- #1
Ruik
- 6
- 0
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.
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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