PX262 - D1 - ladder operators

the hamiltonian operator: $\hat{H} = \frac{{\hat{P}}^{2}}{2 m} + \frac{1}{2} m ω^{2} {\hat{X}}^{2}$
the position, $x$ , can be measured in the units $(\frac{ℏ}{m ω})^{\frac{1}{2}}$ , the momentum units are $(m ℏ ω)^{\frac{1}{2}}$ , and the energy units are $ℏ ω$
the expression for the energy of a given state becomes: $E_{n} = \frac{1}{2} + n$
the hamiltonian operator is: $\hat{H} = \frac{1}{2} ({\hat{P}}^{2} + {\hat{X}}^{2})$
the commutation relation: $[\hat{P}, \hat{X}] = - i$
the ladder operators are defined as:

\begin{aligned} {\hat{R}}_{+} & = (\hat{P} + i \hat{X}) \\ {\hat{R}}_{-} & = (\hat{P} - i \hat{X}) \end{aligned}

\begin{aligned} {\hat{R}}_{+} {\hat{R}}_{-} & = (\hat{P} + i \hat{X}) (\hat{P} - i \hat{X}) \\ = {\hat{P}}^{2} + {\hat{X}}^{2} + i \hat{X} \hat{P} - i \hat{P} \hat{X} \\ = 2 \hat{H} - i [\hat{P}, \hat{X}] \\ {\hat{R}}_{+} {\hat{R}}_{-} & = 2 \hat{H} - 1 \\ similarly, {\hat{R}}_{-} {\hat{R}}_{+} & = 2 \hat{H} + 1 \end{aligned}

the hamiltonian operator can rewritten as: $\hat{H} = \frac{1}{2} ({\hat{R}}_{+} {\hat{R}}_{-} + 1) = \frac{1}{2} ({\hat{R}}_{-} {\hat{R}}_{+} - 1)$
commutation relations of the ladder operators with hamiltonian:

\begin{matrix} [{\hat{R}}_{+}, \hat{H}] = {\hat{R}}_{+} \frac{1}{2} ({\hat{R}}_{-} {\hat{R}}_{+} - 1) - \frac{1}{2} ({\hat{R}}_{+} {\hat{R}}_{-} + 1) {\hat{R}}_{-} = - {\hat{R}}_{+} \\ [{\hat{R}}_{-}, \hat{H}] = + {\hat{R}}_{-} \end{matrix}

applying ${\hat{R}}_{+}$ to some eigenfunction, $ϕ_{n}$ , and working out the energy of the resulting wavefunction: $\hat{H} {\hat{R}}_{+} ϕ_{n} = {\hat{R}}_{+} \hat{H} ϕ_{n} + {\hat{R}}_{+} ϕ_{n} = {\hat{R}}_{+} E_{n} ϕ_{n} + {\hat{R}}_{+} ϕ_{n} = (E_{n} + 1) {\hat{R}}_{+} ϕ_{n}$
this shows that the wavefunction, $\hat{R} ϕ_{n}$ , is an eigenfunction of the hamiltonian operator, with the eigenvalue, $E_{n} + 1$

E_{n} + 1 = n + 1 + \frac{1}{2} = E_{n + 1}

this shows that the result pf applying ${\hat{R}}_{+}$ on $ϕ_{n}$ is $ϕ_{n + 1}$
similarly, $\hat{H} {\hat{R}}_{-} ϕ_{n} = (E_{n} - 1) \hat{R} ϕ_{n} = E_{n - 1} ϕ_{n - 1}$
the two operators help move up and down the ladder of states, where each step is a particular energy eigenstate, hence called ladder operators
if ${\hat{R}}_{-}$ is applied on the ground state, $ϕ_{0}$ should get to zero because it is not possible to move any lower: $0 = {\hat{R}}_{+} {\hat{R}}_{-} ϕ_{0} = (2 \hat{H} - 1) ϕ_{0} = (2 E_{0} - 1) ϕ_{0} ⟹ E_{0} = \frac{1}{2}$
this shows that the lowest energy state has an energy: $E_{0} = \frac{1}{2} ℏ ω$