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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. Let's again look at the most general expression for the energy eigenvalue equation.


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