Wavefunction Analysis
The given wavefunction is $\psi(x) = e^{-(iax+\beta)}$, where $a$ and $\beta$ are real constants.
Schrödinger Equation Application
We use the time-independent Schrödinger equation:
$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi(x) = E\psi(x)$First, calculate the derivatives of the wavefunction:
- First derivative: $\frac{d\psi}{dx} = \frac{d}{dx}(e^{-(iax+\beta)}) = -(ia)e^{-(iax+\beta)} = -(ia)\psi(x)$
- Second derivative: $\frac{d^2\psi}{dx^2} = \frac{d}{dx}(-(ia)\psi(x)) = -(ia)\frac{d\psi}{dx} = -(ia)(-(ia)\psi(x)) = -(a^2)\psi(x)$
Potential Calculation
Substitute the second derivative back into the Schrödinger equation:
$-\frac{\hbar^2}{2m}(-(a^2)\psi(x)) + V(x)\psi(x) = E\psi(x)$Simplify the equation:
$\frac{\hbar^2 a^2}{2m}\psi(x) + V(x)\psi(x) = E\psi(x)$Divide by $\psi(x)$ (since $\psi(x)$ is non-zero):
$\frac{\hbar^2 a^2}{2m} + V(x) = E$Rearrange to solve for the potential $V(x)$:
$V(x) = E - \frac{\hbar^2 a^2}{2m}$Potential Identification
The derived potential $V(x) = E - \frac{\hbar^2 a^2}{2m}$ is a constant value because $E$, $\hbar$, $m$, and the real constant $a$ are all constants.
Comparing this result with the given options:
- Option 1: $V(x) \propto a^2 x^2$ is not constant.
- Option 2: $V(x) \propto e^{-\alpha x}$ is not constant.
- Option 3: $V(x) = 0$ is a constant potential. This matches our derived form where the constant could be zero.
- Option 4: $V(x) \propto \sin(ax)$ is not constant.
Therefore, the only potential that fits the condition is a constant potential, specifically $V(x) = 0$.
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?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) ?
- 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?
