Quantization Light Multimode

Photons

Quantization Light Multimode

Canonical Quantization of Multimode Radiation

Hamiltonian Eigenstates

Total number of photons

Momentum

Field Observables

Heisenberg Formalism of the Free Field

Photoelectric Effect

Canonical quantization of multimode radiation

In the discussion of quantization of a single mode (here) we discovered that in free space the most general solution of the Maxwell’s equation it is a plain wave:

\begin{aligned} \mathbf{E}(\mathbf{r}, t) & = \sum_\lambda E_\lambda(t) \mathbf{e}_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} + \text{c.c.} = \sum_\lambda i \mathcal E_\lambda \alpha_\lambda(t) \mathbf{e}_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} + \text{c.c.}\\ & = \sum_\lambda \mathcal E_\lambda(0) \mathbf{e}_\lambda e^{i(\mathbf{k}_\lambda \cdot \mathbf{r} -\omega_\lambda t)} + \text{c.c.} \\ & = \mathbf{E}^{(+)}(\mathbf{r}, t) + \mathbf{E}^{(-)}(\mathbf{r}, t) \end{aligned}

Also the frequency of oscillation is given by:

\omega_\lambda = c k_\lambda

In this case too as in the the one of the single photon, we impose periodic boundary condition on a volume V_\lambda = L^3; considering a cube with side of length L and taking this length to be is an integer multiple of \mathbf k:

\begin{aligned} & k_{\lambda,x} = 2\pi n_{\lambda,x} \\ & k_{\lambda,y} = 2\pi n_{\lambda,y} \\ & k_{\lambda,z} = 2\pi n_{\lambda,z} \end{aligned}

n_{\lambda,i} can take any positive or negative value in order to span the full space. \omega_\lambda can then take only discrete values:

\omega_\lambda = c\frac{2\pi}{L}\left(n_{\lambda_x}^2 + n_{\lambda_y}^2 + n_{\lambda_z}^2\right)^{\frac{1}{2}}

To define fully a wave vector these three parameters are needed, as well two parameters to describe the polarization, for convenience they will be chosen to be orthogonal between them and to \mathbf k_\lambda.

The energy of the can be written as:

\mathcal H = \frac{\varepsilon_0}{2} \int_{V}\left( \mathbf E^2 + c^2 \mathbf B^2\right) \mathrm d\mathbf r^3

where V = L^3 is the quantization volume.

The result of the calculation is simple:

\mathcal H = 2\varepsilon_0L^3\sum_\lambda \left|\mathcal E_\lambda(t) \right|^2 = \sum_\lambda \mathcal H_\lambda

but it is computed with contributions that cancel out, either because of Kronecker delta, either the electric and magnetic contribution have an opposite sign:

\begin{aligned} & \int_V \mathbf E^2 \mathrm d\mathbf r^3 = L^3 \sum_\lambda \left[ \left|\mathcal E_\lambda(t) \right|^2 + \mathcal E_\lambda(t)\bar{\mathcal E}_\lambda(t) \right] + \text{c.c.}\\ & \int_V \mathbf B^2 \mathrm d\mathbf r^3 = \frac{L^3}{c^2} \sum_\lambda \left[ \left|\mathcal E_\lambda(t) \right|^2 - \mathcal E_\lambda(t)\bar{\mathcal E}_\lambda(t) \right] + \text{c.c.} \end{aligned}

The total energy of radiation is the sum of the energy of each mode without any cross term; so the dynamic of each mode is decoupled and independent.

As for the quantization, it is possible then to proceed as in the single mode case, defining two conjugate variables and verify if the Hamiltonian derived is consistent the Maxwell equations.

Defining:

E_\lambda(t) = i \mathscr E^{(1)}_\lambda \alpha_\lambda(t)

with:

\mathscr E^{(1)}_\lambda = \sqrt{\frac{\hbar \omega_\lambda}{2\varepsilon_0 V_\lambda}}

It results that the Hamiltonian can be again written as:

\mathcal H_\lambda = \hbar \omega_\lambda \left| \alpha_\lambda^2(t) \right|^2 = \frac{\omega_\lambda}{2}\left(Q_\lambda^2 + P_\lambda^2\right)

using the real and imaginary part of \alpha_\lambda(t).

Since the modes are decoupled, the various Q_\lambda and P_\lambda are canonically conjugated variables.

It is the possible to proceed with the canonical quantization, introducing the operators:

\begin{aligned} & \mathbf Q_\lambda = \sqrt{\frac{\hbar}{2}}\left(\mathbf a_\lambda + \mathbf a^\dag_\lambda\right) \\ & \mathbf P_\mu = -i\sqrt{\frac{\hbar}{2}}\left(\mathbf a_\mu - \mathbf a^\dag_\mu\right) \\ & [\mathbf Q_\lambda, \mathbf P_\mu] = i \hbar \delta_{\lambda\mu} \end{aligned}

The Hamiltonian becomes:

\mathbf H_R = \sum_\lambda \frac{\omega_\lambda}{2}\left(\mathbf Q_\lambda^2 + \mathbf P_\lambda^2\right) = \sum_\lambda \frac{\hbar \omega_\lambda}{2}\left(\mathbf a_\lambda^\dag \mathbf a_\lambda + \frac{1}{2}\right)

The subscript R stands for radiation.

This is a central result in quantum optics: quantized radiation, in the absence of charges, can be considered a set of independent quantum harmonic oscillators.

Hamiltonian eigenstates

The eigenstates of a quantum mechanical system form a complete basis for that system; these can also be computed for the radiation, and we will be using what was derived for a single state of radiation:

\mathbf H_R | \psi \rangle = E | \psi \rangle

The starting point is the total Hamiltonian:

\mathbf H_R = \sum_\lambda \frac{\hbar \omega_\lambda}{2}\left(\mathbf a_\lambda^\dag \mathbf a_\lambda + \frac{1}{2}\right)

and the commutator relations:

\left[\mathbf a_i, \mathbf a_j^\dag \right] = i\delta_{ij}

For each mode, the eigenenergies are:

E_{n,\lambda} = \hbar\omega_\lambda \left(n_\lambda + \frac{1}{2}\right), \quad n_\lambda \in \mathbb N

and the following relations hold:

\begin{aligned} & \mathbf N_\lambda | \psi_{n,\lambda} \rangle = n_\lambda | \psi_{n,\lambda} \rangle = | \mathbf n_\lambda \rangle \\ & \mathbf a_\lambda^\dag | \psi_{n,\lambda} \rangle = \sqrt{n_\lambda+1} | \psi_{n_\lambda+1} \rangle \\ & \mathbf a_\lambda | \psi_{n,\lambda} \rangle = \sqrt n | \psi_{n_\lambda-1} \rangle \\ & \mathbf a_\lambda | \psi_{0,\lambda} \rangle = \mathbf a_\lambda | \mathbf 0_\lambda \rangle = 0 \end{aligned}

From these:

| \mathbf n_\lambda \rangle = \frac{\left(\mathbf a_\lambda^\dag\right)^{n_\lambda}}{\sqrt{n_\lambda!}} | \mathbf 0 \rangle

To find the eigenstate of the Hamiltonian, we use the property:

| \psi \rangle = | \mathbf n_1 \rangle \otimes \dots \otimes | \mathbf n_\lambda \rangle \otimes \dots = | n_1, \dots, n_\lambda, \dots \rangle

the state | n_1, \dots, n_\lambda, \dots \rangle, n_\lambda \in \mathbb N is called a Fock state.

So:

\mathbf H_R | \psi \rangle = \sum_\lambda \mathbf H_\lambda | n_1, \dots, n_\lambda , \dots \rangle = \sum_\lambda E_\lambda | n_1, \dots, n_\lambda , \dots \rangle

where the Fock states are a complete basis of the radiation states.

The dimension of this state grow quickly to be very large:

D = \left(N_{\text{max}}\right)^M,\; n_\lambda \le N_{\text{max}}

and M are non-empty modes. There is however a single vacuum state:

| 0_1, \dots, 0_\lambda, \dots \rangle = | \mathbf 0 \rangle

and from this state we can obtain any Fock state:

| n_1,\dots, n_\lambda, \dots \rangle = \frac{\left(\mathbf a^\dag_1\right)^{n_1}}{\sqrt{n_1!}} \dots \frac{\left(\mathbf a^\dag_\lambda\right)^{n_\lambda}}{\sqrt{n_\lambda!}} | \mathbf 0 \rangle

For simplicity, only the non-zero state are listed.

Since all the empty states have a finite energy, a priory the energy is always infinite; it is possible to get rid of these infinities using a process called renormalization, considering the eigenvalue of:

\hbar\omega_\lambda \left(n_\lambda + \frac{1}{2}\right) - E_V

where:

E_V \equiv \sum_\lambda \frac{1}{2} \hbar \omega_\lambda | \mathbf 0 \rangle

is the energy of the vacuum.

Total number of photon

It is possible to define the total number of photons observable as:

\mathbf N = \sum_\lambda \mathbf a^\dag_\lambda \mathbf a_\lambda

The Fock state | n_1,\dots, n_\lambda, \dots \rangle is an eigenstate of the number operator with eigenvalues:

\mathbf N | n_1,\dots, n_\lambda, \dots \rangle = \sum_\lambda n_\lambda | n_1,\dots, n_\lambda, \dots \rangle

which express the number of photons of the state.

Any Fock state which is an eigenvalue of the Hamiltonian is also an eigenvalue of the number of photon observable, but the inverse is not true, as there are eigenvalues of the number of photons observable which are not eigenvalue of the Hamiltonian if the frequency \omega_\lambda of the components is not always the same.

Momentum

The quantization of the linear momentum can be done in a fashion similar to the energy:

\mathbf P_R = \varepsilon_0 \int_{V} \mathbf E \times \mathbf B \; \mathrm d\mathbf r^3

When expanding this and applying the periodic boundary conditions, term cancel out except the square of the modulus of \alpha_\lambda(t) and lead to the simple expression:

\mathbf P_R = \sum_\lambda |\alpha_\lambda(t)|^2 \hbar \mathbf k_\lambda

When the applying the quantization:

\mathbf P_R = \sum_\lambda \hbar \mathbf k_\lambda \left(\mathbf a_\lambda^\dag \mathbf a + \frac{1}{2} \right)

But for each mode \lambda there is a counterpropagating wave \lambda with the opposite \mathbf k vector, so the term \frac{1}{2} cancel and the finale expression is:

\mathbf P_R = \sum_\lambda \hbar \mathbf k_\lambda \mathbf a_\lambda^\dag \mathbf a

Field observables

It is possible to measure the electric field or the magnetic field at a point of space; we can define the quantum observables associated with the corresponding classical observables, expanding the field as function of \alpha_\lambda(t) and it complex conjugate and replacing it with \mathbf a_\lambda and \mathbf a^\dag_\lambda:

\begin{aligned} \mathbf{E}(\mathbf{r}) & = \sum_\lambda i\mathbf{e}_\lambda \mathscr E^{(1)}_\lambda \left(\mathbf a_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} - \mathbf a_\lambda^\dag e^{-i\mathbf{k}_\lambda \cdot \mathbf{r}} \right) \\ \mathbf{B}(\mathbf{r}) & = \sum_\lambda i\frac{\mathbf k_\lambda \times \mathbf{e}_\lambda}{\omega_\lambda} \mathscr E^{(1)}_\lambda \left(\mathbf a_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} - \mathbf a_\lambda^\dag e^{-i\mathbf{k}_\lambda \cdot \mathbf{r}} \right) \end{aligned}

Defining the operators:

\begin{aligned} \mathbf{E}^{(+)}(\mathbf{r}) & = \sum_\lambda i\mathbf{e}_\lambda \mathscr E^{(1)}_\lambda \mathbf a_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} \\ \mathbf{E}^{(-)}(\mathbf{r}) & = \left[ \mathbf{E}^{(+)}(\mathbf{r})\right] ^\dag \end{aligned}

The electric field can be express as the sum of these operators; since they are not Hermitian, they are not physical observables.

The problem with infinities encountered with energies is happening also with the vacuum fluctuation, since the average of the electric field is zero but not the square:

\begin{aligned} & \langle \mathbf 0_\lambda | \mathbf{E}(\mathbf{r}) | \mathbf 0_\lambda \rangle = 0 \\ & \langle \mathbf 0_\lambda | \mathbf{E}^2(\mathbf{r}) | \mathbf 0_\lambda \rangle = \sum_\lambda \left[ \mathscr E^{(1)} \right]^2 \end{aligned}

Since there is an infinite number of modes, the second quantity is infinite. These vacuum fluctuations can be seen in physical phenomena, like the Lamb shift or the Casimir effect and require techniques to get finite values from these infinite series; to obtain them, there is a process similar to the renormalization done for the energy which was developed in the 1940s.

Heisenberg formalism of the free field

The Schrödinger formalism used so far in which the state vector is evolving with time while the operator is not evolving, does not allow to compute the photo detection at two different times; in order to do so, it is necessary to write an operator which is time dependent (the Heisenberg formalism), keeping the state vector constant.

The two are equivalent, and to derive the Heisenberg formalism, it is possible to note that in the Schrödinger formalism the time evolution is described by a unitary operator \mathbf U(t, t_0) and therefore:

\begin{aligned} \langle \mathbf A \rangle (t) & = \langle \psi(t) | \mathbf A | \psi(t) \rangle = \left(\langle \psi(0) | \mathbf U^\dag(t, t_0) \right) | \mathbf A | \left( \mathbf U(t, t_0) | \psi(0) \rangle \right) \\ & = \langle \psi(0) | \left( \mathbf U^\dag(t, t_0) | \mathbf A | \mathbf U(t, t_0) \right) | \psi(0) \rangle = \langle \psi(0) | \mathbf A(t) | \psi(0) \rangle \end{aligned}

Therefore, in the Heisenberg formalism, the operator take a form that evolve with time as:

\mathbf A(t) = \mathbf U^\dag(t, t_0) | \mathbf A | \mathbf U(t, t_0)

The electric field in the Heisenberg formulation take the form:

\mathbf{E}(\mathbf{r}, t) = \sum_\lambda i\mathbf{e}_\lambda \mathscr E^{(1)}_\lambda \left(\mathbf a_\lambda(0) e^{i(\mathbf{k}_\lambda \cdot \mathbf{r} - \omega_\lambda t)} + \mathbf a_\lambda^\dag(0) e^{-(i\mathbf{k}_\lambda \cdot \mathbf{r} - \omega_\lambda t)} \right)

This is the same equation as the semi-classical field with \alpha(0), \bar \alpha(0) replaced by \mathbf a_\lambda(0), \mathbf a_\lambda^\dag(0).

Photoelectric effect

Similarly for a single photon, it is possible to analyze the photoelectric effect, which is the ejection of electrons from matter illuminated by incident radiation under the effect of an frequency \omega higher than a specific threshold.

Using this effect it is possible to detect the number of electrons which are emitted with extremely high precision.

For multimode radiation, the single detection signal probability is similar to the single mode:

\begin{aligned} & \mathrm dP(\mathbf r,t) = w^{(1)}(\mathbf r,t)\,\mathrm dS \,\mathrm dt \\ & w^{(1)}(\mathbf r,t) = s \langle \psi(t) | \mathbf{E}^{(-)}(\mathbf{r})\mathbf{E}^{(+)}(\mathbf{r}) | \psi(t) \rangle = s \left\|\mathbf{E}^{(+)}(\mathbf{r}) | \psi(t) \rangle \right\|^2 \end{aligned}

s is again the sensitivity of the detector; this sensitivity is assumed constant, which is a good assumption if the frequency spectrum is not too broad, otherwise a more precise formula would be:

w^{(1)}(\mathbf r,t) =\left\|\sum_\lambda \mathbf{e}_\lambda \sqrt{s(\omega_\lambda)} \mathscr E^{(1)}_\lambda \mathbf a_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} | \psi(t) \rangle \right\|^2

In this more complete expression, both s and \mathscr E^{(1)}_\lambda are a function of \omega_\lambda.

A quantum efficiency parameter \eta_\lambda can be introduced that takes into account the fact that the detector is not ideal; the perfect detection of 1 require:

s_{0} = \frac{2\epsilon_0 c}{\hbar \omega_\lambda}

as it is multiplied by \mathscr E^{(1)}_\lambda for an ideal detector \omega_\lambda cancels out, and the ideal detection is:

w^{(1)}_0(\mathbf r,t) =\frac{c}{L^3}\left\|\sum_\lambda \mathbf{e}_\lambda \mathbf a_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} | \psi(t) \rangle \right\|^2

For an non ideal detector, we introduce the quantum efficiency \eta_{\omega_\lambda}:

\eta_\lambda = \frac{s_{\omega_\lambda}}{s_0} \le 1

This parameter can be used to express the detection rate:

w^{(1)}(\mathbf r,t) =\frac{c}{L^3}\left\|\sum_\lambda \sqrt{\eta_\lambda} \mathbf{e}_\lambda \mathbf a_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} | \psi(t) \rangle \right\|^2

If the efficiency is not varying too much, it can be considered roughly constant and give a formula often used in quantum optics:

w^{(1)}(\mathbf r,t) =\eta \frac{c}{L^3}\left\|\sum_\lambda \mathbf{e}_\lambda \mathbf a_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} | \psi(t) \rangle \right\|^2

For joint detection, the probability takes the form:

\begin{aligned} \mathrm dP^2(\mathbf r_1,t,\mathbf r_2,t) & = w^{(2)}(\mathbf r_1,t,\mathbf r_2,t)\,\mathrm dS_1 \,\mathrm dt\,\mathrm dS_2 \,\mathrm dt \\ w^{(2)}(\mathbf r_1,t,\mathbf r_2,t)& = s^2\langle \psi(t) | \mathbf{E}^{(-)}(\mathbf{r_1}) \mathbf{E}^{(-)}(\mathbf{r_2}) \mathbf{E}^{(+)}(\mathbf{r_2}) \mathbf{E}^{(+)}(\mathbf{r_1}) | \psi(t) \rangle \\ & = s^2\left\| \mathbf{E}^{(+)}(\mathbf{r_2}) \mathbf{E}^{(+)}(\mathbf{r_1})| \psi(t) \rangle\right\|^2 \end{aligned}

As in the single mode case, this is not just the product of the probability of the two single detections; this is different from the classical description in which this equality holds.

We can then use the Heisenberg formalism to write the expression for the single and double detection, which now allows to have two detections at two different times.

In case of w^{(1)} there is not much change:

w^{(1)}(\mathbf r,t) = s \langle \psi(0) | \mathbf{E}^{(-)}(\mathbf{r}, t)\mathbf{E}^{(+)}(\mathbf{r},t) | \psi(0) \rangle = s \left\|\mathbf{E}^{(+)}(\mathbf{r}, t) | \psi(0) \rangle \right\|^2

It is possible to express now w^{(2)}:

\begin{aligned} w^{(2)}(\mathbf r_1,t_1,\mathbf r_2,t_2)& = s^2\langle \psi(0) | \mathbf{E}^{(-)}(\mathbf{r_1}, t_1) \mathbf{E}^{(-)}(\mathbf{r_2},t_2) \mathbf{E}^{(+)}(\mathbf{r_2},t_2) \mathbf{E}^{(+)}(\mathbf{r_1},t_1) | \psi(0) \rangle \\ & = s^2\left\| \mathbf{E}^{(+)}(\mathbf{r_2},t_2) \mathbf{E}^{(+)}(\mathbf{r_1}, t_1)| \psi(0) \rangle\right\|^2 \end{aligned}

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