Physical Background

Fock States and the Size of the Computational Basis

After the quantization of the Electro-Magnetic fields, which is roughly mode expansion of these fields, it’s possible to describe the state of the system by Fock (number) states. These states describe the number of photons in each mode (which is nonnegative integer). Example for $M=5$ modes and $N=3$ photons state is $|\psi\rangle=|00102\rangle$, which has 1 photon in 3rd mode and 2 photons in 5th mode. At our case each mode represents a specific waveguide in the array. Considering the bosonic nature of photons, the counting of states of (exactly) $N$ indistinguishable photons in $M$ distinguishable modes produce – $$ \tilde{N_b} = {M+N-1 \choose N}$$ where $\tilde{N_b}$ denotes the number of equiphoton states. For example - for $M=20$, $N=2$ exist $\tilde{N_b}=210$ photon states. For a general superposition of Fock states, which has at most $N_{max}$ photons in a single state, the computational basis for calculations is the union of all states which have at most $N_{max}$ photons. Its size is - $$ N_b = \sum_{n=0}^{N_{max}}{M+n-1 \choose n}$$ Here $N_b$ denotes the size of the computational basis. For $M=20$, $N=2$ exist $N_b={20+0-1 \choose 0}+{20+1-1 \choose 1}+{20+2-1 \choose 2}=1+20+210=231$ photon states. Notice that in general case the number of states, $N_b$, is much greater than the number of distinguished modes in waveguide array – $M$.

Hamiltonian of the Waveguide Array and Unitary Propagation

Consider a waveguide array with nearest neighbour coupling. The propagation and the coupling between the waveguides in such an array can be modeled by the following Hamiltonian - $$ H = \beta \sum_{n=1}^M a_n^\dagger a_n +\sum_{n=1}^Mc_{n,n+1}a^\dagger_{n+1}a_n +c_{n,n-1} a^\dagger_{n-1}a_n$$ Here $\beta$ is the propagation constant of the waveguides (taken as identical) and the $c_{i,j}$ coefficients are the coupling coefficients between adjacent waveguides. Use of Heisenberg equation of motion for $a_n^\dagger$ leads to the following system of ODE's - $$i\frac{\partial a_n ^\dagger}{\partial z}=\beta a_n^\dagger+c_{n,n+1} a_{n+1}^\dagger+c_{n,n-1} a_{n-1}^\dagger$$ The solution for $a_n ^\dagger(z)$, as shown in [1], is given by - $$a_n^\dagger(z) =\sum_{l=1}^M U_{n,l}(z) a_l^\dagger(z=0) \\ U_{n,l}(z) = (e^{iHz})_{n,l}$$

Notice that this solution aplies to any general Hamiltonian, and not only to that of nearest neighbour coupling, which I used as an example.

Intensity and Correlation Calculation

Using the Heisenberg representation, $a_n^\dagger(z)$ are all we need for intensity and correlation measurements. The intensity measured at the exit of waveguide $n$ is proportional to the expectation value of the photon number at that waveguide - $$\boxed{I_n(z) \propto APN(n,z) =\langle \psi_0 | a_n^\dagger(z) a_n(z) |\psi_0\rangle}$$

$APN(n,z)$ denotes the average photon number at waveguide $n$ at specified $z$ value. Normalized average photon number can be interpreted as the probability to measure (at least) a single photon at each spesific waveguide.

The second order correlation between waveguides $q$ and $r$ defined as [2] - $$\boxed{\Gamma_{q,r} = \langle \psi_0 | a_q^\dagger(z) a_r^\dagger(z) a_r(z) a_q(z) |\psi_0\rangle}$$

Normalized correlation can be interpreted as the probability to measure simultaniously (at least) a single photon at waveguide $q$ and (at least) a single photon at waveguide $r$. To stress the obvious - as we cannot measure non-zero intensity with 0 photons (state) - we cannot measure non-zero correlations with only single photon states.



[1] Yaron Bromberg, Yoav Lahini, Roberto Morandotti, and Yaron Silberberg, "Quantum and Classical Correlations in Waveguide Lattices", Phys. Rev. Lett. 102, 253904 (2009)

[2] Roy J. Glauber, "The Quantum Theory of Optical Coherence", Phys. Rev. 130, 2529 (1963)

© 2016 Alexander Cheplev, Faculty of Physics, Technion

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Last update:20.02.2016