$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 4.15}$

a) Show that the operator $B$ of Example 4.3.4 is of the form $Q\otimes I$ where $Q$ is a $(2\times 2)$-Hermitian operator.

b) Show that any operator of the form $Q\otimes I$, where $Q$ is a $(2\times 2)$-Hermitian operator and $I$ is the $(2\times 2)$-identity operator, specifies a measurement of a two-qubit system. Describe the subspace decomposition associated with such an operator.

c) Describe the subspace decomposition associated with an operator of the form $I\otimes Q$ where $Q$ is a $(2\times 2)$-Hermitian operator and $I$ is the $(2\times 2)$-identity operator, and give a high-level description of such measurements.

Not 100% sure but I'll try :)

(1)Part a)

(2)Part b)If $Q$ is hermitian, then $Q = Q^+$, hence:

with $q_1, q_2$ real. Then:

(3)is also Hermitian, hence it specifies a measurement on a 2-quibit system.

$I$ has only one eigenvalue (equal to 1) hence the eigenvalues of $Q \otimes I$ are the eigenvalues of $Q$. If $Q$ has only one eigenvalue, then its subspace decomposition is trivial ($V = V \oplus 0$, where $V$ is the entire space for a 1-qubit system). In the non-trivial case where $Q$ has two distinct eigenvalues (with eigenspaces $S_1, S_2$), these are also the eigenvalues of $Q \otimes I$ and the eignespaces of $Q \otimes I$ are $S_1 \otimes V, S_2 \otimes V$. Hence $Q \otimes I$ is a measurement of the 1st qubit.

(4)Part c)Assuming $Q$ hermitian as in point b):

is also Hermitian, hence it specifies a measurement on a 2-quibit system.

As in point b), the non-trivial case is when $Q$ has two distinc eigenvalues (with eigenspaces $S_1, S_2$): these are also the eigenvalues of $I \otimes Q$ and the eignespaces of $I \otimes Q$ are $V \otimes S_1, V \otimes S_2$. Hence $I \otimes Q$ is a measurement of the 2nd qubit.

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