# Rotation which diagonalizes the Hamiltonian

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?

• Oct 1, 2020 at 6:37

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.
• 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). Oct 1, 2020 at 20:09
• 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! Oct 1, 2020 at 20:09
• 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. Oct 1, 2020 at 20:30
• Yes, the problem was chosen easy to just illustrate the principle. You'd have to be more creative if the initial vector were $\hat z - \hat x$ ! Oct 1, 2020 at 20:34
• @Frobenius. Depends on what one understands by "this". Of course any vector may be rotated to $\hat z$. Creative means non-mess-aversive. Oct 2, 2020 at 0:27

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
$$$$H\boldsymbol{=}\alpha\left(\sigma_{y}\boldsymbol{-}\mathrm \sigma_{x}\right)\,,\quad \alpha\boldsymbol{=}\dfrac{\sqrt{2}\epsilon}{\hbar} \tag{01}\label{01}$$$$ From the bijection between hermitian traceless matrices and real 3-vectors (discussed in paragraph ''The reasoning'' of aforementioned REFERENCE) the representative real 3-vector of this Hamiltonian is $$$$\mathbf{h}\boldsymbol{=}\alpha\left(\mathbf{j}\boldsymbol{-}\mathbf{i}\right) \tag{02}\label{02}$$$$ 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 $$$$H'\boldsymbol{=}c\,\sigma_{z} \,,\quad c\in \mathbb{R} \tag{03}\label{03}$$$$ Above expression is justified because not only $$\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 $$$$\mathbf{h'}\boldsymbol{=}c\,\mathbf{k} \tag{04}\label{04}$$$$ If the transformation must be a rotation then the vector $$\mathbf{h'}$$ of equation \eqref{04} will be the image of the vector $$\mathbf{h}$$ of equation \eqref{02} so $$$$\Vert\mathbf{h'}\Vert\boldsymbol{=}\Vert\mathbf{h}\Vert \quad \boldsymbol{\Longrightarrow} \quad c\boldsymbol{=}\sqrt{2}\,\alpha \tag{05}\label{05}$$$$ 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 $$SU(2)$$ $$$$\boxed{\:\:U_{\mathbf{n} ,\theta}\boldsymbol{=} \cos\frac{\theta}{2}\boldsymbol{-}i(\mathbf{n} \boldsymbol{\cdot} \boldsymbol{\sigma})\sin\frac{\theta}{2}\boldsymbol{=}\dfrac{\sqrt{2}}{2}\left(I\boldsymbol{-}i\,\dfrac{\sigma_x\boldsymbol{+}\sigma_y}{\sqrt{2}}\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{08}\label{08}$$$$

It could be verified easily, using the properties of Pauli matrices, that $$U_{\mathbf{n} ,\theta}$$ diagonalizes the Hamiltonian $$H$$, that is $$$$U_{\mathbf{n} ,\theta}\,H\,U^{*}_{\mathbf{n} ,\theta}\boldsymbol{=}H' \tag{09}\label{09}$$$$ or explicitly $$$$\dfrac{\sqrt{2}}{2}\left(I\boldsymbol{-}i\,\dfrac{\sigma_x\boldsymbol{+}\sigma_y}{\sqrt{2}}\right)\,\left(\sigma_y\boldsymbol{-}\sigma_x\vphantom{\dfrac{\sigma_x\boldsymbol{+}\sigma_y}{\sqrt{2}}} \right)\,\dfrac{\sqrt{2}}{2}\left(I\boldsymbol{+}i\,\dfrac{\sigma_x\boldsymbol{+}\sigma_y}{\sqrt{2}}\right)\boldsymbol{=}\sqrt{2}\,\sigma_z \tag{10}\label{10}$$$$

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 $$$$\boxed{\:\:U_{\mathbf{m} ,\phi}= \cos\frac{\phi}{2}-i(\mathbf{m} \boldsymbol{\cdot} \boldsymbol{\sigma})\sin\frac{\phi}{2}=\dfrac{i}{2}\left( \sigma_x-\sigma_y-\sqrt{2}\sigma_z \right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\: } \tag{12}\label{12}$$$$ Again, it could be verified easily, using the properties of Pauli matrices, that $$U_{\mathbf{m} ,\phi}$$ diagonalizes the Hamiltonian $$H$$, that is $$$$U_{\mathbf{m} ,\phi}\,H\,U^{*}_{\mathbf{m} ,\phi}\boldsymbol{=}H' \tag{13}\label{13}$$$$ or explicitly $$$$\left[\dfrac{i}{2}\left(\sigma_x-\sigma_y-\sqrt{2}\sigma_z \right)\right]\,\left(\sigma_y\boldsymbol{-}\sigma_x\vphantom{\dfrac{\sigma_x\boldsymbol{+}\sigma_y}{\sqrt{2}}} \right)\,\left[\boldsymbol{-}\dfrac{i}{2}\left(\sigma_x-\sigma_y-\sqrt{2}\sigma_z \right)\right]\boldsymbol{=}\sqrt{2}\,\sigma_z \tag{14}\label{14}$$$$

• 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. Oct 3, 2020 at 9:10
• @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. Oct 3, 2020 at 9:21
• 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. Oct 3, 2020 at 9:49