For two-qubit systems, do we have $\langle 01|01\rangle = \langle 0|0\rangle\langle 1|1\rangle$?

Question

I am new to quantum computing and I want to know the following: If I have a 2 qubit system in state e.g. $\left|01\right>$ and I want to calculate the probability of measuring e.g. $\left<01\right|$ I can write it as following: $|\left<01|01\right>|^2$. Now is the notation $\left<01|01\right>$ equal to this notation $\left<0|0\right>\left<1|1\right>$?

Answer 1

Yes.

In general, given two Hilbert space $H_1$ and $H_2$ with inner product $\langle \cdot |\cdot \rangle_1$ and $\langle \cdot | \cdot \rangle_2$ then for $u_1, v_1 \in H_1$ and $u_2,v_2 \in H_2$ we have that

$$\langle u_1 \otimes u_2 | v_1 \otimes v_2 \rangle = \langle u_1|v_1\rangle_1 \cdot \langle u_2|v_2\rangle_2$$

Answer 2

A more general rule is called the mixed-product property: $$ (A \otimes B) (C \otimes D) = AC \otimes BD, $$ which holds for any matrices $A,B,C,D$ of such sizes that we can form the matrix products $AC$ and $BD$.