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9.3 Eigenfunctions of \hat{H} through ladder operations

If we start with the ground state we would expect that we can’t go any lower,

\hat{a}{u}_{0}(y) = 0.
(9.17)

This can of course be checked explicitly,

\begin{eqnarray} \hat{a}{e}^{−{y}^{2}∕2 }& =& {1\over \sqrt{2}}\left (\hat{y} + {d\over dy}\right ){e}^{−{y}^{2}∕2 } %& \\ & =& {1\over \sqrt{2}}\left (y{e}^{−{y}^{2}∕2 } − y{e}^{−{y}^{2}∕2 }\right )%& \\ & =& 0 %&(9.18) \\ \end{eqnarray}

Quiz Can you show that {ϵ}_{0} = 1∕2 using the operators \hat{a}?

Once we know that {ϵ}_{0} = 1∕2, repeated application of Eq. (9.15) shows that {ϵ}_{n} = n + 1∕2, which we know to be correct from our previous treatment.

Actually, once we know the ground state, we can now easily determine all the Hermite polynomials up to a normalisation constant:

\begin{eqnarray}{ u}_{1}(y)& ∝& {a}^{†}{e}^{−{y}^{2}∕2 } %& \\ & =& {1\over \sqrt{2}}\left (\hat{y} − {d\over dy}\right ){e}^{−{y}^{2}∕2 } %& \\ & =& {1\over \sqrt{2}}\left (y{e}^{−{y}^{2}∕2 } + y{e}^{−{y}^{2}∕2 }\right )%& \\ & =& \sqrt{2}y{e}^{−{y}^{2}∕2 }. %&(9.19) \\ \end{eqnarray}

Indeed {H}_{1}(y) ∝ y.

From math books we can learn that the standard definition of the Hermite polynomials corresponds to

{H}_{n}(y){e}^{−{y}^{2}∕2 } = {(\sqrt{2})}^{n}{\left (\hat{{a}}^{†}\right )}^{n}{e}^{−{y}^{2}∕2 }.
(9.20)

We thus learn {H}_{1}(y) = 2y and {H}_{2}(y) = (4{y}^{2} − 2).

Question: Prove this last relation.

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