Errors and losses impact on planar integrated photonic circuits fidelity

Abstract

Every manufacturing or synthesis process suffers from possible errors or inaccuracies. Thus, this article presents a theoretical study on how such imperfections affect the efficiency of an planar integrated photonic circuit. Three sources of error were analyzed: absorption and scattering phenomena, inaccuracies in the directional couplers physical dimensions and uncertainties in the phase shifters implementation. As a result, graphs that indicate the behaviour of the photonic circuit efficiency, described by a quantity called fidelity, in function of the source of errors were obtained. These graphs allow to deduce scenarios where, even under the action of such imperfections, its fidelity still presents high values.

Appendices

Appendix A: Circuit matrix for the five-dimensional case

To write the matrix that describes the unitary operation performed on an input state by the photonic circuit shown in Fig. 1, we must first decompose it into simpler sequential operations. In this case, we will decompose it into six operations. The first of these is represented by the five parallel phase shifters located at the beginning of the circuit. This first operation can be described by the following matrix:

$$\begin{aligned} \hat{T}_1=\left( \begin{array}{ccccc} e^{{i\phi _1}} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad e^{{i\phi _2}} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad e^{i\phi _3} &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad e^{i\phi _4} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad e^{i\phi _5} \\ \end{array}\right) . \end{aligned}$$

(A1)

The other operations are performed by vertical arrangements of directional couplers. Sometimes we have a coupler between the first two paths and another coupler acting between the third and fourth paths, sometimes we have a coupler between the second and third paths and another coupler acting between the last two paths. Thus, these operations can be written in matrix form as follows

$$\begin{aligned} \hat{T}_2=\left( \begin{array}{ccccc} e^{i\phi _6}\sqrt{r_1} &{}\quad e^{i\phi _6}\sqrt{t_1} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ \sqrt{t_1} &{}\quad -\sqrt{r_1} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad e^{i\phi _7}\sqrt{r_2} &{}\quad e^{i\phi _7}\sqrt{t_2} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad \sqrt{t_2} &{}\quad -\sqrt{r_2} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ \end{array}\right) , \end{aligned}$$

(A2)

$$\begin{aligned} \hat{T}_3=\left( \begin{array}{ccccc} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad e^{i\phi _8}\sqrt{r_3} &{}\quad e^{i\phi _8}\sqrt{t_3} &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \sqrt{t_3} &{}\quad -\sqrt{r_3} &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad e^{i\phi _9}\sqrt{r_4} &{}\quad e^{i\phi _9}\sqrt{t_4} \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \sqrt{t_4} &{}\quad -\sqrt{r_4} \\ \end{array}\right) , \end{aligned}$$

(A3)

$$\begin{aligned} \hat{T}_4=\left( \begin{array}{ccccc} e^{i\phi _{10}}\sqrt{r_5} &{}\quad e^{i\phi _{10}}\sqrt{t_5} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ \sqrt{t_5} &{}\quad -\sqrt{r_5} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad e^{i\phi _{11}}\sqrt{r_6} &{}\quad e^{i\phi _{11}}\sqrt{t_6} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad \sqrt{t_6} &{}\quad -\sqrt{r_6} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ \end{array}\right) , \end{aligned}$$

(A4)

$$\begin{aligned} \hat{T}_5=\left( \begin{array}{ccccc} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad e^{i\phi _{12}}\sqrt{r_7} &{}\quad e^{i\phi _{12}}\sqrt{t_7} &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \sqrt{t_7} &{}\quad -\sqrt{r_7} &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad e^{i\phi _{13}}\sqrt{r_8} &{}\quad e^{i\phi _{13}}\sqrt{t_8} \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \sqrt{t_8} &{}\quad -\sqrt{r_8} \\ \end{array}\right) , \end{aligned}$$

(A5)

$$\begin{aligned} \hat{T}_6=\left( \begin{array}{ccccc} e^{i\phi _{14}}\sqrt{r_9} &{}\quad e^{i\phi _{14}}\sqrt{t_9} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ \sqrt{t_9} &{}\quad -\sqrt{r_9} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad e^{i\phi _{15}}\sqrt{r_{10}} &{}\quad e^{i\phi _{15}}\sqrt{t_{10}} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad \sqrt{t_{10}} &{}\quad -\sqrt{r_{10}} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ \end{array}\right) . \end{aligned}$$

(A6)

Once these matrices that describe the simplest sequential operations have been determined, the matrix that determines the circuit as a whole will be given by

$$\begin{aligned} \hat{A}=\hat{T}_{6}\cdot \hat{T}_{5}\cdot \hat{T}_{4}\cdot \hat{T}_{3}\cdot \hat{T}_{2}\cdot \hat{T}_{1}. \end{aligned}$$

(A7)

Appendix B: Details about fidelity for the two-dimensional case

If we consider the photonic circuit for $N=2$, we will see that it consists of only one directional coupler and one phase shifter. Thus, the matrices representing the ideal circuit and the lossy circuit will be given by

$$\begin{aligned} \hat{A}=\left( \begin{array}{cc} e^{i\phi }\sqrt{r} &{}\quad e^{i\phi }\sqrt{t} \\ \sqrt{t} &{}\quad -\sqrt{r} \end{array}\right) ,\quad \hat{B}=\left( \begin{array}{cc} e^{i\phi }\sqrt{\delta r} &{}\quad e^{i\phi }\sqrt{\delta t} \\ \sqrt{\delta t} &{}\quad -\sqrt{\delta r} \end{array}\right) . \end{aligned}$$

(B1)

Remembering that we are considering here only losses due to absorption and scattering of photons. To calculate the fidelity between these two operators, we will use Eq. (14). That way, we will first calculate the values of Tr$(\hat{A}^{\dagger }\hat{B})$ and Tr$(\hat{B}^{\dagger }\hat{B})$, obtaining

$$\begin{aligned} \hat{A}^{\dagger }\hat{B}&=\left( \begin{array}{cc} e^{-i\phi }\sqrt{r} &{}\quad \sqrt{t} \\ e^{-i\phi }\sqrt{t} &{}\quad -\sqrt{r} \end{array}\right) \cdot \left( \begin{array}{cc} e^{i\phi }\sqrt{\delta r} &{}\quad e^{i\phi }\sqrt{\delta t} \\ \sqrt{\delta t} &{}\quad -\sqrt{\delta r} \end{array}\right) \nonumber \\&=\left( \begin{array}{cc} \sqrt{\delta } &{}\quad 0 \\ 0 &{}\quad \sqrt{\delta } \end{array}\right) \quad \Rightarrow \quad \text {Tr}(\hat{A}^{\dagger }\hat{B})=2\sqrt{\delta }, \end{aligned}$$

(B2)

and

$$\begin{aligned} \hat{B}^{\dagger }\hat{B}&=\left( \begin{array}{cc} e^{-i\phi }\sqrt{\delta r} &{}\quad \sqrt{\delta t} \\ e^{-i\phi }\sqrt{\delta t} &{}\quad -\sqrt{\delta r} \end{array}\right) \cdot \left( \begin{array}{cc} e^{i\phi }\sqrt{\delta r} &{}\quad e^{i\phi }\sqrt{\delta t} \\ \sqrt{\delta t} &{}\quad -\sqrt{\delta r} \end{array}\right) \nonumber \\&=\left( \begin{array}{cc} \delta &{}\quad 0 \\ 0 &{}\quad \delta \end{array}\right) \quad \Rightarrow \quad \text {Tr}(\hat{B}^{\dagger }\hat{B})=2\delta .\qquad \end{aligned}$$

(B3)

With the results presented in Eq. (B2) and (B3), we can calculate the fidelity between the operators $\hat{A}$ and $\hat{B}$, obtaining

$$\begin{aligned} F(\hat{A},\hat{B})=\left| \dfrac{2\sqrt{\delta }}{\sqrt{4\delta }}\right| ^2 \quad \Rightarrow \quad F(\hat{A},\hat{B})=1. \end{aligned}$$

(B4)

Thus, for the semi-classical case, we have that the fidelity is always equal to 1 as shown in Fig. 3a. For the quantum case, this result does not apply, as we will show below. Let’s imagine the performance of a two-path photonic circuit in a state of the type $\left| {\psi }\right\rangle =\alpha \left| {0}\right\rangle +\beta \left| {1}\right\rangle$. The final state for the ideal case will be given by

$$\begin{aligned} \left| {\psi ^{\prime }}\right\rangle&=\hat{A}\left| {\psi }\right\rangle =\left( \begin{array}{cc} e^{i\phi }\sqrt{r} &{}\quad e^{i\phi }\sqrt{t} \\ \sqrt{t} &{}\quad -\sqrt{r} \end{array}\right) \cdot \left( \begin{array}{c} \alpha \\ \beta \end{array}\right) \nonumber \\ \Rightarrow \quad \left| {\psi ^{\prime }}\right\rangle&=\left( \begin{array}{c} e^{i\phi }\sqrt{r}\alpha + e^{i\phi }\sqrt{t}\beta \\ \sqrt{t} \alpha -\sqrt{r}\beta \end{array}\right) =\left( \begin{array}{c} \alpha ^{\prime } \\ \beta ^{\prime } \end{array}\right) . \end{aligned}$$

(B5)

For the case where there are photon losses, we have

$$\begin{aligned} \left| {\psi ^{\prime \prime }}\right\rangle&=\hat{B}\left| {\psi }\right\rangle =\left( \begin{array}{cc} e^{i\phi }\sqrt{\delta r} &{}\quad e^{i\phi }\sqrt{\delta t} \\ \sqrt{\delta t} &{}\quad -\sqrt{\delta r} \end{array}\right) \cdot \left( \begin{array}{c} \alpha \\ \beta \end{array}\right) \nonumber \\ \Rightarrow \quad \left| {\psi ^{\prime \prime }}\right\rangle&=\left( \begin{array}{c} e^{i\phi }\sqrt{\delta r}\alpha + e^{i\phi }\sqrt{\delta t}\beta \\ \sqrt{\delta t} \alpha -\sqrt{\delta r}\beta \end{array}\right) =\left( \begin{array}{c} \alpha ^{\prime \prime } \\ \beta ^{\prime \prime } \end{array}\right) . \end{aligned}$$

(B6)

Note that we can rewrite the case with losses as

$$\begin{aligned} \left| {\psi ^{\prime \prime }}\right\rangle =\sqrt{\delta }\left( \begin{array}{c} \alpha ^{\prime } \\ \beta ^{\prime } \end{array}\right) . \end{aligned}$$

(B7)

Therefore, the probability of detecting a photon at the outputs of this device will be $|\alpha ^{\prime \prime }|^2=\delta |\alpha ^{\prime }|^2$ and $|\beta ^{\prime \prime }|^2=\delta |\beta ^{\prime }|^2$. This implies a reduction in the probability of the photon being detected at any output by a factor $\delta$. Furthermore, there is a possibility that the photon will not be detected, that is, there is a probability that the photon will be absorbed or scattered during its passage through the circuit. This probability is equal to $1-\delta$. On the other hand, the fidelity between these two final states can be calculated using the following equation

$$\begin{aligned} F_q(\left| {\psi ^{\prime }}\right\rangle ,\left| {\psi ^{\prime \prime }}\right\rangle )=|\left\langle {\psi ^{\prime }|\psi ^{\prime \prime }}\right\rangle |^2 \nonumber \\ \Rightarrow \quad F_q(\left| {\psi ^{\prime }}\right\rangle ,\left| {\psi ^{\prime \prime }}\right\rangle )&=\delta . \end{aligned}$$

(B8)

Appendix C: Unbalanced Coefficients Within the Device

In carrying out this brief analysis, we will consider a photonic circuit for the case $N=5$, and for simplicity, we will set $r=t=1/2$ for all directional couplers and $\phi =0$ for all phase shifters. For an initial state of the form $\left| {\phi }\right\rangle =\sum _{i=1}^{5}a_i\left| {i}\right\rangle$, after the operation of the first column of directional couplers in the ideal circuit case, we will have the following state

$$\begin{aligned} \left| {\phi }\right\rangle =\dfrac{1}{\sqrt{2}}\begin{pmatrix} a_1+a_2 \\ a_1-a_2 \\ a_3+a_4 \\ a_3-a_4 \\ a_5\sqrt{2} \end{pmatrix}. \end{aligned}$$

(C1)

After the operation of the first column of directional couplers in the circuit with losses, we will have

$$\begin{aligned} \left| {\phi '}\right\rangle =\dfrac{1}{\sqrt{2}}\begin{pmatrix} (a_1+a_2)\sqrt{\delta } \\ (a_1-a_2)\sqrt{\delta } \\ (a_3+a_4)\sqrt{\delta } \\ (a_3-a_4)\sqrt{\delta } \\ a_5\sqrt{2} \end{pmatrix}. \end{aligned}$$

(C2)

Comparing Eqs. (C1) and (C2), we see that in the state $\left| {\phi '}\right\rangle$, only four out of the five coefficients are multiplied by the factor $\sqrt{\delta }$. This creates an imbalance among the coefficients, as the pairwise ratios between the coefficients will not always result in the same value. This imbalance becomes more evident when analyzing the states after passing through the second column of directional couplers. For the lossless case, we have

$$\begin{aligned} \left| {\varphi }\right\rangle =\dfrac{1}{\sqrt{2}}\begin{pmatrix} a_1+a_2 \\ (a_1-a_2+a_3+a_4)/\sqrt{2} \\ (a_1-a_2-a_3-a_4)/\sqrt{2} \\ a_5+(a_3-a_4)/\sqrt{2} \\ -a_5+(a_3-a_4)/\sqrt{2} \end{pmatrix}. \end{aligned}$$

(C3)

For the case with losses, however

$$\begin{aligned} \left| {\varphi '}\right\rangle =\sqrt{\dfrac{\delta }{2}}\begin{pmatrix} a_1+a_2 \\ (a_1-a_2+a_3+a_4)\sqrt{\delta /2} \\ (a_1-a_2-a_3-a_4)\sqrt{\delta /2} \\ a_5+(a_3-a_4)\sqrt{\delta /2} \\ -a_5+(a_3-a_4)\sqrt{\delta /2} \end{pmatrix}. \end{aligned}$$

(C4)

Now comparing Eqs. (C3) and (C4), is possible to see that there is a greater variety of factors being multiplied to the quantum states coefficients. The first coefficient is multiplied by a factor of $\sqrt{\delta }$, while the second and third coefficients are multiplied by $\delta$. Different factors are multiplied to different parts of the fourth and fifth coefficients. All this imbalance leads to changes in the probabilities of finding a photon at the input of a subsequent directional coupler and this error propagation within the photonic circuit alters the final quantum state, moving it away from the ideal case and reducing the operation fidelity.