We are going to look at a problem

V(x)=\begin{cases}V_L &x\leq x_L\\ \text{$v$\quad s.t.\quad$v<V_{max}$} & x_L<x\leq x_R\\ V_R &x>x_R\end{cases}

A particle is fired from the left. Describe the $J(x)$ of the particle in all states

Reflecting state

Extends Flow of Probability, Stationary States 1d

This is when

V_L<E<V_R

Let $x_R>x_L$

Let wave function $\phi(x)$ be

\phi(x) = \begin{cases} A e^{ik_L x} + B e^{-ik_L x} & x < x_L \\ C e^{-b_R x} & x > x_R \end{cases}

If $x_L<x<x_R$ , then $\phi(x)$ is whatever solves TISE -- it doesn't really matter.

Let's examine the 1D stationary state in the reflecting state where $E<V_L, V_R$

In each region $V$ is constant, the TISE

-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2}+V\phi=E\phi

\frac{d^2\phi}{dx^2}=\frac{2m(V-E)}{\hbar^2}\phi

Let

\lambda^2 \triangleq \frac{2m(V - E)}{\hbar^2}\triangleq c, \qquad\Rightarrow\qquad \lambda \triangleq \pm\frac{\sqrt{2m(V - E)}}{\hbar}\triangleq \pm\sqrt{c}

\frac{d^2\phi}{dx^2} = \lambda^2\,\phi

This is an ODE. To solve we let

\phi(x)=e^{\pm \lambda x}

Let's try the negative case first. Let $\phi(x)\triangleq e^{-\lambda x}$

\frac{d^2}{dx^2}e^{-\lambda x} = \frac{d}{dx}\big(-\lambda\,e^{-\lambda x}\big) = (-\lambda)(-\lambda)\,e^{-\lambda x} = (-\lambda)^2 e^{-\lambda x}

\Rightarrow\quad \frac{d^2\phi}{dx^2} = \lambda^2\,\phi

\Rightarrow\quad (-\lambda)^2 e^{-\lambda x} = \lambda^2\,(e^{-\lambda x})

true

Let's try the positive case $\phi(x)\triangleq e^{-\lambda x}$

\frac{d^2}{dx^2}e^{+\lambda x} = \frac{d}{dx}\big(\lambda e^{+\lambda x}\big) = (\lambda)(\lambda)e^{+\lambda x} = \lambda^2 e^{+\lambda x}

\Rightarrow\quad (+\lambda)^2 e^{\lambda x} = \lambda^2\,(e^{\lambda x})

also works.

The general solution is therefore

\phi(x) = \alpha\,e^{+\lambda x} + \beta\,e^{-\lambda x}

Solving For Cases

Case 1

Let's consider $E>V_L$

Recall that

\lambda^2 \triangleq \frac{2m(V - E)}{\hbar^2}\triangleq c

Let $V=V_L$

V_L-E<0

\lambda^2 \triangleq \frac{2m(V_L - E)}{\hbar^2}<0

\lambda^2<0

which means $\lambda$ must be imaginary. Let $k_L$ be real and positive

k_L^2 \triangleq -\frac{2m(V_L-E)}{\hbar^2} > 0 \quad\Rightarrow\quad \lambda^2 = -k_L^2

\lambda = \pm\sqrt{-k_L^2} = \pm\sqrt{-1}\,\sqrt{k_L^2} = \pm i\,k_L

sub back into

\phi(x) = e^{\pm i k_L x} \quad\Rightarrow\quad \phi(x) = A e^{i k_L x} + B e^{-i k_L x}

Case 2

Consider $E<V_R$

V_R-E>0

\lambda^2>0

so it must be real. Let $b_R$ be real and positive

b_R^2 \triangleq \frac{2m(V_R - E)}{\hbar^2} > 0 \quad\Rightarrow\quad \lambda = \pm\sqrt{b_R^2} = \pm b_R

sub back into

\phi(x) = D\,e^{+b_R x} + C\,e^{-b_R x}

as $x$ goes to $+\infty$ then $e^{+b_Rx}$ diverges unless $\alpha=0$

\phi(x) = C\,e^{-b_R x}

Working out Probability current

Working it out Probability Current $J$

Let

\omega\triangleq \phi^*\frac{\partial \phi}{\partial x}

\omega^*=\omega = -(\omega -\omega^*)=-2iIm(\omega)

So from the definition of J

J(x,t) \triangleq \frac{i\hbar}{2m}\left(\frac{\partial\psi^*}{\partial x}\psi - \psi^*\frac{\partial\psi}{\partial x}\right)

= \frac{i\hbar}{2m}(w^* - w)

= \frac{i\hbar}{2m}\big(-2i\,\mathbb{Im}(w)\big) = \frac{\hbar}{m}\mathbb{Im}(w)

= \frac{\hbar}{m}\,\mathbb{Im}\!\left(\phi^*\frac{d\phi}{dx}\right)

Case 1

Let's consider $E>V_L$

\frac{d}{dx}\phi = ik_L A e^{ik_L x} - ik_L B e^{-ik_L x}

\phi^* = A^* e^{-ik_L x} + B^* e^{ik_L x}

\omega=\phi^* \frac{d}{dx}\phi = ik_L\Big(|A|^2 - |B|^2 - A^* B e^{-2ik_L x} + A B^* e^{2ik_L x}\Big)

sub back to get

= \frac{\hbar}{m}\,\mathbb{Im}\!\left(ik_L|A|^2-ik_L|B|^2\right)

= \frac{\hbar k_L}{m}\left(|A|^2-|B|^2\right)

Case 2

Let's consider $E<V_R$

w = \phi^*\frac{d}{dx}\phi = (C^*e^{-b_R x})(-b_R C e^{-b_R x}) = -b_R|C|^2 e^{-2b_R x}

J = \frac{\hbar}{m}\,\mathbb{Im}\big(-b_R|C|^2 e^{-2b_R x}\big)

= \frac{\hbar}{m}(0) = 0

Combining

That means the probability current $J$ is this

J(x,t) = \begin{cases} \dfrac{\hbar k_L}{m}\big(|A|^2 - |B|^2\big) & x < x_L \\[2mm] 0 & x > x_R \end{cases}

The physical reason is that the right side is a dead end tail that transmit to nothing, so in steady state all incoming probabikity must be reflected back. Incoming flux is outgoing flux. $|A|^2=|B|^2$ must be true per Flow of Probability.

Scattering State

This is when

E>V_L, V_R

and we already know from Probability Current Problem 1 that

\phi(x) = \begin{cases} A e^{ik_L x} + B e^{-ik_L x} & x < x_L \\ C e^{ik_R x} + De^{-ik_Rx} & x > x_R \end{cases}

Note that $x<x_L$ and $x>x_R$ is like that because $E>V_L, V_R$

Because we're describing a particle fired from the left only, there is nothing coming in from right to left so $D=0$ .

J(x,t) = \begin{cases} \frac{\hbar k_L}{m}\big(|A|^2 - |B|^2\big) & x < x_L \\ \frac{\hbar k_R}{m}|C|^2 & x > x_R \end{cases}

Reflection wise Transmission Reflection Coefficients:

A is incident, B is reflected, C is transmitted

R = \frac{|B|^2}{|A|^2}

T = \frac{k_R|C|^2}{k_L|A|^2}, \qquad R + T = 1

$T>0$ even if $E<V_{max}$ which means a particle with less energy than the potential usually gets stuck. In quantum physics, it is possible for it to transmit through this barrier. This is known as quantum tunnelling

$R>0$

Bound State

This is when

E<V_L, V_R

\phi(x) = \begin{cases} A e^{b_R x} + Be^{-b_Rx} & x < x_L \\ D e^{b_R x} + Ce^{-b_Rx} & x > x_R \end{cases}

Left region, if $x\rightarrow -\infty$ then the $B$ term will blow up so $B=0$ Right region, if $x\rightarrow \infty$ then the $D$ term will blow up so $D=0$

\phi(x) = \begin{cases} A e^{b_R x}& x < x_L \\ Ce^{-b_Rx} & x > x_R \end{cases}

A bound state must decay to zero on both sides.

This means that energy is quantized for bound.

E \in \{E_0, E_1, E_2, \dots\}

this is because, as counterpoint if $E$ is continuous, a wavefunction DNE in between the energies. this is because the left tail $Ae^{b_Lx}$ decays to 0 (is Evanescent). Integrate the middle. You'd need a right tail that also decays so

\frac{\frac{d \phi}{dx}(x_R)}{\phi(x_R)}=-b_R

must be true. If it isn't true then $D\neq 0$ needs to be drue but that blows it up.