If a system's Hamiltonian is non-interacting, does it mean that the system is not entangled? My book is generally being quite unclear about something.
So firstly I know that if the system is not entangled, we can write its state as $|ab\rangle=|a\rangle|b\rangle$ (if we understand the product is actually a tensor product). If it is entangled, we cannot do this.
It states that in general the composite system has Hamiltonian $H_{ab}=H_a+H_b+H_{int}$. Then it does some maths and works out that if $H_{int}=0$, the systems will remain unentangled provided they begin unentangled, otherwise the systems have to be entangled.
Somewhere else it says to consider two non-interacting subsystems (which I interpret as $H_{int}=0$) so that $|ab\rangle=|a\rangle|b\rangle$ - this disagrees with the above in that we haven't ruled out the prospect of the system being initially entangled, which would mean we couldn't write $|ab\rangle=|a\rangle|b\rangle$.
So my question is, which of the books two statements are correct? I.e, does $H_{int}=0$ imply the system is not entangled, or does it imply that the system is not entangled only if it is not entangled at time zero?
 A: You might be misunderstanding entangled states. An entangled state, is a state which we cannot express as a factorised tensor product, i.e it cannot be cast in this form:
$$ |\psi_A\rangle \otimes |\psi_B\rangle \equiv |\psi_A\psi_B\rangle. $$
Consider the state 
$$ |\psi\rangle = \frac{1}{2}\Big (|0\rangle_A|0\rangle_B + |0\rangle_A|1\rangle_B + |1\rangle_A|0\rangle_B + |1\rangle_A|1\rangle_B \Big). $$ 
This state can be factorised, i.e it can be expressed into a tensor product of two other states:
$$ \to |\psi\rangle_A \otimes  |\psi\rangle_B = \frac{1}{\sqrt 2}\Big (|0\rangle_A + |1\rangle_A \Big ) \otimes \Big (|0\rangle_B + |1\rangle_B \Big). $$
Now consider the Bell state:
$$ |\Phi ^{+}\rangle = \frac {1}{\sqrt 2} \Big (|0\rangle_A|0\rangle_B + |1\rangle_A |1\rangle_B \Big) $$ 
There is no way to decompose this state now into the tensor product of two states (try it!), hence $|\Phi ^{+}\rangle$ is said to be entangled.
Bear in mind that I have suppressed tensor notation when I wrote down the basis states, i.e technically I should have written for example:
$$ |\psi\rangle = \frac{1}{2}\Big (|0\rangle_A \otimes |0\rangle_B + |0\rangle_A \otimes |1\rangle_B + |1\rangle_A \otimes |0\rangle_B + |1\rangle_A \otimes |1\rangle_B \Big), $$
but do not confuse products of states with entanglement. Entanglement exists when a system cannot be separated into two subsystems, therefore you can only consider them together and describe them using an entangled state. In your case, as you yourself has said, $|ab\rangle$ can be decomposed into $|a\rangle \otimes |b\rangle$ so the states are not entangled and the system can be described in terms of two independent subsystems.
