Problem with quantum tomography on two qubits

Question

With reference to question on how to do quantum tomography on two qubits, I would like to ask you for help again. I tried to do the tomography on state

\begin{equation}\psi=\frac{1}{2}\begin{pmatrix}1 \\ i \\-1 \\-i\end{pmatrix}\end{equation}

This state can be prepared by application of $HX$ on first qubit and $SH$ on second one, both qubits were in state $|0\rangle$ at the beginning.

To do measurement of the state by the tomography, I evaluated eigenvectors of all observables present in two qubits tomogaphy and created measuring circuits.

Here is a list of observables, their eigenvectors and respective eigenvalues (please note that I omitted normalization in majority of cases because of simplicity):

• Observable $X\otimes X$:
  • $-1$: $\begin{pmatrix}0 & 1 & -1 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}1 & 0 & 0 & -1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & 0 & 1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & 1 & 0\end{pmatrix}^T$
• Observable $Y\otimes Y$:
  • $-1$: $\begin{pmatrix}0 & 1 & -1 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}1 & 0 & 0 & 1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & 0 & -1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & 1 & 0\end{pmatrix}^T$
• Observable $Z\otimes Z$:
  • $-1$: $\begin{pmatrix}0 & 1 & 0 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 0 & 1 & 0\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & 0 & 0\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 0 & 0 & 1\end{pmatrix}^T$
• Observable $X\otimes Y$:
  • $-1$: $\begin{pmatrix}0 & 1 & i & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}1 & 0 & 0 & -i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & 0 & i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & -i & 0\end{pmatrix}^T$
• Observable $X\otimes Z$:
  • $-1$: $\begin{pmatrix}1 & 0 & -1 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 1 & 0 & 1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & 0 & -1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & 1 & 0\end{pmatrix}^T$
• Observable $Y\otimes X$:
  • $-1$: $\begin{pmatrix}0 & 1 & -i & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}1 & 0 & 0 & -i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & 0 & i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & i & 0\end{pmatrix}^T$
• Observable $Y\otimes Z$:
  • $-1$: $\begin{pmatrix}1 & 0 & -i & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 1 & 0 & i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & 0 & -i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & i & 0\end{pmatrix}^T$
• Observable $Z\otimes X$:
  • $-1$: $\begin{pmatrix}1 & -1 & 0 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 0 & 1 & 1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 0 & 1 & -1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 1 & 0 & 0\end{pmatrix}^T$
• Observable $Z\otimes Y$:
  • $-1$: $\begin{pmatrix}1 & -i & 0 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 0 & 1 & i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 0 & 1 & -i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & i & 0 & 0\end{pmatrix}^T$
• Observable $I\otimes X$:
  • $-1$: $\begin{pmatrix}1 & -1 & 0 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 0 & 1 & -1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 0 & 1 & 1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 1 & 0 & 0\end{pmatrix}^T$
• Observable $I\otimes Y$:
  • $-1$: $\begin{pmatrix}1 & -i & 0 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 0 & 1 & -i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 0 & 1 & i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & i & 0 & 0\end{pmatrix}^T$
• Observable $I\otimes Z$:
  • $-1$: $\begin{pmatrix}0 & 1 & 0 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 0 & 0 & 1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 0 & 1 & 0\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 0 & 0 & 1\end{pmatrix}^T$
• Observable $X\otimes I$:
  • $-1$: $\begin{pmatrix}1 & 0 & -1 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 1 & 0 & -1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & 0 & 1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & 1 & 0\end{pmatrix}^T$
• Observable $Y\otimes I$:
  • $-1$: $\begin{pmatrix}1 & 0 & -i & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 1 & 0 & -i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & 0 & i\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & i & 0\end{pmatrix}^T$
• Observable $Z\otimes I$:
  • $-1$: $\begin{pmatrix}0 & 0 & 1 & 0\end{pmatrix}^T$
  • $-1$: $\begin{pmatrix}0 & 0 & 0 & 1\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}1 & 0 & 0 & 0\end{pmatrix}^T$
  • $1$: $\begin{pmatrix}0 & 1 & 0 & 0\end{pmatrix}^T$

Then I realized that basis generated by eigenvectors of observables are repeating (note that I "invented" names for some basis for easier referencing further):

• z-basis: $Z\otimes Z$, $I\otimes Z$ and $Z\otimes I$
• Bell basis: $X\otimes X$ and $Y\otimes Y$
• "Imaginary" Bell basis: $X\otimes Y$ and $Y\otimes X$
• "A" basis: $X\otimes Z$ and $X\otimes I$
• "B" basis: $Y\otimes Z$ and $Y\otimes I$
• "C" basis: $Z\otimes X$ and $I\otimes X$
• "D" basis: $Z\otimes Y$ and $I\otimes Y$

After that I created for each basis a circuit for measuring in the basis, or in other words the circuit for converting basis states to z-basis ones. These circuits are here:

Bell basis

Imaginary Bell basis

A basis

B basis

C basis

D basis

Having this I created a map between basis states and z-basis and assigned each state in z-basis either value -1 or 1 in the following way

\begin{matrix} \mathrm{observable} & -1 & 1\\ X\otimes X & |11\rangle, |10\rangle & |00\rangle, |01\rangle\\ Y\otimes Y & |11\rangle, |00\rangle & |10\rangle, |01\rangle\\ Z\otimes Z & |01\rangle, |10\rangle & |00\rangle, |11\rangle\\ X\otimes Y & |01\rangle, |10\rangle & |00\rangle, |11\rangle\\ X\otimes Z & |10\rangle, |11\rangle & |01\rangle, |00\rangle\\ Y\otimes X & |11\rangle, |10\rangle & |00\rangle, |01\rangle\\ Y\otimes Z & |10\rangle, |11\rangle & |01\rangle, |00\rangle\\ Z\otimes X & |10\rangle, |11\rangle & |01\rangle, |00\rangle\\ Z\otimes Y & |10\rangle, |11\rangle & |01\rangle, |00\rangle\\ I\otimes X & |10\rangle, |01\rangle & |11\rangle, |10\rangle\\ I\otimes Y & |10\rangle, |01\rangle & |11\rangle, |00\rangle\\ I\otimes Z & |01\rangle, |11\rangle & |10\rangle, |00\rangle\\ X\otimes I & |10\rangle, |01\rangle & |00\rangle, |11\rangle\\ Y\otimes I & |10\rangle, |01\rangle & |11\rangle, |00\rangle\\ Z\otimes I & |10\rangle, |11\rangle & |00\rangle, |01\rangle\\ \end{matrix}

When I calculated a density operator estimation and did spectral decomposition of that I really obtained state $\psi$ among eigenstates and its respective eigenvalue was the highest one. However, there was also one negative eigenvalue. So, as the density operator is a positive one, it seems that there is some mistake in my calculation.

To sum it all up:

1. I calculated eigenvalues and eigenvectors of all observables
2. I created circuits for measuring a quantum state in bases given by eigenvectors obtained in the first step
3. I assigned each eigenvector its "mirror" in z-basis after measuring by circuits from step 2
4. I prepared state $\psi$
5. I did a measurement of state $\psi$ in all bases obtained in step 1 with help of circuits from step 2
6. Based on measured probabilities, I calculated estimation of $\mathrm{Tr}(A\rho)$, where $A$ is an observable
7. Then, I calculated estimation of density operator $\rho$
8. Finnaly, I found eigenvalues and eigenvector of $\rho$. State $\psi$ should be one of eigenvectors, its respective eigenvalue should be close to 1 as pure state $\psi$ was measured.

I did all calculations in MatLab and I used matrix representation of quantum gates to simulate both state preparation and measurement in different bases, so the calculations are exact and there is no noise.

My question: is my above described approach right?

I really appreciate any help. Thank you in advance.

Answer 1

I find it a little tough to understand your calculations directly. I am especially confused by the circuit diagrams in your question; why they are there and what you are using them for.

If you are performing calculations on theoretical data (without noise), then I feel you can make do with an easier approach for quantum state tomography. As per my answer on your previous question (that you linked to in this question), the idea behind QST is to reconstruct a unknown state $\rho$ from the expectation values of a set of measurement observables.

When I calculated a density operator estimation and did spectral decomposition of that I really obtained state ψ among eigenstates and its respective eigenvalue was the highest one. However, there was also one negative eigenvalue. So, as the density operator is a positive one, it seems that there is some mistake in my calculation.

What exactly do you mean by 'I really obtained the state $\psi$ among eigenstates (of what?) and its respective eigenvalue was the highest one (among whose)?

Do you mean that the estimated density operator that you obtained had more than one nonzero eigenvalues? If you are just performing exact calculations with matrices, this should not be the case. (If you are incorporating statistical noise into your calculations then this will definitely happen).

Negative eigenvalues can occur in your estimation of the density matrix if there is statistical noise. This will occur in actual experiments, so people have been addressing it for quite some time now. This paper offers a way of combating negative eigenvalues.

Now, if you are performing your calculations solely on pure states $\rho = |\psi\rangle\langle\psi|$ and if you are calculating the expectation value for an observable $M$ as:

\begin{equation} \langle M\rangle = \rm{Tr}\big[M\rho\big] = \langle\psi|M|\psi\rangle, \end{equation}

then reconstructing $\rho$ via the equation \begin{equation} \hat{\rho}= \sum_{P_{i}\in\mathcal{P}^{2}}\rm{Tr}\big[P_{i}\rho\big]P_{i}, \end{equation}

with $\mathcal{P}^{2}$ the two-qubit Pauli group, should be a perfect reconstruction and therefore give only one nonzero eigenvalue. So then it would indicate an error in your calculations.

If you can add some information on how exactly you are performing these calculations (are you doing them by hand, by code, are you actually simulating the quantum circuits etc.) then I might be able to help more. Please feel free to contact me directly.