Field Quantization. Annihilation and Creation operators

Taking up the Hamiltonian equation from Field Quantization. Hamiltonian for a single-mode field

\begin{equation} H = \frac{1}{2} \left( p^2 + \omega^2 q^2  \right) \end{equation}

Where $q$ and $p$ are the canonical variables. To make a quantum approach  we just replace them with the respective operator $\hat{q}$ and $\hat{p}$, which must satisfy the canonical commutation relation:

\begin{equation} \left[ \hat{q},\hat{p} \right]=i\hbar \hat{I} \equiv i\hbar \end{equation}

With these operators the Hamiltonian transforms into the operator:

\begin{equation} \hat{H} = \frac{1}{2} \left( \hat{p} ^2 + \omega^2 \hat{q} ^2  \right)  \end{equation}

Here is convenient to introduce the non-Hermitian annihilation $\hat{a}$ and creation $\hat{a} ^\dagger$ operators given by:

\begin{equation} \hat{a} = (2 \hbar \omega)^{-1/2} (\omega \hat{q}+ i\hat{p}) \end{equation} \begin{equation} \hat{a} ^\dagger = (2 \hbar \omega)^{-1/2} (\omega \hat{q}- i\hat{p}) \end{equation}

These satisfies the commutation relation: 

\begin{equation}  \left[ \hat{a},\hat{a} ^\dagger \right] =  \hat{a} \hat{a} ^\dagger -  \hat{a}^\dagger \hat{a}= 1 \end{equation}

Noting that

\begin{align*} \hat{a} ^\dagger \hat{a} &= (2 \hbar \omega)^{-1/2} (\omega \hat{q}- i\hat{p})(2 \hbar \omega)^{-1/2} (\omega \hat{q}+ i\hat{p}) =  \frac{1}{2 \hbar \omega}\left[  \omega^2\hat{q}^2 + i\omega\hat{q}\hat{p}-i\omega\hat{p}\hat{q}+\hat{p}^2  \right] \\ \\ &=  \frac{1}{2 \hbar \omega}\left[  \omega^2\hat{q}^2 + i\omega \left[ \hat{q},\hat{p} \right]+\hat{p}^2  \right] =  \frac{1}{2 \hbar \omega}\left[  \omega^2\hat{q}^2 - \omega\hbar +\hat{p}^2  \right] = \frac{1}{\hbar \omega} \frac{1}{2} \left( \hat{p}^2 +  \omega^2\hat{q}^2 \right)  - \frac{1}{2} \\ \\ &= \frac{1}{\hbar \omega}\hat{H}-\frac{1}{2} \end{align*}

So, the Hamiltonian operator can be defined as:

\begin{equation} \hat{H} = \hbar\omega \left(  \hat{a} ^\dagger \hat{a} + \frac{1}{2} \right) \end{equation}

Now we can describe the energy eigenvalues. We denote $\vert n \rangle$ as energy eigenstate of the single mode field with eigenvalue $E_n$ such that:

\begin{equation} \hat{H}\vert n \rangle =  \hbar\omega \left(  \hat{a} ^\dagger \hat{a} + \frac{1}{2} \right) \vert n \rangle = E_n \vert n \rangle \end{equation}

The operator $\hat{a}^\dagger \hat{a}$ has a great importance and is called number operator $\hat{n}$. We will generate the eigenvalue equations for $\hat{a}^\dagger \vert n \rangle$ and $\hat{a} \vert n \rangle$ to appreciate the effect of the operators in the energy level and understand why they are called that. Multiplying Eq. (8) by  $\hat{a}^\dagger$ we obtain:

\begin{equation*} \hbar\omega \left( \hat{a} ^\dagger \hat{a} ^\dagger \hat{a} + \frac{1}{2}\hat{a} ^\dagger \right) \vert n \rangle =  E_n \hat{a}^\dagger \vert n \rangle \end{equation*}

\begin{equation*} \hbar\omega \left( \hat{a} ^\dagger \left(\hat{a}\hat{a}^\dagger -1 \right) + \frac{1}{2}\hat{a} ^\dagger \right) \vert n \rangle =  E_n \hat{a}^\dagger \vert n \rangle \end{equation*}

\begin{equation*} \hbar\omega \left( \hat{a} ^\dagger\hat{a}\hat{a}^\dagger + \frac{1}{2}\hat{a} ^\dagger \right) \vert n \rangle =   E_n  \hat{a}^\dagger \vert n \rangle +  \hbar\omega  \hat{a}^\dagger \vert n \rangle \end{equation*}

\begin{equation*} \hbar\omega \left( \hat{a} ^\dagger\hat{a} + \frac{1}{2} \right) \left(\hat{a}^\dagger \vert n \rangle \right)=  \left( E_n + \hbar\omega \right) \left( \hat{a}^\dagger \vert n \rangle \right) \end{equation*}

\begin{equation} \Rightarrow \hat{H} \left(\hat{a}^\dagger \vert n \rangle \right)=  \left( E_n + \hbar\omega \right) \left( \hat{a}^\dagger \vert n \rangle \right) \end{equation}

This is the equation for the eigenstate  $\hat{a}^\dagger \vert n \rangle$, it has energy eigenvalue of $E_n + \hbar \omega$. We can interpret this as if the operator $\hat{a}^\dagger$ created one quantum of energy $\hbar \omega$, that is why it is called as the creation operator. In the same way, we can multiply Eq. (8) by $\hat{a}$ to obtain the equation for the eigenstate $\hat{a}\vert n \rangle$:

\begin{equation} \hat{H} \left(\hat{a} \vert n \rangle \right)=  \left( E_n - \hbar\omega \right) \left( \hat{a} \vert n \rangle \right) \end{equation}

Analogous to the previous case, we can interpret it as if the operator $\hat{a}$ eliminated or annihilated one quantum of energy $\hbar \omega$, that is why it is called as the annihilation operator. 

For the states $\hat{a}\vert n \rangle$, $\hat{a}^\dagger \vert n \rangle$ and $\hat{n} \vert n \rangle$ we have:

\begin{equation}\hat{a}\vert n \rangle = c_n \vert n-1 \rangle \end{equation}

\begin{equation}\hat{a}^\dagger \vert n \rangle = d_n \vert n+1 \rangle \end{equation}

\begin{equation}\hat{n} \vert n \rangle = n \vert n \rangle \end{equation}

Where $c_n$ and $d_n$ are constants. The number of states must be normalized, i.e. $\langle n \vert n \rangle = 1$. The inner product of  $\hat{a}\vert n \rangle$ with itself is:

\begin{equation} (\langle n \vert \hat{a}^\dagger)(\hat{a}\vert n \rangle) = \langle n \vert \hat{a}^\dagger \hat{a} \vert n \rangle = \langle n \vert \hat{n} \vert n \rangle = n \langle n \vert n \rangle = n \end{equation}

Also

\begin{equation} (\langle n \vert \hat{a}^\dagger)(\hat{a}\vert n \rangle) = \langle n-1 \vert c_n^* c_n  \vert n-1 \rangle = \vert c_n \vert ^2 \langle n-1 \vert n-1 \rangle = \vert c_n \vert ^2 \end{equation}

Thus $c_n = \sqrt{n}$ and

\begin{equation} \boxed{ \hat{a}\vert n \rangle = \sqrt{n} \vert n-1 \rangle} \end{equation}

In the same way, the inner product of  $\hat{a}^\dagger \vert n \rangle$ with itself is:

\begin{equation} (\langle n \vert \hat{a})(\hat{a}^\dagger \vert n \rangle) = \langle n \vert \hat{a} \hat{a}^\dagger \vert n \rangle = \langle n \vert (\hat{a}^\dagger\hat{a}+1) \vert n \rangle = \langle n \vert \hat{n} \vert n \rangle +  \langle n \vert  n \rangle =  n + 1\end{equation}

Also

\begin{equation} (\langle n \vert \hat{a})(\hat{a}^\dagger \vert n \rangle) = \langle n+1 \vert d_n^* d_n \vert n+1 \rangle = \vert d_n \vert ^2 \langle n+1 \vert n+1 \rangle = \vert d_n \vert ^2 \end{equation}

Thus $d_n = \sqrt{n+1}$ and

\begin{equation}\boxed{ \hat{a}^\dagger \vert n \rangle = \sqrt{n+1} \vert n+1 \rangle} \end{equation}

Acknowledgement

I would like to thank Dr. Carlos Herman Wiechers Medina for their support, this post is based on the notes from the Quantum Optics class.