Quantum Optics
Free Field Quantization
Quantization of single mode electromagnetic field
Maxwell’s equations: \[\nabla \cdot \vec{E} = 0\] \[\nabla \cdot \vec{B} = 0\] \[\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}\] \[\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}\]
where \(\vec{E} = \vec{E}(\vec{r}, t)\ ;\ \vec{B} = \vec{B}(\vec{r}, t)\)
Field polarization along x-axis: \[\vec{E} = E_x(z, t) \hat{x}\ \]
Single mode field: \[ \vec{E} = \vec{E}(z, t) = \sqrt{\frac{\hbar \omega}{2 \epsilon_0 V}} q(t)\sin(kz) \hat{x}\ \]
where \(q(t)\) has dimensions of length, \(V\) is the volume of the cavity and \(\omega_m = cm\pi/L\) where \(c\) is the speed of light. Also
\[ \vec{B} = \vec{B}(z, t) = \frac{\mu_0\epsilon_0}{k}\sqrt{\frac{\hbar \omega}{2 \epsilon_0 V}}\ p(t)\cos(kz) \hat{y}\ \]
where \(p(t) = \dot{q}(t)\) has dimensions of momentum.
We can write the Hamiltonian of the field as:
\[ H = \frac{1}{2} \int_V dV (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) = \frac{1}{2}(p^2 + \omega^2 q^2) \]
1D Harmonic Oscillator
The Hamiltonian of a 1D harmonic oscillator is given by:
\[ H = \frac{1}{2}(p^2 + \omega^2 q^2) \]
where \(p\) and \(q\) are the momentum and position of the classical oscillator.
Quantization of 1D Harmonic Oscillator
\[q\rightarrow \hat{q}\ ;\ p\rightarrow \hat{p}\]
we have the commutation relation: \[ [\hat{q}, \hat{p}] = i\hbar \]
The Hamiltonian of the 1D quantum harmonic oscillator is given by:
\[ \hat{H} = \frac{1}{2}(\hat{p}^2 + \omega^2 \hat{q}^2) \] where \(\hat{p}, \hat{q}\) are Hermitian operators.
Creation and Annihilation Operators
The creation and annihilation operators are defined as:
\[ \hat{a} = \frac{1}{\sqrt{2\hbar\omega}} (\omega \hat{q} + i\hat{p}) \] \[ \hat{a}^\dagger = \frac{1}{\sqrt{2\hbar\omega}} (\omega \hat{q} - i\hat{p}) \]
The operators are non-Hermitian.
also \[ \hat{q} = \sqrt{\frac{\hbar}{2\omega}} (\hat{a} + \hat{a}^\dagger) \] \[ \hat{p} = i\sqrt{\frac{\hbar\omega}{2}} (\hat{a}^\dagger - \hat{a}) \]
The commutation relation is given by:
\[ [\hat{a}, \hat{a}^\dagger] = 1 \]
Putting the above relations in previous equation we get:
\[ \hat{E_x} = \sqrt{\frac{\hbar \omega}{\epsilon_0 V}} (\hat{a} + \hat{a}^\dagger) \sin(kz) \] \[ \hat{B_y} = \sqrt{\frac{\hbar \omega}{\epsilon_0 V}} (\hat{a}^\dagger - \hat{a}) \cos(kz) \]
where \(\mathcal{E} = \sqrt{\frac{\hbar \omega}{\epsilon_0 V}}, \mathcal{B} = \sqrt{\frac{\hbar \omega}{\epsilon_0 V}}\) are the field per photon.
The Hamiltonian of the field is given by:
\[ \hat{H} = \frac{1}{2}(\hat{p}^2 + \omega^2 \hat{q}^2) = \hbar \omega (\hat{a}^\dagger \hat{a} + \frac{1}{2}) \]
Evolution of the Field
The time evolution of the field is given by:
\[ i\hbar \frac{\partial}{\partial t}\hat{a} = [\hat{a}, \hat{H}] \] \[ \implies \frac{\partial}{\partial t}\hat{a} = -i\omega \hat{a} \]
The solution of the above equation is given by:
\[ \hat{a}(t) = \hat{a}(0) e^{-i\omega t} \] \[ \hat{a}^\dagger(t) = \hat{a}^\dagger(0) e^{i\omega t} \]
\[\hat{E_x} = \mathcal{E}(\hat{a}(0)e^{-i\omega t} + \hat{a}^\dagger(0)e^{i\omega t}) \sin(kz)\]
Fock States
The Fock states are the eigenstates of the number operator \(\hat{a}^\dagger \hat{a}\).
\[ \hat{a}^\dagger \hat{a} |n\rangle = n |n\rangle \] where \(n\) is the number of photons in the state \(|n\rangle\) and \(\hat{a}^\dagger \hat{a} = n\) is the photon number operator in the mode.