By definition, Hadamard transformation (acting on a qubit) maps the unit vector in the $Y$ axis direction of the Bloch Sphere ($S^2$) to its negative, equivalent to a rotation of $\pi$ rad around $X+Z$ axis. I understand it pictorially.
I have trouble showing this explicitly using the matrix representation of Hadamard transformation $H= \frac{1}{\sqrt{2}}\begin{bmatrix} {1}&{1}\\ {1}&{-1} \end{bmatrix}$for a qubit $\hat y$.
This qubit $\hat y$ (the unit vector in the $Y$ axis direction of the Bloch Sphere) has $\theta=\frac{\pi}{2}$ and $\phi=\frac{\pi}{2}$, where these angles are the ones that define the state of a qubit($\lvert \psi\rangle=cos\,(\frac{\theta}{2})\lvert 0\rangle+e^{i\phi}\,sin\,(\frac{\theta}{2})\lvert 1\rangle$). Therefore, $\hat y = \frac{1}{\sqrt2}\lvert 0\rangle + e^{i\frac{\pi}{2}}\frac{1}{\sqrt2}\lvert 1\rangle= \frac{1}{\sqrt2}\lvert 0\rangle + \frac{i}{\sqrt2} \lvert 1\rangle$. Now acting the Hadamard transformation matrix on it we have:
$\begin{equation} H\cdot\hat y=\frac{1}{\sqrt{2}}\begin{bmatrix} {1}&{1}\\ {1}&{-1} \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt2}\\\frac{i}{\sqrt2}\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1+i\\1-i\end{bmatrix}\end{equation}$.
In contrast, we expected the result of $H\cdot\hat y$ be instead $-\hat y$ by the pictorial definition of Hadamard transformation. What am I missing?
Any help regarding the mistake that I'm making here is appreciated.
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$\begingroup$ What state were you expecting, and how does the output you get actually differ from it? Of what operators is your expected output an eigenstate? Does the obtained output obey that same relation? What's the overlap between the two? What happens if you factor out the ugly phase? $\endgroup$– Emilio PisantyCommented May 30, 2018 at 22:06
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$\begingroup$ Hi @EmilioPisanty. As in the answer below, the expected state is $-\hat y= \frac{1}{\sqrt2}\lvert 0\rangle - \frac{i}{\sqrt2} \lvert 1\rangle$. It differs by a factor $\frac{1+i}{2}$ from what I obtained above. And of course, since these two are multiples, both are eigenstates of the same set of operators. How could I see this myself? $\endgroup$– MathistCommented May 30, 2018 at 23:54
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You have to take out a factor $\frac{1+i}{\sqrt{2}}$ -- the action on the Bloch sphere is only up to a phase. Also, not that the qubit vector corresponding to $-\hat y$ is $(|0\rangle-i|1\rangle)/\sqrt{2}$.
answered May 30, 2018 at 22:01
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$\begingroup$ Thanks @Norbert. Is the fact that the action on the Bloch sphere is up to a phase because of the rotational symmetry of the sphere itself? (i.e. we could take any two directions to be two of the axis) $\endgroup$– MathistCommented May 30, 2018 at 23:57
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$\begingroup$ @Mathophile-Mathochist No. It's just that the mapping of a qubit state (a unit vector in $\mathbb C^2$ to the Bloch sphere (a unit vector in $\mathbb R^3$) does not keep track of the overall phase. (After all, just by parameter counting one degree of freedom must be lost.) $\endgroup$ Commented May 31, 2018 at 0:03