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.

  • $\begingroup$ 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 $|1\rangle,~|1\rangle$? $\endgroup$ – Eric Apr 21 at 13:29
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    $\begingroup$ @Eric Yes. The first one is just a property of linearity. The second one is coming from the the fact that $(A \otimes B) (v \otimes u) = Av \otimes Bu$. $\endgroup$ – KAJ226 Apr 21 at 14:57
  • $\begingroup$ is there a reference(s) to your claims? $\endgroup$ – Quantum Guy 123 Jul 5 at 23:19
  • $\begingroup$ @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. $\endgroup$ – Markus Heinrich Jul 6 at 9:56
  • $\begingroup$ 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. $\endgroup$ – Quantum Guy 123 Jul 6 at 14:46

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