Question

The Hamiltonian for a quantum particle of mass \(m\) is given below, where \(\omega < \Omega\). The Schrodinger equation for this system can be solved exactly using the orthogonal transformation \(x = \dfrac{x_1 - x_2}{\sqrt2}\) and \(y = \dfrac{x_1+x_2}{\sqrt2}\).
\[ H = -\dfrac{\hbar^2}{2m}\left[\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2}\right] + \dfrac12 m\Omega^2(x^2+y^2) + m\omega^2 xy \]The ground state energy of this system is

• A. \(\dfrac{\hbar}{2}\Big[\sqrt{\Omega^2-\omega^2} + \sqrt{\Omega^2+\omega^2}\Big]\)
• B. \(\hbar\Big[\sqrt{\Omega^2-\omega^2} + \sqrt{\Omega^2+\omega^2}\Big]\)
• C. \(\dfrac{\hbar}{2}\Big[\sqrt{\Omega^2+\Omega\omega} + \sqrt{\Omega^2-\Omega\omega}\Big]\)
• D. \(\hbar\Big[\sqrt{\Omega^2+\Omega\omega} - \sqrt{\Omega^2-\Omega\omega}\Big]\)

Hint:

Rewrite \(x^2+y^2\) and \(xy\) in terms of \(x_1,x_2\) using the given rotation, then group terms to see two separate uncoupled oscillators.

Answer 1

The Correct Option is A

Step 1: Understand the coupling in the Hamiltonian.
The Hamiltonian describes a particle moving in two dimensions with an isotropic potential \(\tfrac12 m\Omega^2(x^2+y^2)\) plus an extra cross term \(m\omega^2 xy\) that mixes \(x\) and \(y\). Because of this cross term, \(x\) and \(y\) are not independent oscillators as written, so we cannot just read off two separate frequencies yet. The question tells us that rotating to a new pair of coordinates \(x_1, x_2\) removes the mixing.

Step 2: Rewrite the potential in the new coordinates.
We are given \(x = \dfrac{x_1-x_2}{\sqrt2}\) and \(y = \dfrac{x_1+x_2}{\sqrt2}\). First find \(x^2+y^2\):
\[ x^2+y^2 = \frac{(x_1-x_2)^2 + (x_1+x_2)^2}{2} = \frac{2x_1^2+2x_2^2}{2} = x_1^2+x_2^2 \]
This is expected because a rotation of coordinates never changes the total distance from the origin. Next find \(xy\):
\[ xy = \frac{(x_1-x_2)(x_1+x_2)}{2} = \frac{x_1^2-x_2^2}{2} \]

Step 3: Rewrite the kinetic term.
A rotation between two coordinate pairs is an orthogonal transformation, and the sum of second derivatives (the Laplacian) is unchanged by any orthogonal (rotation) transformation. So
\[ \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial x_2^2} \]

Step 4: Assemble the Hamiltonian in the new coordinates.
\[ H = -\frac{\hbar^2}{2m}\left[\frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial x_2^2}\right] + \frac12 m\Omega^2(x_1^2+x_2^2) + m\omega^2\left(\frac{x_1^2-x_2^2}{2}\right) \]
Group the \(x_1\) terms and the \(x_2\) terms separately:
\[ H = \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_1^2} + \frac12 m(\Omega^2+\omega^2)x_1^2\right] + \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_2^2} + \frac12 m(\Omega^2-\omega^2)x_2^2\right] \]
This is now exactly two independent, uncoupled simple harmonic oscillators.

Step 5: Read off the two frequencies and the ground state energy.
A one-dimensional oscillator \(-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}+\tfrac12 m\omega_0^2 x^2\) has frequency \(\omega_0\) and ground state energy \(\tfrac12\hbar\omega_0\). Comparing, the two frequencies here are \(\omega_1 = \sqrt{\Omega^2+\omega^2}\) (from the \(x_1\) oscillator) and \(\omega_2 = \sqrt{\Omega^2-\omega^2}\) (from the \(x_2\) oscillator); the condition \(\omega < \Omega\) given in the question keeps \(\omega_2\) real. The total ground state energy of two independent oscillators just adds:
\[ E_0 = \frac12\hbar\omega_1 + \frac12\hbar\omega_2 = \frac{\hbar}{2}\Big[\sqrt{\Omega^2+\omega^2}+\sqrt{\Omega^2-\omega^2}\Big] \]

Step 6: Why the other options are wrong.
Option (B) has the right square-root terms but is missing the \(\tfrac12\) factor that comes from adding two ground state energies of \(\tfrac12\hbar\omega\) each. Options (C) and (D) mix up the frequencies with \(\Omega\omega\) products under the square root instead of \(\omega^2\), which does not come from this cross term at all.

Final Answer:
The ground state energy is \(\dfrac{\hbar}{2}\big[\sqrt{\Omega^2-\omega^2}+\sqrt{\Omega^2+\omega^2}\big]\), option (A).\[ \boxed{E_0 = \frac{\hbar}{2}\left[\sqrt{\Omega^2-\omega^2}+\sqrt{\Omega^2+\omega^2}\right]} \]