# Probability of measuring one qubit from the state of two qubits

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?

• 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? – Adam Zalcman Feb 7 at 17:19

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 $$|1\rangle \langle 1| = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}$$

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