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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?

Mauricio
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bsaoptima
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1 Answers1

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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!

hft
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