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9.2 The operators \hat{a} and \hat{{a}}^{†}.

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}}^{†},

\hat{a} = {1\over \sqrt{2}}\left (y + {d\over dy}\right ),\kern 2.77695pt \kern 2.77695pt \hat{{a}}^{†} = {1\over \sqrt{2}}\left (y − {d\over dy}\right ).
(9.4)

Since

{d\over dy}(yf(y)) = y {d\over dy}f(y) + f(y),
(9.5)

or in operator notation

{d\over dy}\hat{y} =\hat{ y} {d\over dy} +\hat{ 1},
(9.6)

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

\begin{eqnarray} \hat{a}\hat{{a}}^{†}& =& {1\over 2}\left (\hat{{y}}^{2} − {{d}^{2}\over d{y}^{2}} +\hat{ 1}\right ),%& \\ {\hat{a}}^{†}\hat{a}& =& {1\over 2}\left (\hat{{y}}^{2} − {{d}^{2}\over d{y}^{2}} −\hat{ 1}\right ).%&(9.7) \\ \end{eqnarray}

If we define the commutator

[\hat{f},\hat{g}] =\hat{ f}\hat{g} −\hat{ g}\hat{f},
(9.8)

we have

[\hat{a},\hat{{a}}^{†}] =\hat{ 1}.
(9.9)

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

\begin{eqnarray} \left ({\hat{a}}^{†}\hat{a} + {1\over 2}\right )u(y)& =& ϵu(y),%& \\ \left (\hat{a}{\hat{a}}^{†}−{1\over 2}\right )u(y)& =& ϵu(y).%&(9.10) \\ \end{eqnarray}

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

\left (\hat{a}{\hat{a}}^{†}\hat{a} + {1\over 2}\hat{a}\right )u(y) = ϵ\hat{a}u(y).
(9.11)

If we just rearrange some brackets, we find

\left (\hat{a}{\hat{a}}^{†} + {1\over 2}\right )\hat{a}u(y) = ϵ\hat{a}u(y).
(9.12)

If we now use

\hat{a}{\hat{a}}^{†} =\hat{ {a}}^{†}\hat{a} −\hat{ 1},
(9.13)

we see that

\left ({\hat{a}}^{†}\hat{a} + {1\over 2}\right )\hat{a}u(y) = (ϵ − 1)\hat{a}u(y).
(9.14)

Question: Show that

\left ({\hat{a}}^{†}\hat{a} + {1\over 2}\right )\hat{{a}}^{†}u(y) = (ϵ + 1){\hat{a}}^{†}u(y).
(9.15)

We thus conclude that (we use the notation {u}_{n}(y) for the eigenfunction corresponding to the eigenvalue {ϵ}_{n})

\begin{eqnarray} \hat{a}{u}_{n}(y)& ∝& {u}_{n−1}(y),%& \\ \hat{{a}}^{†}{u}_{ n}(y)& ∝& {u}_{n+1}(y).%&(9.16) \\ \end{eqnarray}

So using \hat{a} we can go down in eigenvalues, using {a}^{†} we can go up. This leads to the name lowering and raising operators (guess which is which?).

We also see from (9.15) that the eigenvalues differ by integers only!

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