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I'm having some trouble understanding the Bloch representation of qubits in some cases.

The canonical representation $\cos(\psi/2) |0\rangle + \sin(\psi/2)e^{i\theta}|1\rangle$ has the first coefficient, $\cos(\psi/2)$, always a nonnegative real number. But the Pauli-Y gate is defined to operate on the computational basis $|0\rangle$ and $|1\rangle$ with the matrix

$$ Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, $$ which when applied to $|1\rangle$ yields $-i|0\rangle$. Now this is not in the canonical representation. My questions is: should we normalize the phase to get $|0\rangle$? and does this means that $-i|0\rangle=|0\rangle$?

Miguel
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1 Answers1

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The state vector
$$ |\Psi\rangle=\cos \psi/2 |0\rangle + \sin \psi/2 ~e^{i\theta} |1\rangle = \begin{pmatrix} \cos \psi/2 \\ e^{i\theta} \sin \psi/2 \end{pmatrix}$$ defines a pure state density matrix through its projection operator, $$\bbox[yellow]{ |\Psi\rangle \langle \Psi | = \begin{pmatrix} \cos^2 \psi/2 & \sin \psi/2 ~ \cos\psi/2 ~e^{-i\theta} \\ \sin \psi/2 ~ \cos\psi/2 ~e^{i\theta} & \sin^2 \psi/2 \end{pmatrix}=\rho }~. $$ Note the manifest invariance under over-all rephasing of $|\Psi\rangle$.

The general principles' expression of this idempotent hermitean density matrix is also, evidently, $$ \rho=\frac{1}{2}(1\!\! 1 + \hat n \cdot \vec \sigma) , $$ with $\hat n = (\sin \psi \cos \theta, \; \sin \psi \sin \theta, \; \cos \psi)^T. $

It is now obvious that $-i|0\rangle$ corresponds to the same point of the sphere as $|0\rangle$, its rephasing by an over-all angle angle of $\pi/2$.

On the Bloch sphere, Y has sent (rotated by π/2) the x-axis (1,0,0) to the z-axis (0,0,1).

Cosmas Zachos
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