Operators acting on a single subsystem within a combined system's state

Question

I was reading over combined systems and operators acting on a single system within the combined system, and am confused by the math.

For example, we have individual qubit states for subsystems $A$ and $B$ that, as a combined system, produce the state:

\begin{align} A\otimes B &= \begin{bmatrix} a_{0}b_{0} \\ a_{0}b_{1} \\ a_{1}b_{0} \\ a_{1}b_{1} \end{bmatrix} \end{align}

We have an operator $O$ that acts on subsystem $A$. To express this on the combined state, it is simply (where $I$ is the identity matrix of the same dimension as $A$ and $B$) \begin{align} (O\otimes I)(A\otimes B) \end{align}

But why is this? I've been reading a text that says this combined operator changes only the coefficients $a_0$ and $a_1$, while leaving $b_0$ and $b_1$ unchanged. But I don't see that. I see this operation as changing whatever the value of the product of $a_i$ and $b_j$ is.

Answer 1

Welcome to SE! Another way of saying what your text claims is $$(O\otimes I)(A\otimes B)=(OA)\otimes B.$$ To prove this to yourself, you could write a general $O$, say $O=\begin{pmatrix}w&x\\y&z\end{pmatrix}$, and then just evaluate both sides of the above equation directly for $A\equiv\begin{pmatrix}a_0\\a_1\end{pmatrix}$ and $B\equiv\begin{pmatrix}b_0\\b_1\end{pmatrix}$ (i.e., for the LHS take the tensor product $O\otimes I$ first, then multiply $A\otimes B$ by this, and for the RHS multiply $A$ by $O$ first, then take the tensor product with $B$).

You will find that the equation indeed holds; this is the sense in which $A$ "just acts on" the state of the first qubit.