The Schr枚dinger equation for the quantum- mechanical harmonic oscillator in
Eq. (40.44) may be written as $$ -\frac{\hbar^{2}}{2 m} \psi^{\prime
\prime}+\frac{1}{2} m \omega^{2} x^{2} \psi=E \psi $$ where
\(\omega=\sqrt{k^{\prime} / m}\) is the angular frequency for classical
oscillations. Its solutions may be cleverly addressed using the following
observations: The difference in applying two operations \(A\) and \(B,\) in
opposite order, called a commutator, is denoted by \([A, B]=A B-B A\). For
example, if \(A\) represents differentiation, so that \(A f=\frac{d f}{d x},\)
where \(f=f(x)\) is a function, and \(B\) represents multiplication by \(x,\) then
\([A, B]=\left[\frac{d}{d x}, x\right] f=\frac{d}{d x}(x f)-x \frac{d f}{d x}
.\) Applying the product rule for differentiation on the first term, we
determine \(\left[\frac{d}{d x}, x\right] f=f,\) which we summarize by writing
\(\left[\frac{d}{d x}, x\right]=1 .\) Similarly, consider the two operators
\(a_{+}\) and \(a_{-}\) defined by \(a_{\pm}=1 / \sqrt{2 m \hbar \omega}\left(\mp
\hbar \frac{d}{d x}+m \omega x\right)\)
(a) Determine the commutator \(\left[a_{-}, a_{+}\right]\).
(b) Compute \(a_{+} a_{-\psi}\) in terms of \(\psi\) and \(\psi^{\prime \prime},\)
where \(\psi=\psi(x)\) is a wave function. Be mindful that \(\frac{d}{d x}(x
\psi)=\psi+x \psi^{\prime} .\) (c) The Schr枚dinger equation above may be
written as \(H \psi=E \psi,\) where \(H\) is a differential operator. Use your
previous result to write \(H\) in terms of \(a_{+}\) and \(a_{-}\)
(d) Determine the commutators \(\left[H, a_{\pm}\right] .\) (e) Consider a wave
function \(\psi_{n}\) with definite energy \(E_{n},\) whereby \(H \psi_{n}=E_{n}
\psi_{n} .\) If we define a new function \(\psi_{n+1}=a_{+} \psi_{n},\) then we
can readily determine its energy by computing \(H \psi_{n+1}=H a_{+}
\psi_{n}=\left(a_{+} H+\left[H, a_{+}\right]\right) \psi_{n}\) and comparing
the result with \(E_{n+1} \psi_{n+1}\). Finish this calculation by inserting
what we have determined to find \(E_{n+1}\) in terms of \(E_{n}\) and \(n .\) (f) We
can prove that the lowest-energy wave function \(\psi_{0}\) is annihilated by
\(a_{-},\) meaning that \(a_{-} \psi_{0}=0 .\) Using your expression for \(H\) in
terms of \(a_{\pm},\) determine the ground-state energy \(E_{0}\). (g) By applying
\(a_{+}\) repeatedly, we can generate all higher states. The energy of the \(n\)
th excited state can be determined by considering \(\mathrm{Ha}_{+}^{n}
\psi_{0}=E_{0} \psi_{0}\). Using the above results, determine \(E_{n}\). (Note:
We did not have to solve any differential equations or take any derivatives to
determine the spectrum of this theory.)
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