Commit 5c8de38c authored by juandaanieel's avatar juandaanieel
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Corrected plots for wkb section

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......@@ -22,7 +22,66 @@ common.configure_plotting()
- Write down the energy of a bound state using the WKB approximation for three different potential configurations.
## Potential Wells
An immediate consequence of the connection formulas is that bound state energies can be approximated for several potential wells as depicted in the following plot.
An immediate consequence of the connection formulas is that bound state energies can be approximated for several potential wells. Let us remember that in a region where $E>V(x)$, $\psi(x)$ decays exponentially. Therefore, $\psi(x)$ is bounded to be in the region where $E > V(x)$.
### Potential with no vertical walls
![No walls](figures/no_wall.svg)
Consider the case with a region $x_1(E)< x< x_2(E)$ where $E < V(x)$. Here, the wavefunction decays beyond the turning points, we know that it is proportional to a sine function. That is,
$$
\sin \left( \frac{i}{\hbar} \int_{x}^{x_2} p(x')dx' + \frac{\pi}{4}\right) = \pm \sin \left( \frac{i}{\hbar} \int_{x_1}^{x} p(x')dx' + \frac{\pi}{4}\right),
$$
The extra $\pm$ sign is included as a shift of $n\pi$. The arguments at both sides are equal, that is,
$$
\frac{i}{\hbar} \int_{x}^{x_2} p(x')dx' + \frac{\pi}{4}= -\frac{i}{\hbar} \int_{x_1}^{x} p(x')dx' - \frac{\pi}{4} + n \pi
$$
We sum the limits of the two integrals, and we conclude,
$$
\int_{x_1}^{x_2} p(x') dx' = \left( n - \frac{1}{2} \right)\pi \hbar
$$
### Potential with one vertical wall
![One wall](figures/one_wall.svg)
Conside the case shown above with one vertical wall. Observe that one of the turning points is fixed such that $\psi(x_1)=0$ and $x_1$ does not depend on $E$. Only the right turning point is a function of $E$, i.e. $x_2 = x_2(E)$. Here one can easily derive a condition for the bound state energies by considering the wavefunction in the region $x_1 < x < x_2$. That is,
$$
\psi_{WKB}(x) \approx \sin\left( \frac{1}{\hbar}\int_x^{x_2} p(x') dx' + \frac{\pi}{4} \right).
$$
Given the boundary condition, we require the sine to vanish, then, the argument must be an integer multiple of $\pi$. So we conclude,
$$
\int_{x_1}^{x_2} p(x') dx' = \left( n - \frac{1}{4} \right)\pi \hbar
$$
### Potential with two vertical walls
![Two walls](figures/two_wall.svg)
Finally, consider the case of a quantum well with vertical walls. In this case, none of the turning points depend on energy. The wavefunction inside the hard walls is given as,
$$
\psi_{WKB}(x) = \frac{1}{\sqrt{p(x)}} \left( A e^{\frac{i}{\hbar} \int_{x_1}^x p(x')dx'} + B e^{-\frac{i}{\hbar} \int_{x_1}^x p(x')dx'} \right).
$$
We apply the boundary condition at $x_1=0$, and we find that,
$$
A = - B \rightarrow \psi_{wkb}(x) = \frac{1}{\sqrt{p(x)}} \sin \left( \frac{i}{\hbar} \int_{x_1}^x p(x')dx' \right).
$$
Then, we apply the second boundary condition, and we find
$$
\psi_{WKB}(x_2) = \frac{1}{\sqrt{p(x)}} \sin \left( \frac{i}{\hbar} \int_{x_1}^{x_2} p(x')dx' \right).
$$
So we conclude,
$$
\int_{x_1}^{x_2} p(x') dx' = n\pi \hbar
$$
!!! summary "Bohr-Sommerfeld quantization rule"
[This rule](https://en.wikipedia.org/wiki/Old_quantum_theory) determines that closed orbits can only be occupied by quantized states. That is,
$$
2 \int_{x_1}^{x_2} p(x) dx = \oint pdx = n h.
$$
## Summary
```python inline
import matplotlib.pyplot as plt
......@@ -64,62 +123,10 @@ for ax in axes:
plt.legend(prop={'size': 14})
fig.show()
```
One can observe that regardless the shape of the potential, all these cases have the following characteristics:
* For a hard wall located at $x_1$, the wavefunction vanishes, i.e. $\psi(x_1)=0$.
* For a turning point at $x_2$, after which $E>V(x)$, $\psi(x)$ decays exponentially.
Observe that the turning points are a function of energy, i.e. $x_1 = x_1(E)$ and $x_2 = x_2(E)$.
### Potential with one vertical wall
One can easily derive a condition for the bound state energies by considering the wavefunction in the region $x_1 < x < x_2$. That is,
$$
\psi_{WKB}(x) \approx \sin\left( \frac{1}{\hbar}\int_x^{x_2} p(x') dx' + \frac{\pi}{4} \right).
$$
Given that the wavefunction vanishes at the hard wall, we have $\psi_{WKB}(x_1) = 0$. In order for the sine to vanish, the argument must be an integer multiple of $\pi$. So we conclude,
$$
\int_{x_1}^{x_2} p(x') dx' = \left( n - \frac{1}{4} \right)\pi \hbar
$$
### Potential with no vertical walls
Since the wavefunction decays beyond the turning points, we know that it is proportional to a sine function in the region $x_1 < x < x_2$. That is,
$$
\sin \left( \frac{i}{\hbar} \int_{x}^{x_2} p(x')dx' + \frac{\pi}{4}\right) = \pm \sin \left( \frac{i}{\hbar} \int_{x_1}^{x} p(x')dx' + \frac{\pi}{4}\right),
$$
The extra $\pm$ sign is included as a shift of $n\pi$. The arguments at both sides are equal, that is,
$$
\frac{i}{\hbar} \int_{x}^{x_2} p(x')dx' + \frac{\pi}{4}= -\frac{i}{\hbar} \int_{x_1}^{x} p(x')dx' - \frac{\pi}{4} + n \pi
$$
We sum the limits of the two integrals, and we conclude,
$$
\int_{x_1}^{x_2} p(x') dx' = \left( n - \frac{1}{2} \right)\pi \hbar
$$
### Potential with two vertical walls
The wavefunction inside the hard walls is given as,
$$
\psi_{WKB}(x) = \frac{1}{\sqrt{p(x)}} \left( A e^{\frac{i}{\hbar} \int_{x_1}^x p(x')dx'} + B e^{-\frac{i}{\hbar} \int_{x_1}^x p(x')dx'} \right).
$$
We apply the boundary condition at $x_1=0$, and we find that,
$$
A = - B \rightarrow \psi_{wkb}(x) = \frac{1}{\sqrt{p(x)}} \sin \left( \frac{i}{\hbar} \int_{x_1}^x p(x')dx' \right).
$$
Then, we apply the second boundary condition, and we find
$$
\psi_{WKB}(x_2) = \frac{1}{\sqrt{p(x)}} \sin \left( \frac{i}{\hbar} \int_{x_1}^{x_2} p(x')dx' \right).
$$
So we conclude,
$$
\int_{x_1}^{x_2} p(x') dx' = n\pi \hbar
$$
!!! summary "Bohr-Sommerfeld quantization rule"
[This rule](https://en.wikipedia.org/wiki/Old_quantum_theory) determines that closed orbits can only be occupied by quantized states. That is,
$$
2 \int_{x_1}^{x_2} p(x) dx = \oint pdx = n h.
$$
## Summary
The previous finding can be summarized in the following formula for bound state energies
The three cases depicted here can be summarized in the following formula for bound state energies
$$
\int_{x_1}^{x_2} p(x') dx' = (n-\nu)\pi \hbar.
$$
Here $\nu=0, 1/2, 1/4$ for the different potential wells. The correspondence can be observed in the initial plot.
Here $\nu=0, 1/2, 1/4$ for the different potential wells.
......@@ -18,11 +18,13 @@ title: The WKB approximation
## Classical intuition
![Classical particle](figures/classical_scat.svg)
The WKB approximation is used to describe a particle of mass $m$ in a one-dimensional (1D) potential with some initial velocity. To call some intuition, let us recall the classical case: consider a particle of energy $E$ moving in a potential $V(x)$. In this situation, wthe total energy of the particle is conserved, and can be at every point $x$ be separated into kinetic and potential energy as $E = E_\text{kin}(x) + V(x)$.
If $E > V(x)$, then the particle will always have an energy larger than the potential at any point $x$. Hence, there will always be some non-zero kinetic energy at every point $x$. As a consequence, the particle will continuously move in one direction (determined by the initial velocity). This happens irrespective of the potential shape: for example, it does not matter whether the potential is smooth or changes abruptly.
If $E > V(x)$, as shown in the upper plots, then the particle will always have an energy larger than the potential at any point $x$. Hence, there will always be some non-zero kinetic energy at every point $x$. As a consequence, the particle will continuously move in one direction (determined by the initial velocity). This happens irrespective of the potential shape: for example, it does not matter whether the potential is smooth or changes abruptly.
If $E < V(x)$, will reach a maximum position $x_0$ such that $E = V(x_0)$ and the kinetic energy becomes zero. Then, the particle will be reflected. Again, this will happen for any potential shape.
If $E < V(x)$, as shown in the lower plots, will reach a maximum position $x_0$ such that $E = V(x_0)$ and the kinetic energy becomes zero. Then, the particle will be reflected. Again, this will happen for any potential shape.
In summary, in classical mechanics the global motion of a particle in a 1D potential is only determined by the value of the total energy compared to the potential $V(x)$.
......
......@@ -51,8 +51,8 @@ def discretize(x, y, step):
n_points = len(x)
n_new = int(max(x)/step)
d = int(n_points/n_new)
for i in range(n_new):
z[i*d:(i+1)*d] = y[i*d]
for i in range(0, n_new, 2):
z[i*d:(i+2)*d] = y[(i+1)*d]
return z
x = np.linspace(-1, 1, 200)*np.pi/2
......@@ -62,11 +62,11 @@ y = np.sin(x)
y1 = np.heaviside(x1-1, 0 ) + np.heaviside(x1-2, 0 ) + np.heaviside(x1-3, 0 )
y2 = (5-np.sin(x2*5*np.pi))
y3 = (5-np.sin(x1*5*np.pi))
z1 = discretize(x, y, step=0.3)
z1 = discretize(x, y, step=0.1)
fig, ax = plt.subplots(nrows=1, ncols=2, figsize=(12, 4))
ax[0].plot(x, y, label='smooth', color='black')
ax[0].plot(x, z1, label=r'$discrete$', color='red')
ax[0].plot(x, y, label='smooth', color='red', linewidth=3, zorder=100)
ax[0].plot(x, z1, label=r'$discrete$', color='black')
ax[0].set_ylabel(r'$V(x)$',fontsize=18)
ax[1].plot(x1, y1, label=r'V(x)', color='black')
......
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