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 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 $\Psi(x, t)$ isA particle is subjected to a potential
$V(x) = \begin{cases} \infty, & x \le 0 \\ V_0, & a \le x \le b \\ 0, & \text{elsewhere} \end{cases}$
Here, $a > 0$ and $b > a$. If the energy of the particle $E < V_0$, which one of the following schematics is a valid quantum mechanical wavefunction ($\Psi$) for the system?- The wavefunction for a particle is given by the form $e^{-(iax+\beta)}$, where $a$ and $\beta$ are real constants. In which one of the following potentials $V(x)$, the particle is moving?
A particle of mass $m$ is moving in the potential
$V(x) = \begin{cases} V_0 + \frac{1}{2}m\omega_0^2x^2, & x > 0, \\ \infty, & x \le 0, \end{cases}$
Figures P, Q, R and S show different combinations of the values of $\omega_0$ and $V_0$.
$E_j^{(P)}, E_j^{(Q)}, E_j^{(R)}$ and $E_j^{(S)}$ with $j = 0, 1, 2, ...$, are the eigen-energies of the $j$-th level for the potentials shown in figures P, Q, R and S, respectively. Which of the statement is/are true?- 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) ?
