Schrödinger equation

Time dependent equation

\[ \mathrm{i} \hbar \frac{\partial}{\partial t} \varphi(x) = \left( -\frac{\hbar^2}{2m} \frac{\mathrm{d^2}}{\mathrm{d}x^2} + V(x) \right) \varphi(x) \]

Time independent equation

\[ \left( -\frac{\hbar^2}{2m} \frac{\mathrm{d^2}}{\mathrm{d}x^2} + V(x) \right) \varphi(x) = E \varphi(x) \]

Weak form with test function \(v\):

\[ \frac{\hbar^2}{2m}(\nabla \varphi, \nabla v) + V(x)(\varphi, v) = E (\varphi, v) \]

Potential well

With

\[\begin{split} V(x) = \begin{cases} 0, &-L/2<&x&<L/2 \\ V_0 &\text{outside} \end{cases} \end{split}\]

we assemble a solution composed of

\[\begin{split} \varphi(x) = \begin{cases} \varphi_1, &-L/2<&x \\ \varphi_2, &-L/2<&x&<L/2 \\ \varphi_3, &&x&>L/2 \end{cases} \end{split}\]

let

\[ k = \frac{\sqrt{2 m E}}{\hbar}, \quad k^` = \frac{\sqrt{2 m (V_0 - E)}}{\hbar} \text{and} \quad \alpha = \frac{\sqrt{2 m (V_0 - E)}}{\hbar} \]

Inside the potential well

For inside the potential well this leads to

\[ \frac{\mathrm{d^2}}{\mathrm{d}x^2} \varphi(x) = - k^2 \varphi(x) \]

which can be solved using

\[ \varphi_2 = A \sin(k x) + B \cos(k x) \]

Outside the potential well

and outside the potential well for unbound solutions, i.e. \(E>V_0\)

\[ \frac{\mathrm{d^2}}{\mathrm{d}x^2} \varphi_{1/3}(x) = -{k^`}^2 \varphi_{1/3}(x) \]

which can similary be solved using

\[ \varphi_{1/3} = C \sin(k^` x) + D \cos(k^` x) \]

and bound solutions, i.e. \(E<V_0\)

\[ \frac{\mathrm{d^2}}{\mathrm{d}x^2} \varphi_{1/3}(x) = \alpha^2 \varphi_{1/3}(x) \]

solved by

\[ \varphi_1 = \mathrm{e}^{-F x} + \mathrm{e}^{G x} \quad \text{and} \quad \varphi_3 = \mathrm{e}^{-H x} + \mathrm{e}^{I x} \]

Bound states

We find for the bound states, i.e. states where we assume that \(\lim_{x\to\pm\inf}\varphi(x)=0\), that the complete wavefunction simplifies to

\[\begin{split} \varphi(x) = \begin{cases} \mathrm{e}^{G x}, &-L/2<&x \\ A \sin(k x) + B \cos(k x), &-L/2<&x&<L/2 \\ \mathrm{e}^{-H x}, &&x&>L/2 \end{cases} \end{split}\]

as the solutions need to be continous and differentiable, i.e.

\[ \varphi_1(-L/2) = \varphi_2(-L/2) \quad \varphi_2(L/2) = \varphi_3(L/2) \]

and

\[ \left.\frac{\mathrm{d}\varphi_1}{\mathrm{d}x}\right|_{x=-L/2} = \left.\frac{\mathrm{d}\varphi_2}{\mathrm{d}x}\right|_{x=-L/2} \quad \text{and} \quad \left.\frac{\mathrm{d}\varphi_2}{\mathrm{d}x}\right|_{x=L/2} = \left.\frac{\mathrm{d}\varphi_3}{\mathrm{d}x}\right|_{x=L/2} \]

which leads to \(A=0\) and \(G=H\) for the symmetric case and \(B=0\) and \(G=-H\) for the assymetric case.

this leads for the symmetric case to the conditions

\[\begin{split} H \mathrm{e}^{-\alpha L/2} = B \cos(kL/2) \text{ and } -\alpha H \mathrm{e}^{-\alpha L/2} = - k B \sin(kL/2) \\ \Rightarrow \alpha = k \tan(kL/2) \end{split}\]

and for the assymetric case to

\[\begin{split} H \mathrm{e}^{-\alpha L/2} = B \sin(kL/2) \text{ and } -\alpha H \mathrm{e}^{-\alpha L/2} = k B \cos(kL/2) \\ \Rightarrow \alpha = - k \cot(kL/2) \end{split}\]

with \(u=\alpha L/2\) and \(v=kL/2\) and using \(u^2=u_0^2-v^2\) with \(u_0^2=mL^2V_0/2\hbar^2\) we can simplify both to

\[\begin{split} \sqrt{u_0^2-v^2} = \begin{cases} v \tan v, &\text{for the symmetric case} \\ -v \cot v, &\text{for the asymmetric case} \end{cases} \end{split}\]

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import numpy as np
import matplotlib.pyplot as plt

u0 = np.sqrt(20)
v = np.linspace(0,5,10000)
yc = np.sqrt(u0**2-v**2)
plt.ylim(0,u0+1)
plt.plot(v, yc)

y = v*np.tan(v)
y = np.where(y>-10,y,np.nan)
plt.plot(v, y, label='symmetric')
idx_s = np.argwhere(np.nan_to_num(np.diff(np.sign(y - yc)),-1))[:,0]
plt.plot(v[idx_s], y[idx_s], 'ro')
print(v[idx_s])

y = -v*1/np.tan(v)
plt.plot(v, np.where(y>-10,y,np.nan), label='asymmetric')
idx_a = np.argwhere(np.nan_to_num(np.diff(np.sign(y - yc)),-1))[:,0]
plt.plot(v[idx_a], y[idx_a], 'ro')
print(v[idx_a])

plt.legend()
plt.show()

[1.28012801 3.72687269]
[2.53775378 3.14131413]