# Measuring first qubit of Bell state

Let's consider the following Bell state:

$$\lvert \Phi^+\rangle = \frac{1}{\sqrt{2}} (\lvert00\rangle + \lvert11\rangle)$$

What would happen if I measure the first qubit in the standard basis and keep the other one intact? Is the following computation correct?

$$\langle 0I\lvert\Phi^+\rangle\langle\Phi^+\lvert0I\rangle = \frac{1}{2}\lvert0\rangle\langle0\lvert$$

I is the two dimensional identity matrix. And $$\langle 1I\lvert\Phi^+\rangle\langle\Phi^+\lvert1I\rangle = \frac{1}{2}\lvert1\rangle\langle1\lvert$$

Shouldn't have I gotten the probability of measurement instead of states?

• Possible duplicate of Density matrix after measurement on density matrix – ahelwer Jun 11 at 5:50
• Can someone explain how a question about measuring Bell states is a duplicate of a question about maximally mixed states? There's a little step in there that's not necessarily obvious and that little step can sometimes be the difference between a duplicate and not-a-duplicate to the person asking the question – Mithrandir24601 Jun 20 at 9:31

Recall the measurement postulate (as applied to projective measurements): A measurement is described by a set of projectors $$P_i$$ satisfying $$\sum_iP_i=1$$. If a state $$|\psi\rangle$$ is being measured, the outcome $$i$$ is obtained with probability $$p_i=\langle\psi|P_i|\psi\rangle$$, and if outcome $$i$$ is obtained, the state after measurement is $$P_i|\psi\rangle/\sqrt{p_i}$$.
The problem is how to describe the projections. In your case, you want to project the first qubit on the standard basis, and do nothing on the second qubit. So, that means you need $$P_0=|0\rangle\langle 0|\otimes1,\qquad P_1=|1\rangle\langle 1|\otimes1,$$ because we remember that "do nothing to a qubit" is described by the identity operation.
So, what's the probability of getting the answer 0? $$\langle\Phi^+|P_0|\Phi^+\rangle=\frac{1}{2},$$ and the two-qubit state after the measurement is $$\sqrt{2}P_0|\Phi^+\rangle=|00\rangle$$