It's very hard to see what exactly you're asking so I want to provide a bit of a broader context: Entanglement is not a second-class citizen of the country that is quantum mechanics. She does not need permission to appear, or "special forces" to create her, or any such thing. In fact I will go one step further to say, we wish she did! If we could limit her scope of appearance, so that she only appeared when we performed some rare sorcery, we would have a quantum computer by now that could do more than prove that with high probability, fifteen is three times five.
She is the twin sister of coherence, the “waviness” that we started QM in order to explain in the first place. In fact they never appear in the same location at the same time, so much so that many physics students expect us teachers to dramatically reveal that they were the same all along, that coherence is really “self-entanglement” or some such, and this is why when you entangle two systems their internal coherence disappears. (I don’t know how to define this well enough to satisfy them with a theorem.)
Coherence says that a quantum bit can be in a superposition of states, $a|0\rangle + b |1\rangle$. If you have two bits, one of which is in the above state, the other is just known to be $0$, then the joint state of the whole system is$$\big(a|0\rangle + b |1\rangle\big)\otimes\big(|0\rangle\big)=a|00\rangle + b |10\rangle.$$
Now almost any interaction which couples these two bits will entangle them, even if the interaction itself has a purely classical description. For example the interaction $(u,v)\mapsto(u, u\text{ xor } v)$ is a typical classical computer operation that can be realized I quantum mechanics as the unitary transform$$\begin{align}
|00\rangle&\mapsto|00\rangle\\
|01\rangle&\mapsto|01\rangle\\
|10\rangle&\mapsto|11\rangle\\
|11\rangle&\mapsto|10\rangle\\
\end{align}$$
And in QM we call this the “CNOT” gate. It has a completely classical description but when you happen to apply it to the above state which has a coherent bit for $u$, you will find the entangled state, $$\operatorname{CNOT}_{1\to 2}\Big(\big(a|0\rangle + b |1\rangle\big)\otimes\big(|0\rangle\big)\Big)=a|00\rangle + b |11\rangle.$$
That state is entangled. Notice how entanglement in this notation “looks like” the coherence looked, that $a|x\rangle+b|y\rangle$ pattern.
What has happened is a sort of failure of transitivity, or something: we have three things playing together: classical reversible logic gates (in this case xor), composing two systems and decomposing them later (the role of $\otimes$), and coherence. By itself, logic gates play well with composition: this is how computers have scaled to terabytes. By itself coherence plays nice with composition. Logic gates by themselves play nice with coherence. But you put all three together and you discover that they don’t all play perfectly well together: you may start with a composition and then be unable to cleanly decompose it into $|x\rangle\otimes|y\rangle$ after.
So the problem in general with our quantum computers, is actually exactly of this form. Our 0 and 1 bits, have some sort of different physical interaction with the atoms around them. In other words, the bits are not fully isolated from the atoms around them, and each bit’s interaction with its environment is slightly different depending on whether the bit is in the zero state or the one state.
Quantum mechanics says that that's a very bad thing. Like, QM doesn't make value judgments directly: but our goal was to make the bits coherent, and any entanglement with any outside environment--including any atoms that we don't immediately have under control--reduces the coherence of the system, and transfers it to the entanglement of the system plus environment. So QM translates any differential interaction, no matter how classically describable, to entanglement and then it tells us that this will make it progressively harder to achieve the goal we set for ourselves as these little entanglements sap away more and more of our coherence.
