# How to know what eigenvalue corresponds to measurements of individual qubits in a multiqubit system?

I'm working through the book "Introduction to the Theory of Quantum Information Processing" by Bergou and Hillary, and I've encountered a scenario that I'm not sure how to approach. In chapter 3, problem 1, you're essentially supposed to show how a 2 qubit quantum system can violate Wigner's inequality, which is given as

$$p(a_1=1, b_2=1)+p(a_2=1,b_3=1)\ge p(a_1=1,b_3=1),$$ where $$a_i$$ and $$b_j$$ are some observables, whose results can either be $$-1$$ or $$1$$ and are are anticorrelated for for $$i=j$$ (that is if $$a_1$$ measures 1, $$b_1$$ measures -1, and vice versa).

The problem instructs to show this with the operators $$a_1=\frac{1}{2}\sigma_z+\frac{\sqrt{3}}{2}\sigma_x,$$ $$a_2=\sigma_z,$$ $$a_3=\frac{1}{2}\sigma_z- \frac{\sqrt{3}}{2}\sigma_x,$$ where $$\sigma_z$$ and $$\sigma_x$$ are the respective Pauli matrices, and similarly for $$b_{1,2,3}$$; and the initial state of the system being the bell state $$|\Omega_-\rangle=\frac{1}{\sqrt{2}}(|0\rangle_a |1\rangle_b+|1\rangle_a |0\rangle_b).$$ The general method of approaching this is fairly obvious to me, you just need to take the "combined" operators $$a_1b_2, a_2b_3$$, and $$a_1b_3$$ and find the probabilities of getting a 1 for each qubit.

Take $$a_1b_2$$, $$a_1b_2=(\frac{1}{2}\sigma_{za}+\frac{\sqrt{3}}{2}\sigma_{xa})\otimes\sigma_{zb},$$ which in matrix form (if my understanding is correct) is, $$a_1b_2=\frac{1}{2} \begin{pmatrix} 1 & 0 & \sqrt{3} & 0 \\ 0 & -1 & 0 & -\sqrt{3} \\ \sqrt{3} & 0 & -1 & 0 \\ 0 & -\sqrt{3} & 0 & 1 \\ \end{pmatrix},$$ which of course has the eigenvalues $$\lambda_1=1,\lambda_2=1,\lambda_3=-1,\lambda_4=-1$$ as expected since we desire those to be the only outcomes. Then I find the the probability of the outcomes in the usual manner, with $$|\lambda_n\rangle$$ being the eigenstate corresponding to the n-th eigenvalue, the probability is $$p(\lambda_n)=|\langle \lambda_n|\Omega_-\rangle|^2$$

However, my problem is that I'm not familiar with how to determine which eigenvalue/eigenstate corresponds to what outcomes for the individual qubits. I would think that $$\lambda_1$$ and $$\lambda_2$$ correspond to the outcomes where the measurement is the same for both qubits, either $$1$$ for $$-1$$ for both qubits, while $$\lambda_3$$ and $$\lambda_4$$ correspond to the outcomes where they're different, $$-1$$ for qubit $$a$$ and $$1$$ for qubit $$b$$ or $$1$$ for $$a$$ and $$-1$$ for $$b$$, but I have no justification for that.

How is this determined? How do I know which eigenstate of an operator in the basis of the full system corresponds to what outcomes for the individual qubits?

Edit: As has been pointed out, I transcribed the problem incorrectly from the book, $$|\Omega_-\rangle$$ should have a minus sign in the expression instead of a plus sign (actually it should also be called $$|\Phi_-\rangle$$ to match the book's notation). However, at this point, several answers have already been given, and I will have marked one as correct based on the original statement of my post. Thank you to those who realized my mistake and pointed it out.

• If you still don't have an answer 1 day before the bounty period ends, you can ping me and let me know. Nov 5, 2023 at 16:09
• – hft
Nov 6, 2023 at 17:44

If you consider the state

$$\Omega_- = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle).\tag{1}$$

you will find that there is NO violation. Let me show this. Note that the approach I will use for solving this problem can be applied to any choice of state, and this ultimately answers your question.

Given that Alice and Bob are set to measure different operators, each having two outcomes (+1 and -1), the simplest method to determine the probability for each measurement outcome is to first identify the eigenvectors corresponding to each operator individually. After this, one can construct the respective probability distributions. Here's how this can be done:

Let's define a general measurement operator $$m_i$$ for $$i \in \{1,2,3\}$$, where $$m$$ stands for $$a$$ (Alice) or $$b$$ (Bob).

Measurement operator: $$m_1 = \frac{1}{2}\sigma_z + \frac{\sqrt{3}}{2}\sigma_x.$$

It can be shown that the eigenvector corresponding to the eigenvalue $$+1$$ is

$$|m^{+}_1\rangle = \frac{\sqrt{3}}{2}|0\rangle + \frac{1}{2}|1\rangle.\tag{2}$$

Measurement operator: $$m_2 = \sigma_z.$$

This one's straightforward! Since it is already diagonal in the computational basis, the eigenvector corresponding to the eigenvalue is just: $$|m_2^{+}\rangle = |0\rangle\tag{3}$$

Measurement operator: $$m_3 = \frac{1}{2}\sigma_z - \frac{\sqrt{3}}{2}\sigma_x.$$

In a manner analogous to $$m_1$$, the eigenvector corresponding to the eigenvalue $$+1$$ for this operator is:

$$|m^{+}_3\rangle = -\frac{\sqrt{3}}{2}|0\rangle + \frac{1}{2}|1\rangle.\tag{4}$$

Now, here's the fun part: when you're ready to crunch out those probabilities, you've got options! You can either go with the matrix representation or stick with our good ol' ket-bra notation.

ket-bra notation:

$$p(a_1=1, b_2=1)=|\langle a^{+}_1 b^{+}_2|\Omega_1\rangle|^2 = \frac{1}{8},\tag{5}$$ $$p(a_2=1, b_3=1)=|\langle a^{+}_2 b^{+}_3|\Omega_1\rangle|^2 = \frac{1}{8},\tag{6}$$ $$p(a_1=1, b_3=1)=|\langle a^{+}_1 b^{+}_3|\Omega_1\rangle|^2 = 0,\tag{7}$$ and, in fact, there is no violation as $$p(a_1=1, b_2=1)+p(a_2=1, b_3=1) > p(a_1=1, b_3=1).\tag{8}$$

However, if you consider the other Bell-state:

$$|\Psi_-\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle),\tag{9}$$

then we see a violation since

$$p(a_1=1, b_3=1)=\operatorname{tr}(M\rho M^{\dagger}) = \frac{3}{8}.\tag{10}$$

Let me just tell you how you can carry out the calculations using a matrix notation:

Matrix notation:

First, note that our state is given by:

$$\rho:= |\Psi_-\rangle \langle \Psi_-| = \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right). \tag{11}$$

Now, defining the following measurement operator

$$M_1:=|a^{+}_1\rangle\langle a^{+}_1|\otimes|b^{+}_2\rangle\langle b^{+}_2| =\left( \begin{array}{cccc} \frac{3}{4} & 0 & \frac{\sqrt{3}}{4} & 0 \\ 0 & 0 & 0 & 0 \\ \frac{\sqrt{3}}{4} & 0 & \frac{1}{4} & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right), \tag{12}$$

one can show that

$$p(a_1=1, b_2=1)=\operatorname{tr}(M\rho M^{\dagger}) = \frac{1}{8}.\tag{13}$$

The other matrices can be similarly reconstructed.

• Welcome to our community! I have added equation numbers because even if you are not referring to any equations in this post, someone else might benefit from being able to say "Eq. 4 in this QCSE post" rather than "the equation for the third measurement operator in this QCSE post". Nov 6, 2023 at 17:05
• Thx @user1271772!
– Alex
Nov 6, 2023 at 17:34

From your current description, it seems that you have found each eigenvector $$\{|v_1\rangle,...,|v_4\rangle\}$$ and its associated eigenvalue $$\lambda_1,...,\lambda_4$$. These eigenvalues represent the product of the eigenvalues of each subspace $$a$$ and $$b$$, and, because of the degeneracy of the eigenvalues you cannot tell them apart directly.

Your intuition is right when you say that $$\lambda_1=\lambda_2=1$$ correspond to either $$\lambda_a=\lambda_b=1$$ or $$\lambda_a=\lambda_b=-1$$, because the eigenvalues of the tensor-product of two operators will be the product of the original operators' eigenvalues. You can easily find the value of $$\lambda_a$$ or $$\lambda_b$$ by applying a single operator (either $$a$$ or $$b$$) to your eigenvector (the eigenvector of a tensor-product $$a\otimes b$$ is also eigenvector of $$I\otimes b$$ and $$a\otimes I$$).

Here is an example with $$|v_1\rangle = \frac{1}{2}\begin{pmatrix} \sqrt{3} \\ 0 \\ 1 \\ 0 \end{pmatrix}$$. We know from our eigen-decomposition that $$\lambda_1 = 1$$ :

$$a_1|v_1\rangle = a_1\otimes I |v_1\rangle= \frac{1}{2}\begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 & 0 \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{\sqrt{3}}{2} \\ 0 & 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \end{pmatrix}\begin{pmatrix} \sqrt{3} \\ 0 \\ 1 \\ 0 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} \sqrt{3} \\ 0 \\ 1 \\ 0 \end{pmatrix} = 1|v_1\rangle,$$

From this you can conclude that $$\lambda_1$$ is associated with the eigenvalue $$\lambda_{a1}=1$$ which in turn implies that $$\lambda_{b1}=1$$. You can check this for the remaining eigenvalues as well as for the remaining operators.

This should allow you to conclude your proof.