Question 705025

SCQHARD

The Hamiltonian for a one-dimensional simple harmonic oscillator is given by $H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$ . The harmonic oscillator is in the state $|\psi\rangle = \frac{1}{\sqrt{1+\lambda^2}}(|1\rangle + \lambda e^{i\theta}|2\rangle)$ , where $|1\rangle$ and $|2\rangle$ are the normalised first and second excited states of the oscillator and $\lambda, \theta$ are positive real constants. If the expectation value $\langle\psi|x|\psi\rangle = \beta \sqrt{\frac{\hbar}{m\omega}}$ , the value of $\beta$ is:

(A)

$\frac{1}{\sqrt{2}(1+\lambda^2)}$

(B)

$\frac{\sqrt{2}\lambda \cos\theta}{1+\lambda^2}$

(C)

$\frac{2\lambda \cos\theta}{1+\lambda^2}$

(D)

$\frac{\lambda^2 \cos\theta}{1+\lambda^2}$

Detailed Solution

Using the operator $x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)$ , where $a|n\rangle = \sqrt{n}|n-1\rangle$ and $a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$ . For state $|\psi\rangle = C(|1\rangle + \lambda e^{i\theta}|2\rangle)$ with $C = (1+\lambda^2)^{-1/2}$ , calculating $\langle\psi|x|\psi\rangle$ results in $\beta = \frac{2\lambda \cos\theta}{1+\lambda^2}$ .

Question 705025

Detailed Solution

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