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Bell State Simulation

I have some problems to grasp the interpretation of the Gottesman-Knill theorem. If the first qubit is measured, since $\mathcal{Z} \otimes \mathcal{I}$ does not commute with all the stabilizers, the first qubit should $0$ or $1$ with equal probability. Same thing with the second one. But we should expect to have only $|00\rangle$ or $|11\rangle$. How can we have that?

Does it make sense to measure both qubits at the same time? This could mean that we should have to search for maximally entangled qubits during the simulations? If so, how the proper stabilizer formalism can be used?

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  • $\begingroup$ Hello. "the first qubit should be $0$ or $1$ with equal probability [...] But we should expect to have only $|00\rangle$ or $|11\rangle$.". There is no contradiction in the sense that the first qubit will indeed be $0$ or $1$ with equal probability if you measure $Z \otimes I$, and it will collapse the two-qubit state in $|00\rangle$ or $|11\rangle$ with equal probability. However I am not sure to have really understood what you are asking (it would be good if you can clarify exactly what you are asking). $\endgroup$ Jul 1, 2023 at 23:20
  • $\begingroup$ If I use Gottesman-Knill theorem, the operator $\mathcal{Z} \otimes \mathcal{I}$ should be used to read the first qubit. Since this operator does not commute with both generators, I should conclude that the first qubit is either $0$ or $1$ with probability $\frac{1}{2}$. Fine. Then, if I want to read the second qubit with operator $\mathcal{I} \otimes \mathcal{Z}$, I should conclude that the second qubit is either $0$ or $1$ with probability $\frac{1}{2}$. However, I should have either $|00\rangle$ or $|11\rangle$. Is it clearer? I cannot mesure the qubits independently. $\endgroup$
    – JMark
    Jul 2, 2023 at 0:12

1 Answer 1

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See Nielson and Chuang 10.5.3.

The state just before measurement is stabilized by $\langle X\otimes X, Z\otimes Z\rangle $.

Measuring qubit one in the $Z$ basis, i.e. measuring $Z \otimes I$ works as follows. Since $Z \otimes I$ does not commute with one generator, $X\otimes X$, measurement leads to either

  1. an outcome of $+1$ and the state is stabilized by $\langle Z\otimes I, Z\otimes Z\rangle$, or
  2. an outcome of $-1$ and the state is stabilized by $\langle -Z\otimes I, Z\otimes Z\rangle$.

Now, we measure $I \otimes Z$. Let's look at both cases one by one.

Case 1

Stabilizer group is $\langle Z\otimes I, Z\otimes Z\rangle$. As it turns out $I \otimes Z = (Z\otimes I)(Z\otimes Z)$ is in this group. So the outcome of measurement will be +1 with probability 1.

Case 2

Stabilizer group is $\langle -Z\otimes I, Z\otimes Z\rangle$. In this case, $-I \otimes Z = (-Z\otimes I)(Z\otimes Z)$ is in this group. So the outcome of measurement will be -1 with probability 1.

Hence, the two measurements either yield +1, +1 or -1, -1.

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  • $\begingroup$ I saw where I made my errors. I was updating the stabilizer group after the measurement with $Z \otimes I$ in both cases. The negative sign was overlooked. Thanks. $\endgroup$
    – JMark
    Jul 3, 2023 at 0:33

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