Let's say we got a CX with a Hadamard gate on the control gate and any state at the target gate, will the target necessarily become a superposition of two states?

Best.

Assuming your control is $$|0\rangle$$ to begin with. Then after application of Hadamard, the control is: $$\frac{|0\rangle + |1\rangle}{\sqrt 2}$$.

Now using this as control and applying $$X$$ gate to the target, say $$|0\rangle$$, you get: $$\frac{|0\rangle|0\rangle + |1\rangle |1\rangle}{\sqrt 2}$$

Now, the system is entangled and is in a superposition of the states $$|00\rangle$$ and $$|11\rangle$$. Because it is entangled, it does not make sense to separately talk about the target state and ask if it is in superposition.

If you measure out the target, you will notice it is in a classical mixture (not a quantum superposition) of states $$|0\rangle$$ and $$|1\rangle$$ represented as follows: $$\frac{1}{2}|0\rangle\langle0| + \frac{1}{2}|1\rangle\langle1|$$

And looking at the outcome, you will see a basis state (I assume this is what you mean by simple state): $$|0\rangle$$ or $$|1\rangle$$, so yes, on measuring you will see a simple state.

A similar thing happens for $$|1\rangle$$. Any state that is not an eigenstate of $$X$$ will lead to some sort of entanglement as shown above. However, like @John Garmon said, if you use an eigenstate of $$X$$, say $$|-\rangle$$, on applying $$CX$$ the states becomes: $$\frac{|0\rangle|-\rangle - |1\rangle|-\rangle}{\sqrt 2}$$ $$= \frac{|0\rangle - |1\rangle}{\sqrt 2} |-\rangle$$

And the target state is not entangled, and is a 'simple' state with respect to the $$|+\rangle, |-\rangle$$ basis.

If I am properly understanding what you are asking, then no. If for the control, we start with $$\left|0\right>$$ and apply a Hadamard to it, then we end with our control in an even mixture of $$\left|0\right>$$ and $$\left|1\right>$$ and therefore 50% of the time the CX gate will act like an X gate, and 50% of the time the CX gate will act like the Identity. The key now is looking for target states that act the same way under application of the X gate as application of the Identity. These states are the eigenstates of the X gate, so $$\left|+\right>$$ and $$\left|-\right>$$. If the target input is in either the $$\left|+\right>$$ or $$\left|-\right>$$ state, then the target after the CX gate will NOT be in a superposition of two states.

• let's keep it with $\left|0\right>$ and $\left|1\right>$, what about measurements after completing the computation ? does this render either simple states ? (not superposition of them) ? – user1319236 Jun 16 '19 at 21:44
• Could you please clarify your question. I am not sure what you mean by simple states. – John Garmon Jun 17 '19 at 2:57
• not superposition states, but i got the answer it's 'measurement' that will collapse the superpostion into a simple state namely a $\left|0\right>$ or a $\left|1\right>$ ;) – user1319236 Jun 17 '19 at 4:12
• @JohnGarmon Could you now use the terminology of mixture and probabilities? I know what you're trying to say, but what you literally say is something quite different, and could confuse people. Talking about superposition and probability amplitude would be preferable. – DaftWullie Jun 17 '19 at 6:41