Wavefunction Analysis for Probability Current Density
The question requires calculating the probability current density, $J$, specifically for the real part of the given wavefunction $\psi = e^{ikx} + 2e^{-ikx}$.
Probability Current Density Definition
The probability current density $J$ for a wavefunction $\psi(x, t)$ in one dimension is given by the formula:
$ J = \frac{\hbar}{2mi} \left( \psi^* \frac{d\psi}{dx} - \psi \frac{d\psi^*}{dx} \right) $
A key property is that if a wavefunction $\phi(x)$ is purely real (i.e., $\phi^* = \phi$), its probability current density is zero:
$ J = \frac{\hbar}{2mi} \left( \phi \frac{d\phi}{dx} - \phi \frac{d\phi}{dx} \right) = 0 $
Real Part Wavefunction Calculation
We first determine the real part of the wavefunction $\psi = e^{ikx} + 2e^{-ikx}$.
Using Euler's relations ($e^{i\theta} = \cos\theta + i\sin\theta$ and $e^{-i\theta} = \cos\theta - i\sin\theta$):
$ \psi = (\cos(kx) + i\sin(kx)) + 2(\cos(kx) - i\sin(kx)) $
$ \psi = \cos(kx) + i\sin(kx) + 2\cos(kx) - 2i\sin(kx) $
Combining real and imaginary parts:
$ \psi = (3\cos(kx)) + i(-\sin(kx)) $
The real part is therefore:
$ \text{Re}(\psi) = 3\cos(kx) $
Probability Current Density for Real Part
Let $\phi(x) = \text{Re}(\psi) = 3\cos(kx)$. Since $3\cos(kx)$ is a real function, its probability current density must be zero based on the definition.
$ J_{\text{real part}} = 0 $
Answer Probability Current Density
The probability current density for the real part of the wavefunction is 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?- 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) ?
