# How to perform a projective measurement on one component of a composite system?

For simplicity, let $$|\phi\rangle|\psi\rangle\in\Bbb C^2\otimes\Bbb C^2$$. I know how to compute the projective measurement $$\{P_m\}_m$$ of $$|\phi\rangle|\psi\rangle$$ on $$\Bbb C^2\otimes\Bbb C^2$$, but I wonder how to measure the first component of $$|\phi\rangle$$ of $$|\phi\rangle|\psi\rangle$$ with respect to a projective measurement $$\{P_m\}_m$$ on $$\Bbb C^2$$. And I also wonder will the second component collapses after the measurement? What will be the resulting state? PS. I haven't seen the explaination in N&C's book. A reference is also welcomed.

If you perform a local measurement $$\{P_m\}$$ on the first system only, then the global measurement is given by the projectors $$P_m \otimes \mathbb{I}$$ where $$\mathbb{I}$$ is the identity matrix.

Consequently, if you perform a local measurement on a product state $$|\phi\rangle\otimes|\psi\rangle$$, then the state of the second system is not disturbed as the post-measurement state is simply $$\frac{(P_m \otimes \mathbb{I})( |\phi\rangle \otimes|\psi\rangle)}{\| (P_m \otimes \mathbb{I})(|\phi\rangle\otimes|\psi\rangle)\|_2} =\frac{P_m |\phi\rangle}{\|P_m |\phi\rangle\|_2} \otimes|\psi\rangle.$$

In constrast, if you measure an entangled state, then this not true anymore. For instance, take the well-known 2-qubit Bell state $$|\phi^+\rangle = \frac{1}{\sqrt 2} \big( | 00 \rangle + | 11 \rangle \big).$$ Measuring the first system in the computational basis will collapse this either into $$| 00 \rangle$$ or $$| 11\rangle$$ with probability $$1/2$$ each. Thus, the (reduced) state of the second system depends on the outcome of the measurement on the first one.

• Many thanks. I want to confirm that $P_m \otimes \mathbb{I}\left( \frac{1}{\sqrt 2} \big( | 00 \rangle + | 11 \rangle \big)\right)=P_m \otimes \mathbb{I}(\frac{1}{\sqrt{2}}|00\rangle)+P_m \otimes \mathbb{I}(\frac{1}{\sqrt{2}}|11\rangle)$. Then apply $P_m$ and $\Bbb I$ separately to $|0\rangle,~|0\rangle$ and $|1\rangle,~|1\rangle$?
– Eric
Apr 21, 2021 at 13:29
• @Eric Yes. The first one is just a property of linearity. The second one is coming from the the fact that $(A \otimes B) (v \otimes u) = Av \otimes Bu$. Apr 21, 2021 at 14:57
• is there a reference(s) to your claims? Jul 5, 2021 at 23:19
• @QuantumGuy123 To which comment are you referring? My answer only uses the standard postulates of QM and KAJ226 uses the definition of the tensor product. Jul 6, 2021 at 9:56
• as the OP stated in the original question 'A reference is also welcomed'. I can't speak for them, but I am personally looking for a reference to the method you showed of constructing multi-qubit measurement operators using tensor products of single qubit measurement operators. Jul 6, 2021 at 14:46