The quantum harmonic oscillator is an important toy model for studying bound states, and it will again serve as an important instance in quantum field theory. To study the quantum counterpart of the classical harmonic oscillator, consider the one-dimensional Schrödinger equation \(\mathring{\imath}\)\(\hslash\)\(\left(\vphantom{\partial \,/\, \partial t}\right.\)\(\partial\)\(\,/\,\)\(\partial\)\(t\)\(\left.\vphantom{\partial \,/\, \partial t}\right)\)\(\psi\)\(\left[\vphantom{x ,\, t}\right.\)\(x\)\(,\,\)\(t\)\(\left.\vphantom{x ,\, t}\right]\)\(\;=\;\)\(H\)\(\psi\)\(\left[\vphantom{x ,\, t}\right.\)\(x\)\(,\,\)\(t\)\(\left.\vphantom{x ,\, t}\right]\) with a quadratic potential:
\(H\)\(\;\equiv\;\)\(\dfrac{-\, \hslash \vphantom{ \hslash}^{2}}{2 m}\)\(\dfrac{\mathrm{d} \vphantom{ \mathrm{d}}^{2}}{\mathrm{d} x \vphantom{ x}^{2}}\)\(\,+\,\)\(\dfrac{1}{2}\)\(m\)\(\omega \vphantom{ \omega}^{2}\)\(x \vphantom{ x}^{2}\)\(;\,\)
If we accept such formal manipulation with matrix exponents:
\(\psi\)\(\left[\vphantom{x ,\, t}\right.\)\(x\)\(,\,\)\(t\)\(\left.\vphantom{x ,\, t}\right]\)\(\;=\;\)\(e \vphantom{ e}^{-\, \mathring{\imath} H t \,/\, \hslash}\)\(\psi\)\(\left[\vphantom{x ,\, t \:\leftarrow\: 0}\right.\)\(x\)\(,\,\)\(t\)\(\:\leftarrow\:\)\(0\)\(\left.\vphantom{x ,\, t \:\leftarrow\: 0}\right]\)\(\;\mathtt{for}\;\)\(x\)\(,\,\)\(t\)\(;\,\)
Then time-dependent Schrödinger equation may be reduced to the time-independent equation:
\(\dfrac{\mathrm{d} \vphantom{ \mathrm{d}}^{2}}{\mathrm{d} x \vphantom{ x}^{2}}\)\(\psi\)\(\left[\vphantom{x}\right.\)\(x\)\(\left.\vphantom{x}\right]\)\(\,-\,\)\(\left(\vphantom{\dfrac{m \vphantom{ m}^{2} \omega \vphantom{ \omega}^{2}}{\hslash \vphantom{ \hslash}^{2}} x \vphantom{ x}^{2} \,-\, \dfrac{2 m E}{\hslash \vphantom{ \hslash}^{2}}}\right.\)\(\dfrac{m \vphantom{ m}^{2} \omega \vphantom{ \omega}^{2}}{\hslash \vphantom{ \hslash}^{2}}\)\(x \vphantom{ x}^{2}\)\(\,-\,\)\(\dfrac{2 m E}{\hslash \vphantom{ \hslash}^{2}}\)\(\left.\vphantom{\dfrac{m \vphantom{ m}^{2} \omega \vphantom{ \omega}^{2}}{\hslash \vphantom{ \hslash}^{2}} x \vphantom{ x}^{2} \,-\, \dfrac{2 m E}{\hslash \vphantom{ \hslash}^{2}}}\right)\)\(\psi\)\(\left[\vphantom{x}\right.\)\(x\)\(\left.\vphantom{x}\right]\)\(\;=\;\)\(0\)\(\;\mathtt{for}\;\)\(x\)\(;\,\)
We sketch the method of ladder operator here. Introduce:
\(\alpha \vphantom{ \alpha}_{\pm}\)\(\;\equiv\;\)\(\surd\!\)\(\left(\vphantom{\dfrac{m \omega}{2 \hslash}}\right.\)\(\dfrac{m \omega}{2 \hslash}\)\(\left.\vphantom{\dfrac{m \omega}{2 \hslash}}\right)\)\(\left(\vphantom{x \,\mp\, \dfrac{\hslash}{m \omega} \dfrac{\mathrm{d}}{\mathrm{d} x}}\right.\)\(x\)\(\,\mp\,\)\(\dfrac{\hslash}{m \omega}\)\(\dfrac{\mathrm{d}}{\mathrm{d} x}\)\(\left.\vphantom{x \,\mp\, \dfrac{\hslash}{m \omega} \dfrac{\mathrm{d}}{\mathrm{d} x}}\right)\)\(;\,\)
It is straightforward to check:
\(x\)\(\;=\;\)\(\surd\!\)\(\left(\vphantom{\dfrac{\hslash}{2 m \omega}}\right.\)\(\dfrac{\hslash}{2 m \omega}\)\(\left.\vphantom{\dfrac{\hslash}{2 m \omega}}\right)\)\(\left(\vphantom{\alpha \vphantom{ \alpha}_{-} \,+\, \alpha \vphantom{ \alpha}_{+}}\right.\)\(\alpha \vphantom{ \alpha}_{-}\)\(\,+\,\)\(\alpha \vphantom{ \alpha}_{+}\)\(\left.\vphantom{\alpha \vphantom{ \alpha}_{-} \,+\, \alpha \vphantom{ \alpha}_{+}}\right)\)\(;\,\)
\(\dfrac{\mathrm{d}}{\mathrm{d} x}\)\(\;=\;\)\(\surd\!\)\(\left(\vphantom{\dfrac{m \omega}{2 \hslash}}\right.\)\(\dfrac{m \omega}{2 \hslash}\)\(\left.\vphantom{\dfrac{m \omega}{2 \hslash}}\right)\)\(\left(\vphantom{\alpha \vphantom{ \alpha}_{-} \,-\, \alpha \vphantom{ \alpha}_{+}}\right.\)\(\alpha \vphantom{ \alpha}_{-}\)\(\,-\,\)\(\alpha \vphantom{ \alpha}_{+}\)\(\left.\vphantom{\alpha \vphantom{ \alpha}_{-} \,-\, \alpha \vphantom{ \alpha}_{+}}\right)\)\(;\,\)
\(\alpha \vphantom{ \alpha}_{+}\)\(\alpha \vphantom{ \alpha}_{-}\)\(\,-\,\)\(\alpha \vphantom{ \alpha}_{+}\)\(\alpha \vphantom{ \alpha}_{-}\)\(\;=\;\)\(\,-\,\)\(1\)\(;\,\)
\(H\)\(\;=\;\)\(\hslash\)\(\omega\)\(\left(\vphantom{\alpha \vphantom{ \alpha}_{+} \alpha \vphantom{ \alpha}_{-} \,+\, \dfrac{1}{2}}\right.\)\(\alpha \vphantom{ \alpha}_{+}\)\(\alpha \vphantom{ \alpha}_{-}\)\(\,+\,\)\(\dfrac{1}{2}\)\(\left.\vphantom{\alpha \vphantom{ \alpha}_{+} \alpha \vphantom{ \alpha}_{-} \,+\, \dfrac{1}{2}}\right)\)\(;\,\)
Therefore we have these interesting facts:
\(\;\mathtt{if}\;\)\(\alpha \vphantom{ \alpha}_{-}\)\(\alpha \vphantom{ \alpha}_{+}\)\(\varphi\)\(\;=\;\)\(n\)\(\varphi\)\(,\,\)
\(\;\mathtt{then}\;\)\(\alpha \vphantom{ \alpha}_{-}\)\(\alpha \vphantom{ \alpha}_{+}\)\(\alpha \vphantom{ \alpha}_{+}\)\(\varphi\)\(\;=\;\)\(\left(\vphantom{\alpha \vphantom{ \alpha}_{+} \alpha \vphantom{ \alpha}_{-} \,+\, 1}\right.\)\(\alpha \vphantom{ \alpha}_{+}\)\(\alpha \vphantom{ \alpha}_{-}\)\(\,+\,\)\(1\)\(\left.\vphantom{\alpha \vphantom{ \alpha}_{+} \alpha \vphantom{ \alpha}_{-} \,+\, 1}\right)\)\(\alpha \vphantom{ \alpha}_{+}\)\(\varphi\)\(\;=\;\)\(\left(\vphantom{n \,+\, 1}\right.\)\(n\)\(\,+\,\)\(1\)\(\left.\vphantom{n \,+\, 1}\right)\)\(\alpha \vphantom{ \alpha}_{+}\)\(\varphi\)\(;\,\)
\(\;\mathtt{if}\;\)\(\alpha \vphantom{ \alpha}_{+}\)\(\alpha \vphantom{ \alpha}_{-}\)\(\varphi\)\(\;=\;\)\(n\)\(\varphi\)\(,\,\)
\(\;\mathtt{then}\;\)\(\alpha \vphantom{ \alpha}_{+}\)\(\alpha \vphantom{ \alpha}_{-}\)\(\alpha \vphantom{ \alpha}_{-}\)\(\varphi\)\(\;=\;\)\(\left(\vphantom{\alpha \vphantom{ \alpha}_{-} \alpha \vphantom{ \alpha}_{+} \,-\, 1}\right.\)\(\alpha \vphantom{ \alpha}_{-}\)\(\alpha \vphantom{ \alpha}_{+}\)\(\,-\,\)\(1\)\(\left.\vphantom{\alpha \vphantom{ \alpha}_{-} \alpha \vphantom{ \alpha}_{+} \,-\, 1}\right)\)\(\alpha \vphantom{ \alpha}_{-}\)\(\varphi\)\(\;=\;\)\(\left(\vphantom{n \,-\, 1}\right.\)\(n\)\(\,-\,\)\(1\)\(\left.\vphantom{n \,-\, 1}\right)\)\(\alpha \vphantom{ \alpha}_{-}\)\(\varphi\)\(;\,\)
Meanwhile, the ground state can be solved by either the the method of series, or by letting \(\alpha \vphantom{ \alpha}_{-}\)\(\psi\)\(\;=\;\)\(0\), to be:
\(\psi \vphantom{ \psi}_{0}\)\(\left[\vphantom{x}\right.\)\(x\)\(\left.\vphantom{x}\right]\)\(\;\equiv\;\)\(\left(\dfrac{m \omega}{\pi \hslash}\right) \vphantom{ \left(\dfrac{m \omega}{\pi \hslash}\right)}^{1 \,/\, 4}\)\(e \vphantom{ e}^{-\, m \omega x \vphantom{ x}^{2} \,/\, 2 \hslash}\)\(;\,\)
And the solutions are given as:
State_energy:
\(\psi \vphantom{ \psi}_{n}\)\(\;\equiv\;\)\(\alpha \vphantom{ \alpha}^{n}_{+}\)\(\psi \vphantom{ \psi}_{0}\)\(,\,\)\(n\)\(\;=\;\)\(0\)\(,\,\)\(1\)\(,\,\)\(2\)\(,\,\)\(\,\dotsc\,\)\(;\,\)
They satisfy:
Level_energy:
\(E \vphantom{ E}_{n}\)\(\;\equiv\;\)\(\left(\vphantom{n \,+\, \dfrac{1}{2}}\right.\)\(n\)\(\,+\,\)\(\dfrac{1}{2}\)\(\left.\vphantom{n \,+\, \dfrac{1}{2}}\right)\)\(\hslash\)\(\omega\)\(,\,\)\(n\)\(\;=\;\)\(0\)\(,\,\)\(1\)\(,\,\)\(2\)\(,\,\)\(\,\dotsc\,\)\(;\,\)
\(H\)\(\psi \vphantom{ \psi}_{n}\)\(\;=\;\)\(E \vphantom{ E}_{n}\)\(\psi \vphantom{ \psi}_{n}\)\(;\,\)
When I learnt about this in the past, I wondered how the expression of \(\alpha\) was motivated. Such method of ladder operator might not be universal to solve all Hamiltonians. Indeed, for this to work, the energy levels must be spaced evenly. Suppose \(E \vphantom{ E}_{n}\)\(\;=\;\)\(n\)\(\hslash\)\(\omega\)\(\,+\,\)\(E \vphantom{ E}_{0}\)\(,\,\)\(n\)\(\;=\;\)\(0\)\(,\,\)\(1\)\(,\,\)\(2\)\(,\,\)\(\,\dotsc\,\), \(H\)\(\varphi\)\(\;=\;\)\(E \vphantom{ E}_{n}\)\(\varphi\), \(H\)\(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{+}\)\(\varphi\)\(\;=\;\)\(E \vphantom{ E}_{n \,+\, 1}\)\(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{+}\)\(\varphi\), then it follows \(H\)\(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{+}\)\(\;=\;\)\(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{+}\)\(H\)\(\,+\,\)\(\hslash\)\(\omega\)\(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{+}\). Since \(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{-}\)\(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{+}\) is diagonal and full ranked in the energy basis, this strongly suggests \(H\)\(\;=\;\)\(D \vphantom{ D}_{1}\)\(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{+}\)\(\tilde{\alpha} \vphantom{ \tilde{\alpha}}_{-}\)\(\,+\,\)\(D \vphantom{ D}_{2}\), where \(D \vphantom{ D}_{2}\) and \(D \vphantom{ D}_{2}\) are diagonal in the energy basis too. The best I can say is that the rest looks now easier to guess, by naïvely trying to factor \(m \vphantom{ m}^{2}\)\(\omega \vphantom{ \omega}^{2}\)\(x \vphantom{ x}^{2}\)\(\,/\,\)\(\hslash \vphantom{ \hslash}^{2}\)\(\,-\,\)\(\mathrm{d} \vphantom{ \mathrm{d}}^{2}\)\(\,/\,\)\(\mathrm{d}\)\(x \vphantom{ x}^{2}\) as something like \(\alpha \vphantom{ \alpha}_{+}\)\(\alpha \vphantom{ \alpha}_{-}\) .
❧ August 29, 2021