9.2: The operators  and †.
- Page ID
- 14803
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
\[\left(\hat{a}^{\dagger} \hat{a}+\frac{1}{2}\right) \hat{a} u(y)=(\epsilon-1) \hat{a} u(y) .\]
Question: Show that
\[\left(\hat{a}^{\dagger} \hat{a}+\frac{1}{2}\right) \hat{a}^{\dagger} u(y)=(\epsilon+1) \hat{a}^{\dagger} u(y) .\]
We thus conclude that (we use the notation \(u_n(y)\) for the eigenfunction corresponding to the eigenvalue \(\epsilon_n\) )
\[\begin{aligned}
\hat{a} u_n(y) & \propto u_{n-1}(y), \\
\hat{a}^{\dagger} u_n(y) & \propto u_{n+1}(y) .
\end{aligned}\]
So using \(\hat{a}\) we can go down in eigenvalues, using \(a^{\dagger}\) we can go up. This leads to the name lowering and raising operators (guess which is which?).
We also see from \(\PageIndex{12}\) that the eigenvalues differ by integers only!