reduced the Hamiltonian of these two coupled harmonic oscillators to the sum of two independent harmonic oscillators, one with frequency ω and mass M and the other of mass μ and frequency $\Omega =\sqrt{\omega ^{2}+k/\mu}$. That is, by introducing the
CM and relative motion variables, we have managed to eliminate the coupled term from the Hamiltonian.
The energy levels of this two-oscillator system can be inferred at once from the suggestive Hamiltonians of (4.331):

$E_{n_{1}n_{2}} =\hbar \omega \left(n_{1}+\frac{1}{2}\right) +\hbar \Omega \left(n_{2}+\frac{1}{2}\right)$.          (4.332)

The states of this two-particle system are given by the product of the two states $|N〉=|n_{1}〉 |n_{2}〉$; hence the total wave function, $\psi _{n} (X,x)$, is equal to the product of the center of mass wave function, $\psi _{n_{1}} (X)$, and the wave function of the relative motion, $\psi _{n_{2}} (x):\psi _{n} (X,x)=\psi _{n_{1}} (X)\psi _{n_{2}} (x)$.

Note that both of these wave functions are harmonic oscillator functions whose forms can be found in (4.172):

$\psi _{n} (X,x)=\frac{1}{\sqrt{\pi }\sqrt{2^{n_{1}}2^{n_{2}} n_{1}!n_{2}!x_{0_{1} }x_{0_{2} } } } e^{-X^{2}/2x^{2}_{0_{1} } }e^{-x^{2}/2x^{2}_{0_{2} } } H_{n_{1}} \left(\frac{X}{x_{0_{1} } } \right) H_{n_{2}} \left(\frac{X}{x_{0_{2} } } \right)$,       (4.333)

where $n=(n_{1},n_{2}),x_{0_{1} }=\sqrt{\hbar /(M\omega )}$, and $x_{0_{2} }=\sqrt{\hbar /(\mu \Omega )}$.