How does a Hadamard gate impact the initial/previous values of a Qubit? [closed]

Question

I've been studying Quantum Computing and one thing that intrigued me is:

given a qubit q1 with an initial value x, when I apply a Hadamard gate on it, then it goes to superposition, so the probability of it collapsing to 1 or 0 is 50%.

So, does it mean that I just don't care about the previous value x at all anymore? Also, if that's in the middle of a circuit, that kind of cancels everything that happened before... For example, in the circuit for teleportation, we have a H gate on Qubit one before measuring it:

In this case, the first qubit goes to superposition, meaning that now it has a 50/50 chance of being 0 or 1. Is this the right interpretation of this concept?

Same thing happens for Deutsch-Jozsa:

We measure on the "H basis", as explained in the example. So, both qubits go in superposition, making their outcome depend on a probability. I didn't get it how these gates help on the calculation.

Answer 1

I think you might just be misunderstanding the Hadamard gate slightly. In the computational basis, $H= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}$ maps $|0\rangle \leftrightarrow |+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|1\rangle \leftrightarrow |-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$ since the Hadamard gate is its own inverse.

In words, the Hadamard gate sends computational basis states to equal superposition states (with phase/sign determined by the input state), and sends superposition states back to the computational basis depending on the phase between $|0\rangle$ and $|1\rangle$.

Finally, to clarify, the Hadamard gate is unitary, you can always undo it to get "x" back.