Say you run qubit $A = (|0⟩ + |1⟩) / √2$ and qubit $B = |0⟩$ through a CNOT gate. What is the state of qubit $B$ afterwards?

Question

I am new to the weeds of quantum computing and this question is probably pretty elementary.

Say you run qubit A = (|0⟩ + |1⟩) / √2 and qubit B = |0⟩ through a CNOT gate. What is the state of qubit B afterwards?

Here is how I have tried to reason my way to an answer:

1. The state of the system after the CNOT gate is (|00⟩ + |11⟩) / √2
2. The state of qubit A remains (|0⟩ + |1⟩) / √2
3. Here's where I feel like I'm doing something wrong. My intuition tells me if I have a global state ac|00⟩ + ad|01⟩ + bc|10⟩ + bd|11⟩, I should be able to deconstruct it into the states of the individual qubits to acquire A = a|0⟩ + b|1⟩ and B = c|0⟩ + d|1⟩
4. Using the above intuition and observations, I have a = 1/√2, b = 1/√2, ad = 0, bc = 0 ac = 1/√2, and bd = 1/√2. I'm looking for c and d.
5. This system of equations is inconsistent. By the first four equations, it must be the case that c = d = 0. But by the last two, that can't be the case.

Where did my thinking go wrong?

Answer 1

1. This is correct.
2. This is not correct. Particle A does not have a definite state anymore. Neither does particle B. Previously, you could write the two-particle state as $\frac{|0\rangle+|1\rangle}{\sqrt 2}|0\rangle$, so the state factored and you could talk about "the state of A" and "the state of B". After the CNOT, this is no longer true. You can no longer talk about the "state of A", you can only talk about the state of the SYSTEM. This is because A and B are entangled.

Punchline: There are some two-particle states that cannot be represented as $(a|0\rangle+b|1\rangle)(c|0\rangle+d|1\rangle)$. The state of your A and B particles after applying the CNOT is one of them.