3
$\begingroup$

I wanted to ask how do you implement a circuit that finds the non-diagonal values of the density matrix of a quantum state on IBM Q?

$\endgroup$
3
$\begingroup$

Basically you add measurements in different bases by applying gates before the (Z-basis) measurement.

See here the standard implementation:

https://github.com/Qiskit/qiskit-ignis/blob/3c59f82c11e87c071bc7e84240b50e2aa995281f/qiskit/ignis/verification/tomography/basis/paulibasis.py#L31

| improve this answer | |
$\endgroup$
3
$\begingroup$

Density matrix of single qubit state can be estimated based on this formula

\begin{equation} \rho = \frac{\text{tr}(\rho)I+\text{tr}(X\rho)X+\text{tr}(Y\rho)Y+\text{tr}(Z\rho)Z}{2}, \end{equation}

where $X$, $Y$, $Z$ are Pauli matrices.

Obviously $\mathrm{tr}(\rho) = 1$.

Terms $\mathrm{tr}(X\rho)$, $\mathrm{tr}(Y\rho)$ and $\mathrm{tr}(Z\rho)$ can be estimated by measuring a quantum state in different bases:

  • In $z$ basis you simply measure the state
  • In $x$ basis Hadamard gate has to be applied before measurement
  • In $y$ basis $S^\dagger$ gate followed by Hadamard gate have to be applied before measurement

Value of $\mathrm{tr}(A\rho)$, where $A \in \{X,Y,Z\}$, is given by

$$ \mathrm{tr}(A\rho) = \frac{1}{m}\sum_{i=1}^{m}\lambda_{i}, $$

where $m$ is number of measurements (i.e. shots on IBM Q) and $\lambda_{i}$ is eigenvalue respective to measured state. Since Pauli matrices are Hermitian and unitary, their eigenvalues are -1 and +1. Moreover, eigenvalue -1 is assigned to such eigenstate that after measurement it is mapped to state $|1\rangle$, eigenvalue +1 is mapped to state $|0\rangle$.

Note: You can check this if you calculate eigenvectors and eigenvalues of Pauli matrices and then apply above mentioned gates. Eigenstates should be mapped to $|0\rangle$ and $|1\rangle$.

For practical purposes formula for calculation $\mathrm{tr}(A\rho)$ can be rewritten in terms of measured states probability distribution followingly

$$ \mathrm{tr}(A\rho) = P(|0\rangle) - P(|1\rangle), $$

because +1 is equivalent to measuring $|0\rangle$, hence $\frac{1}{m}\sum_{\lambda = 1}\lambda_i = \frac{\#(\lambda = 1)}{m} = P(\lambda = 1) = P(|0\rangle)$. Similarly for eigenvalue -1.

Note: probabilities are expressed as decimal number, not percentage!

To sum up how to do quantum tomography on IBM Q:

  1. Prepare a qubit in some state
  2. Measure it in $x$, $y$ and $z$ bases
  3. Use probabilities of measuring $|0\rangle$ and $|1\rangle$ for estimation of $\mathrm{tr}(A\rho)$
  4. Calculate density matrix $\rho$ (the first formula in this answer)
| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.