How to compute the unitary from the $\chi$ matrix obtained from QPT

Question

I am trying to do quantum process tomography for one qubit and obtain the unitary for the gates that are applied on the qubit. I have studied the theory on process tomography from mike and ike and the box 8.5 describes a simple procedure for obtaining the chi matrix for our quantum operation.

Now I am a little confused how would I go from the said chi matrix to obtaining the unitary which directly depend upon $E_i$'s.

Note - I am not getting the same chi matrix when I compare my results to qiskit backends but I figured that was happening because of different basis states.

Furthermore I have implemented the above using Qiskit and I dont think I am getting the correct result.

def one_qubit_process_tomography(circuit, samples):

# I, rho_1
qc = QuantumCircuit(1,1)
qc = qc.compose(circuit)

rhod_1 = one_qubit_tomography(qc, samples)

# X, rho_4
qc = QuantumCircuit(1,1)
qc.x(0)
qc = qc.compose(circuit)
rhod_4 =  one_qubit_tomography(qc, samples)

# H, rho_h,  E(|+><+|)
qc = QuantumCircuit(1,1)
qc.h(0)
qc = qc.compose(circuit)
rhod_h = one_qubit_tomography(qc, samples)

# XH, rho_xh,  E(|-><-|)
qc = QuantumCircuit(1,1)
qc.x(0)
qc.h(0)
qc = qc.compose(circuit)
rhod_xh =   one_qubit_tomography(qc, samples) 

# this is E(|0><1|)
rhod_2 = rhod_h - 1j*rhod_xh - (1-1j)*(rhod_1 + rhod_4)/2

# this is E(|1><0|)
rhod_3 = rhod_h + 1j*rhod_xh - (1+1j)*(rhod_1 + rhod_4)/2

# now we will find the chi matrix 
lambdaa = 0.5*np.array([[1,0,0,1],[0,1,1,0],[0,1,-1,0],[1,0,0,-1]])

 #print(rhod_1)
 #print(rhod_2)
 #print(rhod_3)
 #print(rhod_4)

rho = np.zeros([4,4],dtype=np.complex_)
rho[0:2,0:2] = rhod_1
rho[0:2,2:4] = rhod_2
rho[2:4,0:2] = rhod_3
rho[2:4,2:4] = rhod_4

chi = np.matmul(np.matmul(lambdaa, rho), lambdaa)
  #print(rho)


return chi

Answer 1

Not all quantum operations are unitary. A more general type of quantum operation is the completely positive trace-preserving (CPTP) linear map which is the main subject of chapter $8$ of Nielsen & Chuang. The $\chi$ matrix is one of several concrete descriptions of a CPTP map and is the output of the standard quantum process tomography.

A CPTP map with a given $\chi$ matrix corresponds to a unitary if and only if the $\chi$ matrix is rank one. This is very unlikely to be the case for a $\chi$ matrix obtained in experiment due to noise and decoherence. That said, if the $\chi$ matrix does describe a unitary operation then we can obtain it by converting from the $\chi$ matrix representation to a Kraus representation as described near equation $(8.167)$ on page $392$ in Nielsen & Chuang. Unitary quantum operations have a Kraus representation with a single Kraus operator equal to the unitary operator.

Answer 2

As @AdamZalcman has pointed out, the $\chi$ matrix represents a (more general than unitary) Quantum channel. If you were trying to implement a unitary operation, your channel might be close to a unitary operation (depending on how well your system performs).

As the $\chi$ matrix is positive semidefinite, it has a (unique) eigendecomposition $\{\lambda_{i},\vec{v}_{i}\}$, with the eigenvectors forming an orthonormal basis. To obtain the unitary that you were trying to implement, grab that eigenvector $\vec{v}_{max}$ for which its eigenvalue $\lambda$ is the maximum one (over all eigenvalues of $\chi$). Assuming your $\chi$ matrix is in the Pauli basis, your unitary $U_{implemented}$ is then:

$$ U_{implemented} = \sum_{j} (\vec{v}_{max})_{j} P_{j} $$ where $P_{j}$ are the Paulis $\{I,X,Y,Z\}$ (in that order!), assuming you are performing single-qubit process tomography. Using the operators from N&C as defined in your question, it would be $\{I, X, -iY, Z\}$.

If $U_{implemented}$ is exactly the unitary $U_{goal}$ that you were trying to implement, all your noise is completely random, and your process fidelity is $\lambda_{max}$.

If $U_{implemented}$ is not exactly $U_{goal}$, you (also) have a systematic error, which means that your system is performing an extra (unitary) rotation $U_{err} = U_{implemented}U_{goal}^{\dagger}$ and your process fidelity is (strictly) bounded from above by $\lambda_{max}$.