# Materials in Electronics/Confined Particles/1D Finite Wells

We have seen the result of an infinite potential well, but this is not a situation that can exist in real life - a point cannot have infinite potential with respect to another. Let us consider the symmetric finite potential well, right. It is easier mathematically if the well is centred around the origin. As before, the potential in the well (between -L and L) is 0, but outside the well, it is a finite value, V (equal on both sides):
       $V\left(x\right)={\begin{cases}V,\quad x<-L\\0,\quad -L\leq x\leq L\\V,\quad x>L\end{cases}}$
Because the potential outside is finite, an electron could escape from the well even if its potential was less than V (quantum tunnelling), so we expect to see a wavefunction which does not go to zero at the boundaries of the well.
Let's break up our wavefunction into three pieces, one on each side and one inside the well:
       $\psi \left(x\right)={\begin{cases}\psi _{1}\left(x\right),&\quad x<-L\\\psi _{2}\left(x\right),&\quad -L\leq x\leq L\\\psi _{3}\left(x\right),&\quad x>L\end{cases}}$
We will solve each of the wavefunctions separately using Schrödinger's Equation:
[1D Time-Independent Schrödinger's Equation]
       $-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dx^{2}}}\psi \left(x\right)+V(x)\psi =E\psi$

## Inside the Well

Inside the well, the potential is zero, so we can rewrite Schrödinger's Equation as follows:
       $-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dx^{2}}}\psi _{2}\left(x\right)=E\psi _{2}$
Rearranging,
       ${{d^{2}} \over {dx^{2}}}\psi _{2}\left(x\right)=-{{2mE} \over \hbar ^{2}}\psi _{2}\left(x\right)$
Recall that the energy is given by
       $E\,$ $=T\left(x\right)+V\left(x\right)\,$ $={{\hbar ^{2}k^{2}} \over {2m}}$
Substituting for E in our differential equation gives:
       ${{d^{2}\psi _{2}\left(x\right)} \over {dx^{2}}}=-k^{2}\psi _{2}\left(x\right)$
This is now a second-order differential equation with a standard form. The general solution is given by:
[General Solution for ψ_2]
       $\psi _{2}\left(x\right)=A\sin {kx}+B\cos {kx}$
The energies here are
       $E={\frac {k^{2}\hbar ^{2}}{2m}}$
In this general solution, A and B can be any complex numbers, and k can be any real number.

## Outside the Well

We will now look at the region with the wavefunction ψ$_1$, i.e. to the left. The region to the right of the well will have an identical derivation. For the region outside of the box to the left, since the potential is constant, V(x) = V our statement of Schrödinger's Equation becomes:
       $-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dx^{2}}}\psi _{1}\left(x\right)=(E-V)\psi _{1}$
There are two possible families of solutions, depending on whether E is less than V (the particle is bound in the potential well) or E is greater than V (the particle is free).

### Unbound State

For a free particle, E > V, and letting
       $\kappa ^{2}={\frac {2m(E-V)}{\hbar ^{2}}}$
produces
       ${\frac {d^{2}}{dx^{2}}}\psi _{1}\left(x\right)=-\kappa ^{2}\psi _{1}$
with the same solution form as inside the well:
       $\psi _{1}=C\sin(\kappa x)+D\cos(\kappa x)\quad$

### Bound State

This analysis will first focus on the bound state, where E-V. Letting
       $\alpha ^{2}={\frac {2m(V-E)}{\hbar ^{2}}}$
produces
       ${\frac {d^{2}}{dx^{2}}}\psi _{1}\left(x\right)=\alpha ^{2}\psi _{1}\left(x\right)$
where the general solution is exponential:
       $\psi _{1}\left(x\right)=Fe^{-\alpha x}+Ge^{\alpha x}\,\!$
Similarly, for the other region outside the box:
       $\psi _{3}\left(x\right)=He^{-\alpha x}+Ie^{\alpha x}\,\!$
Now in order to find the specific solution for the problem at hand, we must specify the appropriate boundary conditions and find the values for A , B , F , G , H and I that satisfy those conditions.

## Finding Wavefunctions for the Bound State

Solutions to the Schrödinger Equation must be continuous, and continuously differentiable. These requirements are boundary conditions on the differential equations previously derived.
In this case, the finite potential well is symmetrical, so symmetry can be exploited to reduce the necessary calculations, by choosing wavefunctions of well defined parity.
Summarizing the previous sections:
       $\psi \left(x\right)={\begin{cases}\psi _{1}\left(x\right),&\quad x<-L\\\psi _{2}\left(x\right),&\quad -L\leq x\leq L\\\psi _{3}\left(x\right),&\quad x>L\end{cases}}$
where we found the wavefunctions for the bound state, $\psi _{1},\psi _{2}\,\!$ and $\psi _{3}\,\!$ to be:
       $\psi _{1}\left(x\right)=Fe^{-\alpha x}+Ge^{\alpha x}\,\!$
       $\psi _{2}\left(x\right)=A\sin(kx)+B\cos(kx)\quad$
       $\psi _{3}\left(x\right)=He^{-\alpha x}+Ie^{\alpha x}\,\!$
Where
       $k={\frac {\sqrt {2mE}}{\hbar }}$
       $\alpha ={\frac {\sqrt {2m(V-E)}}{\hbar }}$
We see that as x goes to $-\infty$, the F term goes to infinity. Likewise, as x goes to $+\infty$, the I term goes to infinity. As the wave function must be finite for all x, this means we must set F = I = 0, and we have:
       $\psi _{1}\left(x\right)=Ge^{\alpha x}\,\!$
       $\psi _{2}\left(x\right)=A\sin(kx)+B\cos(kx)$
       $\psi _{3}\left(x\right)=He^{-\alpha x}\,\!$
Next, we know that the overall $\psi \,\!$ function must be continuous and differentiable. In other words the values of the functions and their derivatives must match up at the dividing points:
       $\psi _{1}(-L)=\psi _{2}(-L)\,\!$ $\psi _{2}(L)=\psi _{3}(L)\,\!$ ${\frac {d\psi _{1}}{dx}}(-L)={\frac {d\psi _{2}}{dx}}(-L)\,\!$ ${\frac {d\psi _{2}}{dx}}(L)={\frac {d\psi _{3}}{dx}}(L)\,\!$
Firstly, we need to recall that there must exist wavefunction solutions with well defined parity, odd or even. Since the sine function is odd and the cosine function is even, we can search for a wavefunction solution in the well of the form of sine or cosine. We can now analyse the even and odd wavefunctions separately. It will be shown that for bound states, the even an odd solutions always have different energies. This means there is no degeneracy, and thus there are no bound eigen-states which are a linear combination of sine and cosine, and all bound states have well defined parity. Below is a diagram showing an even parity wavefunction against an odd parity wavefunction. These are illustrative only and do not necessarily relate to the given problem, they are meant to confer the symmetry of the wavefunctions only.

### Even Wavefunctions

The wavefunction for the even solution has no odd (sine) components, so the three wavefunctions are now:
       $\psi _{1}\left(x\right)=Ge^{\alpha x}\,\!$
       $\psi _{2}\left(x\right)=B\cos(kx)$
       $\psi _{3}\left(x\right)=He^{-\alpha x}\,\!$
Let us now consider the conditions requiring continuity of the wavefunction at the two "joining points":
       $Ge^{-\alpha L}=B\cos(-kL)\,$
       $He^{-\alpha L}=B\cos(kL)\,$
Since the cosine function is even, the two equations have the same right hand side. So we have
       $Ge^{-\alpha L}=He^{-\alpha L}\,$
       $G=H\,$
The conditions of continuity of the derivatives at the boundaries give us the following:
       $\alpha Ge^{-\alpha L}=-kB\sin(-kL)\,$
       $\alpha He^{-\alpha L}=kB\sin(kL)\,$
Given the equality of G and H, we can now divide any equation from the second conditions by any from the first conditions. I will choose the first equations from each set:
       ${\frac {\alpha Ge^{-\alpha L}}{Ge^{-\alpha L}}}={\frac {-kB\sin(-kL)}{B\cos(-kL)}}$
We now have an expression for α (recall that the tan function is odd, so the negatives cancel):
       $\alpha =k\tan \left(kL\right),$
where
       $k={\frac {\sqrt {2mE}}{\hbar }}$
       $\alpha ={\frac {\sqrt {2m(V-E)}}{\hbar }}$
This can be reduced to the following equations:
       ${\frac {\alpha }{k}}=\tan \left(kL\right),$
       ${\sqrt {\frac {V-E}{E}}}=\tan \left({\frac {\sqrt {2mE}}{\hbar }}L\right)$
An even parity wavefunction exists at the energies (E) that satisfy this equation. This equation is extremely difficult to solve analytically, but can readiliy be solved numerically or graphically. This will be done later. First, we will derive the odd parity wavefunctions.

### Odd Wavefunctions

The wavefunction for the odd solution has no even (cosine) components, so the three wavefunctions are now:
       $\psi _{1}\left(x\right)=Ge^{\alpha x}\,\!$
       $\psi _{2}\left(x\right)=A\sin(kx)$
       $\psi _{3}\left(x\right)=He^{-\alpha x}\,\!$
Let us now consider the conditions requiring continuity of the wavefunction at the two "joining points":
       $Ge^{-\alpha L}=A\sin(-kL)\,$
       $-He^{-\alpha L}=A\sin(kL)\,$
Since sine is an odd function, we can see that
       $Ge^{-\alpha L}=-He^{-\alpha L}\,$
       $G=-H\,$
The conditions of continuity of the derivatives at the boundaries give us the following:
       $\alpha Ge^{-\alpha L}=kA\cos(-kL)\,$
       $\alpha He^{-\alpha L}=kA\cos(kL)\,$
Dividing the first equation from the second by the first equation from the first set (any combination would do), we get:
       ${\frac {\alpha Ge^{-\alpha L}}{Ge^{-\alpha L}}}={\frac {kA\cos(-kL)}{A\sin(-kL)}}$
       $\alpha =k\cot \left(-kL\right)\,$
And because cot is an odd function,
       ${\frac {\alpha }{k}}=-\cot \left(kL\right)$
       ${\sqrt {\frac {V-E}{E}}}=-\cot \left({\frac {\sqrt {2mE}}{\hbar }}L\right)$
An odd parity wavefunction exists whenever the energy satisfies this condition. As with the even wavefunction, this is not a trival equation to solve analytically, but can be solved graphically or numerically.

## Graphical Solution for Energy Levels

Let us quickly restate our objectives here. We wish to find the energies of the bound states of an electron in a finite square well. An energy level exists only when one of the following conditions is satisfied:
[Condition for an Even Wavefunction to Exist]
       ${\frac {\alpha }{k}}=\tan \left(kL\right)$
[Condition for an Odd Wavefunction to Exist]
       ${\frac {\alpha }{k}}=-\cot \left(kL\right)$
We can plot the three functions together on a graph against energy, E. The conditions are satisfied at the crossing points as shown below. We therefore can find the allowed energy levels by reading the energy at the crossings. Before we can plot a graph, we need to choose some properties of our system. Let us set the width of the well to be 0.4nm in total (L=0.2nm) and set the potential barrier,V, to 75eV.
We see that there are six permitted energy states: even wavefunctions at about 2, 17 and 46eV and odd wavefunctions at about 7.5, 30 and 64.5eV. This method provides an easy way to find out how many levels there are and approximately where they are located. While the graphical method is theoretically as accurate as any, it is possible to practically obtain much higher accuracy using purely numerical methods.
Also notice that however small the potential barrier, there will always be at least one even solution, corresponding to the first intersection. The existence of an odd solution depends on the value of V.

## Numerical Methods for Energy Levels

There are many numerical methods that could be used to solve these equations. While the equations cannot be solved exactly, they can be differentiated. This means that we can use Newton-Raphson Iterations, which are typically fast to converge. By using our graphical method to give us a rough idea of the locations of the roots, we can select suitable starting values for the iterations.
The two equations below are the equations to solve using the iteration (the Newton-Raphson method requires a function equal to zero).
       ${\frac {\alpha }{k}}-\tan \left(kL\right)=0$
       ${\frac {\alpha }{k}}+\cot \left(kL\right)=0$
The derivates of these equations are needed for the iteration (unless done automatically by a computer), and these are given below:
       ${\frac {-V}{2E^{2}{\sqrt {\frac {V-E}{E}}}}}-{\frac {Lm}{\hbar {\sqrt {2mE}}}}\sec ^{2}\left({\frac {L{\sqrt {2mE}}}{\hbar }}\right)$
       ${\frac {-V}{2E^{2}{\sqrt {\frac {V-E}{E}}}}}-{\frac {Lm}{\hbar {\sqrt {2mE}}}}\csc ^{2}\left({\frac {L{\sqrt {2mE}}}{\hbar }}\right)$
The table below shows the six levels in this system, the starting value for the iteration, and the energy to 5 decimal places.
| n | Parity | Starting Value/eV | Energy/eV |
| 1 | Even | 2 | 1.89660 |
| 2 | Odd | 8 | 7.56592 |
| 3 | Even | 17 | 16.93946 |
| 4 | Odd | 30 | 29.87241 |
| 5 | Even | 46 | 46.05440 |
| 6 | Odd | 64 | 64.60181 |
Let's compare these to the first six energy levels for an infinite squre well, which we saw how to derive before:
As you can see, the energy levels are lower in the finite well than in the infinite well. This is due to the wavefunction extending beyond the well's boundaries, effectively resulting in a larger box. As a larger box results in lower energy levels, this makes sense.

## Finding the Numerical Wavefunctions

We have found a way to evaluate the energy levels present in a finite potential well, which is usually the only information required, but we can also find the shape of the wavefunction. We already know the approximate shape - it is sinusoidal inside the well and decays exponentially to zero outside it.

### First Energy Level

Let's start at the first (lowest) energy level, which is even. We already defined the wavefunction piecewise as follows:
       $\psi _{1}\left(x\right)=Ge^{\alpha x}\,\!$
       $\psi _{2}\left(x\right)=B\cos(kx)$
       $\psi _{3}\left(x\right)=He^{-\alpha x}\,\!$
Now that we know the energy of the level, we can easily compute k and α. This means that we just have to find G, B and H. We have already shown that G=H. Because we will have to normalise the wavefunction later anyway, we can manipulate the scaling of the functions at this point without consequence. Let us set the value of B to 1. This way we just have to find the value of G (=H) which satisfies the continuity bondary condition:
       $\psi _{3}(L)=\psi _{2}(L)\,\!$
       $Ge^{-\alpha L}=\cos(kL)\,$
Rearranging,
       $G={\frac {\cos(kL)}{e^{-\alpha L}}}$
The values of k and α and E at the first energy level
* E = 1.89660eV = 3.03868 × 10$^{-19}$ J
* k = 7.05548 × 10$^9$
* α= 4.38034 × 10$^{10}$
This gives
* G = 1014.31
We can now plot the wavefunction against position. Note that we have not performed normalisation yet, so the units are arbitrary.
We can see that the wavefunction extends visibly into the classically forbidden zone (blue). If the exact matching of the gradient is hard to fathom, remeber that it was this very condition that led to the definitions for k and α. However, as the calculations leading to this point apply only to valid energies, if you try to plot the wavefunctions for a non-valid energy, you will get a continuous (as the method of obtaining G enforces this) but not smooth graph, implying that the solution does not fit the physical system.
Because we have no simple relationship between the energy and shape of the graph (notice that G changes as a reasonably complex function of E), we can't find an easily expressible normalisation coefficient. However, by integrating the square of the wavefunction, we can still evaluate it:
       $\int _{-\infty }^{\infty }|\psi (x)|^{2}dx=1$
       $\int _{-\infty }^{-L}|\psi _{1}(x)|^{2}dx+\int _{-L}^{L}|\psi _{2}(x)|^{2}dx+\int _{L}^{\infty }|\psi _{3}(x)|^{2}dx=1$
       $\psi _{0}^{2}\left[\int _{-\infty }^{-L}G^{2}e^{2\alpha x}dx+\int _{-L}^{L}\cos ^{2}(kx)dx+\int _{L}^{\infty }G^{2}e^{-2\alpha x}dx\right]=1$
This integral is quite simple, and yields the following normalisation coefficient:
* ψ$_0$=66990.6
Compare this with 70710.7 for an infinite well. Remember to account for the units (often nanometres cause problems) when doing this integral. It makes sense that the normalisation coeffiecient is lower here, as the wavefunction extends further, so has a higher area when the sinusoid is scaled to 1, meaning that it must be multiplied by a smaller number to give a unity probability of finding the particle somewhere.

### Second Energy Level

Now let's consider the second energy level, which is odd. The wavefunction is given by:
       $\psi _{1}\left(x\right)=Ge^{\alpha x}\,\!$
       $\psi _{2}\left(x\right)=A\sin(kx)$
       $\psi _{3}\left(x\right)=He^{-\alpha x}\,\!$
where G = -H. Setting A=1, and proceeding as above, we get the following:
* E = 7.56592eV = 1.21219389 × 10$^{-18}$ J
* k = 1.40919 × 10$^{10}$
* α= 7.99293 × 10$^9$
This gives a G of 1432.53 and leads to the graph below:
To get the normalisation coeffiecient, we put
       $\int _{-\infty }^{\infty }|\psi (x)|^{2}dx=1$
       $\int _{-\infty }^{-L}|\psi _{1}(x)|^{2}dx+\int _{-L}^{L}|\psi _{2}(x)|^{2}dx+\int _{L}^{\infty }|\psi _{3}(x)|^{2}dx=1$
       $\psi _{0}^{2}\left[\int _{-\infty }^{-L}G^{2}e^{2\alpha x}dx+\int _{-L}^{L}\sin ^{2}(kx)dx+\int _{L}^{\infty }G^{2}e^{-2\alpha x}dx\right]=1$
which gives a normalisation coefficent of 66849.7. This is marginally smaller than the coefficient for the first energy level. Since the wavefunction extends further into the potential barrier region (this is not clearly apparent in the graph, but it is just observable), this is to be expected. We can see this pattern continue through all six levels:
| n | Parity | G | ψ$_0$ |
| 1 | Even | 1014.31 | 66990.6 |
| 2 | Odd | 1432.53 | 66849.7 |
| 3 | Even | -1168.56 | 66575.5 |
| 4 | Odd | -615.794 | 66073.7 |
| 5 | Even | 194.185 | 65055.8 |
| 6 | Odd | 25.2653 | 61954.1 |

## Unbound States

There are also permitted energy levels for a particle above the energy of the top of the well, V. These are called unbound states as they do not decay to zero outside the well. However, they are certainly still affected by it. Unbound states are also called continuum states, because unlike the discrete bound states they have a solution for the continuous range of energies which fulfill
       $E>V\,$
Recall the definition for α:
       $\alpha ={\frac {\sqrt {2m(V-E)}}{\hbar }}$
In unbound states, this is imaginary, which means that we cannot use the same method as we could for bound states. If you look at the graphical solution, you can see that the line for α/k ends at V, meaning that there are no more valid bound state energy levels past this point.
We saw near the beginning of this page that the wavefunction for an unbound state outside the well is
       $\psi _{1}\left(x\right)=C\sin(\kappa x)+D\cos(\kappa x)\quad$
where
       $\kappa ={\sqrt {\frac {2m(E-V))}{\hbar ^{2}}}}$
The solution inside the well is the same as for the bound states:
       $\psi _{2}\left(x\right)=A\sin {kx}+B\cos {kx}$
Becaue the potential is symmetric around the origin, there must be solutions of defined parity also for unbound states, so we will choose a basis of odd and even wavefunctions. As before, we know that the overall wavefunction must be continuous and differentiable. In other words the values of the functions and their derivatives must match up at the dividing points:
       $\psi _{1}(-L)=\psi _{2}(-L)\,\!$ $\psi _{1}(L)=\psi _{2}(L)\,\!$ ${\frac {d\psi _{1}}{dx}}(-L)={\frac {d\psi _{2}}{dx}}(-L)\,\!$ ${\frac {d\psi _{1}}{dx}}(L)={\frac {d\psi _{2}}{dx}}(L)\,\!$

### Odd Wavefunctions

For odd wavefunctions, we can ignore the cosine terms for $\psi _{2}$, but not for $\psi _{1}$ and $\psi _{3}$:
       $\psi _{2}\left(x\right)=A\sin {kx}\,\!$
       $\psi _{1}\left(x\right)=C\sin {\kappa x}+D\cos {\kappa x}=C\sin {(\kappa x-\phi )}$
       $\psi _{3}\left(x\right)=E\sin {\kappa x}+F\cos {\kappa x}=C\sin {(\kappa x+\phi )}$
$\psi _{1}$ and $\psi _{3}$ are not odd functions on their own, but the general function $\psi$ defined for all x is odd, noting that for any $x_{0}>L$:
       $\psi _{1}\left(-x_{0}\right)=\psi _{3}\left(x_{0}\right)$
Applying the boundary condition of continuity of the wavefunction, we get the same result at each boundary:
       $C\sin {(\kappa L+\phi )}=A\sin {kL}\,\!$
taking the derivative and requiring continuity at L, we get:
       $\kappa C\cos {(\kappa L+\phi )}=kA\cos {kL}\,\!$
The choice of wavefunctions of a defined parity reduces the problem to tailoring the values of the wavefunction and its derivative only at one point (as the other yields identical equations). Thus we have two equations with three variables. We will choose C, the amplitude of the wavefunction outside the well to be unity. At the end we will normalize $\psi$ to unity over the full 1D finite well.
Dividing the equation derived from continuity of the wavefunction by that of the continuity of the derivative:
       $\tan {(\kappa L+\phi )}={\frac {\kappa }{k}}\tan {kL}$
       $\phi =\arctan {({\frac {\kappa }{k}}\tan {kL}})-\kappa L$
Inserting the phase $\phi$ into the equation of continuity at L, and taking C=1, we get:
       $A={\frac {\sin {(\kappa L+\phi )}}{\sin {kL}}}$
Thus we see that there is an odd wavefunction solution for any energy, well depth, and width. There is a resonance above the well when:
       $\sin {kL}=0,\!$
This is similar to the Fabri Perot resonances in optics.

### Even Wavefunctions

In the same manner as for the odd wavefunctions, we can ignore the sine terms for $\psi _{2}$, but not for $\psi _{1}$ and $\psi _{3}$ for the even wavefunctions:
       $\psi _{2}\left(x\right)=B\cos {kx}\,\!$
       $\psi _{1}\left(x\right)=C\sin {\kappa x}+D\cos {\kappa x}=D\cos {(\kappa x-\phi )}$
       $\psi _{3}\left(x\right)=E\sin {\kappa x}+F\cos {\kappa x}=C\cos {(\kappa x+\phi )}$
Applying the boundary condition of continuity of the wavefunction, we get the same result at each boundary:
       $C\cos {(\kappa L+\phi )}=B\cos {kL}\,\!$
taking the derivative and requiring continuity at L, we get:
       $\kappa C\sin {(\kappa L+\phi )}=kA\sin {kL}\,\!$
Dividing the equation derived from continuity of the wavefunction by that of the continuity of the derivative:
       $\tan {(\kappa L+\phi )}={\frac {k}{\kappa }}\tan {kL}$
       $\phi =\arctan {({\frac {k}{\kappa }}\tan {kL}})-\kappa L$
Inserting the phase $\phi$ into the equation of continuity at L, and taking C=1, we get:
       $A={\frac {\cos {(\kappa L+\phi )}}{\cos {kL}}}$
Thus we see that there is an even wavefunction solution for any energy, well depth, and width. There is a resonance above the well when:
       $\cos {kL}=0,\!$