This is a post for my second year Phys Chem students but it may have tidbits of interest to others.
Particleinabox models are great for seeing how quantum works.
Particleinabox models are not just toy models, either. For example:

The images above show the probability density waves of a surface electron trapped in a “ring” of iron atoms placed on a copper surface by scanning tunneling microscopy (STM). The shapes of these waves are well modelled by an electron trapped in a diskshaped box.

The particleonaring model has also been used by my colleague PierreFrancois Loos to develop insight into electron correlation.

Later in this post I will show how such models can be used to calculate conduction of an electron “through” a series of barriers e.g. for designing nanoelectronic devices.
For students
To understand these notes you need to

Know basic calculus (derivatives and integrals) and you need to understand the basics of complex numbers. If you do not, bail out here, and come back after you know.

Second, you should read the previous super quick intro to quantum mechanics
Scattered through this and the previous post are some questions. Carefully study these questions and their answers. The exam questions will be very similar, if not identical.
The later material on transmission through multiple barriers is not examinable. I present it because it could be useful if you are interested in, for example, modelling energy levels and conduction through molecules (Prof. George Koutsantonis is interested in making such “molecular wires”).
What is a particleinabox problem?
Particle in a box problems are those for a single particle moving in one dimension subject to a piecewise constant potential energy i.e. a potential energy that looks like the following function:
From looking at this picture you can see that the particle experiences different “constant” energies at different points along the axis. In fact, if we wanted to write this potential in “maths” it would be this:
In this case the article seems to encounter a kind of “step barrier”. Of course, other shapes for \( V(x) \) are possible and each leads to a different problem.
Strategy for solving particleinabox problems
The strategy for solving these one dimensional problems is always the same:

Write down the Hamiltonian for the problem,
and hence the Schrodinger equation \( \hat{H}\psi = E\psi \) for the unknown energies \( E \) and unknown wavefunctions \( \psi \).

Write down the general solution in each region where the potential is constant. The general solutions in each region is always of the form \( \psi(x) = A e^{B x}\) where \( A \) and \(B\) are (possibly complex) unknown constants to be found. To save time you can learn these possible cases to be applied in different regions where the potential is constant:

\( \boxed{\psi(x) = A e^{i k x}} \) or \( \boxed{\psi(x) = A e^{i k x} + B e^{i k x}} \) for free or unbound particles.

A free particle is one where \( V(x) = 0 \).

An unbound particle has energy \( E \) greater than the largest value of the potential \( V(x) \) for any \( x \) i.e. the particle has enough energy to overcome any barrier.

A free particle is unbound, but an unbound particle is not necessarily free.

The first solution is for free particle with a well defined momentum, the second solution does not have well defined momentum. See Q12 below.


\( \boxed{\psi(x) = A e^{\alpha x} + B e^{+\beta x}} \) for electrons in a region where the energy of the particle \( E \) is less than the value of the potential. This is called an exponential decay region or barrierpenetration region of the wavefunction.

\( \boxed{\psi(x) = A \sin kx + B \cos kx} \) for electrons where the energy of the particle \( E \) is greater than the value of the potential. This is called an oscillatory region of the wavefunction.


Find the unknown constants in the wavefunction(s) by making sure that the solutions are continuous and smooth  see Rule 1 of quantum mechanics in the previous post. At the end of the problem there will usually only one unknown constant left which can be found by the normalization condition if needed.
A pretty good graphic for illustrating this procedure is shown at the hyperphysics site here.
Q9. What is the basic strategy for solving particle in a box problems? You do not need to list the possible general solutions for each case.
Answer
Write down the Schrodinger equation. Write down the general solutions to the Schrodinger equation in each region where the potential is constant. Find the unknown constants in the different pieces of the wavefunction for each region by makeing sure that it is continuous and smooth.
Q10. Show that the most general solution for a free particle \( \psi(x) = A e^{i k x} + B e^{i k x} \) is the same as the solution for a bound particle \( \psi(x) =C \sin kx + D \cos kx \)
Strategy
From the de Moivre formula \( e^{i k x} = \cos kx + i \sin kx \) it looks as if the first solution can be written using terms which look like those in the second. Therefore use de Moivre in the first and try to obtain the second. We will probably have to use the trig identities \( \cos (kx) = \cos (kx) \) and \( \sin(kx) =  \sin (kx) \).
Answer
The last line is the same as \( C \cos kx + D \sin kx \) provided \( C = (A+B) \) and \( D = (AB) \).
This shows that there is, actually, no mathematical difference between these general solutions. In fact, by choosing the constant \( k \) to be pure complex \( k’ = i k \) we can show that the general bound solutions are also equivalent. The main reason to use these different forms is that they are more convenient.
Free electron in one dimension  otherwise known as the particle not in a box
Now we get to the meat. Note that the meat is not in a box.
Q11. Solve the Schrodinger equation for an electron moving freely in one dimension.
Strategy
The strategy for solving these problems was given above. The Schrodinger equation for the free particle was written down in Q5 in the last post. Basically a free particle means that \( V(x) = 0 \). But before applying the strategy and solving the equation let’s think what answer we might expect. Actually, since the electron can move freely, we would expect that it can have … any energy it likes … as long as it is positive. Also, we expect that an electron moving freely should be located … well, anywhere it likes … meaning to say that the probability of finding it at any position is equally likely.
Answer
From Q5 the Schrodinger equation for the free electron is
From the strategy we know a possible solution of this equation is \( \psi(x) = A e^{i k x} \). Note that the constant \( i k \) is explicitly complex, so that \( k \) is a real number. To find the unknown numbers \( A \) and \(k \) we substitute this general form into the left hand side of the equation and simplify:
This shows that \( \psi(x) = A e^{i k x} \) is indeed a solution of our Schrodinger equation provided that
Q11 is now solved!
Note that, as we expected:

\( E \) is positive regardless of whether the arbitrary constant \( k \) is positive or negative (since the square of an arbitrary number is always positive).

The probability of fiding the electron an any position is \( \psi^{\ast}(x) \psi(x) = A^{\ast} e^{i k x} A e^{i k x} = A^2 \) and does not depend on \( x \) i.e. the probability of finding the particle at one place or another is the same i.e. the particle is located everywhere with equal probability.
Q12. Show that the free electron wavefunction \( \psi = A e^{i k x} \) has momentum \( \hbar k \). Show that the free electron wavefunction \( \psi = A e^{i k x} \) has momentum \( \hbar k \). What is the meaning of these two wavefunctions?
Strategy
According to the Rules of QM, a wavefunction has a certain momentum if it is an eigenfunction of the corresponding operator. Hence we need to show that \( \psi(x) \) is an eigenfunction of the momentum \( \hat{p} \).
Answer
We evaluate \( \hat{p}\psi(x) \) for the wavefunction in Q11
This is an eigenvalue equation with eigenvalue \( \hbar k \). Therefore the momentum of a free electron with wave function \( \psi(x) = e^{i k x} \) is \( \hbar k \).
Likewise:
This is an eigenvalue equation with eigenvalue \( \hbar k \). Therefore the momentum of a free electron with wave function \( \psi(x) = e^{i k x} \) is \( \hbar k \).
Clearly the wavefunction \( \psi(x) = A e^{i k x} \) corresponds to an electron moving with positive momentum (i.e. moving to the right) while \( \psi(x) = A e^{i k x} \) corresponds to an electron with negative momentum (i.e. moving to the left), assuming \( k>0 \).
Note that the unknown constant \( A \) can not be determined by the normalization condition: recall that according to the Rules of QM, normalization only applies to bound states.
Q13. Show that the wavefunction \( \psi = A e^{i k x} + B e^{i k x} \) is also a solution to the free electron Schrodinger equation. Is this a valid wavefunction?
Strategy
Substitute this wavefunction into the free particle Schrodinger equation and see if it is an eigenfunction i.e. see if it obeys \( \hat{H}\psi = E \psi \).
Answer
You can check that \( \psi(x) = B e^{i k x} \) is a solution to the same free electron Schrodinger equation by substitution. In fact, the algebra is identical except that instead of \( k \) the opposite value \( k \) is used. Since both \( A e^{i k x} \) and \( B e^{i k x} \) are separately solutions, their sum must also be i.e.
is also a solution.
Is this a valid solution?
Well, yes.
However, you can show the probability density for this wavefunction is
The last equality follows from noting that the cross terms are complex conjugates. Therefore this wavefunction does not have a constant probability density. There is no explicit rule to forbid such a wavefunction, but it does not quite agree with what we might have expected.
This question illustrates a very important point about quantum mechanics:
If we have more than one solution to the SChrodinger equation with the same energy, then we can take a linear combination of these solutions, like above with arbitrary constants \( A \) and \( B \), and get a different wavefunction with the energy! There are an inifinite number of such wavefunctions corresponding to the same energy  just choose different arbitrary constants. This is the idea behind hybrid atomic orbitals in the valence bond theory of chemical bonding. In the hybrid atomic orbital theory, linear combinations of wavefunctions of equal or nearly equal energy, like the s and p orbitals, are used to explain how atomic orbitals adjust themselves to maximize the bonding to neighbouring atoms around a central atom.
Particle in a box
Q14. Solve the Schrodinger equation for an electron in a box of length \( L \) which starts at the origin (the potential is infinite outside the box).
Strategy
Draw a picture and apply the given strategy above. Note that the the wavefunction \( \psi(x) = 0 \) outside the box; if not we would get infinity appearing in the equation \( \hat{H}\psi(x) = E\psi(x) \) when \(x \) is outside the box.
Answer
The potential \( V(x) \) looks as shown below. Also shown are the solutions which we have yet to find.
According to the strategy we should try wavefunctions of the form
Substituting in the Schrodinger equation we get
So as before we see that \( \psi(x) = A \sin k x + B \cos k x \) is a solution of the Schrodinger equation with
In fact this solution would be acceptable as an answer to the “free particle” problem. The difference is now that we have to apply the boundary conditions on the wavefunction, which are from the picture:

Substituting \( \psi(x) \) in the first equation above determines the value of the unknown constant \( B \) as follows:
Here we used the standard values of \( \sin 0 = 0 \) and \( \cos 0 = 1 \). Therefore we have figured out that \( \psi(x) = A \sin k x \).

Now we apply the second boundary condition and see what we get:
This equation has solutions only when the angle \( k L = n \pi \) radians where \( n = 1, 2, \ldots \). Note that the value \( n = 0 \) leads to a a wavefunction \( \psi(x) = 0 \) which has no area underneath it i.e. it is unnormalizable and therefore doesn’t obey the rules for an allowed wavefunction. Likewise although \( n = 1, 2, 3, \ldots \) are also solutions of the above equation, they do not lead to different wavefunction solutions since \( \psi(x) = \sin kx = \sin (kx) \) due to the properties of the sine function.
The last boundary condition equation gives a value for the unknown \( k \) constant which means it determines the allowed energy eigenvalues:
Corresponding to wavefunction \( \psi(x) = A \sin \displaystyle\frac{n \pi}{L} x \).
To find the unknown constant \( A \) we must normalize the wavefunction,
From which we get:
We can evaluate this integral by looking up tables if we like. The important thing is not the answer but how to get it.
Particle in a half infinite box
Q15. Solve the Schrodinger equation for an electron in a box of length \( L \) which starts at the origin. The potential is infinite to the left of the box, and has a given value \( V_0 \) to the right. The energy of the electron is less than \( V_0 \).
Strategy.
 First draw a picture of the potential. It looks similar like the previous question, except that to the right, the height of the box is \( V_0 \) and not infinite.
 The wavefunction to the left of the box is zero as explained in the previous question.
 There are two other wavefunctions \(\psi\sub\s{in} \) and \(\psi\sub\s{out}\) for the inside and outside (right) of the box respectively. These two wavefunctions need to be “joined” at the boundary to make a “smooth” wavefunction.
 Inside the box, the wavefunction is the same as before, \( \psi\sub\s{in}(x) = A \sin k x \). This satisfies the boundary condition \( \psi\sub{in}(0) = 0 \).
 Outside the box, according to strategy above, the wavefunction outside the box is \( \psi\sub\s{out}(x) = B e^{\beta x} + C e^{\beta x} \). However the part \( C e^{\beta x } \) gets bigger as \(x \) gets larger and so if constant \( C \) is not zero the wavefunction would have an infinite area underneath it i.e. it would be unnormalizable. Therefore the wavefunction outside the box is \( \psi\sub\s{out}(x) = B e^{\beta x} \).
 Finally, we substitute \(\psi\sub\s{in}\) and \(\psi\sub\s{out}\) into the Schrodinger equation and apply the boundary conditions to get the answer.
Answer
Here is the picture of the potential:
Now let’s substitute in our general solutions for each region.

Exactly as before, we substitute \( \psi\sub\s{in}(x) = A \sin k x \) into the Schrodinger equation to get

Outside the box i.e. for \( 0 < x < L \) we have a slightly different Schrodinger equation:
From the above equation we can see that we need a second derivative of \( \psi\sub\s{out}(x) = B e^{\beta x} \), so lets do that:
Using this result, the Schrodinger equation outside the box becomes:
So we get another equation for the energy,
Now we can equate our two energy formula (the one inside the box and the one outside the box) to get an expression for \( \beta \) in terms of \( E \), and hence \( k \) like this:
This shows that \( \beta \) is determined in terms of the unknown constant \( k \).
Now we need to apply the boundary conditions, which are:
The first condition ensures that the wavefunctions are “unbroken” or “continuous”. The second condition ensures that the derivatives of the two wavefunction match so the join is “smooth”.

The first condition gives

The second condition gives
We can eliminate the constants \( A \) and \( B \) by dividing the second equation by the first. The result is:
This is a difficult equation to solve. In general it has to be solved numerically once we are given a value for \( V_0 \). Before choosing a value let’s simplify the problem a bit by using atomic units i.e. units where we set \( \hbar = m_e = 1 \). Atomic units are widely used in quantum chemistry. Let’s further assume that \( L = 1 \) in these units. Then our equation becomes:
All we need now is a value of \( V_0 \). Let’s take \( V_0 = 18 \). We can now solve this equation numerically.
The easiest way to do it, go to the Wolfram Alpha web site and type in \( \texttt{k cot k = (36  k^2)^(1/2)} \). You will get two solutions: \( k = 2.679, 5.226 \). Cool!
What about if \( V_0 \) is very large? Don’t we expect to get something close to the particleinaninfinitebox problem?
Let’s check. Choose \( V_0 = 500 \) and type into Wolfram Alpha \( \texttt{k cot k = (1000  k^2)^(1/2)} \). You should get the results \( k = 3.05, 6.09, 12.17 \ldots \). These values are indeed close to the solutions \( k = \pi, 2\pi, 3\pi, \ldots \). Everything seems to make sense!
Below are the wavefunctions for a particular case of the half infinite box. You should be able to draw a sketch of these. See how the exponential part in the barrier ouside joins smoothly up with the sin wave part in the box?
Note that the wavefunction is not zero inside the barrier. Therefore, there is a non zero probability for it to be inside the barrier even though the particle does not have enough energy to go in! If the particle were a person, this would be like saying that they could walk inside a wall!
You can also see on these plots the solutions in the case where the energy of the particle is greater than the barrier height \( V_0 \). To get these solutions we have to assume a different general solution in that part and repeat the whole procedure above.
Congratulations!
If you’ve got this far then congratulations! You understand the basics of quantum mechanics.
Tunneling through a barrier
The following is not for examination, but for interest only.
(Hello? Is anyone out there? Or am I the only wierdo around?)
This example illustrates the fact that, not only can a quantum particle exist inside a barrier, it can tunnel right through!
To see this we solve the Schrodinger equation for this potential
This is called a barrier potential as the picture below makes clear.
As before, let’s consider the case \( E<V_0 \) where the particle does not have enough classical energy to penetrate the barrier.
The general solutions in each region have the form:
These are general wavefunctions in the left (L), middle (M) and right (R) regions. Plugging these into the Schrodinger equation we get
Now let’s deal with the boundary conditions. To ensure that the wavefunctions are continuous and smooth:

The boundary conditions at \( x=0 \) are:
It is easier to write this as a matrix equation

The boundary conditions at \(x=a\) are:
The general solution to these equations will always have at least two unknown constants e.g. \(A_1\) describing momentum from the left or \(B_2\) describing momentum from the right. The transmission and reflection characteristic relating the left wavefunction to the right wavefunction are given by the transfer matrix \(\B{P}\) which is defined by
From the above definitions we see that
so that
After a lot of tedious algebra making use of standard formulas for inverting matrices, we get
Note: this transfer matrix method can be generalized to calculate transmission through any series of barriers—in case you ever need to do that for, say, nanotechnology applications.
Of course, you would not do the algebra by hand, but set these equations up in a computer algebra program like Wolfram Alpha
Now for some results.
Assume there is only momentum coming from the left (\(A_1\neq 0\), \(B_1=0 \)) but none on the right (\(B_2=0\), \(B_1\neq 0\)). The transmitted probability flux to the right is then:
Note that the (probability) velocity opertor used above to calculate the probability flux is the real part of \(\hat{v} = \hat{p}/m \). It is just the momentum divided by the mass.
We define the transmission coefficient by
Likewise the reflection coefficient by is defined as
It may be verified (tediously hand, but easily with Wolfram Alpha that:
Plots of the transmission and reflection coefficient are shown in the following graphs (from the highly recommended textbook by Ballentine) assuming \(L=\hbar=1=2m=1\), for \(E=0.16\) and \(E=1.0\) respectively.
The formula for the transmission coefficient is:
The transmission coefficient is plotted below:
Some comments:

\( \psi \) does not simply decrease in the barrier. However, \(\psi\) always decreases

\( \psi \) always has zero slop at the exit on the right hand side because \( \psi\) is constant on the right hand side

The particle can tunnel right through the barrier, as advertised.

The transmission coefficient decreases exponentially with thickness.