# Canonical quantisation harmonic oscillator

I have a question on the canonical quantisation as described at the linked wiki page: https://en.wikipedia.org/wiki/Quantum_field_theory#Canonical_quantisation

we take the displacement of a classical harmonic oscillator described as $$x(t)={\frac {1}{\sqrt {2\omega }}}ae^{-i\omega t}+{\frac {1}{\sqrt {2\omega }}}a^{*}e^{i\omega t},$$

and promote $$x(t)$$ to a linear operator $${\displaystyle {\hat {x}}(t)}$$:

$${\displaystyle {\hat {x}}(t)={\frac {1}{\sqrt {2\omega }}}{\hat {a}}e^{-i\omega t}+{\frac {1}{\sqrt {2\omega }}}{\hat {a}}^{\dagger }e^{i\omega t}.}$$

the coefficients $$a$$ and it's complex conjugate $$a^*$$ are replaced $${\hat a}$$ and $${\hat a}^{\dagger }$$.

Question: why following this instruction $${\hat a}$$ becomes creation operator and $${\hat a}^{\dagger }$$ annihilation opererator and not conversely? is there any physical reason that justifies it based on the plus or minus sign of $$e^{-i\omega t}$$ resp. $$e^{i\omega t}$$? or is it just a "random choice"?

• In quantum mechanics, $$a$$ and $$a^\dagger$$ are ladder operators, in that they raise and lower the number of excitations presence in the harmonic oscillator: $$a^\dagger |0\rangle \propto |1\rangle \quad \Rightarrow \quad a^\dagger |n\rangle = \sqrt{n+1}|n+1\rangle, \\ a|1\rangle \propto |0\rangle \quad \Rightarrow \quad a|n\rangle = \sqrt{n}|n-1\rangle.$$

Each application on $$a$$ on your number-eigenstate $$|n\rangle$$ removes a quantum of excitation ($$\hbar \omega$$ for the harmonic oscillator). The opposite for $$a^\dagger$$.

$$a^\ast$$ is the specific case of $$a^\dagger$$ when $$a$$ is a scalar, not a matrix.

In field theory now, the vacuum $$|0\rangle$$ has no field excitations (i.e. no particles, no waves on the string). I can create one excitation by applying $$a^\dagger$$, like before. Since in the field theory context I am now creating a particle, it's called the creation operator. Same reasoning for the annihilation operator $$a$$.

• In the Heisenberg picture, the time dependence of an operator is given by: $$\dot{A} = \frac{i}{\hbar} [H,A],$$ so: $$\dot{a} = \frac{i}{\hbar}[H, a] = \frac{i}{\hbar}[\hbar \omega \, a^\dagger a,a] = -\mathrm{i}\omega a \quad \Rightarrow \quad a(t) = a(0)e^{-\mathrm{i}\omega t},$$ and

$$\dot{a^\dagger} = \frac{i}{\hbar}[H, a^\dagger] = \frac{i}{\hbar}[\hbar \omega \, a^\dagger a,a^\dagger] = \mathrm{i}\omega a^\dagger \quad \Rightarrow \quad a^\dagger(t) = a^\dagger(0)e^{\mathrm{i}\omega t}.$$

• morally you use the notation $x(t)={\frac {1}{\sqrt {2\omega }}}a(t)+{\frac {1}{\sqrt {2\omega }}}a^{*}(t)$ at first without knowing which exponential $e^{-i\omega t}$ or $e^{i\omega t}$ belong to which coefficient and deduce it by applying derivatives at the equation with commutator from Heisenberg picture? does this association creation operator -> $e^{-i\omega t}$ & annihilation operator -> $e^{i\omega t}$ have an "intuitive" meaning? Commented Sep 22, 2019 at 3:01
• You first start with defining $a$ and $a^\dagger$ in terms of $x$ and $p$, all in time-independent formalism. The you compute the time dependence of the operators with the Heisenberg equation of motion (the commutator thing). If you want an intuitive meaning, the energy associated with a state after applying $a$ is $-\hbar \omega$ (since I can write $-\omega$ in the exponent as $E/\hbar$, because you are removing a quantum of energy $\hbar \omega$. Commented Sep 22, 2019 at 3:05
• thinking about the intuitive meaning you explained I'm not sure if I understand you completely. let consider the case of annihilation operator $\hat {a}$: let $|n\rangle$ our initial state and it becomes $\sqrt{n}|n-1\rangle$. can I see the belonging coefficient $e^{-i\omega t}=e^{-iE/ \hbar t}$ as "energy portion that leaves the system" after applying the annihilation? heuristically justified by energy conservation? or have I misunderstood you? Commented Sep 22, 2019 at 14:58
• My way was just a hand wavy way of remembering which operator has the $-\omega$. An energy eigenstate has a time dependence $\exp(iE/\hbar)$, so you act with $a$ it now becomes $E-\hbar\omega$ in the exponent. Commented Sep 22, 2019 at 15:52