# What is the result of measuring $\sigma_x$ on the state $|01\rangle+|10\rangle$?

I confused about how to calculate the probabilities and getting a certain result of measuring Bell's states with Pauli matrices as the operator. When you measure something, the state involved would be projected onto an eigenstate of the observable.

given $$|\psi\rangle = \frac{1}{\sqrt2} (|01⟩ + |10⟩)$$ as the state and $$\sigma_x = \left[\begin{matrix}0&1\\1&0\\\end{matrix}\right]$$ as the observable.

if the probability is 1/2, how to calculate them actually? What is the state after measurement?

The way to do this first requires rewriting whatever state you have in terms of the eigenstates of the operator in question. $$|0\rangle = \frac{1}{\sqrt{2}} (|+\rangle + |-\rangle)\\ |1\rangle = \frac{1}{\sqrt{2}} (|+\rangle - |-\rangle)$$ so that $$|\psi\rangle$$ becomes \begin{align} |\psi\rangle &= \frac{1}{\sqrt{2}} \Big( \frac{1}{2} (|+\rangle + |-\rangle)(|+\rangle - |-\rangle) + \frac{1}{2} (|+\rangle - |-\rangle)(|+\rangle + |-\rangle) \Big) \\ &= \frac{1}{2\sqrt{2}} (2|++\rangle - 2|--\rangle) \\ &= \frac{1}{\sqrt{2}} (|++\rangle - |--\rangle) \end{align}

Notice that this state still looks entangled in the $$X$$-basis, which is general property that entanglement cannot disappear just by rotating between bases. The second thing is that when you say you want to measure $$X$$, you have written a single-qubit operator, but this a two-qubit state. We can easily make it a two-qubit operator by taking the tensor product $$\text{X on the first qubit} \rightarrow X \otimes I\\ \text{X on the second qubit} \rightarrow I \otimes X$$ Here it actually doesn't matter which qubit of the entangled pair we measure, because their states are maximally correlated in the $$X$$-basis. Since the state's components are now expressed in terms of the eigenstates of either operator above, we can take the norm-squared of the amplitudes of the components of the state to see that we get either $$|+\rangle$$ (+1) or $$|-\rangle$$ (-1) with probability $$1/2$$ on either qubit we choose to measure.

See @KAJ226's post for the probability/measuring, below I'll explain how to get the eigenvectors

For the operators $$Z$$ and $$Y$$ the procedure is the same. Note that the computational state $$|0\rangle$$ and $$|1\rangle$$ are the eigenstates of the $$Z$$ operator, so you don't need to rewrite them. The eigenstates of the $$Y$$ operator are $$|y+\rangle = \frac{1}{\sqrt{2}} (|0\rangle + i|1\rangle)\\ |y-\rangle = \frac{1}{\sqrt{2}} (|0\rangle - i|1\rangle)$$ so that the computational states, rewritten in the $$Y$$-basis, are $$|0\rangle = \frac{1}{\sqrt{2}} (|y+\rangle + |y-\rangle)\\ |1\rangle = \frac{1}{i\sqrt{2}} (|y+\rangle - |y-\rangle)$$

To obtain the eigenvectors of any matrix (operator), you look for solutions of the following equation

$$A|v\rangle = \lambda |v\rangle\\ (A - \lambda I) |v\rangle = 0$$

where $$A$$ is an operator, $$I$$ is the identity matrix, $$|v\rangle$$ is a vector, and $$\lambda$$ is an eigenvalue. This equation has solutions when the determinant of the matrix $$A - \lambda I$$ is 0. As an example, for the $$X$$ operator

$$det(X - \lambda I) = 0\\ det\begin{bmatrix} -\lambda & 1 \\ 1 & -\lambda \end{bmatrix} = 0\\ \lambda^2 - 1 = 0\\ \lambda = \pm 1$$

Knowing the eigenvalues, we plug them back into the original equation to find the eigenvectors. For the first eigenvalue $$\lambda = +1$$

$$A|v\rangle = +1 |v\rangle\\ \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix} a \\ b\end{bmatrix} = \begin{bmatrix} a \\ b\end{bmatrix}$$

which says that $$a=b$$, so that the eigenvectors of eigenvalue +1 are the vectors $$a\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ for some scalar $$a$$. But remember, these vectors are quantum states, which must be normalized, so we can find $$a$$

$$\langle v | v \rangle = 1 = a^* a \begin{bmatrix} 1 & 1\end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = 2|a|^2 \\ \frac{1}{2} = |a|^2$$

and we see that a simple solution is just that $$a = \frac{1}{\sqrt{2}}$$, which is the familiar normalization factor. The state we have found is

\begin{align} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1\end{bmatrix} &= \frac{1}{\sqrt{2}} \bigg( \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \bigg) \\ &= \frac{1}{\sqrt{2}} \big( |0\rangle + |1\rangle \big) \equiv |+\rangle \end{align}

So if an eigenvector of $$X$$ is $$|+\rangle$$, then applying $$X \otimes I$$ to the state $$|+\rangle \otimes |\phi\rangle$$ for arbitrary $$|\phi\rangle$$ yields $$+1 |+\rangle |\phi\rangle$$, which is an eigenvector of eigenvalue +1.

• I would just like to add that for qiskit users, the tensor product would be reversed such that X on the first qubit: $I\bigotimes X$, and X on the second qubit: $X\bigotimes I$. Commented Nov 15, 2020 at 23:45
• Is it the same if we want to measure Y and Z? May I know how you get the eigenstates of the operator. Sorry, I still don't understand how the calculation works. can you explain step by step for measuring the probability of X on the first qubit? What is the state after measurement ? Commented Nov 16, 2020 at 1:06
• I added more to my post Commented Nov 16, 2020 at 11:03
• thank you so much. I understood and convinced about the calculation made. I really appreciate it. Commented Nov 17, 2020 at 4:00

Note that $$\sigma_x = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ has two eigenvectors $$|+ \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ and $$|- \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$ with the corresponding eigenvalues of $$+1$$ and $$-1$$, respectively.

According to the Born rule, the probability that we will get $$+1$$ on the first qubit which correspond to the $$|+\rangle$$ is $$Tr( \rho M)$$, where $$\rho$$ is the density matrix and in this case it is $$\rho = |\psi \rangle \langle \psi|$$, and $$M$$ is the projection operator onto the basis vector corresponding to the measurement outcome of $$+1$$ on the first qubit in this case. Calculating it out explicitly:

$$\rho = |\psi \rangle \langle \psi | = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$

\begin{align} M = |+\rangle \langle+| \otimes I = \bigg[ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} \bigg] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} &= \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \\ &= \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix} \end{align}

And therefore,

$$Tr(\rho M) = Tr\bigg(\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix} \bigg) = \dfrac{1}{2}$$

UPDATE: The state after measurement, $$|\psi_{post}\rangle$$, is going to be \begin{align} |\psi_{post}\rangle = \dfrac{ M |\psi \rangle }{ \sqrt{prob(+1)}} = \dfrac{ \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} }{ \sqrt{ 1/\sqrt{2} } } = \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} = |+ +\rangle \end{align}

Note that $$|++\rangle = |+ \rangle \otimes |+\rangle = \bigg(\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}\bigg) \otimes \bigg(\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}\bigg) = \dfrac{|00\rangle + |01\rangle + |10\rangle + |11\rangle }{2}$$

This can be extended to measurement in $$Y$$ basis as well. But $$\sigma_y = \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}$$ and hence its two eigenvectors are $$|i \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}$$ and $$|-i \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix}$$ with corresponding eigenvalues of $$+1$$ and $$-1$$ respectively. Then the probability to see $$+1$$ on the first qubit can be again calculated by $$Tr(\rho M)$$ but here $$M = |i\rangle \langle i| \otimes I$$ which can be calculated explicitly as

\begin{align} M = |i\rangle \langle i| \otimes I = \bigg[ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i \end{pmatrix} \bigg] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} &= \dfrac{1}{2} \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \\ &= \dfrac{1}{2}\begin{pmatrix} 1 & 0 & -i & 0\\ 0 & 1 & 0 & -i\\ i & 0 & 1 & 0\\ 0 & i & 0 & 1\end{pmatrix} \end{align}

and hence

$$Tr(\rho M) = Tr\bigg(\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & -i & 0\\ 0 & 1 & 0 & -i\\ i & 0 & 1 & 0\\ 0 & i & 0 & 1\end{pmatrix} \bigg) = \dfrac{1}{2}$$

Update 2:

If you want to measure in the $$Z$$ basis then it's trivial as you can already see the answer but we can follow the same procedure for confirmation. $$\sigma_z = \begin{pmatrix} 1 & 0\\ 1 & -1 \end{pmatrix}$$ and it has two eigenvectors $$|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ and $$|1 \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ with two associate eigenvalues $$+1$$ and $$-1$$ respectively. Now to know the probability of getting the first qubit to be measured with $$+1$$ is $$Tr(\rho M)$$ where $$\rho$$ is similar as before, and $$M = |0\rangle \langle 0| \otimes I$$, which works out explicitly as

\begin{align} M = |0\rangle \langle 0| \otimes I = \bigg[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \bigg] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \end{align}

Hence

$$Tr(\rho M) = Tr\bigg(\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \bigg) = \dfrac{1}{2}$$

and the state after collapsed is: \begin{align} |\psi_{post}\rangle = \dfrac{ \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} }{ \sqrt{ 1/\sqrt{2} } } = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = |0 1\rangle \end{align}

This makes sense because we have $$|\psi \rangle = \dfrac{|01\rangle + |10\rangle}{\sqrt{2}}$$ and so if the first qubit is in the state $$|0\rangle$$ after measurement then this means that the state $$|\psi \rangle$$ has collapsed onto the eigenvector $$|01 \rangle$$, hence the reason why our calculation show $$|01\rangle$$ for $$|\psi_{post}\rangle$$. Similarly, if the read out indicates that the first qubit is a $$|1\rangle$$ then $$|\psi_{post}\rangle$$ would have been $$|10\rangle$$.

• Thank you so much for your answer. I really appreciate it. May I know how to write the state after obtaining the result in Dirac notation? how to change from matrix to braket form ya ? Commented Nov 17, 2020 at 3:59
• See my update portion of the answer. Commented Nov 17, 2020 at 4:42
• thank you so much Commented Nov 17, 2020 at 4:49
• if we want to measure the second qubit of the state, using the collapse state, the probability of getting +1 will be 1 while the state after the measurement is still |++>. is this correct? Commented Nov 18, 2020 at 7:14
• can you show the calculation for measuring Z-basis and the state after measurement? I get probability 1/2 with the state after the measurement is 1x4 matrix wich is 0 1 0 0 . is this correct ? Commented Nov 18, 2020 at 7:49