Quantum circuit simulation of linear optics using fermion to qubit encoding

We first explain how a specific form of entangled multi-fermions can effectively behave as multi-bosons. In the second quantization language, the bosonic creation and annihilation operators ${\hat{a}}^{\dagger}_{i}$ and ${\hat{a}}_{i}$ ($i=1,\cdots,M$) obey the following commutation relations:

$\displaystyle[{\hat{a}}_{i},{\hat{a}}^{\dagger}_{j}]=\delta_{ij},\quad[{\hat{a}}_{i},{\hat{a}}_{j}]=[{\hat{a}}^{\dagger}_{i},{\hat{a}}^{\dagger}_{j}]=0,$

(1)

while the fermionic operators ${\hat{b}}^{\dagger}_{i}$ and ${\hat{b}}_{i}$ obey the anti-commutation relations:

$\displaystyle\{{\hat{b}}_{i},{\hat{b}}^{\dagger}_{j}\}=\delta_{ij},\quad\{{\hat{b}}_{i},{\hat{b}}_{j}\}=\{{\hat{b}}^{\dagger}_{i},{\hat{b}}^{\dagger}_{j}\}=0,$

(2)

where $\{\hat{A},\hat{B}\}\equiv\hat{A}\hat{B}+\hat{B}\hat{A}$. The above relations satisfy the Pauli exclusion principle for fermions, which prohibits the superposition of two fermions in the same state. Indeed, we see that ${\hat{b}}^{\dagger}_{i}{\hat{b}}^{\dagger}_{i}|vac\rangle=-{\hat{b}}^{\dagger}_{i}{\hat{b}}^{\dagger}_{i}|vac\rangle=0$ by Eq. (2), where $|vac\rangle$ is a vacuum state. On the other hand, if the fermions have internal degrees of freedom, such as spin, fermionic modes with different internal states can occupy the same spatial mode. By denoting a $K$-dimensional internal degree of freedom as $\mu$ ($\mu=0,\cdots K-1$), a fermionic operator with internal degrees of freedom $\mu$ is defined as ${\hat{b}}^{\dagger\mu}_{i}$ and ${\hat{b}}^{\mu}_{i}$. The anticommutation relations for the operators are as follows:

$\displaystyle\{{\hat{b}}^{\mu}_{i},{\hat{b}}^{\dagger\nu}_{j}\}=\delta_{ij}\delta^{\mu\nu},\quad\{{\hat{b}}^{\mu}_{i},{\hat{b}}^{\nu}_{j}\}=\{{\hat{b}}^{\dagger\mu}_{i},{\hat{b}}^{\dagger\nu}_{j}\}=0.$

(3)

In such a case, the fermions can condensate in the same spatial mode up to $K$. We aim to employ this feature of multi-fermionic states for mimicking the Bose-Einstein condensation (BEC) with the cutoff $K$. Fig. 2 explains the concept of fermionic condensation.

On the other hand, for the fermionic condensation to operate like the BEC, we must properly consider the fundamental differences between bosons and fermions, i.e., the exchange symmetry and antisymmetry indicated in Eqs. (1) and (2). Shchesnovich [19] showed that the interchangeability of entanglement and exchange symmetry can render entangled multi-fermions symmetric under the exchange of spatial modes. Here, we introduce the effective bosonic state of multi-fermions with the condensation limit $K$ in the second quantization language, which offers a more refined explanation than of the first quantization language used in Ref. [19].

Let us consider an $N$-fermionic state,

$\displaystyle{\hat{b}}_{i_{1}}^{\dagger\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}\cdots{\hat{b}}_{i_{N}}^{\dagger\mu_{N}}|vac\rangle,$

(4)

($i_{\alpha}=1,2,\cdots,M$ and $\mu_{\alpha}=1,2,\cdots,K$ for $1\leq\alpha\leq N$). This state is always antisymmetric under the exchange of the total indices $(\mu,i)$. However, if $K\geq N$, we can obtain a symmetric state under the spatial modes $i_{\alpha}$ by suitably superposing fermionic states as follows:

$\displaystyle\frac{1}{\sqrt{N!}}{\hat{b}}_{i_{1}}^{\dagger[\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}\cdots{\hat{b}}_{i_{N}}^{\dagger\mu_{N}]}|vac\rangle$

(5)

(a square bracket $[,]$ on the upper indices means that the indices are antisymmetrized. For the simplest example, ${\hat{b}}_{i_{1}}^{\dagger[\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}]}\equiv{\hat{b}}_{i_{1}}^{\dagger\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}-{\hat{b}}_{i_{1}}^{\dagger\mu_{2}}{\hat{b}}_{i_{2}}^{\dagger\mu_{1}}$). Since the following relation,

	$\displaystyle{\hat{b}}_{i_{1}}^{\dagger[\mu_{1}}\cdots{\hat{b}}_{i_{\alpha}}^{\dagger\mu_{\alpha}}\cdots{\hat{b}}_{i_{\beta}}^{\dagger\mu_{\beta}}\cdots{\hat{b}}_{i_{N}}^{\dagger\mu_{N}]}|vac\rangle$
	$\displaystyle=-{\hat{b}}_{i_{1}}^{\dagger[\mu_{1}}\cdots{\hat{b}}_{i_{\beta}}^{\dagger\mu_{\beta}}\cdots{\hat{b}}_{i_{\alpha}}^{\dagger\mu_{\alpha}}\cdots{\hat{b}}_{i_{N}}^{\dagger\mu_{N}]}|vac\rangle$
	$\displaystyle={\hat{b}}_{i_{1}}^{\dagger[\mu_{1}}\cdots{\hat{b}}_{i_{\beta}}^{\dagger\mu_{\alpha}}\cdots{\hat{b}}_{i_{\alpha}}^{\dagger\mu_{\beta}}\cdots{\hat{b}}_{i_{N}}^{\dagger\mu_{N}]}|vac\rangle,$		(6)

holds for any $\alpha$ and $\beta$ for $1\leq\alpha\leq N$ and $1\leq\beta\leq N$, we have

		$\displaystyle\frac{1}{\sqrt{N!}}{\hat{b}}_{i_{1}}^{\dagger[\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}\cdots{\hat{b}}_{i_{N}}^{\dagger\mu_{N}]}|vac\rangle$
		$\displaystyle=\frac{1}{\sqrt{N!}}{\hat{b}}_{\{i_{1}}^{\dagger[\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}\cdots{\hat{b}}_{i_{N}\}}^{\dagger\mu_{N}]}|vac\rangle$		(7)

(a brace $\{,\}$ on the lower indices on the right hand side denotes that the indices are symmetrized. For the simplest example, ${\hat{b}}_{\{i_{1}}^{\dagger\mu_{1}}{\hat{b}}_{i_{2}\}}^{\dagger\mu_{2}}\equiv{\hat{b}}_{i_{1}}^{\dagger\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}+{\hat{b}}_{i_{2}}^{\dagger\mu_{1}}{\hat{b}}_{i_{1}}^{\dagger\mu_{2}}$). Therefore, we can consider Eq (5) to be an effective $N$-boson state with the condensation limit $K$.

As a simple example, when $N=2$, Eq. (5) becomes

$\displaystyle\frac{1}{\sqrt{2}}\big{(}{\hat{b}}_{i_{1}}^{\dagger\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}-{\hat{b}}_{i_{1}}^{\dagger\mu_{2}}{\hat{b}}_{i_{2}}^{\dagger\mu_{1}}\big{)}|vac\rangle.$

(8)

By exchanging the mode indices $i_{1}$ and $i_{2}$, we have

	$\displaystyle\frac{1}{\sqrt{2}}\big{(}{\hat{b}}_{i_{2}}^{\dagger\mu_{1}}{\hat{b}}_{i_{1}}^{\dagger\mu_{2}}-{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}{\hat{b}}_{i_{1}}^{\dagger\mu_{1}}\big{)}|vac\rangle$
	$\displaystyle=\frac{1}{\sqrt{2}}\big{(}-{\hat{b}}_{i_{1}}^{\dagger\mu_{2}}{\hat{b}}_{i_{2}}^{\dagger\mu_{1}}+{\hat{b}}_{i_{1}}^{\dagger\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}\big{)}|vac\rangle$
	$\displaystyle=\frac{1}{\sqrt{2}}\big{(}{\hat{b}}_{i_{1}}^{\dagger\mu_{1}}{\hat{b}}_{i_{2}}^{\dagger\mu_{2}}-{\hat{b}}_{i_{1}}^{\dagger\mu_{2}}{\hat{b}}_{i_{2}}^{\dagger\mu_{1}}\big{)}|vac\rangle.,$		(9)

where the second line is obtained by changing the order of fermionic operators.

Since the antisymmetrical entanglement of the fermions is essential for effective bosonic states to behave like bosons, the exchange symmetry of the state must be preserved under evolutions. In other words, if we want to simulate the bosonic scattering process with fermions, the transformation operators of fermions must preserve the antisymmetrical entanglement. We observe that some transformation operators satisfy this restriction. We first consider a bosonic operator $T$ of the following form:

$\displaystyle T=\exp\big{[}it(\sum_{j,k=1}^{M}\Phi_{jk}{\hat{a}}_{j}^{\dagger}{\hat{a}}_{k})\big{]},$

(10)

where $t$ is the evolution time and $\Phi_{jk}\in\mathbb{C}$. We note that

$\displaystyle\sum_{jk}\Phi_{jk}{\hat{a}}_{j}^{\dagger}{\hat{a}}_{k}$

(11)

behaves as the Hamiltonian of the given system by setting $\Phi_{jk}=\Phi^{*}_{kj}$. Then, the transformation of ${\hat{a}}_{i}^{\dagger}$ under $T$ is given by

	$\displaystyle T{\hat{a}}^{\dagger}_{i}T^{\dagger}=$	$\displaystyle\sum_{j}\exp(it\Phi^{*})_{ij}{\hat{a}}_{j}^{\dagger}$
	$\displaystyle\equiv$	$\displaystyle\sum_{j}u_{ij}{\hat{a}}_{j}^{\dagger},$		(12)

where $\Phi$ is a Hermitian matrix whose elements are $\Phi_{ij}$ and $\sum_{j}u_{ij}u_{kj}^{*}=\delta_{ik}$. In the fermionic system, the corresponding operator $T_{f}$ is expressed as follows:

$\displaystyle T_{f}=\exp(it\sum_{\mu}\sum_{j,k}\Phi_{jk}\hat{b}_{j}^{\dagger\mu}{\hat{b}}_{k}^{\mu}),$

(13)

which gives

$\displaystyle T_{f}{\hat{b}}^{\dagger\mu}_{i}T_{f}^{\dagger}=\sum_{j}u_{ij}{\hat{b}}_{j}^{\dagger\mu}.$

(14)

Then, the state Eq. (5) evolves via $T_{f}$ as follows:

	$\displaystyle|\Psi\rangle_{f}=$	$\displaystyle\frac{1}{\sqrt{N!}}\sum_{k_{1},\cdots,k_{N}}u_{\{j_{1}}^{k_{1}}\cdots u_{j_{N}\}}^{k_{N}}{\hat{b}}^{\dagger[\mu_{1}}_{k_{1}}\cdots{\hat{b}}^{\dagger\mu_{N}]}_{k_{N}}|vac\rangle$
	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{N!}}\sum_{k_{1},\cdots,k_{N}}u_{\{j_{1}}^{\{k_{1}}\cdots u_{j_{N}\}}^{k_{N}\}}{\hat{b}}^{\dagger[\mu_{1}}_{\{k_{1}}\cdots{\hat{b}}^{\dagger\mu_{N}]}_{k_{N}\}}|vac\rangle.$

The second line of the above equation shows that the transformed state is a linear combination of effective multi-boson states, which itself is an effective multi-boson state. In a more general form, we see that any number-conserving Hamiltonian looks like $H=\sum_{jk}\Phi_{jk}{\hat{a}}_{j}^{\dagger}{\hat{a}}_{k}+\mathrm{c.c.}$.

Finally, we check whether the measurement of the state Eq. (5) that evolves with Eq. (10) is effectively bosonic, i.e., the scattering probability is proportional to the absolute square of the transformation matrix permanent. Suppose first that we postselect terms without bunching, irrespective of what the internal states of the particles are. Without loss of generality, we can assume the boson number distribution vector as follows:

$\displaystyle{\vec{n}}=(\underbrace{1,1,\cdots,1}_{N},\underbrace{0,\cdots,0)}_{M-N}.~{}(M\geq N)$

(16)

Then, the scattering probability is given with a projector $E=\sum_{\mu_{1}\cdots\mu_{N}}({\hat{b}}^{\dagger\mu_{1}}_{1})\cdots{\hat{b}}_{N}^{\dagger\mu_{N}}|vac\rangle\langle vac|{\hat{b}}^{\mu_{N}}_{N}\cdots{\hat{b}}^{\mu_{1}}_{1}$ as follows:

	$\displaystyle P=$	$\displaystyle{\rm Tr}(E\rho_{f})$
	$\displaystyle=$	$\displaystyle\sum_{\mu_{1},\cdots,\mu_{N}}\langle vac|{\hat{b}}^{\mu_{N}}_{N}\cdots{\hat{b}}^{\mu_{1}}_{1}|\Psi\rangle\langle\Psi|_{f}{\hat{b}}^{\dagger\mu_{1}}_{1}\cdots{\hat{b}}_{N}^{\dagger\mu_{N}}|vac\rangle.$		(17)

Using the relation,

	$\displaystyle{\hat{b}}_{i_{1}}^{\mu_{1}}\cdots{\hat{b}}_{i_{N}}^{\mu_{N}}{\hat{b}}_{k_{1}}^{\dagger[\nu_{1}}\cdots{\hat{b}}_{k_{N}}^{\dagger\nu_{N}]}|vac\rangle$
	$\displaystyle=\delta^{\{k_{1}}_{i_{1}}\cdots\delta^{k_{N}\}}_{i_{N}}\delta^{[\nu_{1}}_{\mu_{1}}\cdots\delta^{\nu_{N}]}_{\mu_{N}}|vac\rangle,$		(18)

we have

$\displaystyle P\sim|{\textrm{perm}}(u)|^{2},$

(19)

where $u$ is an $N\times N$ matrix whose entries are $u_{ij}$ and ${\textrm{perm}}(u)$ denotes the permanent of $u$, as expected for a bosonic systems with $T$ [24, 11]. If the postselected states permit bunching, the probability becomes proportional to the permanent of the submatrix of $u$ as expected [25, 26, 27, 28].

Since a fermionic state of the form indicated in Eq. (5) can simulate a linear scattering of bosons, we conclude that digital quantum computers can also simulate the same system using the JW transformation. Before explaining how we actually organize quantum circuits and algorithms for such a simulation, we first review the JW transformation, which maps fermions to qubits [20].

In the JW transformation, qubit states $|0\rangle$ and $|1\rangle$ correspond to the empty and occupied states of fermions for a given mode, i.e., the following isomorphism should hold:

		$\displaystyle\textrm{$N$ qubit state }|{\vec{n}}\rangle=|n_{1},\cdots,n_{N}\rangle\quad(n_{j}=0,1)$
		$\displaystyle\quad\cong\quad\textrm{$N$ fermionic state }({\hat{b}}^{\dagger}_{1})^{n_{1}}\cdots({\hat{b}}^{\dagger}_{1})^{n_{N}}|vac\rangle.$		(20)

The left and right hand side denotes an $N$-qubit state and an $N$-fermionic state, respectively, and $\cong$ represents that the two sides are in a correspondence relationship with each other. For this relationship to hold, there must be operators acting on the $N$-qubit system that play the roles of creation and annihilation operators. Indeed, we can construct such operators by combining the Pauli operators $X_{j},Y_{j}$ and $Z_{j}$ ($j=1,\cdots L$), i.e., ${\hat{b}}_{j}^{\dagger}(X,Y,Z)\cong{\hat{b}}_{j}^{\dagger}$ and ${\hat{b}}_{j}(X,Y,Z)\cong{\hat{b}}_{j}$.

We can see that $|{\vec{n}}\rangle$ and $b^{\dagger}_{j}(X,Y,Z)$ must satisfy the following conditions:

• •

  If $n_{j}=0$, then ${\hat{b}}_{j}|{\vec{n}}\rangle=0$

• •

  If $n_{j}=1$, then ${\hat{b}}_{j}|{\vec{n}}\rangle=(-1)^{s_{\vec{n}}^{j}+1}|n_{1},\cdots,n_{j}\oplus 1,\cdots,n_{L}\rangle$ where $s_{\vec{n}}^{j}\equiv\sum_{k=1}^{j-1}n_{k}$. Note that $(-1)^{s_{\vec{n}}^{j}+1}$ comes from the anticommutation property of the creation-annihilation operators.

It can easily be verified that

		$\displaystyle{\hat{b}}_{j}(X,Y,Z)\equiv(\otimes_{k=1}^{j-1}Z_{k})\otimes\sigma^{-}_{j},$
		$\displaystyle{\hat{b}}^{\dagger}_{j}(X,Y,Z)\equiv(\otimes_{k=1}^{j-1}Z_{k})\otimes\sigma^{+}_{j}$		(21)

($\sigma^{+}\equiv|1\rangle\langle 0|$ and $\sigma^{-}\equiv|0\rangle\langle 1|$) satisfy the above conditions. One can also check that Eq. (II.2) satisfies the anticommutation relations, i.e., $\{{\hat{b}}_{j},{\hat{b}}_{k}^{\dagger}\}=\delta_{jk}$ and $\{{\hat{b}}_{j}^{\dagger},{\hat{b}}_{k}^{\dagger}\}=\{{\hat{b}}_{j},{\hat{b}}_{k}\}=0$. The state transformation of Eq. (II.2) and operator transformations in Eq. (II.2) define the JW transformation for the digital simulation of fermionic systems.

By combining the JW transformation and the results of Section II.1, we can see that $N$ bosons in $M$ modes can be simulated with $NM$ qubits (see Fig. 3). To impose this correspondence, consider an $MN$-qubit state

$\displaystyle|(n_{1}^{1},\cdots,n_{1}^{N}),(n_{2}^{1},\cdots,n_{2}^{N}),\cdots,$

$\displaystyle(n_{M}^{1},\cdots,n_{M}^{N})\rangle$

(22)

where $n_{i}^{\mu}=0,1$ and each bracket $(n_{i}^{1},n_{i}^{2},\cdots,n_{i}^{N})$ denotes the state of a bundle of $N$ qubits.

If $n_{i}^{\mu}=1$, then it is considered in the fermion picture that a fermion exists in the $i$th mode with internal state $j$. Any state of this kind can be generated from $|vac\rangle\cong|\underbrace{00\cdots 0}_{N\times M}\rangle$ with the creation operators as follows:

		$\displaystyle{\hat{b}}^{\dagger 1}_{1}=\sigma^{+},$
		$\displaystyle{\hat{b}}^{\dagger 2}_{1}=Z\otimes\sigma^{+},$
		$\displaystyle\quad\vdots$
		$\displaystyle{\hat{b}}^{\dagger N}_{1}=\underbrace{Z\otimes\cdots\otimes Z}_{N-1}\otimes\sigma^{+},$
		$\displaystyle\quad\vdots$
		$\displaystyle{\hat{b}}^{\dagger N}_{M}=\underbrace{Z\otimes\cdots\otimes Z}_{NM-1}\otimes\sigma^{+}.$		(23)

Now we can express an effective multi-boson state described in Eq. (5), which is entangled as antisymmetric under the internal states in the qubit space. As an example, consider the case with $N$ bosons when all bosons from 1 to $N$ are in different modes with respect to each other. Using Eq. (II.1), such a state can be expressed as $\frac{1}{\sqrt{N!}}{\hat{b}}^{\dagger[1}_{1}\cdots{\hat{b}}^{\dagger N]}_{N}|vac\rangle$. By defining $\chi_{i}=(\underbrace{0,\cdots,0,\overset{\textrm{$i$th}}{1},0,\cdots,0}_{N})$ and $\chi_{0}=(\underbrace{0,0,\cdots,0}_{N})$, the state can be expressed in the $NM$ qubit space as follows:

$\displaystyle\frac{1}{\sqrt{N!}}\sum_{\rho\in S_{N}}sgn(\rho)|\chi_{\rho(1)},\chi_{\rho(2)},\cdots,\chi_{\rho(N)},\chi_{0},\cdots,\chi_{0}\rangle,$

(24)

where $S_{N}$ is the permutation group On the other hand, if all the bosons are in the same mode, e.g., the first mode, the state can be written as follows:

	$\displaystyle\frac{1}{\sqrt{N!}}{\hat{b}}^{\dagger[1}_{1}\cdots{\hat{b}}^{\dagger N]}_{1}|vac\rangle$	$\displaystyle={\hat{b}}^{\dagger 1}_{1}\cdots{\hat{b}}^{\dagger N}_{1}|vac\rangle$
		$\displaystyle\cong|(\underbrace{1,1,\cdots,1}_{N}),\chi_{0},\cdots,\chi_{0}\rangle.$		(25)

For the case with $N=2$ and $M=3$, Eq. (24) takes the following form:

	$\displaystyle\frac{1}{\sqrt{2}}\Big{(}|\chi_{1},\chi_{2},\chi_{0}\rangle-|\chi_{2},\chi_{1},\chi_{0}\rangle\Big{)}$
	$\displaystyle=\frac{1}{\sqrt{2}}\big{(}|10,01,00\rangle-|01,10,00\rangle\big{)},$		(26)

which corresponds to the bosonic state ${\hat{a}}^{\dagger}_{1}{\hat{a}}^{\dagger}_{2}|vac\rangle$, while Eq. (II.2) becomes $|11,00,00\rangle$, which corresponds to the bosonic state $\frac{1}{\sqrt{2}}({\hat{a}}^{\dagger}_{1})^{2}|vac\rangle$.

Since Eqs. (22) and (II.2) represent a mapping from bosonic systems to qubits, we can digitally simulate multi-boson systems with the following process:

1. 1.

   Preparation of the initial state: We first need to prepare the initial states of the form shown in Eq. (II.1), which can be achieved by adopting one of the known antisymmtrization algorithms, e.g., those in Refs. [29, 30]. On the other hand, we can find optimal algorithms for the states with small $N$ case-by-case.

2. 2.

   Evolution: The unitary operations can be executed by substituting Eq. (II.2) into the Hamiltonian operator of Eq. (11).

3. 3.

   Measurement: While the order of the excited states is unimportant, the number of excited states in each bundle is crucial because it determines the distributions of boson numbers. For example, if $N=3$, $(100)$, $(010)$, and $(001)$, in all cases a mode has one particle with different internal state. Nevertherless, we only record that one of three qubit states in the bundle is excited. Eq. (II.1) represents such a measurement process.

Since two qubits can represent a bosonic mode with a maximal photon number of two, our protocol needs four qubits here.

We will now simulate the HOM dip (see, e.g., [31] for a pedagogic review) with a two-dimensional internal degree of freedom that creates distinguishability. By denoting the internal state of bosons as $s$ $(=0,1)$, the creation and annihilation operators are written as ${\hat{a}}^{\dagger}_{is}$ and ${\hat{a}}_{is}$ with $[{\hat{a}}_{is},{\hat{a}}^{\dagger}_{jr}]=\delta_{ij}\delta_{sr}$. Then, an $N$-boson state ${\hat{a}}_{i_{1}s_{1}}^{\dagger}{\hat{a}}_{i_{2}s_{2}}^{\dagger}\cdots{\hat{a}}_{i_{N}s_{N}}^{\dagger}|vac\rangle$ ($i_{\alpha}\in\{1,\cdots,N\}$, $s_{\beta}\in\{0,1\}$ for $\alpha,\beta\in\{1,\cdots,N\}$) can effectively be expressed as a fermionic state as follows:

$\displaystyle\frac{1}{\sqrt{N!}}{\hat{b}}_{i_{1}s_{1}}^{\dagger[\mu_{1}}{\hat{b}}_{i_{2}s_{2}}^{\dagger\mu_{2}}\cdots{\hat{b}}_{i_{N}s_{N}}^{\dagger\mu_{N}]}|vac\rangle.$

(34)

Therefore, the general initial state for the HOM dip with two photons can be written as follows:

$\displaystyle|\Psi\rangle_{i}={\hat{a}}^{\dagger}_{1|s\rangle}{\hat{a}}^{\dagger}_{2|r\rangle}|vac\rangle\cong\frac{1}{\sqrt{2}}{\hat{b}}_{1|s\rangle}^{\dagger[1}{\hat{b}}_{2|r\rangle}^{\dagger 2]}|vac\rangle,$

(35)

where $|s\rangle$ and $|r\rangle$ are the general internal states of the form $\zeta|0\rangle+\xi|1\rangle$ ($\zeta,\xi\in\mathbb{C}$ and $|\zeta|^{2}+|\xi|^{2}=1$). To simulate this type of HOM dip, we need eight qubits, which are displayed in Fig. 6. Each qubit corresponds to the particle states $(i,\mu,s)$ as indicated in the figure.