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=====Solving for $\alpha$ and $\beta$ for the 2-well Case (10 minutes)=====

  * In general, the two-well system Hamiltonian is represented in matrix notation as

$$H \; = \; \left[\begin{array}{cc}
\alpha & \beta \\
\beta & \alpha \\ \end{array}\right] \; \; . $$

But, how do we explicitly find what $\alpha$ and $\beta$ are?

First, let's find alpha.  We will start by looking at the operation

$$H\vert 1 \rangle \; = \; \left[\begin{array}{cc}
\alpha & \beta \\
\beta & \alpha \\ \end{array}\right]\left[\begin{array}{c}
1 \\
0 \\ \end{array}\right] \; =  \; \left[\begin{array}{c}
\alpha \\
\beta \\ \end{array}\right]. $$

Changing this operation into function notation will give:

$$H \phi(x-a) \; = \; \alpha \phi(x-a) + \beta \phi (x-2a) \; \; .$$

Multiplying both sides by $\int \phi^{*}(x-a)$ will then give:


$$\int \phi^{*}(x-a)H \phi(x-a) \; = \; \int \phi^{*}(x-a)\alpha \phi(x-a) + \int \phi^{*}(x-a)\beta \phi (x-2a) \; \; .$$

We can then pull out $\alpha$ and $\beta$ from the integrals because they are constants.

$$\int \phi^{*}(x-a)H \phi(x-a) \; = \; \alpha\int \phi^{*}(x-a) \phi(x-a) + \beta\int \phi^{*}(x-a) \phi (x-2a) \; \; .$$

The basis states are orthogonal to each other, so evaluating the integral next to $\beta$ will give zero while the integral next to $\alpha$ will equal 1.  Doing so shows that

$$\alpha \; = \; \int \phi^{*}(x-a)H \phi(x-a) \; \; . $$

Note that in bra-ket notation, the expression looks like

$$\alpha \; = \; \langle 1 \vert H \vert 1 \rangle \; \; . $$

  * Performing the same operation beginning with the expression $H\vert 2 \rangle$ would show us that

$$\beta \; = \; \int \phi^{*g}(x-a) \, H \, \phi^{g}(x-2a)dx \; \; . $$

Notice that $\beta$ is an overlap function.  As such, if the potential wells are very far apart, we can expect that the value for $\beta$ will be very small.  In either case, it is expected that $\beta < \alpha$.

  * To explicitly perform these calculations and find values for $\alpha$ and $\beta$, we would need to find the expression for both basis states, insert into the integral that

$$H=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} + V_{atom}\left(x-a\right) + V_{atom}\left(x-2a\right) \; \; ,$$

and evaluate the integral over all space.
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