What happens if I measure only the first qubit of a Bell state?

Question

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?

Answer 1

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 $$ \sqrt{2}P_0|\Phi^+\rangle=|00\rangle $$

(I think all that is actually missing from your calculation is a trace over all the systems, but there's some of the intermediate steps that are not quite right, at least in terms of notation, even though mathematically you appear to have done the right thing.)