How to prove that antipodal points on the Bloch sphere are orthogonal?

Question

I started by assuming two antipodal states $$ |(\theta,\psi)\rangle = \cos\dfrac{\theta}{2}|0\rangle + \sin\dfrac{\theta}{2}e^{i\psi}|1\rangle\\ |(\theta+\pi,\psi+\pi)\rangle= \cos\dfrac{\theta+\pi}{2}|0\rangle + \sin\dfrac{\theta+\pi}{2}e^{i(\psi+\pi)}|1\rangle $$ and then try to take the inner product of them. However, the math doesn't check out (as in the result I get $\ne0$). What is wrong with my deduction?

Answer 1

In spherical coordinates antipodal point to $(\theta,\psi)$ is $(\theta+\pi,\psi)$, not $(\theta+\pi,\psi+\pi)$

Answer 2

In the most widespread convention, the Bloch sphere uses $\theta = 0$ radians latitude to indicate the north pole $|0\rangle$, $\theta = \pi$ to refer to the south pole $|1\rangle$ and $\theta = \pi/2$ to refer to the equator, which includes the superpositions $(1+i)/\sqrt 2$ and $(1-i)/\sqrt 2$ as well as $i|1\rangle$ and $-i|1\rangle$.

If two great circles are distinct and intersect at a given point they will intersect exactly one other time at the opposite point on the sphere and such that a ray passing through both points will also pass through the center of the sphere. Given this, the difference in latitude between one point and the equator will be of equal magnitude but opposite sign relative to the other, with the equator being at $\pi/2$. As such, the latitude of the antipode to $(\theta,\psi)$ will be $\pi-\theta$. As there are $2\pi$ radians of longitude to a sphere, the opposite of a given longitude with be $\pi+\psi$.

This being the case, coordinates for the antipode $(\theta,\psi)$ is $(\pi-\theta,\pi+\psi)$. Trigonometric identities will be sufficient for the rest.

Answer 3

Attaching my answer for finding the antipodal points on the Bloch sphereution in an alternate way

R Sridhar