Complementing a Qubit in the Bloch-Sphere

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In summary, the paper states that the complement of a qubit is impossible to flip, but that you can get back to the original representation by multiplying a state by ##e^{ix}##.
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Ruik
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Hi :-)

for my master thesis I'm working with qubits in the Bloch-sphere representation ##|q\rangle = cos(\frac{\theta}{2})|0\rangle + e^{i\phi}sin(\frac{\theta}{2})|1\rangle##.
Side question: why is only the second amplitude complex?

But let's move to my main question. I need to know how the complement qubit is written in this representation; so the qubit which is placed at the opposite side of the bloch sphere. There is a paper about this problem, that states that it is impossible to flip an unknown qubit ideally but that's not what I want. There they say, that the function complementing a qubit would be
##NOT(\alpha |0\rangle + \beta |1\rangle)= \beta^*|0\rangle - \alpha^*|1\rangle##

That would lead to:
##|q^\perp\rangle = e^{-i\phi}sin(\frac{\theta}{2})|0\rangle - cos(\frac{\theta}{2})|1\rangle##

Now, how do I get back to the representation where the first factor is a Cosinus function an the second factor is the only complex factor?
Is there a representation like ##|q^\perp\rangle = cos(\frac{\theta'}{2})|0\rangle + e^{i\phi'}sin(\frac{\theta'}{2})|1\rangle## with a mapping ##(\theta,\phi) \rightarrow (\theta',\phi')## ?

Looking forward to your answers!

Edit: I found the word for what I mean: Antipodes xD For a given point on the Blochsphere I want the representation of the antipode of this point.
 
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Only one of the terms needs a phase factor because only relative phase factors matter. Keeping track of both can be mathematically convenient, and does matter when entangled with other qubits (because your global phase factors become relative as more cases are added), but is not necessary in this simple case.

Because global phase factors don't matter, you can always multiply a quantum state by ##e^{ix}## for any ##x## you want. So ##|q^\perp\rangle = e^{-i\phi}sin(\frac{\theta}{2})|0\rangle - cos(\frac{\theta}{2})|1\rangle## is congruent to ##|q^\perp\rangle = sin(\frac{\theta}{2})|0\rangle - e^{i\phi}cos(\frac{\theta}{2})|1\rangle##.

There is no operation that takes the complement of a state. Quantum operations are always unitary, but you can't represent the complement operation as a unitary matrix.

I'm pretty sure that assuming you can perform the complement gives you super powers, like erasing qubits and performing FTL communication, but I can't seem to recall how to make that happen.
 
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The fact that I can always multiply a quantum state by ##e^{ix}## is very interesting and useful, thanks!

I came to another solution, by taking the 3D-Koordinates of a point on the Bloch sphere ##
\left(\begin{array}{c}
sin(\theta)cos(\phi)\\
sin(\theta)sin(\phi)\\
cos(\theta)\\\end{array}\right)
##

negating them to ##
\left(\begin{array}{c}
-sin(\theta)cos(\phi)\\
-sin(\theta)sin(\phi)\\
-cos(\theta)\\\end{array}\right) =
\left(\begin{array}{c}
sin(\theta+\pi)cos(\phi)\\
sin(\theta+\pi)sin(\phi)\\
cos(\theta+\pi)\\\end{array}\right)
##
wich lead me to ##|q^\perp\rangle=cos\left(\frac{\theta}{2}+\frac{\pi}{2}\right)|0\rangle+e^{i\phi}sin\left(\frac{\theta}{2}+\frac{\pi}{2}\right)|1\rangle##
wich is the same solution if you switch ##sin## and ##cos## functions and add a phase factor of ##e^{i\pi}=-1##

But I'm not yet sure if the global phase is really irrelevant for me. On the one hand I'm working with two qubits and apply an operation on both of them, on the other hand, in the beginning they are totally unentangled. Can I take these two pure states and ignore their initial global phase because they are pure at that moment, or is the global phase really just ignorable, if I work with one qubit all the time?

Maybe you know some literature about global and relative phase? I don't think I really have a mental concept of what a phase really is yet.
 

FAQ: Complementing a Qubit in the Bloch-Sphere

What is a qubit?

A qubit is the basic unit of quantum information and is a two-state quantum system. It can exist in a superposition of both states, unlike classical bits which can only exist in one of two states (0 or 1).

What is the Bloch sphere?

The Bloch sphere is a geometric representation of the state of a single qubit. It is a three-dimensional sphere with the poles representing the two basis states (0 and 1) and all other points representing the possible superposition states.

How do you complement a qubit in the Bloch sphere?

To complement a qubit in the Bloch sphere, you can either rotate it by 180 degrees around the x-axis or invert its state, flipping it from 0 to 1 or vice versa.

What is the significance of complementing a qubit in the Bloch sphere?

Complementing a qubit in the Bloch sphere is important in quantum computing as it allows for operations such as logical NOT and phase-flipping, which are essential for performing calculations and creating quantum algorithms.

Can a qubit exist in a state other than 0 or 1 on the Bloch sphere?

Yes, a qubit can exist in a superposition of states on the Bloch sphere, meaning it can exist in any point on the sphere other than the poles representing 0 and 1. This allows for a larger range of possible states and makes quantum computing more powerful than classical computing.

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