A one dimensional harmonic oscillator is in the superposition of number states, $ |n\rangle $, given by $ |\Psi\rangle = \frac{1}{2} |2\rangle + \frac{\sqrt{3}}{2} |3\rangle $. The average energy of the oscillator in the given state is ________ $ \hbar \omega $.
To find the average energy of a quantum harmonic oscillator in the state $|\Psi\rangle=\frac{1}{2}|2\rangle+\frac{\sqrt{3}}{2}|3\rangle$, we start by using the formula for the expectation value of energy, given by:
$\langle E \rangle = \langle \Psi | \hat{H} | \Psi \rangle$, where $\hat{H}$ is the Hamiltonian operator for the harmonic oscillator.
The energy eigenvalues for the number states $|n\rangle$ are $E_n=(n+\frac{1}{2})\hbar \omega$.
Thus, $E_2=(2+\frac{1}{2})\hbar\omega=\frac{5}{2}\hbar\omega$ and $E_3=(3+\frac{1}{2})\hbar\omega=\frac{7}{2}\hbar\omega$.
The average energy is then evaluated as:
$\langle E \rangle = \left(\frac{1}{2}\right)^2 \langle 2 | \hat{H} | 2 \rangle + \left(\frac{\sqrt{3}}{2}\right)^2 \langle 3 | \hat{H} | 3 \rangle$.
Simplifying further:
$\langle E \rangle = \frac{1}{4}E_2 + \frac{3}{4}E_3$.
Substituting the energy values, we have:
$\langle E \rangle = \frac{1}{4} \times \frac{5}{2}\hbar\omega + \frac{3}{4} \times \frac{7}{2}\hbar\omega$.
Calculating each term separately:
$\frac{1}{4} \times \frac{5}{2} = \frac{5}{8}$ and $\frac{3}{4} \times \frac{7}{2} = \frac{21}{8}$.
Adding these results, $\langle E \rangle = \frac{5}{8}\hbar\omega + \frac{21}{8}\hbar\omega = \frac{26}{8}\hbar\omega = \frac{13}{4}\hbar\omega$.
Therefore, the average energy of the oscillator in the given state is $3.25\hbar\omega$.
This value falls within the provided range of 3.2 to 3.3, confirming its correctness.
Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator
- The energy $E$ and degeneracy $d$ of the second excited state of a three-dimensional, isotropic quantum harmonic oscillator with angular frequency $\omega$ are
- A particle of mass $ m $ is in a potential $ V(x) = \frac{1}{2}m\omega^2x^2 $ for $ x > 0 $ and $ V(x) = \infty $ for $ x \leq 0 $, where $ \omega $ is the angular frequency. The ratio of the energies corresponding to the lowest energy level to the next higher level is
- Young's double slit experiment is performed using a beam of $C_{60}$ (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with $C_{70}$ molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4th bright fringe for $C_{60}$ will correspond to the $n^{th}$ bright fringe for $C_{70}$. What is the value of $n$ (rounded off to the nearest integer) ?
- Consider a particle in a two dimensional infinite square well potential of side $L$, with $0 \le x \le L$ and $0 \le y \le L$. The wavefunction of the particle is zero only along the line $y = \frac{L}{2}$, apart from the boundaries of the well. If the energy of the particle in this state is $E$, what is the energy of the ground state?
The wavefunction of a particle in an infinite one-dimensional potential well at time $t$ is
$\Psi(x, t) = \sqrt{\frac{2}{3}} e^{-iE_1t/\hbar}\psi_1(x) + \frac{1}{\sqrt{6}} e^{i\pi/6}e^{-iE_2t/\hbar}\psi_2(x) + \frac{1}{\sqrt{6}} e^{i\pi/4}e^{-iE_3t/\hbar}\psi_3(x)$
where $\psi_1, \psi_2$ and $\psi_3$ are the normalized ground state, the normalized first excited state and the normalized second excited state, respectively. $E_1, E_2$ and $E_3$ are the eigen-energies corresponding to $\psi_1, \psi_2$ and $\psi_3$, respectively. The expectation value of energy of the particle in state $\Psi(x, t)$ is
