# 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? Commented Feb 7, 2021 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
Commented Feb 7, 2021 at 21:46