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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?

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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.

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  • $\begingroup$ 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: $X\bigotimes I$. $\endgroup$ Nov 15 '20 at 23:45
  • $\begingroup$ 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 ? $\endgroup$ Nov 16 '20 at 1:06
  • $\begingroup$ I added more to my post $\endgroup$
    – chrysaor4
    Nov 16 '20 at 11:03
  • $\begingroup$ thank you so much. I understood and convinced about the calculation made. I really appreciate it. $\endgroup$ Nov 17 '20 at 4:00
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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 $|10\rangle$.

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  • $\begingroup$ 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 ? $\endgroup$ Nov 17 '20 at 3:59
  • $\begingroup$ See my update portion of the answer. $\endgroup$
    – KAJ226
    Nov 17 '20 at 4:42
  • $\begingroup$ thank you so much $\endgroup$ Nov 17 '20 at 4:49
  • $\begingroup$ 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? $\endgroup$ Nov 18 '20 at 7:14
  • $\begingroup$ 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 ? $\endgroup$ Nov 18 '20 at 7:49

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