Bloch rotation matrix anomaly

Mike Ulrey Apr 28, 2023

By computing a simple Bloch rotation matrix by (what should be) two equivalent methods, we find a slight discrepancy.

Create a list of the Pauli matrices.

In[]:=

σs={σ1,σ2,σ3}=Normal[QuantumOperator[#]["Matrix"]]&/@{"PauliX","PauliY","PauliZ"};

Create a particular Bloch rotation in two different ways -- by using1. The definition

R

n

(θ)=Exp[-θ/2

n

.

σ

]

, and2. A linear combination of Pauli operators

n

.{QuantumOperator[{"RX",θ}],QuantumOperator[{"RY",θ}],QuantumOperator[{"RZ",θ}]}

.

bRotate1[θ_,n_]:=MatrixExp-

θ

2

n.σs;bRotate2[θ_,n_]:=n.{QuantumOperator[{"RX",θ}],QuantumOperator[{"RY",θ}],QuantumOperator[{"RZ",θ}]};

Now find the matrix representing each of the rotations.

rotateMat1=bRotate1θ,

1

3

{1,1,1}//FullSimplify;rotateMat2=ComplexExpandNormalbRotate2θ,

1

3

{1,1,1}["Matrix"];

Finally compare the rotation matrices. Note they are slightly different in the diagonal entries. Why??

In[]:=

MatrixForm/@{rotateMat1,rotateMat2}%/.{θ->90°}//Simplify

Out[]=



Cos θ 2 - Sin θ 2  3	- (1+)Sin θ 2  3
(1-)Sin θ 2  3	Cos θ 2 + Sin θ 2  3

,

3 Cos θ 2 - Sin θ 2  3	- (1+)Sin θ 2  3
(1-)Sin θ 2  3	3 Cos θ 2 + Sin θ 2  3



Out[]=



1 2 -  6	- 1+ 6
1- 6	1 2 +  6

,

3- 6	- 1+ 6
1- 6	3+ 6

