Canonical quantisation harmonic oscillator

Question

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"?

Answer 1

• 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

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