9.2: The operators  and †.

In a previous chapter I have discussed a solution by a power series expansion. Here I shall look at a different technique, and define two operators \(\hat{a}\) and \(\hat{a}^{\dagger}\),

\[\hat{a}=\frac{1}{\sqrt{2}}\left(y+\frac{d}{d y}\right), \quad \hat{a}^{\dagger}=\frac{1}{\sqrt{2}}\left(y-\frac{d}{d y}\right) .\]

Since

\[\frac{d}{d y}(y f(y))=y \frac{d}{d y} f(y)+f(y)\]

or in operator notation

\[\frac{d}{d y} \hat{y}=\hat{y} \frac{d}{d y}+\hat{1}\]

(the last term is usually written as just 1 ) we find

\[\begin{array}{l}
\hat{a} \hat{a}^{\dagger}=\frac{1}{2}\left(\hat{y}^2-\frac{d^2}{d y^2}+\hat{1}\right) \\
\hat{a}^{\dagger} \hat{a}=\frac{1}{2}\left(\hat{y}^2-\frac{d^2}{d y^2}-\hat{1}\right)
\end{array}\]

If we define the commutator

\[[\hat{f}, \hat{g}]=\hat{f} \hat{g}-\hat{g} \hat{f}\]

we have

\[\left[\hat{a}, \hat{a}^{\dagger}\right]=\hat{1}\]

Now we see that we can replace the eigenvalue problem for the scaled Hamiltonian by either of

\[\begin{aligned}
\left(\hat{a}^{\dagger} \hat{a}+\frac{1}{2}\right) u(y) & =\epsilon u(y) \\
\left(\hat{a} \hat{a}^{\dagger}-\frac{1}{2}\right) u(y) & =\epsilon u(y)
\end{aligned}\]

By multiplying the first of these equations by \(\hat{a}\) we get

\[\left(\hat{a} \hat{a}^{\dagger} \hat{a}+\frac{1}{2} \hat{a}\right) u(y)=\epsilon \hat{a} u(y) .\]

If we just rearrange some brackets, we find

\[\left(\hat{a} \hat{a}^{\dagger}+\frac{1}{2}\right) \hat{a} u(y)=\epsilon \hat{a} u(y) .\]

If we now use

\[\hat{a} \hat{a}^{+}=\hat{a}^{\dagger} \hat{a}-\hat{1},\]

we see that

\[\left(\hat{a}^{\dagger} \hat{a}+\frac{1}{2}\right) \hat{a} u(y)=(\epsilon-1) \hat{a} u(y) .\]

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