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 )} .