Commit b02b1107 authored by Han-Miru Kim's avatar Han-Miru Kim
Browse files

QM w03

parent 49daa00b
......@@ -329,11 +329,21 @@ Setup numbering system and define a function that creates environment.
\defboxenv{dfn}{Definition}{orange!50}
\defboxenv{thm}{Theorem}{blue!40}
\defboxenv{lem}{Lemma}{blue!20}
\defboxenv{xmp}{Example}{black!20}
\defboxenv{exr}{Exercise}{black!50}
\defboxenv{xmp}{Example}{green!50!blue!30!white}
\defboxenv{exr}{Exercise}{green!30!blue!50!white}
\defboxenv{rem}{Remark}{red!20}
#+END_SRC
* Quotes
"Physiker haben ein besseres Leben als Mathematiker, die Paranoid sind."
- Gaberdiel 2021-09-28
"Wenn dieser Ausdruck nicht mer da ist, dann ist er nicht mehr da."
- Gaberdiel 2021-09-30
"Die Chemikar sagen einfach: 'Das ist doch kein Problem weil es funktioniert'.
Ich will euch nicht dazu zu bringen, wie Chemiker zu denken."
- Gaberdiel 2021-10-05
"Das E ist das E ist das E"
- Gaberdiel 2021-10-05
......@@ -308,4 +308,49 @@ Moreover, since $E$ is real valued, we have
\end{align*}
The Schrödinger equation easily generalizes to higher dimensions. We get
\begin{align*}
i \hbar \frac{\del }{\del t} \Psi(\vec{x},t) = H \Psi(\vec{x},t), \quad \text{for} \quad H = - \frac{\hbar^{2}}{2m} \Delta + V(\vec{x},t)
\end{align*}
and the TISE stays the same, where $H$ has a different domain
\begin{align*}
\psi: \R^{3} \to \C, \quad H: L^{2}(\R^{3}) \to L^{2}(\R^{3})
\end{align*}
There is a connection between the TISE and the time-independent Hamilton-Jacobi equation.
We make the Ansatz
\begin{align*}
\Psi(\vec{x},t) = \psi(\vec{x}) e^{- \frac{i E}{\hbar}t}, \quad \psi(\vec{x}) = A(\vec{x}) e^{i \frac{S(\vec{x})}{\hbar}}
\end{align*}
Calculating the Laplacian $\Delta \psi(\vec{x})$ yields
\begin{align*}
\Delta \psi(x) = (\Delta A) e^{- \frac{S(x)}{\hbar}} + \frac{2i}{\hbar} (\nabla S) (\nabla A) e^{- \frac{S(x)}{\hbar}} + \frac{i}{\hbar}(\Delta S) A e^{i \frac{S(x)}{\hbar}}
- \frac{1}{\hbar^{2}}(\nabla S)^{2} A e^{i \frac{S(x)}{\hbar}}
\end{align*}
Entering this into the TISE
\begin{align*}
\left(
- \frac{\hbar^{2}}{2m} \Delta + V(x)
\right)
\psi(x)
= E \psi(x)
\end{align*}
and separating real and imaginary parts, we obtain
\begin{align*}
\frac{(\nabla S)^{2}}{2m} + (V(x) - E)
&=
\frac{\hbar^{2}}{2m} \frac{\Delta A}{A}
\\
(\nabla A) (\nabla S) + \frac{A}{2} \Delta S
&= 0
\end{align*}
If we think of $\hbar$ as some parameter that describes the deformation of classical physics\footnote{See \url{https://ncatlab.org/nlab/show/classical+limit}}, we can consider the limit $\hbar \to 0$
and obtain the time-independent Hamilton-Jacobi equation
\begin{align*}
\frac{(\nabla S)^{2}}{2m} + V(x)
= H(x, \nabla S) = E
\end{align*}
\subsection{Energy Eigenstates}
Starting with the TISE $H \psi = E \psi$, we can think of $H$ as some infinite matrix, as it's just a linear operator.
Since a diagonalisable $n \times n$ matrix has $n$ solutions, one would assume that, if $H$ were diagonalisable, we would find that $H \psi = E \psi$ has $n = \infty$ many solutions.
What we can do is add a parameter and solve $H \psi_n = E_n \psi_n$ instead and hope to find all (and hopefully unique) solutions to this.
If $\psi_n$ form a basis of the solution space, then we can say that any other solution to the Schrödinger equation is a linear combination of the basis vectors, i.e.
\begin{align*}
\psi(x) = \sum_{n \in \N} a_n \psi_n(x)
\end{align*}
and in order to develop it over time, we instead develop the $\psi_n$ over time and find that
\begin{align*}
\Psi(x,t) = \sum_{n \in \N} a_n \psi_n(x) \exp\left(
- \frac{i E_n}{\hbar}t
\right)
\end{align*}
Then $\Psi(x,t)$, being a linear combination of solutions of the TISE, should also be a solution for the boundary problem.
What is nice is that we only have to solve for the $\psi_n$ once, and then for any other problem, we can reuse the $\psi_n$.
\subsection{Energy measurements}
For now, let's assume that the $(\psi_n)_{n \in \N}$ form an orthonormal basis of Eigenfunctions of $H \psi = E_n \psi_n$.
Another assumption we often make is that there is no \textbf{degeneracy}, i.e. that $E_n \neq E_m$ for $n \neq m$.
Orthonormalcy means that
\begin{align*}
\int_\R dx \psi_n^{\ast}(x) \psi_m(x) = \delta_{nm}
\end{align*}
In fact, it is possible to prove this if we assume that there is no degeneracy.
\begin{proof}[No degeneracy $\implies$ ONB]
Since they satisfy the schrödinger equation, we have
\begin{align*}
E_n \int_{\R} dx \psi_n^{\ast}(x) \psi_m(x)
&=
\int_{\R} dx \psi_n^{\ast}(x) H \psi_m(x)\\
&=
- \frac{\hbar^{2}}{2m} \int_{\R} dx \psi_n^{\ast}(x) \psi_m''(x)
+
\int_\R dx \psi_n^{\ast}V(x) \psi_m(x)
\end{align*}
Since $V(x)$ is real-valued, we can write the integrand of the second term as $(V(x) \psi_n(x))^{\ast} \psi_m(x)$
Then by doing partial integration of the first term, we get
\begin{align*}
- \frac{\hbar^{2}}{2m}
\int_{\R} dx \psi_n^{\ast}(x) \psi_m''(x)
&=
\ldots
\end{align*}
in the end, we obtain
\begin{align*}
E_m \int_{\R} dx \psi_n^{\ast}(x) \psi_m(x) = \int_{\R} dx (H \psi_n)^{\ast} \psi_m(x) = E_n^{\ast} \int_{\R} \psi_n^{\ast}(x) \psi_m(x)
\end{align*}
in particular, if $n = m$, we see that $E_n = E_n^{\ast}$ so the energy is real.
Subtracting both sides, we see that
\begin{align*}
(E_m - E_n) \int_{\R} dx \psi_n^{\ast}(x) \psi_m(x) = 0
\quad \text{for} \quad m \neq n
\end{align*}
\end{proof}
This shows that the general solution to the SE can always be written in the form
\begin{empheq}[box=\bluebase]{align*}
\Psi(x,t) = \sum_{n \in \N} a_n \psi_n(x) \exp\left(
- \frac{i E_n}{\hbar}t
\right)
\end{empheq}
Now, let's calculate the expectation value for the energy.
It is given by
\begin{align*}
\scal{H}(\Psi)
= \braket{\Psi|H|\Psi}
&= \int_\R \Psi(x,t)^{\ast} H \Psi(x,t)\\
&=
\sum_{m \in \N} \sum_{n \in \N}
a_m^{\ast} a_n
\exp\left(
- \frac{i (E_n- E_m)}{\hbar} t
\right)
\underbrace{
\int_{\R} dx \psi_m^{\ast}(x) \overbrace{H \psi_n(x)}^{= E_n \psi_n(x)}}_{= E_n \delta_{mn}}\\
&=
\sum_{n \in \N} \abs{a_n}^{2} E_n
\end{align*}
This shows that the possible measurements of a system are exactly the Eigenvalues of the Hamiltonian $H$.
One way to interpret this is to say that the values $E_n$ are the possible values of energy and the coefficient $\abs{a_n}^{2}$ determines the probability that $E = E_n$.
This gives us another postulate of wave mechanics.
\begin{enumerate}
\setcounter{enumi}{3}
\item The possible values for measurements correspond to the Eigenvalues of the Hamiltonian.
\item The probability to measure the energy $E_n$ is given by
\begin{align*}
\abs{\braket{\psi_n|\Psi}}^{2}
= \abs{\int dx\psi_n^{\ast} \Psi}^{2}
= \abs{\sum_{m} \underbrace{\braket{\psi_n|a_m \psi_m \exp\left(
- \frac{i E_m t}{\hbar}
\right)}}_{= \delta_{nm}}
}^{2}
=
\abs{a_n}^{2}
\end{align*}
\end{enumerate}
Let's assume we measure some energy $E = E_n$ and we let the system ``idle'', then we should expect that over time, its energy should not change.
If we were to measure the energy again some time later, we would expect to measure the same energy $E = E_n$ again, so the probability $\abs{a_n}^{2} = 1$ and for all other $m \neq 0$, we have $a_m = 0$.
This means that \emph{after the measurement}, the wave function must be given by
\begin{align*}
\Psi = A \psi_n \exp\left(- \frac{i E_n t}{\hbar}
\right)
\end{align*}
In other words: After the measurement, the wave function \emph{collapses} into the state $n$.
This diverges from our physical intuition of measureing things.
Looking at a ball doesn't really change its position or momentum.
But one way to expect this collapse of the wave function is that in order to \emph{measure} its position, we need to shine some light on it and have it bounce back to us.
But this photon will carry some momentum and therefore bouncing off of some particle will of course change its trajectory.
There are many types of interpretations about the collapse of wavefunctions, but mathematically speaking, they shouldn't really affect how we move forwards.
\section{Example of simple systems}
Before we formalize quamtum mechanics, we will build some intuition by solving simple, one-dimensional examples.
\subsection{Particle in a Box}
We consider a particle that is trapped in a 1D Box, say the interval $[0,a]$.
This can be described by the potential
\begin{align*}
V(x) =
\left\{\begin{array}{ll}
0 & 0 \leq x \leq a\\
\infty & \text{ otherwise}
\end{array} \right.
\end{align*}
We start solving the TISE by looking for a basis of Eigenfunctions $\psi$ satisfying $H \psi = E \psi$.
Since the potential outside of the box is infinite, we need that $\psi(x) = 0$, or else the particle would have infinite energy.
Inside of the box, the TISE reads
\begin{align*}
H \psi =
- \frac{\hbar^{2}}{2m} \frac{d^{2}}{d x^{2}} = E \psi
\end{align*}
The most general solutions for this is given by
\begin{align*}
\psi(x) =
\left\{\begin{array}{ll}
A \cosh\left(
\sqrt{2m \abs{E} \frac{x}{\hbar}}
\right)
+
B \sinh\left(
\sqrt{2m \abs{E} \frac{x}{\hbar}}
\right)
& \text{ if } E < 0\\
A + Bx & \text{ if } E = 0
\\
A \cos\left(
\sqrt{2m \abs{E} \frac{x}{\hbar}}
\right)
+
B \sin\left(
\sqrt{2m \abs{E} \frac{x}{\hbar}}
\right)
&
\text{ if } E > 0
\end{array} \right.
\end{align*}
Now we look for boundary conditions.
Since we are taking a second derivative, we need that the $\psi$ and its derivative must at least be continuous.
In particular, continuity implies $\psi(0) = 0$, so in all three cases, we have $A = 0$.
The condition $\psi(a) = 0$ means that
in the cases $E \leq 0$, we find $B = 0$,
so if $E \leq 0$, we only have the trivial solution, which cannot be normalized, so we discard it.
This leaves us with the case $E > 0$.
In order for the sine to vanish at $x = a$, we get the condition
$\sqrt{2m \abs{E_n}} = \frac{n \pi \hbar}{a}$,
so the possible values for Energy are
\begin{align*}
E_n = \frac{n^{2} \pi^{2} \hbar^{2}}{2 m a^{2}}
\quad \text{for} \quad n \in \N
\end{align*}
and the non-trivial solutions are
\begin{align*}
\psi_n = B_n \sin\left(
\frac{n \pi x}{a}
\right)
\end{align*}
in order to determine the constant $B_n$, we use the normalisation condition
\begin{align*}
1 = \abs{B_n}^{2} = \int_0^{a} dx \sin^{2} \left(
\frac{n \pi x}{a}
\right)
= \frac{1}{2} a \abs{B_n}^{2}
\end{align*}
which is solved for $B_n = \sqrt{\frac{2}{a}}$.
We note the following
\begin{rem}[]
\begin{enumerate}
\item The energies are discrete
\item $E = 0$ is not a valid solution!
So the particle cannot sit still, which is an example of the Heisenberg uncertainty principle in action. We will prove it later.
\item The Energy corresponds to the number of roots of the function $\abs{\psi}$.
\end{enumerate}
\end{rem}
\subsection{Finite square well}
We consider a particle in a potential
\begin{align*}
V(x) =
\left\{\begin{array}{ll}
- V & \text{ if } x \in [-a,a]\\
0 & \text{ otherwise}
\end{array} \right.
\end{align*}
Again, we look for an Eigenbasis of solutions $H \psi = E \psi$.
Moreover, we want our particle to be \textbf{bounded}, i.e. $\abs{\psi(x)}^{2} \to 0$ as $\abs{x} \to \infty$.
\begin{itemize}
\item[$x > a$] We are solving $- \frac{\hbar^{2}}{2m} \frac{d^{2}}{d x^{2}} \psi(x) = E \psi(x)$.
This has the general solution
\begin{align*}
\psi(x) = A e^{i \kappa x} + D e^{- i \kappa x} \quad \text{where} \quad E = - \frac{\hbar^{2} \kappa^{2}}{2m} \implies \kappa = \frac{\sqrt{-em E}}{\hbar}
\end{align*}
in order to obtain a bounded solution, we need $\kappa \in \R_{> 0}$ and thus $E < 0$.
And to ensure that it can be normalized, we need $A = 0$, so
\begin{align*}
\psi(x) = D e^{- i \kappa x}
\end{align*}
\item[$x < a$] Here we get the same as in the case $x > a$, but since $x < 0$, we get
\begin{align*}
\psi(x) = A e^{i \kappa x}
\end{align*}
\item[$-a \leq x \leq a$]
Since $V(x) = -V$, the TISE reads
$- \frac{\hbar^{2}}{2m} \frac{d^{2}}{d x^{2}} \psi = (E + V) \psi$.
with general solution
\begin{align*}
\psi(x) = B e^{i k x} + C e^{- i k x} \quad \text{and} \quad V + E = - \frac{\hbar^{2} \kappa^{2}}{2m} \implies k = \sqrt{\frac{2m (V + E)}{\hbar^{2}}}
\end{align*}
\end{itemize}
Now we solve the boundary and continuity conditions.
But solving this is quite tedious and would require lots of calculations.
One trick we can use is to make use of \textbf{symmetry}.
Note that the Hamiltonian $H$ is invariant under the transformation $x \mapsto -x$.
Therefore,
\begin{align*}
\psi(-x) = \psi(x) &\implies H \psi(-x) = H \psi(x)
\\
\psi(-x) = \psi(x) &\implies H \psi(-x) = -H \psi(x)
\end{align*}
so the hamiltonian maps even solutions to even ones, and odd ones to odd ones.
Therefore, it suffices to find all even solutions and all odd solutions.
\subsubsection*{Even solutions}
The even solutions require $A = D$ and $B = C$, so we find.
\begin{align*}
\psi(x) =
\left\{\begin{array}{ll}
A e^{- \kappa x} & \text{ if } x > a\\
2B\cos kx & \text{ if } -a \leq x \leq a\\
A e^{\kappa x} & \text{ if } x < -a
\end{array} \right.
\end{align*}
since the function is even, it's enough to consider the continuity conditions at $x = a$, as the ones for $x = -a$ follow immediately. We see
\begin{align*}
\psi \text{ continuous:} \quad
&A e^{- \kappa x} = 2B \cos ka
\frac{d \psi}{d x}\text{ continuous:} \quad
&\psi(x) = \kappa A e^{- \kappa x} = 2B \sin ka
\end{align*}
dividing both equations, we get
\begin{align*}
\kappa = k \tan(k a)
\end{align*}
and by the characterisation for $k, \kappa$ from beore, we get
\begin{align*}
\kappa^{2} + \kappa^{2} = \frac{2mV}{\hbar^{2}}
\end{align*}
For convenience, it's nice to introduce the variables $\eta = \kappa a$ and $\xi = ka$, giving us
\begin{align*}
\eta = \xi \tan \xi \quad \text{and} \quad \eta^{2} + \xi^{2} = \frac{2ma^{2}}{V}\hbar^{2}
\end{align*}
Plotting this in an $f(\xi) = \eta$ graph, we will intersect the graph of $\tan$ and a circle.
And we will do this with the restriction that $\xi, \eta > 0$.
Graphically, one sees that there is always going to be $1$ solution, no matter how small the radius of the circle is.
And the radius gets larger gets larger, there are more and more solutions.
With some calculations, we can find that if $N$ marks the number of solutions, then
\begin{align*}
(N-1)^{2} \pi^{2} < \frac{2 m a^{2} V}{\hbar^{2}} < N^{2} \pi^{2}
\end{align*}
\subsubsection*{Odd solutions}
Similar to the even case, one quickly finds that $A = -D$ and $B = -C$.
The Ansatz is
\begin{align*}
\psi(x) =
\left\{\begin{array}{ll}
-A e^{-\kappa x} & \text{ if }x > a\\
2iB \sin(kx) & \text{ if }-a < x < a\\
A e^{\kappa x} & \text{ if }x < -a
\end{array} \right.
\end{align*}
And with a similar calculation as before, we get
\begin{align*}
\eta = - \xi \cot \xi \quad \text{and} \quad \eta^{2} + \xi^{2} = \frac{2ma^{2}V}{\hbar^{2}}
\end{align*}
Unlike earlier, there is no odd solution if
\begin{align*}
\frac{2ma^{2}V}{\hbar^{2}} < \frac{\pi^{2}}{4}
\end{align*}
This corresponds to our remark earlier, where we showed that the smallest energy value corresponds to the solution with the last amount of roots, which must produce an even function.
\begin{rem}[]
If we constrast this with the naive approach of simply solving the ODE numerically, then one finds that simply guessing $\kappa$ will generate a solution that explodes as $x \to \infty$.
We \emph{need} to use symmetry to find out that the spectrum of values for $\kappa, k$ is discrete.
\end{rem}
What if we consider solutions that unbound?
The following example gives a nice intuition behind them.
\subsection{Step potential}
The potential is given by
\begin{align*}
V(x) =
\left\{\begin{array}{ll}
0 & \text{ if }x \leq 0\\
V_0 & \text{ if }x \geq 0\\
&
\end{array} \right.
\end{align*}
And as always, we solve for eigenfunctions to the TISE $H \psi = E \psi$.
In the region $x < 0$, the general solution is again
\begin{align*}
\psi_k(x) = A e^{\frac{ikx}{\hbar}} + B e^{\frac{-ikx}{\hbar}} \quad \text{where} \quad k = \sqrt{2m E}
\end{align*}
Again, normalisation is only possible if $k \in \R$, and so $E > 0 $.
The term with coefficient $A$ describes a wave moving from the left side, and $B$ a wave reflecting off the potential.
And the reflection probability is given by $R = \abs{\frac{B}{A}}^{2}$.
In the region $x \geq 0$, the TISE reads
\begin{align*}
\frac{d^{2}}{d x^{2}} \psi = - \frac{2m}{\hbar^{2}} (E - V_0) \psi
\end{align*}
Classically, we expect a difference for the case $E < V_0$, where the particle bouncess off the potential, and for the case $E > V_0$, where the particle can go ``over'' the potential.
\subsubsection*{Case I: $E > V_0$}
Here, the TISE is
\begin{align*}
- \frac{\hbar^{2}}{2m} \frac{d^{2}}{d x^{2}}\psi = (E - V_0) \psi
\end{align*}
and the general solution is
\begin{align*}
\psi(x) = C e^{\frac{ilx}{\hbar}} + D e^{- \frac{ilx}{\hbar}} \quad \text{where} \quad l = \sqrt{2m (E - V_0)}
\end{align*}
But the term with coefficent $D$ would describe a particle coming from the right side, which we do not want.
So we are only looking for solutions with $D = 0$.
The continuity condition at $x = 0$ for $\psi$, $\frac{d }{d }\psi$ gives us
\begin{align*}
A + B = C \quad \text{and} \quad ik(A - B) = i l C \implies (A-B) = \frac{l}{k}C
\end{align*}
solving this is easy. In the end, one finds
\begin{align*}
A = \frac{1}{2} C(1 + \frac{l}{k}), \quad B = \frac{1}{2}C(1 - \frac{l}{k})
\end{align*}
and the reflection and transmission probability is
\begin{align*}
R &= \abs{\frac{B}{A}}^{2} = \frac{(1 - \tfrac{l}{k})^{2}}{(1 + \tfrac{l}{k})^{2}}\\
T &= \frac{l}{k} \abs{\frac{C}{A}}^{2} = \frac{l}{k} \frac{4}{(1 + \tfrac{l}{k})^{2}}
\end{align*}
indeed, one verifies that $T + R = 1$, which is expected.
For a sanity check, lets consider the case where $E \gg V_0$. Then we have that $l \sim k$, so most of the wave is transmitted and ``makes it trough'' the barrier.
And if $E$ is just barely bigger than $V_0$, $l \sim 0$ so almost nothing makes it through.
\textbf{Case II:} $E \leq V_0$
Classically, the particle would not make it past the barrier.
The general solution in the region $x \geq 0$ would be
\begin{align*}
\psi(x) = C e^{- \frac{\kappa x}{\hbar}} + D e^{\frac{\kappa x}{\hbar}} \quad \text{where} \quad \kappa = \sqrt{2m(V_0 - E)}
\end{align*}
since the exponent is real, we have to discard solutions with coefficient $D \neq 0$, or else it would explode as $x \to \infty$.
Solving the boundary conditions yield
\begin{align*}
A = \frac{1}{2} C(1 + \frac{i \kappa}{k}), \quad
B = \frac{1}{2} C(1 - \frac{i \kappa}{k})
\end{align*}
Calculating the probability of reflection, we find that $R = \abs{\frac{B}{A}}^{2} = 1$.
But nevertheless, one finds that $\psi \neq 0$ in the region $x \geq 0$ and instead only finds exponential decay.
However, this does not change anything about the probability current, because since the wave-function is real in that region, we have
\begin{align*}
J = \left(
\psi^{\ast} \del_x \psi - \psi \del_x \psi^{\ast}
\right)
= 0 \quad \text{ for} \quad x > 0
\end{align*}
In the limit case $V_0 \to \infty$, we also have $\kappa \to 0$ and thus the wave-function is identitally zero in this region.
\subsection{Normalisation and Wavepackets}
Consider the general eigenvector to the TISE for a free particle $\psi_k(x) = e^{\frac{i k x}{\hbar}}$, $k = \sqrt{2m E}$.
Such wave-functions are not square-integrable and cannot be normalized!
But we don't care. These are just the basis vectors of our solution space. An actual solution is a linear combination of these eigenvectors
\begin{align*}
\Psi(x,t) = \sum_{k} a_k \psi_k(x) e^{- \frac{i E_k t}{\hbar}} \quad \text{for} \quad E_k = \frac{k^{2}}{2m}
\end{align*}
or, in the continuous case
\begin{align*}
\Psi(x,t) = \int dk\ f(k) \psi_k(x) e^{- \frac{i E_k t}{\hbar}}
\end{align*}
The normalisation conditions then becomes
\begin{align*}
\int dx \abs{\Psi(x,t)}^{2}
&=
\frac{1}{2 \pi}
\int dx \int dk \int d \tilde{k} f(k) e^{\frac{ikx}{\hbar}} e^{- \frac{i E_kt}{\hbar}}
f^{\ast}(\tilde{k}) e^{- \frac{i \tilde{k}x}{\hbar}} e^{\frac{i E_{\tilde{k}}t}{\hbar}}
\\
&=
\frac{1}{2 \pi}
\int d \tilde{k} \int d k f(k) f^{\ast}(\tilde{k}) e^{- \frac{t}{\hbar}(E_{\tilde{k}} - E_k)} \underbrace{\int dx e^{\frac{ix}{\hbar}(k - \tilde{k})}}_{= 2 \pi \delta(k - \tilde{k})}
\\
&=
\int dk \abs{f(k)}^{2} \stackrel{!}{=}1
\end{align*}
so the normalisation condition of the wave-function is not a condition about the form of the eigenvectors of the TISE, but rather a condtion about \emph{how these eigenvectors are combined}.
\subsection{Finite square well, positive energies}
We come back to the potential
\begin{align*}
V(x) = \left\{\begin{array}{ll}
-V & \text{ if } x \in [-a,a]\\
0 & \text{ otherwise}
\end{array} \right.
\end{align*}
and consider solutions to $H \psi = E \psi$ for $E > 0$.
Setting
\begin{align*}
k = \frac{\sqrt{2 m E}}{\hbar}, \quad l = \frac{\sqrt{2 m (E +V)}}{\hbar}
\end{align*}
the general solutions are
\begin{align*}
\psi(x) =
\left\{\begin{array}{ll}
A e^{ilx} + B^{-ilx} & \text{ for } -a < x < a\\
Ce^{ikx} + r^{-ikx} & \text{ for } \abs{x} x < 0\\
te^{ikx} + F^{-ikx} & \text{ for } \abs{x} x > 0
\end{array} \right.
\end{align*}
since we assume that the particle comes form the left side, and not from the right side, we set $F = 0$.
Here, $r$ corresponds to the amplitude of the reflected, and $t$ of the transmitted wave.
The continuity conditions for $\psi,\frac{d \psi}{d x}$ at $x = a$ give
\begin{align*}
\psi(x)|_{x = a}&: \qquad Ae^{ila} = \frac{1}{2} t e^{ila}(1 + \tfrac{l}{2})
\\
\psi'(x)|_{x = a}&: \qquad Be^{-ila} = \frac{1}{2} t e^{ila}(1 - \tfrac{l}{2})
\end{align*}
and at $x = -a$: we get
\begin{align*}
e^{-ika} + re^{ika} &= A e^{-ila} + B^{ila}\\
e^{-ika} - re^{ika} &= \frac{l}{k}\left(A e^{-ila} - B^{ila}\right)
\end{align*}
in the end, we find
\begin{align*}
t = \frac{
e^{-2ika}
}{
\cos(2la) - i \frac{l^{2} + k^{2}}{2lk} \sin(2la)
}
\end{align*}
which means that the transmission probability is
\begin{align*}