Rotations in Bloch Sphere, and Free Parameters of a Qubit

modulus
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This question is mostly about group theory but I would like to understand it in the context of qubits rotating in a Bloch Sphere.

What my understanding of things are right now:
In the rotation Lie Group ##SO(3)##, we have three free parameters (##\frac{n(n-1)}{2}##), and this is also why we end up with three generators for the group. The two dimensional complex matrix representation of ##SO(3)## (which is also a representation to the ##SU(2)## group) also has three generators, namely the Pauli matrices.

What the problem is about:
I can't understand how these facts relate to the number of free parameters in the 'representation space' on which these matrix representations act. For example, the representation space for the 3x3 matrices representing ##SO(3)## would be the column vectors spanned by the standard basis ##e_1, e_2, e_3##.

How it relates to qubits:
If we consider a general complex 2-vector (in Dirac notation):
\begin{equation}
r_{1}e^{\iota\phi_{1}}{|{0}>} + r_{2}e^{\iota\phi_{2}}{|{1}>}
\end{equation}
and use the idea that the global phase of the state is of no significance, we end up with a three parameter representation:
\begin{equation}
r_{1}{|{0}>} + r_{2}e^{\iota\phi^{'}}{|{1}>}
\end{equation}
the free parameters being ##r_1, r_2## and ##\phi^{'}##. Now, if we add the condition of the normalization of the qubit, we get another constraint on these three parameters, and we end up with inly two free parameters. But this is enough to place the qubit exactly on the Bloch Sphere - ##\theta## is the azimuthal angle and ##\phi## is the polar angle. (the surface of the Bloch Sphere is fixed, so we only need two angular spherical coordinates and no ##r##).

What the problem is:
All single qubit quantum gates are unitary and thus belong to the group ##U(2)## which has three free parameters. If the quantum gates are to perform rotations on these qubits, shouldn't the qubits - being the representation space of the group of quantum gates - have three parameters (not two)? Is it possible that the Bloch Sphere can't represent all possible qubits?
 
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modulus said:
This question is mostly about group theory but I would like to understand it in the context of qubits rotating in a Bloch Sphere.

What my understanding of things are right now:
In the rotation Lie Group ##SO(3)##, we have three free parameters (##\frac{n(n-1)}{2}##), and this is also why we end up with three generators for the group. The two dimensional complex matrix representation of ##SO(3)## (which is also a representation to the ##SU(2)## group) also has three generators, namely the Pauli matrices.

What the problem is about:
I can't understand how these facts relate to the number of free parameters in the 'representation space' on which these matrix representations act. For example, the representation space for the 3x3 matrices representing ##SO(3)## would be the column vectors spanned by the standard basis ##e_1, e_2, e_3##.

How it relates to qubits:
If we consider a general complex 2-vector (in Dirac notation):
\begin{equation}
r_{1}e^{\iota\phi_{1}}{|{0}>} + r_{2}e^{\iota\phi_{2}}{|{1}>}
\end{equation}
and use the idea that the global phase of the state is of no significance, we end up with a three parameter representation:
\begin{equation}
r_{1}{|{0}>} + r_{2}e^{\iota\phi^{'}}{|{1}>}
\end{equation}
the free parameters being ##r_1, r_2## and ##\phi^{'}##. Now, if we add the condition of the normalization of the qubit, we get another constraint on these three parameters, and we end up with inly two free parameters. But this is enough to place the qubit exactly on the Bloch Sphere - ##\theta## is the azimuthal angle and ##\phi## is the polar angle. (the surface of the Bloch Sphere is fixed, so we only need two angular spherical coordinates and no ##r##).

What the problem is:
All single qubit quantum gates are unitary and thus belong to the group ##U(2)## which has three free parameters. If the quantum gates are to perform rotations on these qubits, shouldn't the qubits - being the representation space of the group of quantum gates - have three parameters (not two)? Is it possible that the Bloch Sphere can't represent all possible qubits?
Qubits in a pure state are points on the Block sphere. Two parameters needed to fix a point on the sphere. Qubits in possibly mixed states are points *in* the "Bloch ball". Three parameters. A unitary operation works on qubits by rotating the sphere. Three parameters needed: two for the direction *about which* the rotation takes place, and one more for the amount of the rotation.
 
Three parameters. A unitary operation works on qubits by rotating the sphere. Three parameters needed: two for the direction *about which* the rotation takes place, and one more for the amount of the rotation.

This is precisely what is troubling me. We need three parameters to define the rotation, and yet the object we are rotating has only two free parameters. Shouldn't the number of free parameters for both be the same? A greater number of parameters would imply a greater number of objects; if the number of possible rotations exceeds the number of possible vectors you can rotate, aren't some of the rotations redundant?

Also, given an axis and angle of rotation, I know how to construct a rotation in 3-dimensional space (and its corresponding 3x3 matrix), but the corresponding construction in the Bloch Sphere and it's relation to a 2x2 unitary matrix is unclear.

Qubits in a pure state are points on the Block sphere. Two parameters needed to fix a point on the sphere. Qubits in possibly mixed states are points *in* the "Bloch ball".

How is it that we represent mixed states using only three parameters - wouldn't you require three parameters for every contributing pure state?
 
Why should a description of how to modify a state have the same number of parameters as the state itself?

Consider the space of functions that operate on an ##n## bit number. There are way, way more than ##n## (the number of parameters for the space) or ##2^n## (the size of the state space) such functions. Instead, there's ##(2^n)^{(2^n)}## of them!

A point on the surface of a sphere takes two parameters to describe. Rotations take three parameters to describe. The operations have more degrees of freedom than the state. So what? The operations are still all distinct. They're still all linear and unitary. I don't see the problem.
 
Strilanc said:
Consider the space of functions that operate on an ##n## bit number. There are way, way more than ##n## (the number of parameters for the space) or ##2^n## (the size of the state space) such functions. Instead, there's ##(2^n)^{(2^n)}## of them!

Okay, I understand now. There is no reason for the representation space and the group's representation to be required to have the same number of parameters. I don't know why I felt that was necessary.

Thanks for the example, I really needed it.
 
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