Commit 7baf86fe authored by Michael Wimmer's avatar Michael Wimmer
Browse files

rewrite tunneling section

parent dcce4161
......@@ -15,9 +15,10 @@ common.configure_plotting()
Before the start of this lecture, you should be able to:
- Use the connection formulas at the two kinds of possible turning points.
- Describe what happens when a wave encounters a barrier.
- Filp and move limits of integration.
- Describe the behavior of tunneling through a rectangular barrier
- Write down the WKB wave function
- Apply the connection formulas to connect the WKB wave functions across a turning point
- Rewrite integrals
!!! summary "Learning goals"
......@@ -29,37 +30,58 @@ common.configure_plotting()
## Physical intuition
## Setting: transport problem
Having looked at bound states before, we will now focus on quite the opposite situation: the case where far away from from the potential the energy is *larger* than the potential, i.e. $E>V(x)$ for $|x|\rightarrow \infty$. In such a case is *open*, we will not have bound states but there will be eigenstates for a continuous range of energies. Far away from some central potential, the eigenstates can be written as a superposition of plane waves. One particular choice for the eigenstate is to choose to have an incoming wave only one side (this is always possible), as shown in the following sketch:
![tunneling](figures/tunneling.svg)
As discussed before, the WKB approximation allows us to study any smooth potential. Opposite to the potential wells discussion, one can consider the case where far away from from the potential, i.e. $x > x_2$ and $x < x_1$>, $E > V(x)$. Depending on the potential around $x_1 < x < x_2$, we describe a transport or a tunneling process. A tunneling process is depicted in the picture above. Such processes are determined by the coefficients of the incident, reflected and transmitted waves. The solution of the Schrodinger equation away from the potential is,
Assuming that the potential becomes constant for some point away from the central potential, the wave function can be written as
$$
\psi(x) = \left\{ \begin{array}{cc}
A e^{ikx} + B e^{-ikx}, & x <0,\\
F e^{ikx}, & x >L.
\end{array}
\right.
\psi(x) = \left\{\begin{array}{cc}
A e^{ik_L x} + B e^{-ik_L x} & x \rightarrow -\infty,\\
F e^{ik_R x}, & x \rightarrow \infty.
\end{array}\right.
$$
From quantum mechanics, recall that the tunneling probability is given as,
where $k_L = \sqrt{\frac{2m}{\hbar^2} (E-V(x\rightarrow -\infty))}$ and $k_R =\sqrt{\frac{2m}{\hbar^2} (E-V(x\rightarrow \infty))}$ are the wave vectors on the left and right side. These can be different if the potential left and right are not equal.
In such a setting, the interesting question is how much of the incoming wave is transmitted to the right, and how much reflected to the left. In fact, you solved exactly such a system already in your first quantum mechanics class when you computed the tunnel probability through a square barrier! With WKB, we can do approximate solutions now for any kind of potential.
The central quantity of interest is the tunneling probability $T$. As in the example of the square tunnel barrier, it is generally defined as
$$
t = \left| \frac{T}{A} \right|^2 \frac{v_R}{v_L}.
T = \left| \frac{F}{A} \right|^2 \frac{v_R}{v_L},
$$
The factor involving the velocities comes the conservation of probability current. It does not appear if the energy at both sides is the same.
where $v_L$ and $v_R$ are the velocities to the left and the right of the central potential. The reson why the velocity is entering is that transmission probability is with respect to the probability *current* - the question is how much of the incident current goes through or is reflected.
## Transport
It is possible to get rid of the awkward velocity term by a small redefinition: We can instead write
$$
\psi(x) = \left\{\begin{array}{cc}
\frac{A}{\sqrt{v_L} e^{ik_L x} + \frac{B}{v_R} e^{-ik_L x} & x \rightarrow -\infty,\\
\frac{F}{v_R} e^{ik_R x}, & x \rightarrow \infty.
\end{array}\right.
$$
The point is that with this convention, the waves to the left and right have a constant current if e.g. the coefficients $A$ and $F$ are the same. In fact, we can immediately see that for the WKB wave functions we have already chosen such a normalization:
$$
\psi(x)_{E > V(x)} \sim \frac{A}{\sqrt{p(x)}}e^{\frac{i}{\hbar} \int_{x_0}^{x} p(x') dx'} + \frac{B}{\sqrt{p(x)}}e^{-\frac{i}{\hbar} \int_{x_0}^{x} p(x') dx'}\,,
$$
where $\frac{1}{\sqrt{p(x)}} \sim \frac{1}{\sqrt{v}}$. The tunnel probability is thus simply given by
$$
T = \left| \frac{F}{A}\right|^2\,.
$$
Let us start by considering the case $E > V(x)$. In such case, the wave packet is always transmitted. Therefore, $t=1$. However, the wavefunction interacts with the potential. The amplitude changes as it moves along. This behaviour can be explained using current conservation. The probability current is,
## $E > V(x)$: Reflectionless transport
Let us start with the simplest case, where $E > V(x)$ for all $x$. For that case we already previously wrote down the [WKB wave function](wkb_wf.md) for all $x$. In particular, if we only consider an incoming wave from the left, we obtain
$$
\begin{split}
j =& \frac{\hbar}{m}\text{Im} \left( \psi^* \frac{d \psi}{dx}\right),\\
\frac{d \psi}{dx} &= \frac{1}{\sqrt{p(x)}} e^{ -\frac{1}{\hbar} \int_{x_0}^x p(x') d x'}\left( \pm \frac{i}{\hbar}p(x) \right),\\
&\longrightarrow j = \pm\frac{1}{m}.
\end{split}
\psi_\text{WKB}(x) \sim \frac{1}{\sqrt{p(x)}}e^{\frac{i}{\hbar} \int_{x_0}^{x} p(x') dx'}\,.
$$
We see here that there is no reflected wave, so the transmission probablity is $T=1$. Indeed, we argued initially that for a smooth potential there is no backreflection, so no surprises here!
We have found that the current is constant. When the particle is closer to the potential, the density must increase. Similarly, when the particle is far from the potential, the density must decrease, which means that the particle is more spreaded. Such phenomena can be observed in the following animation.
It can still be interesting to think about how the wavefunction interacts with the potential. When [deriving the WKB wave function intuitively](wkb_wf.md#intuitive-derivation), we used current conservation to derive the amplitude of the wave function. In particular,
$$
j \sim A(x)^2 \frac{p(x)}{m}\,,
$$
where $A(x)$ was the amplitude of the wave function. We thus see that for regions where the momentum $p(x)$ decreases, i.e. for larger $V(x)$, the amplitude needs to increase. Conversely, when $p(x)$ increases, the amplitude needs to decrease. This phenomenon can be observed in the following animation:
```python inline
from wkb import wkb_static_animation
......@@ -70,9 +92,11 @@ anim = wkb_static_animation(x, E=1.5)
display(HTML(common.to_svg_jshtml(anim)))
```
## Tunneling
## Intuitive derivation for the tunneling case
In a tunneling problem, the wavefunction decays exponentially across the barrier. The case of a rectangular barrier is shown in the animation below. One can observe how the magnitude of the transmitted wave decreases exponentially as the barrier height increases.
The more interesting case is when there is some central part where $V(x) > E$, as then we will have finite reflection and a transmission $T<1$.
In this section, we will first give a intuitive, but not rigorous derivation of the tunneling probablity in WKB approximation. To this end, let us first consider the familiar case of a rectangular barrier. We know that in this case the wave function under the barrier consists of both an exponentially decaying and an exponentially growing part: $\psi(x) = C e^{-\kappa x} + D e^{+\kappa x}$. In the following interactive picture, you can explore how the wave function changes as a function of barrier height:
```python inline
from wkb import tunnel_animation
......@@ -90,23 +114,35 @@ mags = np.linspace(0.035, 0.1, 20)
anim = tunnel_animation(mags=mags, potential_f=pot_step)
display(HTML(common.to_svg_jshtml(anim)))
```
We observe that as the barrier gets higher, the wave function in the central part becomes more and more dominated by the exponentially decaying part, $\psi(x) \sim C e^{-\kappa x}$. Even more,
the wave function to the right is suppressed due to this dominance of the decaying part. To a good approximation the wave function to the right of the barrier is then suppressed by $e^{-\kappa L}$ where $L$ is the barrier length. As a consequence, we can estimate the tunnel probability to be
$$
T \sim \left(e^{-\kappa L}\right)^2 = e^{-2 \kappa L}\,.
$$
We can use the same argument now for a tunnel barrier that has some smooth variation instead of a constant height:
To solve for the wavefunction in the region $x_1=0 < x < x_2=L$, we can use the WKB approximation. To account for an arbitrary potential shape, recall that the WKB wavefunction is,
$$
\psi_{WKB}(x) = \frac{1}{\sqrt{|p(x)|}} \left( C e^{ -\frac{1}{\hbar} \int_{0}^x |p(x')| d x'} + D e^{ \frac{1}{\hbar} \int_{0}^x |p(x')| d x'} \right).
\psi_{WKB}(x) \sim \frac{1}{\sqrt{|p(x)|}} \left( C e^{ -\frac{1}{\hbar} \int_{0}^x |p(x')| d x'} + D e^{ \frac{1}{\hbar} \int_{0}^x |p(x')| d x'} \right).
$$
Finally one needs to perform a wavefunction matching at the boundaries $x=0$ and $x=L$. However, here we will consider the limiting case of a high and long barrier. In such case, there will be only a decaying component in the wavefunction in the barrier region. The incident and transmitted waves will remain, but the amplitude of the latter will be much smaller. In particular, it will decrease by a factor of,
In the limit of a high barrier and/or long barrier, we can again approximate the wave function under the barrier by only the decaying part:
$$
\left| \frac{T}{A} \right| \sim e^{-\frac{1}{\hbar}\int_0^L |p(x')|dx'}.
\psi_{WKB}(x) \sim \frac{C}{\sqrt{|p(x)|}} e^{ -\frac{1}{\hbar} \int_{0}^x |p(x')| d x'}\,.
$$
Therefore, the tunneling amplitude is proportional to,
Hence, the WKB tunneling probability is
$$
t = \left| \frac{T}{A} \right|^2 \frac{v_R}{v_L} \sim e^{-2\gamma},\qquad \gamma = \frac{1}{\hbar}\int_0^L |p(x')|dx'.
T = \sim e^{-2\gamma},\qquad \gamma = \frac{1}{\hbar}\int_0^L |p(x')|dx'. \tag{1}\label{eq:wkb_tunnel}
$$
This formula is in fact one of the most famous results of the WKB formulation, and very useful to estimate tunneling probablities
## Derivation of the WKB tunneling probability for a smooth barrier
### WKB formula for tunneling
The previous derivation lacks mathematical rigor. Using the connection formulas, we *can* actually solve the full problem in WKB approximation - and as we will see below we will exactly recover the WKB tunneling probability \eqref{eq:wkb_tunnel} in the limit of a high and wide barrier.
The tunneling amplitude can be found using the WKB approximation, and the result corresponds to the one derived above. In particular, we'll use the transfer matrix method. Let us recall that the WKB wavefunction for propagating waves is,
To this end, we'll use the transfer matrix method. Let us recall that the WKB wavefunction for propagating waves is,
$$
\psi(x) = \frac{1}{\sqrt{p(x)}} \left[ A \exp\left( \frac{i}{\hbar} \int_{x}^{x_1} p(x') d x'\right) + B \exp\left( -\frac{i}{\hbar} \int_{x}^{x_1} p(x') d x'\right)\right].
$$
......@@ -278,14 +314,15 @@ F
\end{array}
\right) .
$$
Here we have omitted the terms that do not correspond to the tunneling amplitude. We calculate the transmission probability as,
Here we have omitted the terms that do not correspond to the tunneling amplitude. We can thus read off the tunneling probability:
$$
\begin{split}
T &= \left| \frac{E}{A}\right|^2 = \frac{1}{|\frac{i e^{-\gamma }}{4}-i e^{\gamma }|} = \frac{1}{e^\gamma - e^{-\gamma}/4} \\
&= \frac{e^{-2\gamma}}{1+e^{-2\gamma}/4}\approx e^{-2\gamma}.
T &= \left| \frac{E}{A}\right|^2 = \frac{1}{|\frac{i e^{-\gamma }}{4}-i e^{\gamma }|}\\
& = \frac{1}{e^\gamma - e^{-\gamma}/4}\\
&= \frac{e^{-2\gamma}}{1+e^{-2\gamma}/4}
\end{split}
$$
We have used the approximation for a wide and high tunnel barrier. In this way, we have recovered the result for the tunneling probability found in the previous section. That is,
In the limit of a high and/or wide tunnel barrier $\gamma$ will be large and hence $e^{-2\gamma} \ll 1$. The denominator of the tunneling probability can thus be approximated as $1 + e^{-2\gamma}/4 \approx 1$, and we recover the result for the tunneling probability found in the previous section. That is,
$$
T \approx e^{-2\gamma},\qquad \gamma = \frac{1}{\hbar}\int_{x_1}^{x_2} |p(x')|dx'.
$$
......
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment