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I am new to quantum information and I am trying to work on some problems but I have confused myself over a qubit problem. I have the state of two qubits $|\psi\rangle_{AB}=a_{00}|00\rangle+a_{01}|01\rangle+a_{10}|10\rangle+a_{11}|11\rangle$ and $\sum_{j,k} |a_{j,k}|^2=1$. If I am to measure qubit B in the basis {$|0\rangle_{B},|1\rangle_{B}$}, what is the probability of getting $|1\rangle_B$?

I am not familiar with measuring only one of the qubits. From my understating, the entire state will not collapse after the measurement, and only one of the subsystems will collapse. Am I wrong?

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  • $\begingroup$ Welcome to QCSE! Do the replies to this post answer your question? If not, perhaps the "Remark on measuring composite states" in this answer does? $\endgroup$ Feb 7 at 17:19
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If we have the state $|\psi \rangle = a_{00}|00\rangle +a_{01}|01\rangle +a_{10}|10\rangle +a_{11}|11\rangle $ then the probability of the second qubit being in the state $|1\rangle$ is the probability of the state $|\psi \rangle$ having $|1\rangle$ on the second qubit. In this case, it is from the states $|01\rangle$ and $|11\rangle$. So The probability of measuring the second qubit in the state $|1\rangle$ is $\bigg| a_{01} \bigg|^2 + \bigg| a_{11} \bigg|^2 $.

Similarly, the probability of the second qubit of the state $|\psi\rangle$ being measured in the state $|0\rangle$ is then $\bigg| a_{00} \bigg|^2 + \bigg| a_{10} \bigg|^2 $.

You can work this out explicitly as well. First, we have

$$ |\psi \rangle = a_{00}|00\rangle +a_{01}|01\rangle +a_{10}|10\rangle +a_{11}|11\rangle = \begin{pmatrix} a_{00} \ \ \\ a_{01} \\ a_{10} \\ a_{11} \end{pmatrix} $$ since we taken $|0\rangle = \begin{pmatrix} 1 \\ 0\end{pmatrix}$ and $|1\rangle = \begin{pmatrix} 0 \\ 1\end{pmatrix}$. The computational basis. And now we are looking for the probability that the second qubit is in the state $|0\rangle$ and ignore the first qubit, then the measurement $M$ can be described as

$$ M = I \otimes |0\rangle \langle 0 | = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}= \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} $$ where $I$ corresponds to the identity operator, and $|0\rangle \langle 0|$ corresponds to the outer product operation.

And so according to Born's rule the probability to measure the second qubit in the state $|0 \rangle$ is

$$ \langle \psi | M | \psi \rangle = \begin{pmatrix} a_{00}^* & a_{01}^* & a_{10}^* & a_{11}^* \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} a_{00} \ \ \\ a_{01} \\ a_{10} \\ a_{11} \end{pmatrix} = |a_{00}|^2 + |a_{10}|^2 $$ here $a^*$ indicates the conjugate of $a$ and hence $a^*a = |a|^2$.

Now if you want to construct $M$ for the second qubit being measured in the state $1\rangle$ and not measuring the first qubit then you can do construct it as $M = I \otimes |1\rangle \langle 1|$. Where $|1\rangle \langle 1| = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} $

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  • $\begingroup$ This clears things a lot. Thank you. $\endgroup$
    – user14766
    Feb 7 at 21:46

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