Chapter 3 - Physics of Quantum Information

Rotations of Bloch Vectors

As shown above, the solution to the Schrodinger equation for the density operator is (see Eq.(3.11))

$\rho (t)=U(t)\rho (0)U^{\dagger }(t).\,\!$

In general an open system will evolve according to

$\rho =U\rho _{0}U^{\dagger },\,\!$

whether or not the time dependence is explicitly taken into account. When the density operator is represented using the Bloch vector, the vector is rotated by the unitary transformation. This is seen through an explicit calculation.

There are two ways to see this. One is to simply act with the matrices in the Euler angle parameterization in Section C.5.1 one each of the Pauli matrices to show that indeed,

	$U{\vec {m}}\cdot {\vec {\sigma }}U^{\dagger }=\sum _{ij}m_{i}R_{ij}\sigma _{j}.\,\!$	(3.40)

This is easily seen to be a standard rotation matrix. (See for example http://en.wikipedia.org/wiki/Rotation_matrix.)

Another way to do this is to take

$\rho _{0}={\frac {1}{2}}(\mathbb {I} +{\vec {m}}\cdot {\vec {\sigma }}),\,\!$

as in Eq.(3.33). (Recall ${\vec {m}}\cdot {\vec {sigma}}=\sum _{i}m_{i}\sigma _{i}\,\!$.) Now act on $\rho$ with $U\,\!$ as given in Section C.5.1 by the so-called adjoint action $\rho =U\rho _{0}U^{\dagger }\,\!$,

	$\rho =[\mathbb {I} \cos(\theta /2)-i{\vec {n}}\cdot {\vec {\sigma }}\sin(\theta /2)]{\frac {1}{2}}(\mathbb {I} +{\vec {m}}\cdot {\vec {\sigma }})[\mathbb {I} \cos(\theta /2)+i{\vec {n}}\cdot {\vec {\sigma }}\sin(\theta /2)].\,\!$	(3.41)

To do this calculation explicitly, it helps (but is not necessary) to use the following identity,

	$\sum _{i}\epsilon _{ijk}\epsilon _{ilm}=\delta _{jl}\delta _{km}-\delta _{jm}\delta _{kl}.\,\!$	(3.42)

Then, if one only considers the non-trivial part of the density operator, ${\vec {m}}\cdot {\vec {\sigma }}\,\!$, the result is

	$e^{-i{\vec {n}}\cdot {\vec {\sigma }}\theta /2}{\vec {m}}\cdot {\vec {\sigma }}e^{-i{\vec {n}}\cdot {\vec {\sigma }}\theta /2}={\vec {m}}\cdot {\vec {\sigma }}\cos(\theta )+({\vec {n}}\cdot {\vec {m}})({\vec {n}}\cdot {\vec {\sigma }})(1-\cos(\theta ))+({\vec {n}}\times {\vec {m}})\cdot {\vec {\sigma }}\sin(\theta ),\,\!$	(3.43)

or

	${\begin{aligned}e^{-i{\vec {n}}\cdot {\vec {\sigma }}\theta /2}{\vec {m}}\cdot {\vec {\sigma }}e^{-i{\vec {n}}\cdot {\vec {\sigma }}\theta /2}&={\frac {1}{2}}{\vec {m}}\cdot {\vec {\sigma }}\cos(\theta )+{\frac {1}{2}}({\vec {n}}\cdot {\vec {m}})({\vec {n}}\cdot {\vec {\sigma }})\cos(\theta )+({\vec {n}}\cdot {\vec {m}})({\vec {n}}\cdot {\vec {\sigma }})\\&\;\;\;\;+({\vec {n}}\times {\vec {m}})\cdot {\vec {\sigma }}\sin(\theta )+{\frac {1}{2}}[({\vec {n}}\times {\vec {m}})\times {\vec {n}}]\cdot {\vec {\sigma }}\cos(\theta )\end{aligned}}\,\!$	(3.44)

where

	$({\vec {n}}\times {\vec {m}})\cdot {\vec {\sigma }}=\sum _{ijk}\epsilon _{ijk}n_{i}m_{j}\sigma _{k}.\,\!$	(3.45)

Therefore, the result of the action of $U\,\!$ is to produce, from ${\vec {m}}\,\!$, the vector

	${\vec {m}}^{\prime }={\vec {m}}\cos(\theta )+({\vec {n}}\cdot {\vec {m}}){\vec {n}}(1-\cos(\theta ))+({\vec {n}}\times {\vec {m}})\sin(\theta ).\,\!$	(3.46)

This equation can be interpreted as follows. We consider three components of the vector, the part along the axis of rotation and the two parts in the plane perpendicular to the axis of rotation. The part of the vector along the axis of rotation $({\vec {m}}\cdot {\vec {n}}){\vec {n}}\,\!$ does not change. The parts perpendicular to $({\vec {m}}\cdot {\vec {n}}){\vec {n}}\,\!$ change just like a vector rotated in a plane, but these parts are rotated in the plane perpendicular to the rotation axis and sitting at the end of the vector $({\vec {m}}\cdot {\vec {n}}){\vec {n}}\,\!$. It takes a bit of geometry and vector algebra to show this is the case.