Diagonalization of 2x2 Hermitian matrices using Wigner D-Matrix

[Max1]

TL;DR Summary

A Hermitian 2x2 matrix can be diagonalized by a similarity transform with a $\mathrm{SU}(2)$ matrix. This $\mathrm{SU}(2)$ can be represented by a 2x2 Wigner D-Matrix depending on three real parameters $(\alpha,\beta,\gamma)$. Is there an explicit formula for the parameters?

Motivation:

Due to the spectral theorem a complex square matrix $H\in \mathbb{C}^{n\times n}$ is diagonalizable by a unitary matrix iff $H$ is normal ($H^\dagger H=HH^\dagger$). If H is Hermitian ($H^\dagger=H$) it follows that it is also normal and can hence be diagonalized by a unitary transformation. That means
$$
U^\dagger HU=D
$$
with $D$ a diagonal matrix with the eigenvalues of $H$ on its diagonal. According to [General form for 2x2 unitary matrices] a unitary matrix can also be written as
$$
U=\mathrm{e}^{i\phi/2}S
$$
with $S \in \mathrm{SU}(2)$ (the group of two dimensional unitary matrices with $\mathrm{det}(S)=1$). The phase cancels anyway
$$
U^\dagger HU=\mathrm{e}^{-i\phi/2}S^\dagger H\mathrm{e}^{i\phi/2}S=S^\dagger HS\,.
$$
So a unimodular ($\mathrm{det}(S)=1$) unitary matrix S exists, which diagonalizes H. Following Sakurai "Modern Quantum Mechanics", any element of $\mathrm{SU}(2)$ can be represented by
$$
D(\mathbf{n},\phi)=\exp(-i\phi\,\mathbf{n}\cdot \pmb{\sigma}/2)=S
$$
with $\mathbf{n}\in \mathbb{R}^{3\times3}$, $\phi\in [0,4\pi)$ and $\pmb{\sigma}$ denoting the vector of Pauli matrices. From a physical point of view $D(\mathbf{n},\phi)$ can be interpreted as a rotation of a two-component spinor $\chi$ around the $\mathbf{n}$-axis by the angle $\phi$. The direction of $\mathbf{n}$ can be described by an azimuthal angle and a polar angle. Hence, in total three real parameters are necessary to describe $D(\mathbf{n},\phi)$. According to Sakurai page 179 the rotation in spin-space can also be represented by
$$
D(\alpha,\beta,\gamma)=
\begin{pmatrix}
\exp^{-i(\alpha+\gamma)/2}\cos(\beta/2) & -\exp^{-i(\alpha-\gamma)/2}\sin(\beta/2)\\
\exp^{i(\alpha-\gamma)/2}\sin(\beta/2) & \exp^{i(\alpha+\gamma)/2}\cos(\beta/2)
\end{pmatrix}
$$
and especially any $S\in SU(2)$ can be represented in the upper way. This representation is also known as the Wigner D-Matrix (with $j=1/2$ in this case, compare Wikipedia "Wigner D-Matrix").

Question:

I would like to diagonalize a given hermitian 2x2 Matrix $H$. How do I obtain the three angles $(\alpha,\beta,\gamma)$? I would like to have a forumla which gives the angles explicitly in terms of the matrix elements. I need this to diagonalize a Hamiltonian in spin space which does not commute with $S_\mathrm{z}$ and hence has some off-diagonal elements in $S_\mathrm{z}$ representation.

Motivation 2:

A real symmetric matrix ($A=A^\mathrm{T}$) can be diagonalized by an orthogonal matrix ($Q^\mathrm{T}=Q^{-1}$). Any orthgonal 2x2 matrix can be written as
$$
Q=
\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}\,.
$$
A real symmetric matrix is necessarily of the form
$$
A=
\begin{pmatrix}
a & c\\
c & a
\end{pmatrix}
$$
by calculating the eigenvectors and equating the matrix of eigenvectors with $Q$ one finds
$$
\theta=
\begin{cases}
\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right) & \text{if}\, a-b>0 \land c>0 \\
\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right)+\pi/2 & \text{if}\, a-b<0 \neq c \\
\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right)+\pi & \text{if}\, a-b>0 \land c<0 \\
\end{cases}
$$
In the general case and for the special cases
$$
\theta=
\begin{cases}
0 &\text{if}\, c=0 \land a-b>0 \\
\pi/2 &\text{if}\, c=0 \land a-b<0 \\
\pi/4 &\text{if}\, c>0 \land a-b=0 \\
3\pi/4 &\text{if}\, c<0 \land a-b=0
\end{cases}\,.
$$

I tried obtaining the three angles in the SU(2) case in the same way, but did not succeed in finding a nice and short representation as in the case above. Can anyone give me some pointers on where to find an explicit representation?

 

[New Simplicio]

Don't stuff three angles into a 2X2 matrix. Use Dirac Gamma Matrices.

 

[king vitamin]

Any Hermitian 2x2 matrix may be written
$$
H = a \, \mathbb{I} + b_x \sigma^x + b_y \sigma^y + b_z \sigma^z,
$$
where $a, b_x, b_y, b_z$ are real, and the $\sigma$'s are the Pauli matrices. The relation between these four constants and the four components of $H$ is very easy to derive. Now, without loss of generality, let's take $a = 0$, since the matrix $S$ that diagonalizes $H$ trivially diagonalizes $H - a \mathbb{I}$ as well. So we just need $S$ to diagonalize
$$
H = \mathbf{b} \cdot \pmb{\sigma}.
$$
Here, I have defined the vector $\mathbf{b} = (b_x,b_y,b_z)$.

We now consider conjugating $H$ by $S$:
$$
S^{\dagger} H S = \exp(i\phi\,\mathbf{n}\cdot \pmb{\sigma}/2) \left( \mathbf{b} \cdot \pmb{\sigma} \right) \exp(-i\phi\,\mathbf{n}\cdot \pmb{\sigma}/2).
$$
But expressions of this form are evaluated explicitly in Sakurai, section 3.2 (in my copy anyways... my page numbers don't seem to agree with your so maybe you'll have to find it yourself). In particular, the final result is
$$
S^{\dagger} H S = \sum_{i,j = x,y,z} \sigma_i R_{i,j} b_j,
$$
where $R_{i,j}$ is the ordinary SO(3) rotation matrix which rotates the vector $\mathbf{b}$ counterclockwise about the axis $\mathbf{n}$ by an angle $\phi$. If we want $S^{\dagger}HS$ to be diagonal, clearly we want the right-hand side to only contain $\sigma^z$, so we want the rotation which rotates $\mathbf{b}$ to the vector $|\mathbf{b}| \hat{z}$.

Hopefully this is enough to get you going. The key point is that the Hamiltonian can be written as a vector in spin space, and then you take the rotation which aligns the spin Hamiltonian along the diagonal direction (which is the $z$ direction by convention).

 

[Max1]

Thanks for the replies and sorry for not responding sooner,

as stated in king vitamins answer I expressed the Hamiltonian as a vector in spin space. The Hamiltonian $H=\mathbf{b}\cdot\pmb{\sigma}$ can then be diagonalized by a similarity transform with the matrix of eigenvectors of $\mathbf{b}\cdot\pmb{\sigma}$. As demonstarted at the end of Sakurai section 3.2, the eigenvectors can easily be found by rotating the eigenvectors $|\sigma_\mathrm{z};\pm\rangle$ of $(0,0,1)^\mathrm{T}\cdot\pmb{\sigma}=\sigma_\mathrm{z}$ to the eigenvectors $|\pmb{\sigma}\cdot\mathbf{b};\pm\rangle$ of $\pmb{\sigma}\cdot\mathbf{b}$. This can be done by a counterclockwise rotation around the $\mathrm{y}$-Axis by the azimuthal angle $\theta$ followed by a counterclockwise rotation by the polar angle $\varphi$ around the $\mathrm{z}$-Axis. The eigenvectors are then given by
$$
|\pmb{\sigma}\cdot\mathbf{b};\pm\rangle=
\exp(-\mathrm{i}\sigma_\mathrm{z}\phi/2)
\exp(-\mathrm{i}\sigma_\mathrm{y}\theta/2)
|\sigma_\mathrm{z};\pm\rangle
$$
In the Pauli two-component formalism the resulting eigenvectors can be written as
$$
|\sigma_\mathrm{z};+\rangle=
\begin{pmatrix}
\cos(\theta/2)\mathrm{e}^{-\mathrm{i}\phi/2}\\
\sin(\theta/2)\mathrm{e}^{\mathrm{i}\phi/2}\\
\end{pmatrix}
$$
and
$$
|\sigma_\mathrm{z};-\rangle=
\begin{pmatrix}
-\sin(\theta/2)\mathrm{e}^{-\mathrm{i}\phi/2}\\
\cos(\theta/2)\mathrm{e}^{\mathrm{i}\phi/2}\\
\end{pmatrix}\,.
$$
Hence the matrix of eigenvectors is given by
$$
S(\theta,\phi)=
\begin{pmatrix}
\cos(\theta/2)\mathrm{e}^{-\mathrm{i}\phi/2}&-\sin(\theta/2)\mathrm{e}^{-\mathrm{i}\phi/2}\\
\sin(\theta/2)\mathrm{e}^{\mathrm{i}\phi/2}&\cos(\theta/2)\mathrm{e}^{\mathrm{i}\phi/2}\\
\end{pmatrix}
$$
The angles can be obtained by the usual formulas for spherical coordinates, Wikipedia "Spherical coordinate system":
$$
\theta=\arccos{b_\mathrm{z}/|\mathbf{b}|},\;\;\;\;\phi=\operatorname{atan2}(b_\mathrm{y},b_\mathrm{z})\,.
$$
Just to make sure I also diagonalized
$$
\pmb{\sigma}\cdot\mathbf{n}=
\begin{pmatrix}
n_3&& n_1-\mathrm{i}n_2\\
n_1+\mathrm{i}n_2&&-n_3
\end{pmatrix}
$$
with $\mathbf{n}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)^\mathrm{T}$ expressed in spherical coordinates and obtained the same eigenvectors (up to phase and normalization) in the end. As stated in the posts before the diagonalform of the Hamiltonian is given by:
$$
D=S^{\dagger}HS
$$
As stated in king vitamin's answer any Hermitian matrix can be written as $H=\mathbb{I}a+\pmb{\sigma}\cdot\mathbf{b}$. Hence, as far as I can see, two real angles are enough to parametrize a unitary matrix diagonalizing any 2x2 Hermitian matrix. Is this really true or did I miss something? Obviously I'm chosing a particular phase of the eigenvectors with only two real angles, but that does not matter for diagonalization.

Further question: If I define the set of matrices which diagonalize $H$ as an equivalence class, with each matrix in the class that gives the same $D$. Then my particular matrix $S(\theta,\phi)$ would be a representative of some class. Will I get some kind of well known group if I consider the set of all these eqivalence classes, which diagonalize a Hermitian 2x2 matrix, with matrix multiplication as group operation?

 

[king vitamin]

As stated in king vitamin's answer any Hermitian matrix can be written as H=Ia+σσ⋅bH=Ia+σσ⋅bH=\mathbb{I}a+\pmb{\sigma}\cdot\mathbf{b}. Hence, as far as I can see, two real angles are enough to parametrize a unitary matrix diagonalizing any 2x2 Hermitian matrix. Is this really true or did I miss something?

Yes, I would say this is correct.

Further question: If I define the set of matrices which diagonalize HHH as an equivalence class, with each matrix in the class that gives the same DDD. Then my particular matrix S(θ,ϕ)S(θ,ϕ)S(\theta,\phi) would be a representative of some class. Will I get some kind of well known group if I consider the set of all these eqivalence classes, which diagonalize a Hermitian 2x2 matrix, with matrix multiplication as group operation?

This is a great question, and I can't guarantee that I'll answer it in complete detail. To restate your question, if we consider the set of matrices which diagonalize $H$,
$$
M = \{ S \in SU(2) | D = S^{\dagger} H S \},
$$
you have shown that $M \in SU(2)$ is two-dimensional, whereas it is known that $SU(2)$ is three-dimensional. Therefore we know that $M \neq SU(2)$. So what is $M$?

A starting point is to compare your equation for a general element of $SU(2)$,
$$
D(\alpha,\beta,\nu)=
\begin{pmatrix}
\exp^{-i(\alpha+\nu)/2}\cos(\beta/2) & -\exp^{-i(\alpha-\nu)/2}\sin(\beta/2)\\
\exp^{i(\alpha-\nu)/2}\sin(\beta/2) & \exp^{i(\alpha+\nu)/2}\cos(\beta/2)
\end{pmatrix}
$$
with your equation for a particular (not general!) element of $M$,
$$
S(\theta,\phi)=
\begin{pmatrix}
\cos(\theta/2)\mathrm{e}^{-\mathrm{i}\phi/2}&-\sin(\theta/2)\mathrm{e}^{-\mathrm{i}\phi/2}\\
\sin(\theta/2)\mathrm{e}^{\mathrm{i}\phi/2}&\cos(\theta/2)\mathrm{e}^{\mathrm{i}\phi/2}\\
\end{pmatrix}.
$$
By inspection, we immediately see
$$
S(\theta,\phi) = D(\phi,\theta,0).
$$

Ok, the above will help us. We now use some intuition: after we have applied your particular $S$, which has made $H$ proportional to $\sigma^z$, what further $SU(2)$ transformations can we do which keep $H$ diagonal? The answer is clearly any rotation about the $z$-axis. That is, we expect a general matrix which diagonalizes $H$ takes the form
$$
\tilde{S}(\theta,\phi,\gamma) = S(\theta,\phi) e^{-i \gamma \sigma^z/2}.
$$
Evaluating this explicitly, we can quickly show that
$$
\tilde{S}(\theta,\phi,\gamma) = D(\phi - \gamma,\theta,\gamma),
$$
so this really does construct a general element of $SU(2)$. Therefore, the matrices $D(\phi - \gamma,\theta,\gamma)$, where $\theta$ and $\phi$ are fixed but $\gamma$ is arbitrary, is the full set of $SU(2)$ matrices which diagonalize $H$.

The formal mathematical statement is that the set of matrices of $SU(2)$ which diagonalize $H$ correspond to elements of $SU(2)$ modulo the $U(1)$ factor $e^{- i \gamma \sigma^z/2}$ multiplied on the right-hand side. That is, the space of matrices we seek is the so-called right coset
$$
M = SU(2)/U(1).
$$
Since $U(1)$ is not a normal subgroup of $SU(2)$, this coset is not a group. So a product of two matrices within $M$ will not in general be an element of $M$. (This is very reasonable if you think about this geometrically.) This manifold is topologically the two-sphere (and a quick Google tells me that mathematicians have proven that no Lie group can have this topology, which is consistent with our finding).

I'm sure someone more mathematically savvy than myself can give more properties of $M$, but this is about the limit of what I can say.