How to compute the average value $\langle X_1 Z_2\rangle$ for a two-qubit system?

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  1. Show that the average value of the observable $X_1Z_2$ in a two-qubit system measured in the state $(|00\rangle + |11\rangle)/\sqrt{2}$ is zero.

How would we approach this question? I understand that $X_1$ means $\sigma_1$ acting on the first qubit, and $Z_2$ means $\sigma_3$ acting on the second qubit. I also know that the average value is given by $\left<\psi\vert M\vert \psi\right>$ (the inner product of $\psi$ with $M \psi$). I know how to solve similar problems for a single qubit system, however what confuses me here is what is the vector representation of the states $\left| 00\right>$ and $\left| 11\right>$ and how is this related to the vector representation of $\left| 0\right>$ and $\left| 1\right>$? Also, what is the matrix form of $M$ in this case? Is it the tensor product of $\sigma_1$ applied to the first qubit and $\sigma_3$ applied to the second qubit?

Source of the question

Alk

Posted 2018-04-10T18:24:33.557

Reputation: 163

Possible duplicate of What does it mean for two qubits to be entangled?

– Mithrandir24601 – 2018-04-10T19:51:22.430

4Yes, $X_1 Z_2 = \sigma_1\otimes \sigma_3$ and $|00\rangle=|0\rangle\otimes|0\rangle$. You stated everything you need to solve this! – M. Stern – 2018-04-10T20:15:46.163

1I'm voting to close this question as off-topic because the question arises from a trivial notational confusion. – M. Stern – 2018-04-10T20:19:23.393

3@M.Stern It may be reasonable to be confused about notation and ask about it on stackexchange. I've had multiple students confused by both of the issues in this question every time I've taught. – Jalex Stark – 2018-04-11T02:38:37.043

Answers

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I suggest two different ways of trying to solve this, which will give you experience of different bits of the formulation of Quantum Information Theory. I'll give examples that are closely related to the question you asked, but are not what you asked so that you still get the value of answering the question yourself.

Long-hand Method

Represent the kets as vectors, the Pauli matrices as matrices, explicitly perform the tensor products, and multiply everything out. So, we represent $$ |0\rangle\equiv\left(\begin{array}{c} 1 \\ 0 \end{array}\right)\qquad|1\rangle\equiv\left(\begin{array}{c} 0 \\ 1 \end{array}\right) $$ To calculate the tensor product, such as $|01\rangle$, we do $$ |01\rangle=|0\rangle\otimes|1\rangle\equiv\left(\begin{array}{c}1\times\left(\begin{array}{c} 0 \\ 1 \end{array}\right)\\ 0\times \left(\begin{array}{c} 0 \\ 1 \end{array}\right) \end{array}\right)=\left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right) $$ Remember that $\langle 01|$ is just the Hermitian conjugate of this, $$ \langle 01|=\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \end{array}\right) $$ Then you do something similar for the operators. For example, $$X_1X_2=\sigma_1\otimes\sigma_1=\left(\begin{array}{cc} 0\times \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) & 1\times \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \\ 1\times \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) & 0\times \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \end{array}\right)=\left(\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right). $$ Once you have all of this, you simply multiply it out: $$ \langle 01|X_1X_2|01\rangle=\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \end{array}\right)\left(\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right)\left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right)=0. $$

Shorter Method

(With experience, this method lets you perform this calculation just by looking at it! Of course, when giving an answer, I don't recommend that; you should always justify your answer.)

Remember to think of each term $|01\rangle$ as $|0\rangle\otimes|1\rangle$. So, when you also write $X_1X_2=\sigma_1\otimes\sigma_1$, you see that $$ (\sigma_1\otimes\sigma_1)|0\rangle\otimes|1\rangle=(\sigma_1|0\rangle)\otimes(\sigma_1|1\rangle) $$ i.e. when everything is just tensor products, individual terms match up. Now, hopefully you know the action of $\sigma_1$ and $\sigma_3$ on the basis states: $$ \sigma_1|0\rangle=|1\rangle\qquad \sigma_1|1\rangle=|0\rangle \qquad \sigma_3|0\rangle=|0\rangle \qquad \sigma_3|1\rangle=-|1\rangle $$ Thus, $$ (\sigma_1|0\rangle)\otimes(\sigma_1|1\rangle)=|1\rangle\otimes |0\rangle=|10\rangle $$ One can then easily observe that a state such as $(|01\rangle+|10\rangle)/\sqrt{2}$ is acted on by $X_1X_2$ to give $$ X_1X_2(|01\rangle+|10\rangle)/\sqrt{2}=(|01\rangle+|10\rangle)/\sqrt{2},\tag{1} $$ the same state. So it is clear that the inner product of the state with itself is 1: $$ (\langle 01|+\langle 10|)X_1X_2(|01\rangle+|10\rangle)/2=1. $$ On the other hand, had the outcome in Eq. (1) been a different one of the four Bell states, because we know the Bell states form an orthonormal basis, the expectation value would be 0.

DaftWullie

Posted 2018-04-10T18:24:33.557

Reputation: 35 722

1

Notationally, $|00\rangle = |0\rangle \otimes |0\rangle$. A basic property of tensor products is that inner products split like so: $$(\langle 0|_A \otimes \langle 0|_B) O_A\otimes O_B (|0\rangle_A \otimes |0\rangle_B) = \langle0|O_A|0\rangle \cdot \langle0|O_B|0\rangle.$$

Jalex Stark

Posted 2018-04-10T18:24:33.557

Reputation: 349