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.
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- 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?
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 $\Psi(x, t)$ is
