# Understanding the Hadamard gate and what is meant by qubits [duplicate]

I'm trying to understand the Qiskit documentation in order to see if there are differences in notation from my quantum mechanics lecture notes.

The Hadamard Gate transforms $$|0 \rangle$$ into $$|+ \rangle = \frac{|0 \rangle + |1 \rangle}{\sqrt{2}}$$ and $$|1 \rangle$$ into $$|- \rangle = \frac{|0 \rangle - |1 \rangle}{\sqrt{2}}$$. From what I understand, a qubit is a superposition of $$|0 \rangle$$ and $$|1 \rangle$$ but what is the difference between the superpositions $$|+ \rangle$$ and $$|- \rangle$$ in terms of computation and interpretation?

• They differ in phase, which is as much important as the amplitude. Jan 28 at 17:43
• @TristanNemoz yes it does, thank you! Jan 30 at 17:07

The Hadamard Gate transforms $$|0 \rangle$$ into $$|+ \rangle = \frac{|0 \rangle + |1 \rangle}{\sqrt{2}}$$ and $$|1 \rangle$$ into $$|- \rangle = \frac{|0 \rangle - |1 \rangle}{\sqrt{2}}$$. From what I understand, a qubit is a superposition of $$|0 \rangle$$ and $$|1 \rangle$$...
what is the difference between the superpositions $$|+ \rangle$$ and $$|- \rangle$$ in terms of computation and interpretation?
In a sense, those two state are as different as they can be; they are orthogonal to each other: $$\langle - |+\rangle = \frac{1}{2}\left(\langle 0|-\langle 1|\right)\left(|0\rangle+|1\rangle\right) = \frac{1}{2}\left(1 + 0 -0 -1\right) = 0$$
Also, just like the $$|0\rangle$$ and $$|1\rangle$$ states are eigenstates of the $$\hat Z$$ matrix (Pauli Z-matrix), with eigenvalues +1 and -1. So too the $$|+\rangle$$ and $$|-\rangle$$ state are eigenstates of the $$\hat X$$ matrix (Pauli X-matrix), with eigenvalues +1 and -1, respectively. For example: $$\hat X|+\rangle = \left(\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}\right)\left(\begin{matrix}1/\sqrt{2}\\1/\sqrt{2}\end{matrix}\right) = \left(\begin{matrix}1/\sqrt{2}\\1/\sqrt{2}\end{matrix}\right) =|+\rangle$$ $$\hat X|-\rangle = \left(\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}\right)\left(\begin{matrix}1/\sqrt{2}\\-1/\sqrt{2}\end{matrix}\right) = \left(\begin{matrix}-1/\sqrt{2}\\1/\sqrt{2}\end{matrix}\right) = -\left(\begin{matrix}1/\sqrt{2}\\-1/\sqrt{2}\end{matrix}\right) = - |-\rangle$$
For further help interpreting the meaning of these states recall that, for a spin-1/2 particle, a rotation about the y axis by an angle $$\theta$$ is represented by the matrix: $$e^{-i\theta\hat Y/2} = \cos(\theta/2) - i \hat Y \sin(\theta/2)\;,$$ where $$\hat Y$$ is the Pauli-Y matrix. (Note: Sometimes $$\hat Y$$ is defined as $$-i$$ times the Pauli-Y matrix, but here we define it as the unmodified Pauli-Y matrix.)
If you rotate a z-eigenstate into a x-eigenstate, you perform a rotation about the y axis by $$\pi/2$$. And, lo and behold, plugging into the above equation with $$\theta = \pi/2$$ shows that the Hadamard Gate implements exactly this rotation!