I am trying to understand what is the affect of Hadamard gate on qubit. So far, I understand that applying $H$ gate on qubit, puts the qubit into superposition state where the probability of the qubit being in one of the definite state is 50%.
Now I apply hadamard gate on $|0\rangle$ using IBM Qiskit composer. I now take a measurement of the qubit after the $H$ gate. I can understand that the probabilities graph which says 50% of qubit being in $|0\rangle$ or $|1\rangle$ state. But why is the state in the bloch sphere plot always showing $|1\rangle$.Also, the statevector too shows $|1\rangle$.
My theory was that, maybe it shows 50% of the time in $|1\rangle$ state when measured. So, what I see is part of that probability. But I have tested it many times, and every time the statevector and the state in the Bloch sphere plot shows as $|1\rangle$.
Does it mean that, applying the $H$ gate on $|0\rangle$ and measuring it the qubit, inverts the output?If so, why so?
Could anyone please explain this?
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1$\begingroup$ If you try several times, you will notice that it is not always being $|1\rangle$. Sometime $|1\rangle$, sometimes $|0\rangle$, as it should be 50/50. $\endgroup$– GuangliangCommented Mar 18, 2022 at 13:00
1 Answer
First of all, Q-sphere in IBM composer is not the same as Bloch sphere. While Bloch sphere allows to graphically represent single-qubit state (the state is described by two angles $\theta$ and $\varphi$ similarly to geographic coordinates), Q-sphere shows basis states some general state consists of. North and south poles of Q-sphere corresponds to states $|0\dots0\rangle$ and $|1\dots1\rangle$, respectivelly. Going from north to south means that number of $1$s is increasing while number of $0$s is decreasing. There is equal number of $0$s and $1$s in states placed on the equator. A color of the state correspond to its phase.
In you particular case, you should see two states on the Q-spehere, namely $|0\rangle$ and $|1\rangle$. You see only $|1\rangle$, however, this is caused by the measurement gate. In other words, you see one of possible results after measuring your circuit. Just try to remove the measurement and you will see that your state is composed of $|0\rangle$ and $|1\rangle$.