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 Orthogonality Condition
Two quantum states, $\Psi_1$ and $\Psi_2$, are considered orthogonal if their inner product equals zero. The inner product is denoted as $\langle \Psi_1 | \Psi_2 \rangle$.
The harmonic oscillator states $\phi_0$, $\phi_1$, and $\phi_2$ are given as normalized and mutually orthogonal. This means $\langle \phi_i | \phi_j \rangle = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta ($\delta_{ij} = 1$ if $i=j$, and $\delta_{ij} = 0$ if $i \neq j$).
Calculating the Inner Product
The states $\Psi_1$ and $\Psi_2$ are defined as:
- $\Psi_1 = \phi_0 - 2\phi_1 + 3\phi_2$
- $\Psi_2 = \phi_0 - \phi_1 + \alpha\phi_2$
We calculate the inner product $\langle \Psi_1 | \Psi_2 \rangle$ using linearity:
$ \langle \Psi_1 | \Psi_2 \rangle = \langle (\phi_0 - 2\phi_1 + 3\phi_2) | (\phi_0 - \phi_1 + \alpha\phi_2) \rangle $Expanding this expression and applying the orthogonality and normalization conditions:
$ \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 $Substituting $\delta_{ij}$ values:
$ \langle \Psi_1 | \Psi_2 \rangle = (1) - (0) + \alpha(0) \\ - 2(0) + 2(1) - 2\alpha(0) \\ + 3(0) - 3(0) + 3\alpha(1) $Simplifying the expression:
$ \langle \Psi_1 | \Psi_2 \rangle = 1 + 2 + 3\alpha $ $ \langle \Psi_1 | \Psi_2 \rangle = 3 + 3\alpha $Determining the Value of Alpha
For orthogonality, $\langle \Psi_1 | \Psi_2 \rangle = 0$. Setting the result to zero:
$ 3 + 3\alpha = 0 $Solving for $\alpha$:
$ 3\alpha = -3 $ $ \alpha = \frac{-3}{3} $ $ \alpha = -1 $The value of $\alpha$ required for $\Psi_2$ to be orthogonal to $\Psi_1$ is -1.
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) ?
