# Why is the projection operator corresponding to $\tilde M$ given by $P_m\otimes I_B$?

Nielsen and Chuang, Chapter 2 (Box 2.6):

Suppose $M$ is any observable on a system $A$, and we have some measuring device which is capable of realizing measurements of $M$. Let $\tilde M$ denote the corresponding observable for the same measurement, performed on the composite system $AB$. Our immediate goal is to argue that $\tilde M$ is necessarily equal to $M \otimes I_B$. Note that if the system $AB$ is prepared in the state $|m\rangle |\psi\rangle$, where $|m\rangle$ is an eigenstate with an eigenvalue $m$ and $|\psi\rangle$ is any state of $B$, then the measuring device must yield the result $m$ for the measurement, with probability one. Thus, if $P_m$ is the projector onto the $m$ eigenspace of the observable $M$, then the corresponding projector for $\tilde M$ is $P_m \otimes I_B$. We therefore have

$\tilde M = \sum_{m} m P_m\otimes I_B = M \otimes I_B$

Could someone please explain me why the projection operator is $P_m\otimes I_B$? A proof or at least an example which illustrates the motivation behind the formula would be helpful. It has not been explained clearly in the textbook.

Thus, if $P_m$ is the projector onto the $m$ eigenspace of the observable $M$

Let $\left|m_1\right> \ldots \left|m_n\right>$ be a basis of the $m$ eigenspace of the observable $M$. This means that any state $\left|\psi_m\right>$ which is an eigenstate of $M$ i.e.: $$M\left|\psi_m\right>=m\left|\psi_m\right>$$ can be written as: $$\left|\psi_m\right>=\sum_i \alpha_i \left|m_i\right>$$ for some constants $\alpha_i$.

Now consider a general state $\left|\psi\right>$ which can be written as: $$\left|\psi\right>=\sum_n\sum_i \alpha_{n,i}\left|n_i\right>$$ here $n$ denotes the eigenvalue associated with $M$ and the sum of $i$ is the sum over all orthonormal states with this eigenvalue. The projection operator acts as follows: $$P_M\left|\psi\right>=\frac{1}{\sqrt{\sum_i|\alpha_{m,i}|^2}}\sum_i\alpha_{m,i}\left|m_i\right>$$
where the factor out front is a normalization factor.

You can see here that we have just kept the states with eigenvalue $m$.

then the corresponding projector for $\tilde M$ is $P_m \otimes I_B$. We therefore have

Let us call this corresponding projector $\tilde P_m$. Then we need to ask ourselves; what is the action of $\tilde P_m$?

Well we define it to be the projector which projects a general ket: $$\left|\psi_A\right>\otimes \left|\phi_B\right>$$ onto the $m$ eigenspace of the observable $M\otimes I_B$.

For a general operator $A\otimes B$ the action on a general state is: $$(A\otimes B)(\left|\psi_A\right>\otimes \left|\phi_B\right>)=(A\left|\psi_A\right>)\otimes (B\left|\phi_B\right>)$$ So in our case we have that: $$(M\otimes I_B)(\left|\psi_A\right>\otimes \left|\phi_B\right>)=(M\left|\psi_A\right>)\otimes (I_B\left|\phi_B\right>)$$ It should be evident then that the $m$ eigenspace of $M\otimes I_B$ takes the form: $$(\sum_i \alpha_i \left|m_i\right>)\otimes \left|\phi_B\right>$$ where $\left|\phi_B\right>$ is any ket in $B$. Since then: $$(M\otimes I_B)(\sum_i \alpha_i \left|m_i\right>)\otimes \left|\phi_B\right>=(\sum_i \alpha_i M\left|m_i\right>)\otimes (I_B\left|\phi_B\right>)=(\sum_i \alpha_i m\left|m_i\right>)\otimes \left|\phi_B\right>=m(\sum_i \alpha_i \left|m_i\right>)\otimes \left|\phi_B\right>$$ Like we had before we can write an arbitrary state as: $$(\sum_n\sum_i \alpha_{n,i} \left|n_i\right>)\otimes \left|\phi_B\right>$$ and by definition we need: $$\tilde P_m(\sum_n\sum_i \alpha_{n,i} \left|n_i\right>)\otimes \left|\phi_B\right>=(\sum_i \alpha_{m,i} \left|m_i\right>)\otimes \left|\phi_B\right>$$ So consider the action of $P_m \otimes I_B$ $$(P_m \otimes I_B)(\sum_n\sum_i \alpha_{n,i} \left|n_i\right>)\otimes \left|\phi_B\right>=(P_m\sum_n\sum_i \alpha_{n,i} \left|n_i\right>)\otimes I_B\left|\phi_B\right>$$ $$=(\sum_i \alpha_{m,i} \left|m_i\right>)\otimes \left|\phi_B\right>$$ where we used the action of $P_m$ as found above. Now this holds for all vectors in $\tilde M$ since we have being using general vectors throughout. Thus we must have that: $$\tilde P_m=(P_m \otimes I_B)$$

tl;dr Version

The action of $(P_m \otimes I_B)$ on a general state in $\tilde M$ is: $$(P_m \otimes I_B)(\left|\psi_A\right>\otimes\left|\phi_B\right>)=P_m\left|\psi_A\right>\otimes I_B\left|\phi_B\right>$$ $$\left|\psi_A^m\right>\otimes \left|\phi_B\right>$$ where $\left|\psi_A^m\right>$ is the projection of $\left|\psi_A\right>$ onto the $m$ eigenspace of $m$. This is by definition the action of the corresponding operator in $\tilde M$ and as such $(P_m \otimes I_B)$ is that opeartor.