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If we apply Hadamard gate 2 times on a qubit $|0\rangle$ then we will get $|0\rangle$. So there must be something else which is unique for both $|1\rangle$ and $|0\rangle$ which is constant even in superposition. What is it?

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If we apply Hadamard gate 2 times on a qubit $|0\rangle$ then we will get $|0\rangle$. So there must be something else which is unique for both $|1\rangle$ and $|0\rangle$ which is constant even in superposition. What is it?

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Much as in CMOS we almost always equate a low voltage with binary zero, and high voltage with binary one, and a classical $\mathsf{AND}$ gate as implementing the truth-table by adjusting voltages to be consistent, we have similar requirements in quantum computing to call $\vert 0\rangle$, $\vert 1\rangle$, and the Hadamard gate acting on it in the manner you described. We could have a convention of low-voltage being one and high-voltage being zero; we would have to adjust how we implement $\mathsf{AND}$ appropriately.

The polarity of photons is straightforward to visualize as qubits. If we settle on what we mean by $\vert 0\rangle$ and $\vert 1\rangle$ with respect to the polarity of photons, then a Hadamard gate would be the one that rotates the polarity appropriately.

So if a photon has 20% chance of being in horizontal polarization , can i call it superposition ? Then if a photon has 100% chance to be in horizontal polarization then let's call it |0>, if i apply hadamard operation in this |0> then what it would do? – bipul kalita – 2019-11-09T18:24:58.353

When you equate horizontal polarity with $\vert 0\rangle$, such a Hadamard gate (please capitalize for now) acting on a photon having horizontal polarity would rotate the polarity of the photon by $45^circ$. If you had equated *vertical* polarity with $\vert 0\rangle$ instead, then the Hadamard gate would correspond to rotating the polarity by $135^\circ$. – Mark S – 2019-11-09T18:35:00.087

feeling better now. I think I should forget about "superposition" word, it's better and meaningful when I call it , "a photon with probabilities". So, how much the Hadamard gate will rotate if I apply twice in both |0> and |1> . Because Hadamard is also reversible . – bipul kalita – 2019-11-09T19:13:13.383

Since Hadamard gate is inversion of itself, i.e. $HH=I$ I would expect that the rotation will be zero. However, when Hadamard causes rotation by 45 degrees, two Hadamards applied in row should lead to rotation by 90 degress resulting in changing state $|0\rangle$ to $|1\rangle$. But apparently this is not the case. Could you please explain this? – Martin Vesely – 2019-11-09T22:35:06.643

@MartinVesely you're right, let me think... – Mark S – 2019-11-09T22:43:24.150

"The Hadamard gate can also be expressed as a 90º rotation around the Y-axis, followed by a 180º rotation around the X-axis " found in this page https://www.quantum-inspire.com/kbase/hadamard/ this makes sense now. It works. But we have to imagine it in a 3d sphere.

– bipul kalita – 2019-11-10T05:08:26.920
Convention of what we mean by $\vert 0\rangle$, $\vert 1 \rangle$, and Hadamard gate. – Mark S – 2019-11-09T17:19:53.497

Please help me ,i think I don't understand it well enough. All i just know is mathematical descriptions of some gate [matrix] and they say spin up |1> and spin down |0> and hadamard gate can make them turn into superposition and bla bla bla. But what's happening at physical level ? – bipul kalita – 2019-11-09T17:50:16.197