Quantum operation of fermionic systems and process tomography using Majorana fermion gates

The density matrix of fermion modes is an operator on $\mathcal{H}_{m}$, which can be expressed in the form

$$\rho=\sum_{\bar{n},\bar{n}^{\prime}}\rho_{\bar{n},\bar{n}^{\prime}}|\bar{n}\rangle\langle\bar{n}^{\prime}|,$$

(2)

where $\rho_{\bar{n},\bar{n}^{\prime}}$ are elements of the density matrix. The density matrix must be positive semidefinite and normalised, i.e. $\rho\geq 0$ and $\mathrm{Tr}(\rho)=1$. According to SR, $[C,\rho]=0$, i.e. $\rho_{\bar{n},\bar{n}^{\prime}}=0$ for all $\bar{n}$ and $\bar{n}^{\prime}$ that $(-1)^{|\bar{n}|}\neq(-1)^{|\bar{n}^{\prime}|}$ Cisneros1998 ; Hegerfeldt1997 ; Wick1997 ; Aharonov1967 ; Wick1970 .

Because $C=C^{\dagger}$ and $C^{2}=\openone$, the SR condition can be re-expressed as $\rho=\mathcal{P}(\rho)$, where $\mathcal{P}(\bullet)=\frac{1}{2}\left(\bullet+C\bullet C\right)$. We remark that $\mathcal{P}$ is a projection superoperator, i.e. $\mathcal{P}^{2}=\mathcal{P}$. For all positive semidefinite and normalised density matrices $\rho$, $\mathcal{P}(\rho)$ is a valid state.

The physical process of a quantum system is characterised by a trace-preserving completely-positive map Choi1975 . A completely-positive map can be expressed as $\mathcal{M}(\rho)=\sum_{q}F_{q}\rho F_{q}^{\dagger}$, where $F_{q}$ are Kraus operators on $\mathcal{H}_{m}$ Nielsen2010 . The map is trace-preserving if and only if $\sum_{q}F_{q}^{\dagger}F_{q}=\openone$. A map is unital if $\mathcal{M}(\openone)=\openone$. We note that not all physical processes can be described by a completely-positive map, specifically when the system and environment have initial correlations Jordan2004 .

According to SR, for the physical process of fermion modes, the output state of the map must be valid for all valid input states. Lemma 1 can be proved by noticing that $\mathcal{M}\mathcal{P}(\rho)=\mathcal{P}\mathcal{M}\mathcal{P}(\rho)$ for all density matrices $\rho$. However, this is not the sufficient condition of a valid map on fermion modes, similar to the difference between positive and completely-positive maps. For a valid fermionic map, if it acts on a subset of fermion modes, the corresponding composite map on all the modes should also obey SR.

For a system with $m+p$ modes, the Hilbert space is $\mathcal{H}_{m}\otimes\mathcal{H}_{p}$, and we use A and B to denote two subsystems, respectively. The Fock state of the composed system reads $|{\bar{n},\bar{n}^{\prime}}\rangle_{\rm AB}=|{\bar{n}}\rangle_{\rm A}\otimes|{\bar{n}^{\prime}}\rangle_{\rm B}=A_{{\rm A};\bar{n}}^{\dagger}A_{{\rm B};\bar{n}^{\prime}}^{\dagger}|{{\rm V}}\rangle_{\rm AB}$. Note the order of two operators. According to this definition of Fock states for the composite system, we always have $a_{{\rm AB};i}=a_{{\rm A};i}\otimes\openone_{\rm B}$ if the mode-$i$ is in the first $m$ modes, where $a_{{\rm AB};i}$ is the operator that acts on $\mathcal{H}_{m}\otimes\mathcal{H}_{p}$, and $a_{{\rm A};i}$ is the operator that acts on $\mathcal{H}_{m}$. For a mode-$i$ in the other $p$ modes, the operator $a_{{\rm AB};i}$ cannot be written in the tensor product form with the identity operator on the subsystem A, to be consistent with the anticommutation relation. Although fermion operators are not local in general, parity operators are local, i.e. $C_{\rm A}\otimes\openone_{\rm B}$ is the parity operator of the first $m$ modes, $\openone_{\rm A}\otimes C_{\rm B}$ is the parity operator of the other $p$ modes, and $C_{\rm A}\otimes C_{\rm B}$ is the parity operator of all modes.

The map $\mathcal{M}$ denotes a local process on the first $m$ modes. According to the definition of composite-system Fock states, the map that acts on the composite system can be expressed in the tensor product form $\mathcal{M}_{\rm AB}=\mathcal{M}_{\rm A}\otimes\mathcal{I}_{\rm B}$, where $\mathcal{M}_{\rm A}$ is the map on the operator space of $\mathcal{H}_{m}$, and $\mathcal{I}_{\rm B}$ is the identity map on the operator space of $\mathcal{H}_{p}$. We define $\mathcal{P}_{\rm AB}(\bullet)=\frac{1}{2}\left(\bullet+C_{\rm A}\otimes C_{\rm B}\bullet C_{\rm A}\otimes C_{\rm B}\right)$.

Later, we will show that a valid map according to SR is always local.

According to Choi’s theorem Choi1975 , a map is completely positive if and only if the corresponding Choi matrix is positive semidefinite. For a fermion map $\mathcal{M}$, the Choi matrix is

$\displaystyle{\rm Choi}(\mathcal{M})=\mathcal{M}_{\rm A}\otimes\mathcal{I}_{\rm B}\left(|\Phi\rangle\langle\Phi|\right),$

(4)

where $|{\Phi}\rangle=\sum_{\bar{n}}|{\bar{n}}\rangle_{\rm A}\otimes|{\bar{n}}\rangle_{\rm B}$, and each of A and B has $m$ fermion modes. The map is trace-preserving if and only if $\mathrm{Tr}_{\rm A}\left[{\rm Choi}(\mathcal{M})\right]=\openone_{\rm B}$, and the map is unital if and only if $\mathrm{Tr}_{\rm B}\left[{\rm Choi}(\mathcal{M})\right]=\openone_{\rm A}$.

The state $|{\Phi}\rangle$ is an eigenstate of the parity operator $C_{\rm A}\otimes C_{\rm B}$ with the eigenvalue $+1$, i.e. the total particle number is even. Therefore, if $\mathcal{M}$ is valid,

	$\displaystyle{\rm Choi}(\mathcal{M})$	$\displaystyle=$	$\displaystyle(\mathcal{M}_{\rm A}\otimes\mathcal{I}_{\rm B})\mathcal{P}_{\rm AB}\left(|\Phi\rangle\langle\Phi|\right)$		(5)
		$\displaystyle=$	$\displaystyle\mathcal{P}_{\rm AB}(\mathcal{M}_{\rm A}\otimes\mathcal{I}_{\rm B})\mathcal{P}_{\rm AB}\left(|\Phi\rangle\langle\Phi|\right).$

Because ${\rm Choi}(\mathcal{M})=\mathcal{P}_{\rm AB}\left({\rm Choi}(\mathcal{M})\right)$, we have $[C_{\rm A}\otimes C_{\rm B},{\rm Choi}(\mathcal{M})]=0$. The only if part is proved.

Let $\rho$ be a density matrix of $m+p$ modes. Subsystems A, B and C have $m$ modes, and the subsystem D has $p$ modes. According to the teleportation formalism,

$\displaystyle\mathcal{P}_{\rm AD}(\rho_{\rm AD})=\langle{\Phi}|_{\rm BC}\left[|\Phi\rangle\langle\Phi|_{\rm AB}\otimes\mathcal{P}_{\rm CD}(\rho_{\rm CD})\right]|{\Phi}\rangle_{\rm BC}.$

(6)

Then,

			$\displaystyle(\mathcal{M}_{\rm A}\otimes\mathcal{I}_{\rm D})\mathcal{P}_{\rm AD}(\rho_{\rm AD})$		(7)
		$\displaystyle=$	$\displaystyle\langle{\Phi}|_{\rm BC}\left[{\rm Choi}(\mathcal{M})_{\rm AB}\otimes\mathcal{P}_{\rm CD}(\rho_{\rm CD})\right]|{\Phi}\rangle_{\rm BC}.$

Because $\langle{\Phi}|_{\rm BC}\bullet|{\Phi}\rangle_{\rm BC}=\langle{\Phi}|_{\rm BC}\mathcal{P}_{\rm BC}(\bullet)|{\Phi}\rangle_{\rm BC}$ and $(\mathcal{P}_{\rm BC}\otimes\mathcal{I}_{\rm AD})(\mathcal{P}_{\rm AB}\otimes\mathcal{P}_{\rm CD})=(\mathcal{P}_{\rm AD}\otimes\mathcal{I}_{\rm BC})(\mathcal{P}_{\rm AB}\otimes\mathcal{P}_{\rm CD})$, we have

			$\displaystyle(\mathcal{M}_{\rm A}\otimes\mathcal{I}_{\rm D})\mathcal{P}_{\rm AD}(\rho_{\rm AD})$		(8)
		$\displaystyle=$	$\displaystyle\langle{\Phi}|_{\rm BC}\left[\mathcal{P}_{\rm AB}({\rm Choi}(\mathcal{M})_{\rm AB})\otimes\mathcal{P}_{\rm CD}(\rho_{\rm CD})\right]|{\Phi}\rangle_{\rm BC}$
		$\displaystyle=$	$\displaystyle\mathcal{P}_{\rm AD}(\mathcal{M}_{\rm A}\otimes\mathcal{I}_{\rm D})\mathcal{P}_{\rm AD}(\rho_{\rm AD}).$

Here, we have used Eq. (5), Eq. (7) and ($|{\Phi}\rangle_{\rm BC}$ is an eigenstate of $C_{\rm B}\otimes C_{\rm C}$)

			$\displaystyle\langle{\Phi}|_{\rm BC}\rho_{\rm ABCD}|{\Phi}\rangle_{\rm BC}$		(9)
		$\displaystyle=$	$\displaystyle\langle{\Phi}|_{\rm BC}(\mathcal{P}_{\rm BC}\otimes\mathcal{I}_{\rm AD})(\rho_{\rm ABCD})|{\Phi}\rangle_{\rm BC}.$

The if part is proved.

A valid state of fermion modes is block diagonal according to the parity. For a valid map $\mathcal{M}$, although off-diagonal blocks of the input and output states are zero, the map may have transitions between elements of off-diagonal blocks. An example is the evolution driven by the Hamiltonian $H=\mu a^{\dagger}a$ for only one mode. The corresponding map is $\mathcal{M}(\bullet)=e^{-i\mu a^{\dagger}at}\bullet e^{i\mu a^{\dagger}at}$. These off-diagonal-block transitions do not cause any effect on the output state of the mode. However, the effect can be observed in a multi-mode system, e.g. the evolution from the state $\frac{1}{\sqrt{2}}(|{0,1}\rangle+|{1,0}\rangle)$ to $\frac{1}{\sqrt{2}}(|{0,1}\rangle+e^{-i\mu t}|{1,0}\rangle)$ if the map acts on the first mode.

In general, the parity operator $C$ is not a conserved quantity. For example, the one-mode map $\mathcal{M}(\bullet)=a\bullet a^{\dagger}+a^{\dagger}\bullet a$ is valid and flips the parity. The parity is not conserved when the system and environment exchange particles. The parity is a conserved quantity in unitary processes.

The measurement on a quantum system is described by a set of completely-positive maps $\{\mathcal{E}_{k}\}$. Each map can be expressed as $\mathcal{E}_{k}(\bullet)=\sum_{q}F_{k,q}\bullet F_{k,q}^{\dagger}$, $E_{k}=\sum_{q}F_{k,q}^{\dagger}F_{k,q}$ are POVM operators, and $\sum_{k}E_{k}=\openone$. Given an input state $\rho$, the probability of the measurement outcome $k$ is $\mathrm{Tr}(E_{k}\rho)$, and the output state is $\mathcal{E}_{k}(\rho)/\mathrm{Tr}(E_{k}\rho)$.

For a measurement on fermion modes, maps $\mathcal{E}_{k}$ must be valid according to SR. Because $F_{k,q;{\rm A}}\otimes\openone_{\rm B}|{\Phi}\rangle=\openone_{\rm A}\otimes F_{k,q;{\rm B}}^{\rm T}|{\Phi}\rangle$, where the transpose is in the Fock basis, we have $E_{k;{\rm B}}^{*}=\mathrm{Tr}_{\rm A}[{\rm Choi}(\mathcal{E}_{k})]$. For a valid map, $[C_{\rm A}\otimes C_{\rm B},{\rm Choi}(\mathcal{E}_{k})]=0$, then $[C_{\rm B},E_{k;{\rm B}}^{*}]=0$. In the Fock basis, $C_{\rm B}$ is real. Therefore, $[C_{\rm B},E_{k;{\rm B}}]=0$, i.e. $[C,E_{k}]=0$.

The only if part has been proven. Now we prove the if part. Given a valid POVM operator $E_{k}$, there exists a completely-positive map $\mathcal{E}_{k}(\bullet)=\sqrt{E_{k}}\bullet\sqrt{E_{k}}$. Because $E_{k}$ is positive semidefinte and block diagonal, $\sqrt{E_{k}}$ is also positive semidefinte and block diagonal. We have $[C,\sqrt{E_{k}}]=0$. Then, $|{\Psi}\rangle=\sqrt{E_{k;{\rm A}}}\otimes\openone_{\rm B}|{\Phi}\rangle$ is an eigenstate of $C_{\rm A}\otimes C_{\rm B}$ with the eigenvalue $+1$. Because ${\rm Choi}(\mathcal{E}_{k})=|\Psi\rangle\langle\Psi|$, $\mathcal{E}_{k}$ is a valid fermion map.

Using the Jordan-Wigner transformation, we can express fermion operators using Pauli operators Ortiz2001 , which can be used to obtain the explicit expressions of fermion operators in the Fock basis. We decompose the Hilbert space of $m$ fermion modes into $m$ subsystems, i.e. $\mathcal{H}_{m}=\mathcal{H}_{1}^{\otimes m}$, and the Hilbert space of each subsystem is two-dimensional. Accordingly, the Fock state $|{\bar{n}}\rangle=\bigotimes_{i}|{n_{i}}\rangle$.

The four Pauli operators of one subsystem are $\sigma^{I}=|0\rangle\langle 0|+|1\rangle\langle 1|$, $\sigma^{X}=|1\rangle\langle 0|+|0\rangle\langle 1|$, $\sigma^{Y}=i|1\rangle\langle 0|-i|0\rangle\langle 1|$ and $\sigma^{Z}=|0\rangle\langle 0|-|1\rangle\langle 1|$. We define the Pauli operator on $m$ subsystems $S_{i}=\sigma^{I\otimes(i-1)}\otimes\sigma^{S}\otimes\sigma^{I\otimes(m-i)}$, which is the operator of $\sigma^{S}=\sigma^{X},\sigma^{Y},\sigma^{Z}$ that acts on the $i$-th subsystem.

In the Fock basis, we can express Majorana fermion operators as Gluza2018

	$\displaystyle c_{2i-1}=a_{i}+a_{i}^{\dagger}=\sum_{\bar{n}}(-1)^{\sum_{j=i+1}^{m}n_{j}}X_{i}|\bar{n}\rangle\langle\bar{n}|,$		(12)
	$\displaystyle c_{2i}=-i(a_{i}-a_{i}^{\dagger})=\sum_{\bar{n}}(-1)^{\sum_{j=i+1}^{m}n_{j}}Y_{i}|\bar{n}\rangle\langle\bar{n}|.$		(13)

Accordingly, the Jordan-Wigner transformation reads

	$\displaystyle c_{2i-1}=X_{i}\prod_{j=i+1}^{m}Z_{j},$		(14)
	$\displaystyle c_{2i}=Y_{i}\prod_{j=i+1}^{m}Z_{j}.$		(15)

We remark that the expression in the Fock basis and the Jordan-Wigner transformation depend on the definition of the Fock state, i.e. the order of fermion operators in $A_{\bar{n}}$, which must be consistent. When using the Fock basis and the Jordan-Wigner transformation, we must take into account all the fermion modes, e.g. including both $m+m$ modes when we compute the Choi matrix.

We can express a completely positive map as

$\displaystyle\mathcal{M}(\bullet)=\sum_{\bar{u},\bar{u}^{\prime}}\chi_{\bar{u},\bar{u}^{\prime}}C_{\bar{u}}\bullet C_{\bar{u}^{\prime}}.$

(19)

The corresponding Choi matrix is

$\displaystyle{\rm Choi}(\mathcal{M})=\sum_{\bar{u},\bar{u}^{\prime}}\chi_{\bar{u},\bar{u}^{\prime}}C_{\bar{u}}|\Phi\rangle\langle\Phi|C_{\bar{u}^{\prime}}.$

(20)

States $C_{\bar{u}}|{\Phi}\rangle$ are orthogonal, because $\langle{\Phi}|C_{\bar{u}}C_{\bar{u}^{\prime}}|{\Phi}\rangle=\mathrm{Tr}(C_{\bar{u}}C_{\bar{u}^{\prime}})$. The particle-number parity of $C_{\bar{u}}|{\Phi}\rangle$ is the parity of $|\bar{u}|$. Therefore, for a valid map, $\chi_{\bar{u},\bar{u}^{\prime}}=0$ for all $\bar{u}$ and $\bar{u}^{\prime}$ with different parities, i.e. $\chi$ is block diagonal. Similar to the case of Pauli operators, the map is completely positive if and only if $\chi\geq 0$; the map is trace-preserving if and only if $\mathrm{Tr}(\chi)=1$.

Now, we can prove that valid maps according to SR are local. We consider four fermion operators $C_{\bar{u}_{1}}$, $C_{\bar{u}_{2}}$, $C_{\bar{u}_{3}}$ and $C_{\bar{u}_{4}}$ that act on a composite system with $m+p$ modes. In the four operators, $C_{\bar{u}_{1}}$ and $C_{\bar{u}_{2}}$ act on the first $m$ modes, and $C_{\bar{u}_{3}}$ and $C_{\bar{u}_{4}}$ act on the other $p$ modes. A process on the first $m$ modes has terms in the form $C_{\bar{u}_{1}}\bullet C_{\bar{u}_{2}}$; and a process on the other $p$ modes has terms in the form $C_{\bar{u}_{3}}\bullet C_{\bar{u}_{4}}$. Note that $\bar{u}_{1}\cdot\bar{u}_{3}=\bar{u}_{2}\cdot\bar{u}_{4}=0$ because these operators act on different modes. Then, the composite process has terms in the form $C_{\bar{u}_{3}}C_{\bar{u}_{1}}\bullet C_{\bar{u}_{2}}C_{\bar{u}_{4}}=C_{\bar{u}_{1}}C_{\bar{u}_{3}}\bullet C_{\bar{u}_{4}}C_{\bar{u}_{2}}$. Here, we have used that $C_{\bar{u}_{1}}$ and $C_{\bar{u}_{2}}$ have the same parity, and $C_{\bar{u}_{3}}$ and $C_{\bar{u}_{4}}$ have the same parity. Therefore, the process on the first $m$ modes and the process on the other $p$ modes are always commutative.

The Majorana transfer matrix of a map reads

	$\displaystyle\mathcal{M}^{\rm m}_{\bar{u},\bar{u}^{\prime}}$	$\displaystyle=$	$\displaystyle 2^{-m}\mathrm{Tr}\left[C_{\bar{u}}\mathcal{M}(C_{\bar{u}^{\prime}})\right]$		(21)
		$\displaystyle=$	$\displaystyle 2^{-m}\sum_{\bar{v},\bar{v}^{\prime}}\chi_{\bar{v},\bar{v}^{\prime}}\mathrm{Tr}\left(C_{\bar{u}}C_{\bar{v}}C_{\bar{u}^{\prime}}C_{\bar{v}^{\prime}}\right).$

The trace is non-zero if and only if $C_{\bar{u}}C_{\bar{v}}C_{\bar{u}^{\prime}}C_{\bar{v}^{\prime}}=\eta\openone$, where $\eta=\pm 1,\pm i$ is a phase factor. Therefore, for a non-zero term, $\bar{u}+\bar{v}+\bar{u}^{\prime}+\bar{v}^{\prime}=\bar{\bf 0}$. Then, we have $|\bar{u}+\bar{u}^{\prime}|=|\bar{v}+\bar{v}^{\prime}|$. Because $|\bar{v}+\bar{v}^{\prime}|$ is even for all non-zero $\chi_{\bar{v},\bar{v}^{\prime}}$, $\mathcal{M}^{\rm m}_{\bar{u},\bar{u}^{\prime}}$ is non-zero only for elements that $|\bar{u}+\bar{u}^{\prime}|$ is even, i.e. the Majorana transfer matrix $\mathcal{M}^{\rm m}$ is also block diagonal.

The map is trace-preserving if and only if $\mathcal{M}^{\rm m}_{\bar{\bf 0},\bar{u}^{\prime}}=\delta_{\bar{\bf 0},\bar{u}^{\prime}}$. The map is unital if and only if $\mathcal{M}^{\rm m}_{\bar{u},\bar{\bf 0}}=\delta_{\bar{u},\bar{\bf 0}}$.

The gate set tomography is a self-consistent process tomography protocol Merkel2013 ; BlumeKohout2013 ; Stark2014 ; Greenbaum2015 ; BlumeKohout2017 ; Sugiyama2018 , which does not require the prior knowledge on the state preparation and measurement. In this section, we take the gate set tomography as an example. Other quantum tomography protocols, e.g. state tomography and measurement tomography, can be applied to fermionic systems in a similar way.

To implement the gate set tomography, we need to prepare a set of linear-independent states $|{\rho_{i}}\rangle\rangle$ and have a set of linearly-independent measurement operators $\langle\langle{E_{k}}|$, where $i,k=1,2,\ldots,4^{m}/2$. We note that $4^{m}/2$ is the maximum number of linearly-independent vectors, limited by dimension of the even subspace. These vectors form two $4^{m}/2$-dimensional matrices $M_{\rm in}=[|{\rho_{1}}\rangle\rangle~{}|{\rho_{2}}\rangle\rangle~{}\cdots~{}|{\rho_{4^{m}/2}}\rangle\rangle]$ and $M_{\rm out}=[\langle\langle{E_{1}}|^{\rm T}~{}\langle\langle{E_{2}}|^{\rm T}~{}\cdots~{}\langle\langle{E_{4^{m}/2}}|^{\rm T}]^{\rm T}$.

The Gram matrix is $g=M_{\rm out}M_{\rm in}$, and each element of the Gram matrix $g_{k,i}=\langle\langle E_{k}|\rho_{i}\rangle\rangle=\mathrm{Tr}(E_{k}\rho_{i})$ can be measured in the experiment, by preparing the state $\rho_{i}$ and measuring the probability of the measurement operator $E_{k}$. If $M_{\rm out}$ is known, we can obtain the prepared states by computing $M_{\rm in}=M_{\rm out}^{-1}g$. The state tomography can be implemented in this way, where $M_{\rm in}$ does not need to be a square matrix and can have any number of columns. Similarly, if $M_{\rm in}$ is known, we can obtain measurement operators by computing $M_{\rm out}=gM_{\rm in}^{-1}$. The measurement tomography can be implemented in this way, where $M_{\rm out}$ may not be a square matrix and can have any number of rows.

We consider a set of maps $\{\mathcal{M}_{j}\}$. The matrix of a map $\mathcal{M}_{j}$ that can be directly measured in the experiment is $\widetilde{M}_{j}=M_{\rm out}\mathcal{M}^{\rm even}_{j}M_{\rm in}$. The element of the matrix is $\widetilde{M}_{j;k,i}=\langle\langle{E_{k}}|\mathcal{M}^{\rm m}_{j}|{\rho_{i}}\rangle\rangle=\mathrm{Tr}[E_{k}\mathcal{M}_{j}(\rho_{i})]$, which can be measured by preparing the state $\rho_{i}$, implementing the map $\mathcal{M}_{j}$ and measuring the probability of the measurement operator $E_{k}$. If $M_{\rm in}$ and $M_{\rm out}$ are know, we can obtain the even block by computing $\mathcal{M}^{\rm even}_{j}=M_{\rm out}\widetilde{M}_{j}M_{\rm out}$, which is the process tomography.

In the gate set tomography, if both $M_{\rm in}$ and $M_{\rm out}$ are unknown, we can guess an estimate of $M_{\rm out}$, which is $\widehat{M}_{\rm out}$, and then compute an estimate of $M_{\rm in}$, which is $\widehat{M}_{\rm in}=\widehat{M}_{\rm out}^{-1}g$. The estimate of the even block can be obtained accordingly, which is $\widehat{M}_{j}=\widehat{M}_{\rm out}^{-1}\widetilde{M}_{j}\widehat{M}_{\rm in}^{-1}$. These estimates are different from true matrices up to a transformation $T=M_{\rm out}^{-1}\widehat{M}_{\rm out}$: $\widehat{M}_{\rm in}=T^{-1}M_{\rm in}$, $\widehat{M}_{j}=T^{-1}\mathcal{M}^{\rm even}_{j}T$ and $\widehat{M}_{\rm out}=M_{\rm out}T$. The gate set tomography is self-consistent in the sense that $\widehat{M}_{\rm out}\widehat{M}_{j_{N}}\cdots\widehat{M}_{j_{2}}\widehat{M}_{j_{1}}\widehat{M}_{\rm in}=M_{\rm out}\mathcal{M}^{\rm even}_{j_{N}}\cdots\mathcal{M}^{\rm even}_{j_{2}}\mathcal{M}^{\rm even}_{j_{1}}M_{\rm in}$ for any sequence of maps.

Such a method for implementing the tomography is called the linear inversion method. An alternative way is based on the maximum likelihood estimation Merkel2013 ; BlumeKohout2013 ; Stark2014 ; Greenbaum2015 ; BlumeKohout2017 ; Sugiyama2018 . Next, we will discuss how to measure the odd block. For simplicity, we assume that $M_{\rm in}$ and $M_{\rm out}$ are known in the following. The gate set tomography can be directly generalised to the case of measuring the odd block.

To measure the odd block of a map $\mathcal{M}$ that acts on $m$ modes, we introduce an ancillary mode (two Majorana fermion modes). The composite system has $m+1$ modes, the Hilbert space is $\mathcal{H}_{m}\otimes\mathcal{H}_{1}$, and we use A and B to denote two subsystems, respectively. Let $\mathcal{M}^{\rm m}_{\rm A}=\mathcal{M}^{\rm even}\oplus\mathcal{M}^{\rm odd}$ be the Majorana transfer matrix of the map on the operator space of $\mathcal{H}_{m}$. Then, the Majorana transfer matrix of the map on the composite system reads $\mathcal{M}^{\rm m}_{\rm AB}=\mathcal{M}^{\rm even}_{\rm AB}\oplus\mathcal{M}^{\rm odd}_{\rm AB}$, where

$\displaystyle\mathcal{M}^{\rm even}_{\rm AB}=\left(\begin{array}[]{cccc}\mathcal{M}^{\rm even}&0&0&0\\ 0&\mathcal{M}^{\rm odd}&0&0\\ 0&0&\mathcal{M}^{\rm odd}&0\\ 0&0&0&\mathcal{M}^{\rm even}\end{array}\right)\begin{array}[]{c}C_{\rm even}\\ iC_{\rm odd}c_{2m+1}\\ iC_{\rm odd}c_{2m+2}\\ iC_{\rm even}c_{2m+1}c_{2m+2}\end{array}$

(30)

is the even block of the map on the composite system, and

$\displaystyle\mathcal{M}^{\rm odd}_{\rm AB}=\left(\begin{array}[]{cccc}\mathcal{M}^{\rm odd}&0&0&0\\ 0&\mathcal{M}^{\rm even}&0&0\\ 0&0&\mathcal{M}^{\rm even}&0\\ 0&0&0&\mathcal{M}^{\rm odd}\end{array}\right)\begin{array}[]{c}C_{\rm odd}\\ C_{\rm even}c_{2m+1}\\ C_{\rm even}c_{2m+2}\\ iC_{\rm odd}c_{2m+1}c_{2m+2}\end{array}$

(39)

is the odd block of the map on the composite system. Here, the Majorana fermion operators denote the basis the matrix, where $C_{\rm even}=\{C_{\bar{u}}\,:\,|\bar{u}|\in{\rm Even}\}$ and $C_{\rm odd}=\{C_{\bar{u}}\,:\,|\bar{u}|\in{\rm Odd}\}$ are $C_{\bar{u}}$ operators on the first $m$ modes with the even and odd parities, respectively.

The even block of the composite system $\mathcal{M}^{\rm even}_{\rm AB}$ can be measured using the quantum tomography, with which we can obtain both $\mathcal{M}^{\rm even}$ and $\mathcal{M}^{\rm odd}$, i.e. the whole Majorana transfer matrix of the map. If we measure the entire even block of the composite system, we will have two copies of each $\mathcal{M}^{\rm even}$ and $\mathcal{M}^{\rm odd}$. Later, we will show that one can only measure one copy of each block without using more ancillary modes, because of the limited operation set in a Majorana fermion quantum computer.

First, we identify the operator space which is the span of states that can be prepared using the limited operation set. We consider $2m$ Majorana fermion modes. The basis of the even-parity operator space is $C_{\rm even}$. All valid states according to SR are in this space. Now, we consider another basis of the space, which is $C_{+}\cup C_{-}$, where $C_{\pm}=\{[\openone\pm(-1)^{m}C]C_{\bar{u}}\,:\,|\bar{u}|\in{\rm Even},~{}u_{2m}=0\}$. We note that $(-1)^{m}CC_{\bar{u}}=\mu_{\bar{u}}C_{\bar{\bf 1}-\bar{u}}$, where $\mu_{\bar{u}}=(-1)^{m}CC_{\bar{u}}C_{\bar{\bf 1}-\bar{u}}=C_{\bar{\bf 1}}C_{\bar{u}}C_{\bar{\bf 1}-\bar{u}}=\pm 1$ is a function of $\bar{u}$. Basis operators in $C_{+}$ and $C_{-}$ are orthogonal and form two $4^{m-1}$-dimensional subspaces. We have $CB=BC=\pm(-1)^{m}B$ if $B\in C_{\pm}$. We focus on the operator subspace ${\rm span}(C_{+})$.

There are at most $4^{m-1}$ linearly-independent states that can be prepared using the limited operation set. We suppose these states are $\{\rho_{i}\,:\,i=1,2,\ldots,4^{m-1}\}$, which obey $C\rho_{i}=\rho_{i}C=(-1)^{m}\rho_{i}$. Therefore, these states form a basis of the subspace ${\rm span}(C_{+})$, i.e. we can express basis operators in terms of states in the form $C_{\bar{u}}+\mu_{\bar{u}}C_{\bar{\bf 1}-\bar{u}}=\sum_{i}\alpha_{\bar{u},i}\rho_{i}$, where $\alpha_{\bar{u},i}$ are real coefficients.

Second, we show that $2m+2$ Majorana fermion modes are sufficient for measuring a map on $2m$ Majorana fermion modes. Let $\overline{C}_{+}$ be the set of basis operators on $2m+2$ Majorana fermion modes that is accessible for the state preparation. We consider four operators

	$\displaystyle G_{\bar{u}}$	$\displaystyle=$	$\displaystyle C_{\bar{u}}+\mu_{\bar{u}}C_{\bar{\bf 1}-\bar{u}}ic_{2m+1}c_{2m+2},$		(40)
	$\displaystyle H_{\bar{u}}$	$\displaystyle=$	$\displaystyle C_{\bar{\bf 1}-\bar{u}}+\mu_{\bar{u}}C_{\bar{u}}ic_{2m+1}c_{2m+2},$		(41)
	$\displaystyle I_{\bar{u}}$	$\displaystyle=$	$\displaystyle C_{\bar{u}}ic_{2m}c_{2m+1}+\mu_{\bar{u}}C_{\bar{\bf 1}-\bar{u}}c_{2m}c_{2m+2},$		(42)
	$\displaystyle J_{\bar{u}}$	$\displaystyle=$	$\displaystyle C_{\bar{\bf 1}-\bar{u}}c_{2m}c_{2m+1}-\mu_{\bar{u}}C_{\bar{u}}ic_{2m}c_{2m+2},$		(43)

where $|\bar{u}|$ is even and $u_{2m}=0$. We have $\overline{C}_{+}=\{G_{\bar{u}},H_{\bar{u}},I_{\bar{u}},J_{\bar{u}}\}$. Focusing on the first term on RHS of each equation, we can find that $C_{\rm even}=\{C_{\bar{u}},C_{\bar{\bf 1}-\bar{u}}\}$ and $C_{\rm odd}=\{C_{\bar{u}}c_{2m},-iC_{\bar{\bf 1}-\bar{u}}c_{2m}\}$. In the measurement, if we only measure operators of the first two blocks in Eq. (30), i.e. $C_{\rm even}$ and $iC_{\rm odd}c_{2m+1}$, the second term of each element in $\overline{C}_{+}$ contributes zero to the measurement result. In this way, we can reconstruct the first two blocks of Eq. (30).