# Heisenberg Equation for Raising and Lowering Operators of the Harmonic Oscillator

For a Schrödinger operator $$O$$ that is independent of time, the Heisenberg equation is $$\dot O_H(t) = i[H, O_H(t)], \tag{1}$$ with solution given by $$O_H(t) = e^{iHt}O_S e^{-iHt}. \tag{2}$$ I want to apply this formalism to the quantum Harmonic Oscillator, with Hamiltonian $$H = \omega a^\dagger a = \omega n,$$ and determine the time evolution of $$a_H(t)$$ and $$a^\dagger_H(t)$$ (in what follows I will omit the subscript $$H$$). Since we have the following commutation relations: $$[n, a] = -a \quad [n, a^\dagger] = a^\dagger,$$ Heisenburgs equation becomes: $$\dot a(t) = i[\omega n, a(t)] = -i\omega a(t) \implies a(t) = e^{-i\omega t}a(t),$$ and similarly, $$\dot a^\dagger(t) = i[\omega n, a^\dagger(t)] = i\omega a^\dagger(t) \implies a^\dagger(t) = e^{-i\omega t}a^\dagger(t).$$ However, I am wondering how to connect this with the general solution given by equation $$(2)$$. For example, if I try to expand equation two for $$a(t)$$, in an attempt to get this solution, I get $$e^{i\omega n t}a(t)e^{-i\omega n t} = e^{i\omega n t}\big(e^{-i\omega nt}a(t) + [a(t), e^{-i\omega nt}]\ \big) = a(t) + e^{i\omega n t}[a(t), e^{-i\omega nt}],$$ but since $$[n(t), [n(t), a(t)]] \not= 0$$, I am not sure how to proceed. Is there a way to connect the solution obtained from Heisenberg's equation to the general solution for the quantum Harmonic oscillator?

One straightforward and informative approach is to compare your 2 expressions for $$a_H(t)$$ in the basis of energy eigenstates. Consider the matrix element \begin{align} \langle m | e^{i H t} a e^{-iHt} | n\rangle &= e^{-i \omega (n-m)t}\langle m| a |n\rangle\;. \end{align} It is easy to see that this is zero unless $$n = m+1$$, in which case the exponential reduces to $$e^{-i\omega t}$$, so clearly this will equal $$\langle m | e^{-\imath \omega t}a|n\rangle$$ for all $$m$$ and $$n$$. Since the energy eigenstates form a complete basis, this implies they are equal as operators.
Having seen how the proof goes in a particular basis, it is easier to see how it must go in a basis free version. The key point is that the lowering operator acting on the state between the two exponentials creates an offset that results in the remaining exponential at the end. So we want to derive something along the lines of $$e^{i\omega t n} a = a e^{i\omega t (n-1)}$$
With that in mind, first we write $$na = an + [n,a] = a(n-1)$$ This implies that, for an analytic function $$f$$, \begin{align} f(n)a &= \sum_m \frac{ f^{(m)}(0)}{m!}n^m a\\ &= a \sum_m \frac{ f^{(m)}(0)}{m!}(n-1)^m\\ &= a f(n-1) \end{align} so that $$e^{i \omega nt}ae^{-i\omega nt} = a e^{i \omega (n-1)t}e^{-i\omega nt} = a e^{-i\omega t}$$