1
$\begingroup$

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?

$\endgroup$
1

2 Answers 2

2
$\begingroup$

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.
$\endgroup$
5
  • $\begingroup$ Thank you, I forgot about the "Euler formula" for Pauli matrices. But is there an intuitive way of doing the same procedure in the reverse order? After all, $\hat{n}$ and $\theta$ are actually the unknowns, whereas here we simply confirmed that $U$ indeed corresponds to a rotation with some given values of $\hat{n}$ and $\theta$ (which I looked up in the answer). $\endgroup$
    – grjj3
    Oct 1, 2020 at 20:09
  • $\begingroup$ Well, the bullet item tells you right away what your best guess for the angle and the axis are supposed to be, and you confirm them! Draw the three vectors suggested--nothing to do with matrices, beyond the fact that you wish to end up on the z-axis, since $\sigma_z$ is the only diagonal Pauli matrix! $\endgroup$ Oct 1, 2020 at 20:09
  • 1
    $\begingroup$ Thank you. Indeed, the diagonalized $\hat H$ is proportional to $\sigma_z$ hence the transformation that diagonalizes $\hat H$ must correspond to a rotation which takes it from $\hat y - \hat x$ to $\hat z$. This is possible only if we rotate $\hat y - \hat z$ around a perpendicular axis through a $90^{\circ}$ angle. $\endgroup$
    – grjj3
    Oct 1, 2020 at 20:30
  • $\begingroup$ Yes, the problem was chosen easy to just illustrate the principle. You'd have to be more creative if the initial vector were $\hat z - \hat x$ ! $\endgroup$ Oct 1, 2020 at 20:34
  • $\begingroup$ @Frobenius. Depends on what one understands by "this". Of course any vector may be rotated to $\hat z$. Creative means non-mess-aversive. $\endgroup$ Oct 2, 2020 at 0:27
2
$\begingroup$

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.

enter image description here

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}

enter image description here

$\endgroup$
3
  • $\begingroup$ Thank you for the detailed and illustrative answer. If I'm not mistaken, the fact that there exists a correspondence between traceless Hermitian 2x2 matrices and vectors is a reflection of the fact that there's an isometric Lie algebra isomorphism between the 3-dimensional Lie algebra of traceless Hermitian 2x2 matrices equipped with the determinant as a norm square, and the 3D space equipped with the standard vector cross product and the standard norm square. Correct me if I'm wrong, but there also exists a similar isomorphism between vectors and traceless anti-Hermitian 2x2 matrices. $\endgroup$
    – grjj3
    Oct 3, 2020 at 9:10
  • $\begingroup$ @grjj3 : I am not expert on this subject so I suggest to post a question in Mathematics Stack Exchange. By the way, I usually try to "discover" things on an elementary level (the trees) and after that to see them in the frame of a higher and more general level (the forest). The bijection was a thought of mine, I didn't find it in textbooks or the web, if any. $\endgroup$
    – Frobenius
    Oct 3, 2020 at 9:21
  • 1
    $\begingroup$ I believe it's merely a result of the fact that the real vector space of all 2x2 traceless Hermitian matrices is spanned by Pauli matrices. And since this three-dimensional vector space of matrices is isomorphic to the Euclidean space $\mathbb{R}^3$ there exists a one-to-one correspondence between ordinary 3D vectors and traceless Hermitian matrices. $\endgroup$
    – grjj3
    Oct 3, 2020 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.