Rotation which diagonalizes the Hamiltonian

Question

I stumbled upon the following question:

Given the Hamiltonian of a spin-$1/2$ particle
$$hat{H}=epsilonbegin{pmatrix} 0 & -e^{ipi/4} 
 -e^{-ipi/4} & 0 end{pmatrix} = frac{2epsilon}{hbar} vec{S} cdot frac{hat{y}-hat{x}}{sqrt{2}}$$
what is the rotation transformation that diagonalizes $hat{H}$? Find the angle of rotation $theta$ and the axis of rotation $hat{n}$.

Finding the matrix which diagonalizes $hat{H}$ is not particularly difficult. For instance,
$$U=frac{1}{sqrt{2}}begin{pmatrix}
1 & -frac{1+i}{sqrt{2}} 
frac{1-i}{sqrt{2}} & 1
end{pmatrix}$$
does the job. However it is then claimed that this matrix corresponds to a transformation through an angle $theta=pi/2$ about $hat{n}=(hat{x}+hat{y})/sqrt{2}$. But I'm not quite sure how this can be immediately inferred from the entries of $U$. Moreover, I don't think that $U$ can be decomposed into a sum of $sigma_x$ and $sigma_y$ Pauli matrices. I thought about directly calculating $mathcal{D(hat{n},theta})=expleft[-frac{i}{hbar}vec{S}cdot left(frac{hat{x}+hat{y}}{sqrt{2}}right)right]$ (rotation operator) to check if it coincides with $U$ in the relevant basis, but it seems too exhausting. Perhaps I miss something trivial?

Cosmas Zachos · Accepted Answer

Just use the standard exponentiation of Pauli matrices, knowing that $vec S =hbar vec sigma /2$ for the doublet representation,  which halves the rotation angles,
$$
e^{-i{piover 4}vec sigma cdot { (hat x + hat y) over sqrt{2}} } = 
 cos (pi /4) -i vec{sigma}cdot frac{(hat x +hat y)}  {sqrt{2}} ~ sin (pi/4) =
 frac{1}{sqrt{2}}  (I -i(sigma_x+sigma_y)/sqrt{2})=U.
$$
It is analogous to Euler's formula.
I gather you have done the diagonalization algebra utilizing the properties of Pauli matrices: hardly any calculation!

It might help your intuition to consider the two orthogonal unit vectors $hat y pm hat x$ and to rotate the second by a right angle around the first: it will take you along $hat z$, so a diagonal $sigma_z$. Conversely, knowing $sigma_z$ is the only diagonal Pauli matrix, how do you rotate $hat y  - hat x$ to $hat z$? Obviously by a π/2 rotation around their cross product as an axis! Draw the figure.

Frobenius · Answer

REFERENCE :  My answer on How does the Hamiltonian changes after rotating the coordinate frame.
$boldsymbol{=!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!=}$
Note : In the following for the unit vectors along the coordinate axes $hat{x},hat{y},hat{z}$ I use the symbols $mathbf{i},mathbf{j},mathbf{k}$ respectively.
$boldsymbol{=!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!==!=!=!=}$
The Hamiltonian is the following hermitian traceless matrix
begin{equation}
Hboldsymbol{=}alphaleft(sigma_{y}boldsymbol{-}mathrm sigma_{x}right),,quad alphaboldsymbol{=}dfrac{sqrt{2}epsilon}{hbar}
tag{01}label{01}
end{equation}
From the bijection between hermitian traceless matrices and real 3-vectors (discussed in paragraph ''The reasoning'' of aforementioned REFERENCE) the representative real 3-vector of this Hamiltonian is
begin{equation}
mathbf{h}boldsymbol{=}alphaleft(mathbf{j}boldsymbol{-}mathbf{i}right)
tag{02}label{02}
end{equation}
as shown in Figure-01.

If the Hamiltonian $H'$ of equation eqref{01} must be transformed to a diagonal one $H'$ then we must have
begin{equation}
H'boldsymbol{=}c,sigma_{z} ,,quad cin mathbb{R}
tag{03}label{03}
end{equation}
Above expression is justified because not only  $sigma_{z}$ is a diagonal hermitian matrix but moreover is traceless ($H'$ must be traceless since trace is invariant under similarity transformations).
To the transformed diagonal hermitian traceless matrix $H'$ there corresponds the representative real 3-vector
begin{equation}
mathbf{h'}boldsymbol{=}c,mathbf{k}
tag{04}label{04} 
end{equation}
If the transformation must be a rotation then the vector $mathbf{h'}$ of equation eqref{04} will be the image of the vector $mathbf{h}$ of equation eqref{02} so
begin{equation}
Vertmathbf{h'}Vertboldsymbol{=}Vertmathbf{h}Vert quad boldsymbol{Longrightarrow} quad cboldsymbol{=}sqrt{2},alpha
tag{05}label{05} 
end{equation}
that is
begin{align}
H'&boldsymbol{=}sqrt{2},alpha,sigma_{z}
tag{06a}label{06a}
mathbf{h'} &boldsymbol{=}sqrt{2},alpha,mathbf{k}
tag{06b}label{06b} 
end{align}
as shown in Figure-01.
The most simple rotation that brings the vector $mathbf{h}$ on vector  $mathbf{h'}$ is around  a unit vector $mathbf{n}$ through an angle $theta$ given by
begin{align}
mathbf{n}&boldsymbol{=}dfrac{mathbf{i}boldsymbol{+}mathbf{j}}{sqrt{2}}
tag{07a}label{07a}
theta &boldsymbol{=}dfrac{pi}{2}
tag{07b}label{07b}  
end{align}
shown in Figure-01. This rotation is represented by the following special unitary matrix $SU(2)$
begin{equation}
boxed{::U_{mathbf{n} ,theta}boldsymbol{=} cosfrac{theta}{2}boldsymbol{-}i(mathbf{n}   boldsymbol{cdot}  boldsymbol{sigma})sinfrac{theta}{2}boldsymbol{=}dfrac{sqrt{2}}{2}left(Iboldsymbol{-}i,dfrac{sigma_xboldsymbol{+}sigma_y}{sqrt{2}}right)vphantom{dfrac{dfrac{a}{b}}{dfrac{a}{b}}}::}
tag{08}label{08}
end{equation}
It could be verified easily, using the properties of Pauli matrices, that $U_{mathbf{n} ,theta}$ diagonalizes the Hamiltonian $H$, that is
begin{equation}
U_{mathbf{n} ,theta},H,U^{*}_{mathbf{n} ,theta}boldsymbol{=}H'
tag{09}label{09} 
end{equation}
or explicitly
begin{equation}
dfrac{sqrt{2}}{2}left(Iboldsymbol{-}i,dfrac{sigma_xboldsymbol{+}sigma_y}{sqrt{2}}right),left(sigma_yboldsymbol{-}sigma_xvphantom{dfrac{sigma_xboldsymbol{+}sigma_y}{sqrt{2}}} right),dfrac{sqrt{2}}{2}left(Iboldsymbol{+}i,dfrac{sigma_xboldsymbol{+}sigma_y}{sqrt{2}}right)boldsymbol{=}sqrt{2},sigma_z
tag{10}label{10} 
end{equation}
Note that as there exist infinitely many rotations that bring the vector $mathbf{h}boldsymbol{=}alphaleft(mathbf{j}boldsymbol{-}mathbf{i}right)$ of equation eqref{02} to the vector $mathbf{h'}boldsymbol{=}sqrt{2},alpha,mathbf{k}$ of equation eqref{06b}, so there are  infinitely many unitary matrices like that of equation eqref{08} which  diagonalize the Hamiltonian $Hboldsymbol{=}alphaleft(sigma_{y}boldsymbol{-}mathrm sigma_{x}right)$ of equation eqref{01}. For example, a rotation around a unit vector $mathbf{m}$ through an angle $phi$ given by
begin{align}
mathbf{m}&boldsymbol{=}dfrac{boldsymbol{-}mathbf{i}boldsymbol{+}mathbf{j}boldsymbol{+}sqrt{2}mathbf{k}}{2}
tag{11a}label{11a}
phi &boldsymbol{=}pi
tag{11b}label{11b}  
end{align}
as shown in Figure-02 diagonalizes the Hamiltonian. The corresponging special unitary matrix is
begin{equation}
boxed{::U_{mathbf{m} ,phi}= cosfrac{phi}{2}-i(mathbf{m}  boldsymbol{cdot}  boldsymbol{sigma})sinfrac{phi}{2}=dfrac{i}{2}left(  sigma_x-sigma_y-sqrt{2}sigma_z right)vphantom{dfrac{dfrac{a}{b}}{dfrac{a}{b}}}:: }
tag{12}label{12}
end{equation}
Again, it could be verified easily, using the properties of Pauli matrices, that $U_{mathbf{m} ,phi}$ diagonalizes the Hamiltonian $H$, that is
begin{equation}
U_{mathbf{m} ,phi},H,U^{*}_{mathbf{m} ,phi}boldsymbol{=}H'
tag{13}label{13} 
end{equation}
or explicitly
begin{equation}
left[dfrac{i}{2}left(sigma_x-sigma_y-sqrt{2}sigma_z right)right],left(sigma_yboldsymbol{-}sigma_xvphantom{dfrac{sigma_xboldsymbol{+}sigma_y}{sqrt{2}}} right),left[boldsymbol{-}dfrac{i}{2}left(sigma_x-sigma_y-sqrt{2}sigma_z right)right]boldsymbol{=}sqrt{2},sigma_z
tag{14}label{14} 
end{equation}

Rotation which diagonalizes the Hamiltonian

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