Schrodinger Wave Equation Questions

Probability current density represent the overall drift of a particle in a particular direction. It is given by $$ \boxed{ \vec{J} = \frac{\hbar}{2im} [ \Psi^* \nabla \Psi - \Psi \nabla \Psi^* ] } $$ where, $\Psi = e^{i(kx-wt)} \\$ $ \Psi^* = e^{-i(kx-wt)} \\$ $\nabla \Psi = ik \Psi \\$ $\nabla \Psi^* = -ik \Psi^* \\$ $\implies \vec{J} = \frac{\hbar}{2im} [ \Psi^* ik \Psi - \Psi(-ik \Psi^*) ] \\$ $\implies \vec{J} = \frac{\hbar}{2im} [2ik] = \frac{\hbar k}{m}$ $\\$ The direction of $\vec{J}$ is along the direction of the momentum of the particle as $J \propto \hbar k $(=momentum). Also, $\frac{\hbar k}{m}$ has dimensions of velocity. So probability can be higher at some time and lower at another time and vice versa.

The transmission coefficient $T$, when $E<V_0$ is given by: $$ T = \bigg[ 1+ \frac{V_0^2 \sinh^2 (Ka)}{4E(V_0-E)} \bigg]^{-1} $$ $V_0$ is the potential barrier height $\\$ $a$ is the width of the barrier $\\$ $E$ is the energy of the particle $\\$ $K = \frac{\sqrt{2m(V_0-E)}}{\hbar}$ $\\$ Substituting the values on the equation, we will get T =0.373. Hence the transmission probability is around 37.3%.

The angular frequency of a single harmonic oscillator is given by: $$ \tag{1} w = \sqrt{\frac{k}{m}} $$ $k$ is the force constant and $m$ is the mass of the oscillator. $\\$ Using Hooke's law, we have $$ \tag{2} F = kx $$, $k$ is the force constant and $x$ is the displacement. $\\$ For a quantum harmonic oscillator, the energy levels are quantized and given by the formula: $$ \tag{3} \boxed{E_n = (n + \frac{1}{2}) \hbar w } $$ Here, $E_n$ is the energy of the $n^{th}$ level, $\hbar$ is the reduced Planck’s constant, $w$ is the angular frequency of the oscillator, and $n$ is the quantum number which can take any non-negative integer value (0, 1, 2, …). $\\$ So, using (1) and (2), we have $$ w = \sqrt{\frac{k}{m}} = \sqrt{\frac{F}{xm}} $$ substituting the values for $F = 10^{-1}, x= 10^{-2}$ and $m=10^{-3}$, we have, $$ w = \frac{10^{-1}}{10^{-2} \times 10^{-3}} = 10^{2} \text{rad/s} $$ Substituting this value in (3), we have $$ E_0 = 5.28 \times 10^{-33} J $$, in the case of zero point energy, n=0

A wave function is a single-valued, continuous and normalized wave function. $\\$ (a) In the first case, $\tan x$ goes to $\infty$ as $x$ goes to $\infty$. So it is not an acceptable wave function $\\$ (b) sine and cosine functions are acceptable as they are single-valued, continuous and square-integrable. Similarly their linear combinations are also accepted.

In order to explain interference effects, it is assumed that with each particle a wave function is associated. Born postulated that if a particle is described by a wave function $\Psi(x,t)$, then $|\Psi(x,t)|^2 dx$ gives the probability of finding the particle within an element $dx$ about the point $x$ at time $t$. Since the probability of finding the particle somewhere must be unity, the wave function must be normalized: $$ \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx = \int_{-\infty}^{\infty} \Psi^* \Psi dx = 1$$ Using Schrodinger equation, we have: $$ \tag{1} \frac{\partial \Psi}{\partial t} = \frac{i \hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2} - \frac{i}{\hbar} V \Psi$$ and $$ \tag{2} \frac{\partial \Psi^*}{\partial t} = -\frac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar} V \Psi^*$$ where $V$ is assumed to be real function. On taking the time derivative of the normalized function, $$ \tag{3} \frac{d}{dt} \bigg[ \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx] \bigg] = \int_{-\infty}^{\infty} \frac{\partial (|\Psi(x,t)|^2)}{\partial t} dx $$ using, $|\Psi|^2 = \Psi^* \Psi$ $$ \frac{\partial |\Psi|^2}{\partial t} = \frac{\partial \Psi^* \Psi}{\partial t} = \Psi^* \frac{\partial \Psi}{\partial t} + \Psi \frac{\partial \Psi^*}{\partial t} $$, using (1) and (2), $$ \frac{\partial |\Psi|^2}{\partial t} = \frac{\partial}{\partial x} \bigg[ \frac{i \hbar}{2m} \Big( \Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x} \Big) \bigg]$$, so (3) becomes, $$ \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi|^2 dx] = \frac{i \hbar}{2m} \Big( \Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x} \Big) \bigg|^{\infty}_{-\infty} $$ $$ = \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi|^2 dx] = 0 $$, so, $$ \int_{-\infty}^{\infty} |\Psi|^2 dx] = \text{constant} $$ So, if $\Psi$ is normalized at $t=0$, then it is normalized at any time $t$. This property preserves the normalization of wave function. Without whihc features of Schrodinger equation would be incompatible.

Expectation value of position can be calculated using $$ \braket{x} = \braket{\Psi ^* | x|\Psi} = \int_{-\infty}^{\infty} \Psi^* x \Psi dx $$ So, using the wavefunction $\psi(x)$, $$ \braket{x} = \int_{-\infty}^{\infty} N exp (\frac{-x^2}{2a} - ikx) \space x \space N exp (\frac{-x^2}{2a} + ikx) dx $$ $$ = N^2 \int_{-\infty}^{\infty} x \space exp (\frac{-x^2}{a}) $$ As we know $x$ is an odd function [because $f(-x) = -f(x)$] and using the standard result $ \int_{-a}^{a} f(x) dx = 0 $, where $f(x)$ is an odd function. So the integrand become zero, thus, $$ \braket{x} = 0 $$ Hence the expectation value of position is zero. $\\$ Note: In Quantum Mechanics, when expectation value of an observable is zero that doesn't necessarily means that every measurement is zero, but it means that average result of the measurement over multiple measurements are zero.

We need to normalize $\Psi = A e^{\gamma |x|}$. which is $$ \begin{matrix} \Psi = A e^{\gamma x} & \text{for} & x \ge 0 \\ \Psi = A e^{-\gamma x} & \text{for} & x < 0 \end{matrix}$$ Now, for $x \rightarrow \infty \implies \Psi \rightarrow \infty \\$ and for $x \rightarrow -\infty \implies \Psi \rightarrow \infty $. that is wave function is not finite in $-\infty \le x \le \infty$. Hence it is not normalizable.