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## Finding the Energy Eigenstates of an N-Well System (40 minutes)

• Now, for the n-well system, we have the n $x$ n Hamiltonian

$$H \; \dot{=} \left[\begin{array}{ccccc} \alpha & \beta & 0 & 0 & \dots\\ \beta & \alpha & \beta & 0 & \\ 0 & \beta & \alpha & \beta & \ddots \\ 0 & 0 & \beta & \alpha & \ddots \\ \vdots & & \ddots & \ddots & \ddots \\ \end{array}\right] \; \; .$$

Now, we want to find both the possible eigenstates of the electron in the potential landscape and the energy of each eigenstate. To do this, let's again use the energy eigenvalue equation

$$H\vert \psi \rangle = E \vert \psi \rangle \; \; ,$$

Where $\vert \psi \rangle$ is a superposition of all of the possible ground states, written as

$$\vert \psi \rangle \; = \; \sum_{i}^{n} \, c_{i}\, \vert \, i \, \rangle \; \dot{=} \; \left[\begin{array}{c} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array}\right] \; \; .$$

Let's try to see if an unknown state with $c_{i}=1$ for all entries is an eigenvector.

$$H \vert \psi \rangle = \left[\begin{array}{ccccc} \alpha & \beta & 0 & 0 & \dots\\ \beta & \alpha & \beta & 0 & \\ 0 & \beta & \alpha & \beta & \ddots \\ 0 & 0 & \beta & \alpha & \ddots \\ \vdots & & \ddots & \ddots & \ddots \\ \end{array}\right] \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ \vdots \\ \end{array}\right] = E\left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ \vdots \\ \end{array}\right] \; \; .$$

Performing the matrix algebra, we find that at the ends of the vector, $E=\alpha + \beta$. But, for every other entry in the resulting expression, $E=\alpha + 2\beta$. So, if this is a very long chain of potential wells, it is close to being an eigenstate. Would a vector with alternating values of +1 and -1 satisfy the eigenvalue equation?

Guessing any further for eigenvectors is going to become very difficult, so let's see what we can find about the eigenenergies. Let's again look at the most general expression for the energy eigenvalue equation.

$$\left[\begin{array}{ccccccc} \alpha c_{1}& \beta c_{2} & 0 & 0 & 0 & 0 & \dots\\ \beta c_{1}& \alpha c_{2}& \beta c_{3}& 0 & 0 & 0 & \\ 0 & \beta c_{2} & \alpha c_{3} & \beta c_{4} & 0 & 0 & \ddots \\ & & & \vdots & & & \\ 0 & 0 & 0 & \beta c_{p-1} & \alpha c_{p} & \beta c_{p+1} & \dots \\ & & & \vdots & & & \\ \end{array}\right] = \left[ \begin{array}{c} E c_{1} \\ E c_{2} \\ E c_{3} \\ \vdots \\ E c_{p} \\ \vdots \\ \end{array} \right] \; \; ,$$

where I have multiplied the Hamiltonian and energy by the eigenvector. Now, we can see that any arbitrary $p$th coefficient (except for the ends) must satisfy

$$E\, c_{p} \; = \; \beta\, c_{p-1} + \alpha\, c_{p} + \beta\, c_{p+1} \; \; .$$

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