In a one-dimensional harmonic oscillator, $\phi_0$, $\phi_1$ and $\phi_2$ are respectively the ground, first and the second excited states. These three states are normalized and are orthogonal to one another. $\Psi_1$ and $\Psi_2$ are two states defined by $$\Psi_1 = \phi_0 - 2\phi_1 + 3\phi_2$$ $$\Psi_2 = \phi_0 - \phi_1 + \alpha\phi_2$$ where $\alpha$ is a constant.
Harmonic Oscillator Energy Expectation Value Calculation
The problem requires calculating the expectation value of energy for the quantum state $\Psi_2$ in a one-dimensional harmonic oscillator. The system's energy eigenstates $\phi_0, \phi_1, \phi_2$ are given as normalized and orthogonal, with corresponding energy eigenvalues $E_n = (n + 1/2)\hbar\omega$.
Understanding Energy Expectation Value
When a quantum state is represented as a linear combination of energy eigenstates, $|\Psi\rangle = \sum_n c_n \phi_n$, where $\phi_n$ are normalized and orthogonal, the expectation value of the energy is calculated using the formula:
$ \langle E \rangle = \frac{\sum_n |c_n|^2 E_n}{\sum_n |c_n|^2} $
Determining the Constant $\alpha$
The value of the constant $\alpha$ is specified as being determined from Q. 52. A common method to determine such constants in quantum mechanics problems is by imposing an orthogonality condition between related states. We assume $\Psi_1$ and $\Psi_2$ are orthogonal, i.e., $\langle \Psi_1 | \Psi_2 \rangle = 0$.
The given states are:
- $\Psi_1 = \phi_0 - 2\phi_1 + 3\phi_2$
- $\Psi_2 = \phi_0 - \phi_1 + \alpha\phi_2$
The orthogonality condition $\langle \Psi_1 | \Psi_2 \rangle$ is evaluated as:
$ \langle \Psi_1 | \Psi_2 \rangle = \langle (\phi_0 - 2\phi_1 + 3\phi_2) | (\phi_0 - \phi_1 + \alpha\phi_2) \rangle $
Using the property $\langle \phi_i | \phi_j \rangle = \delta_{ij}$ for normalized and orthogonal states:
$ \langle \Psi_1 | \Psi_2 \rangle = \langle \phi_0|\phi_0\rangle - \langle \phi_0|\phi_1\rangle + \alpha\langle \phi_0|\phi_2\rangle - 2\langle \phi_1|\phi_0\rangle + 2\langle \phi_1|\phi_1\rangle - 2\alpha\langle \phi_1|\phi_2\rangle + 3\langle \phi_2|\phi_0\rangle - 3\langle \phi_2|\phi_1\rangle + 3\alpha\langle \phi_2|\phi_2\rangle $
$ \langle \Psi_1 | \Psi_2 \rangle = 1 - 0 + 0 - 0 + 2 - 0 + 0 - 0 + 3\alpha(1) = 3 + 3\alpha $
Setting the result to zero for orthogonality:
$ 3 + 3\alpha = 0 $ $ \implies \alpha = -1 $
Calculating Energy Expectation Value for $\Psi_2$
With the determined value $\alpha = -1$, the state $\Psi_2$ is:
$ \Psi_2 = \phi_0 - \phi_1 - \phi_2 $
The coefficients of the energy eigenstates in $\Psi_2$ are $c_0 = 1$, $c_1 = -1$, and $c_2 = -1$.
The energy eigenvalues for the harmonic oscillator are:
- $E_0 = (0 + 1/2)\hbar\omega = \frac{1}{2}\hbar\omega$
- $E_1 = (1 + 1/2)\hbar\omega = \frac{3}{2}\hbar\omega$
- $E_2 = (2 + 1/2)\hbar\omega = \frac{5}{2}\hbar\omega$
The normalization factor, which is $\langle \Psi_2 | \Psi_2 \rangle$, is calculated as:
$ \langle \Psi_2 | \Psi_2 \rangle = |c_0|^2 + |c_1|^2 + |c_2|^2 = |1|^2 + |-1|^2 + |-1|^2 = 1 + 1 + 1 = 3 $
The sum of the weighted energy eigenvalues, $\sum |c_n|^2 E_n$, is:
$ \sum |c_n|^2 E_n = |1|^2 E_0 + |-1|^2 E_1 + |-1|^2 E_2 $ $ = 1 \cdot \left(\frac{1}{2}\hbar\omega\right) + 1 \cdot \left(\frac{3}{2}\hbar\omega\right) + 1 \cdot \left(\frac{5}{2}\hbar\omega\right) $ $ = \left(\frac{1}{2} + \frac{3}{2} + \frac{5}{2}\right)\hbar\omega = \frac{9}{2}\hbar\omega $
The expectation value of the energy for $\Psi_2$ is therefore:
$ \langle E \rangle_{\Psi_2} = \frac{\sum |c_n|^2 E_n}{\sum |c_n|^2} = \frac{\frac{9}{2}\hbar\omega}{3} = \frac{9}{6}\hbar\omega = \frac{3}{2}\hbar\omega $
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) ?
