Measuring entanglement of a rank-2 mixed state prepared on a quantum computer

Abstract

We study the entanglement between a certain qubit and the remaining system in rank-2 mixed states prepared on the quantum computer. The protocol, which we propose for this purpose, is based on the relation of geometric measure of entanglement with correlations between qubits. As a special case, we consider a two-qubit rank-2 mixed state and find the relation of concurrence with the geometric measure of entanglement. On the ibmq-melbourne quantum computer, we measure the geometric measure of entanglement in the cases of 2- and 4-qubit mixed quantum states which consist of Schrödinger cat states. We study the dependence of the value of entanglement on the parameter which defines the weight of pure states. Finally, we determine the concurrence of 2-qubit mixed state.

Appendices

Appendix A: Derivation of concurrence for a two-qubit rank-2 mixed state

In this appendix using Wootters definition (6), we obtain the expression for concurrence of a two-qubit rank-2 mixed state. The two-qubit rank-2 density matrix (1) with $\vert \psi _{\alpha }\rangle =a_{\alpha }\vert 00\rangle +b_{\alpha }\vert 11\rangle$ and its $\tilde{\rho }$ matrix which are spanned by $\vert 00\rangle$, $\vert 01\rangle$, $\vert 10\rangle$, $\vert 11\rangle$ basis have the form

$$\begin{aligned} \rho =\left( \begin{array}{ccccc} \sum _{\alpha }\omega _{\alpha }\vert a_{\alpha }\vert ^2 &{} \quad 0 &{} \quad 0 &{} \quad \sum _{\alpha }\omega _{\alpha }a_{\alpha }b^*_{\alpha }\\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ \sum _{\alpha }\omega _{\alpha }a^*_{\alpha }b_{\alpha } &{} \quad 0 &{} \quad 0 &{} \quad \sum _{\alpha }\omega _{\alpha }\vert b_{\alpha }\vert ^2 \end{array}\right) ,\quad \tilde{\rho }=\left( \begin{array}{ccccc} \sum _{\alpha }\omega _{\alpha }\vert b_{\alpha }\vert ^2 &{} \quad 0 &{} \quad 0 &{} \quad \sum _{\alpha }\omega _{\alpha }a_{\alpha }b^*_{\alpha }\\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ \sum _{\alpha }\omega _{\alpha }a^*_{\alpha }b_{\alpha } &{} \quad 0 &{}\quad 0 &{}\quad \sum _{\alpha }\omega _{\alpha }\vert a_{\alpha }\vert ^2 \end{array}\right) . \nonumber \\ \end{aligned}$$

(A1)

Then, the matrix $\rho \tilde{\rho }$ can be expressed as follows

$$\begin{aligned} \rho \tilde{\rho }=\left( \begin{array}{ccccc} \sum _{\alpha ,\beta }\omega _{\alpha }\omega _{\beta }\left( \vert a_{\alpha }\vert ^2\vert b_{\beta }\vert ^2+a_{\alpha }a^*_{\beta }b^*_{\alpha }b_{\beta }\right) &{} \quad 0 &{} \quad 0 &{} \quad \sum _{\alpha ,\beta }\omega _{\alpha }\omega _{\beta }\left( \vert a_{\alpha }\vert ^2 a_{\beta }b^*_{\beta }+a_{\alpha }b^*_{\alpha } \vert a_{\beta }\vert ^2\right) \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ \sum _{\alpha ,\beta }\omega _{\alpha }\omega _{\beta }\left( a^*_{\alpha } b_{\alpha } \vert b_{\beta }\vert ^2+\vert b_{\alpha }\vert ^2 a^*_{\beta } b_{\beta }\right) &{} \quad 0 &{}\quad 0 &{}\quad \sum _{\alpha ,\beta }\omega _{\alpha }\omega _{\beta }\left( \vert b_{\alpha }\vert ^2\vert a_{\beta }\vert ^2+a^*_{\alpha }a_{\beta }b_{\alpha }b^*_{\beta }\right) \end{array}\right) . \nonumber \\ \end{aligned}$$

(A2)

The eigenvalues of this matrix

$$\begin{aligned} \lambda _{1,2}^2=\left( \sqrt{\sum _{\alpha ,\beta }\omega _{\alpha }\omega _{\beta }\vert a_{\alpha }\vert ^2\vert b_{\beta }\vert ^2}\pm \left| \sum _{\alpha }\omega _{\alpha }a_{\alpha }b^*_{\alpha }\right| \right) ^2\quad \lambda _{3,4}^2=0. \end{aligned}$$

(A3)

Now, substituting $\lambda _i$ in formula (6) we obtain expression (7).

Appendix B: Errors of calculations

Total error $\Delta$ of calculations on a quantum computer consists of a standard error $\Delta _s$, an error of gates $\Delta _g$ and a readout error $\Delta _r$. Standard error is inversely proportional to the square root of number of the shots. In our case, the number of shots is equal to 8192, which makes the standard error very small with respect to other errors.The upper limit of the gate error consists of the sum of the errors of each gate included by the quantum circle (see, for instance, [49]). Using the data from Table 1, we can easily estimate this error for different quantum circles. In the cases of the 4-qubit Schrödinger cat (Fig. 3) and Bell states, these errors are equal to 0.07535 and 0.0290, respectively. Here, we use the fact that the sequential action of single-qubit $\sigma ^x$ and Hadamard gates is implemented using one single-qubit basis gate on a quantum computer. Note that in the case of states projected on the x-axis the errors of the single-qubit operators which provide rotations of qubits should be added. For a 2-qubit state, these rotations are provided by two single-qubit gates. The error, which appears from these gates, is equal to 0.0017. Finally, let us evaluate the readout error. The readout error is presented in Table 1, where only one qubit of the circle is measured. In the case of several qubits, the readout errors of each qubit cannot be added. The more the qubits are measured, the faster the total readout error grows. The readout error also depends on the measuring state. We suggest estimating the readout error of our calculations by using the average value $F=\left( F_0+F_1\right) /2$ of fidelities $F_0$ and $F_1$ of the basis states $\vert \mathbf{0}\rangle$ and $\vert \mathbf{1}\rangle$, respectively. The fidelities of achieving these states on the quantum computer are degraded by the readout error. Thus, the readout error is defined as follows $\Delta _r=1-F$, where $F=\left( \vert a_0\vert ^2+\vert a_1\vert ^2\right) /2$ is the average fidelity, and $\vert a_0\vert$, $\vert a_1\vert$ are separately measured amplitudes which correspond to states $\vert \mathbf{0}\rangle$, $\vert \mathbf{1}\rangle$, respectively. Here, $F_0=\vert a_0\vert ^2$ and $F_1=\vert a_1\vert ^2$. Then, for the 4-qubit and 2-qubit we obtain the readout errors equal to 0.3332 and 0.1773, respectively. In Table 2, we represent the errors which appear on the ibmq-melbourne quantum computer for different quantum circuits.

Table 2 Errors which appear on the ibmq-melbourne quantum computers

Now, using errors presented in Table 2, the errors of the value of entanglement can be obtained. Thus, for mean values (5) which we directly measure on the quantum computer we obtain the following restrictions: $\langle \psi _{\alpha }\vert \Sigma ^a_i\vert \psi _{\alpha }\rangle =\langle \psi _{\alpha }\vert \Sigma ^a_i\vert \psi _{\alpha }\rangle \pm \langle \psi _{\alpha }\vert \Sigma ^a_i\vert \psi _{\alpha }\rangle \Delta ^{(\alpha )}$. Then, if we have two pure states $\vert \psi _1\rangle$, $\vert \psi _2\rangle$ which define the mixed state, we obtain the restrictions for the square of mean value (3) in the form

$$\begin{aligned} \langle \Sigma _i^a\rangle ^2= & {} \langle \Sigma _i^a\rangle ^2\pm \left[ 2\vert \langle \Sigma _i^a\rangle \vert \left( \omega \vert \langle \psi _1\vert \Sigma ^a_i\vert \psi _1\rangle \vert \Delta ^{(1)}+(1-\omega ) \vert \langle \psi _2\vert \Sigma ^a_i\vert \psi _2\rangle \vert \Delta ^{(2)}\right) \right. \nonumber \\&\left. +\left( \omega \vert \langle \psi _1\vert \Sigma ^a_i\vert \psi _1\rangle \vert \Delta ^{(1)}+ (1-\omega ) \vert \langle \psi _2\vert \Sigma ^a_i\vert \psi _2\rangle \vert \Delta ^{(2)}\right) ^2 \right] .\nonumber \\ \end{aligned}$$

(A1)

Finally, substituting these restrictions into expressions (2) and (9) we find the deviations for the values of entanglement caused by the errors which appear on the ibmq-melbourne quantum computer.