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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?

JMark
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1 Answers1

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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.

Abdullah Khalid
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