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GATE 2026 PH – Question 39

Quantum Mechanics · one dimensional potentials: step potential, finite rectangular well, tunnelling from a potential barrier, particle in 1,2,3-dimensional box, particle in single and double delta function potentials, 1,2,3 dimensional harmonic oscillator: concept of degeneracy · 2 marks · Multiple choice

The Hamiltonian for a quantum particle of mass $m$ is given below, where $\omega<\Omega$. The Schrödinger equation for this system can be solved exactly using the orthogonal transformations: $x=(x_1-x_2)/\sqrt2$ and $y=(x_1+x_2)/\sqrt2$. $$H=-\frac{\hbar^2}{2m}\left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right]+\frac12m\Omega^2(x^2+y^2)+m\omega^2xy$$ The ground state energy of this system is

  1. $\frac\hbar2[\sqrt{\Omega^2-\omega^2}+\sqrt{\Omega^2+\omega^2}]$
  2. $\hbar[\sqrt{\Omega^2-\omega^2}+\sqrt{\Omega^2+\omega^2}]$
  3. $\frac\hbar2[\sqrt{\Omega^2+\Omega\omega}+\sqrt{\Omega^2-\Omega\omega}]$
  4. $\hbar[\sqrt{\Omega^2+\Omega\omega}-\sqrt{\Omega^2-\Omega\omega}]$

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Correct answer: (A) $\frac\hbar2[\sqrt{\Omega^2-\omega^2}+\sqrt{\Omega^2+\omega^2}]$

Explanation

The Hamiltonian has the cross term $m\omega^2xy$. Go to normal coordinates by the rotation given: $x=(x_1-x_2)/\sqrt2$ and $y=(x_1+x_2)/\sqrt2$.

**Potential energy in the new coordinates.** With $x^2+y^2=x_1^2+x_2^2$ and $xy=\dfrac{x_2^2-x_1^2}{2}$:
$$V=\tfrac12m\Omega^2(x_1^2+x_2^2)+\tfrac12m\omega^2(x_2^2-x_1^2)=\tfrac12m(\Omega^2-\omega^2)x_1^2+\tfrac12m(\Omega^2+\omega^2)x_2^2 .$$

The kinetic energy is unchanged by an orthogonal rotation, so the problem separates into two independent harmonic oscillators with angular frequencies
$$\omega_1=\sqrt{\Omega^2-\omega^2},\qquad\omega_2=\sqrt{\Omega^2+\omega^2}.$$

**Ground-state energy** = the sum of the two zero-point energies:
$$E_0=\tfrac12\hbar\omega_1+\tfrac12\hbar\omega_2=\frac\hbar2\left[\sqrt{\Omega^2-\omega^2}+\sqrt{\Omega^2+\omega^2}\right]\quad(\text{option A}).$$