Rotation which diagonalizes the Hamiltonian

Question

I stumbled upon the following question:

Given the Hamiltonian of a spin-$1/2$ particle $$\hat{H}=\epsilon\begin{pmatrix} 0 & -e^{i\pi/4}\\ -e^{-i\pi/4} & 0 \end{pmatrix} = \frac{2\epsilon}{\hbar} \vec{S} \cdot \frac{\hat{y}-\hat{x}}{\sqrt{2}}$$

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 $\mathcal{D(\hat{n},\theta})=\exp\left[-\frac{i}{\hbar}\vec{S}\cdot \left(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\right)\right]$ (rotation operator) to check if it coincides with $U$ in the relevant basis, but it seems too exhausting. Perhaps I miss something trivial?

Answer 1

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, $$ e^{-i{\pi\over 4}\vec \sigma \cdot { (\hat x + \hat y) \over \sqrt{2}} } = \cos (\pi /4) -i \vec{\sigma}\cdot \frac{(\hat x +\hat y)} {\sqrt{2}} ~ \sin (\pi/4)\\ = \frac{1}{\sqrt{2}} (I -i(\sigma_x+\sigma_y)/\sqrt{2})=U. $$

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.

Answer 2

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} H\boldsymbol{=}\alpha\left(\sigma_{y}\boldsymbol{-}\mathrm \sigma_{x}\right)\,,\quad \alpha\boldsymbol{=}\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{=}\alpha\left(\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 c\in \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} \Vert\mathbf{h'}\Vert\boldsymbol{=}\Vert\mathbf{h}\Vert \quad \boldsymbol{\Longrightarrow} \quad c\boldsymbol{=}\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{=} \cos\frac{\theta}{2}\boldsymbol{-}i(\mathbf{n} \boldsymbol{\cdot} \boldsymbol{\sigma})\sin\frac{\theta}{2}\boldsymbol{=}\dfrac{\sqrt{2}}{2}\left(I\boldsymbol{-}i\,\dfrac{\sigma_x\boldsymbol{+}\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(I\boldsymbol{-}i\,\dfrac{\sigma_x\boldsymbol{+}\sigma_y}{\sqrt{2}}\right)\,\left(\sigma_y\boldsymbol{-}\sigma_x\vphantom{\dfrac{\sigma_x\boldsymbol{+}\sigma_y}{\sqrt{2}}} \right)\,\dfrac{\sqrt{2}}{2}\left(I\boldsymbol{+}i\,\dfrac{\sigma_x\boldsymbol{+}\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 $\mathbf{h}\boldsymbol{=}\alpha\left(\mathbf{j}\boldsymbol{-}\mathbf{i}\right)$ 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 $H\boldsymbol{=}\alpha\left(\sigma_{y}\boldsymbol{-}\mathrm \sigma_{x}\right)$ 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}= \cos\frac{\phi}{2}-i(\mathbf{m} \boldsymbol{\cdot} \boldsymbol{\sigma})\sin\frac{\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_y\boldsymbol{-}\sigma_x\vphantom{\dfrac{\sigma_x\boldsymbol{+}\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}