# Unitary transformations as rotations

Considering for simplicity only real $$n$$-by-$$n$$ Hamiltonian matrix, $$H_n$$. Can unitary transformations, $$S$$, diagonalizing such a Hamiltonians, $$S^T H S = \Lambda,$$ be always represented as rotations? We clearly can do this for $$n=2$$: $$S=\begin{bmatrix} \cos\phi & \sin\phi \\ -\sin\phi &\cos\phi\end{bmatrix},$$ and for $$n=3$$ (since the corresponding rotation matrices in 2d and 3d are the general orthogonal matrices). However I am not sure how to describe a rotation in an arbitrary number of dimensions. Also, what is exactly a rotation in $$n$$-dimensions? Can we always represent it in terms of an infinitesimal generator as a rotation about an axis, $$R=e^{\theta K},$$ where $$K^T=-K$$?

Remark The answers might be obvious for people with a strong background in the group theory. However, my background is not in high-energy physics, and I have never studied continuous groups. Thus, I would appreciate down-to-earth explanations or references where I could find such explanations.

• Because your $H$ is an observable, the restriction that it be Hermitian reduces to it being a symmetric real matrix. Now, it is a fundamental result of finite-dimension linear algebra that every real, symmetric endomorphism can be diagonalised by means of orthogonal matrices. $$\Lambda = R^THR$$ $$R\in O\left(n\right)$$ So my guess would be no. You cannot guarantee that this $R$ is necessarily part of $SO\left( n \right)$ --the part that's identity-connected. Give me some time to find a counterexample --or maybe someone abler than me--, and maybe we'd have a satisfactory answer. Mar 8, 2021 at 11:45
• @joigus Do I understand you correctly: $R\in O(n)$ is already by definition a rotation, but it cannot always be represented as a composition of infintesimal rotations and/or a rotation about a single axis? Mar 8, 2021 at 12:42
• An element of $O\left( n \right)$ is not necessarily an element of $SO\left( n \right)$. It is modulo determinant's sign. Look at point 1 in @Qmechanic 's answer. I think they clinch the case. It's true that you can always find this $O\left(3 \right)$ element. But a simple re-arrangement of the basis order gives you what OP wanted. Mar 8, 2021 at 13:27

1. It is well-known that real symmetric $$n\times n$$ matrices are diagonalizable by orthogonal matrices $$\in O(n)$$. In fact, they are diagonalizable by special orthogonal matrices (=rotations) $$\in SO(n)$$, because we can always include/remove a reflection with negative determinant.

2. The Lie group $$SO(n)$$ of $$n$$-dimensional rotations are classified in e.g. this & this Phys.SE posts.

3. OP's last question is answered affirmatively by the fact that the exponential map $$\exp: so(n)\to SO(n)$$ is surjective, cf. e.g. this Math.SE post.

• Say your $H$ is made diagonal by, $$R^{T}HR$$ but $\det R=-1$. $\Lambda$ being, $$\Lambda=\left(\begin{array}{ccc} E_{1} & 0 & 0\\ 0 & E_{2} & 0\\ 0 & 0 & E_{3} \end{array}\right)$$ You may not care what the ordering is, so as the disconnected part to the identity you can obtain by $O^{-}\left(3\right)=P\times SO\left(3\right)$, with $P$ any inversion, you can set $\Lambda$ as, $$\Lambda=\left(\begin{array}{ccc} E_{2} & 0 & 0\\ 0 & E_{1} & 0\\ 0 & 0 & E_{3} \end{array}\right)$$ Now $R \in SO\left(3\right)$. But there's a price to pay, as ordering changes. Mar 8, 2021 at 14:11

Given a vector space $$V$$ equipped with an inner product $$\langle \cdot,\cdot \rangle:V\times V \rightarrow \mathbb R$$ and an ordered basis $$\{\hat e_n\}$$, a rotation is defined as a linear transformation $$R:V\rightarrow V$$ which preserves the inner product (i.e. $$\langle R(x),R(y)\rangle = \langle x,y\rangle$$) and the orientation of the space. Such transformations are represented as matrices $$r$$ such that $$r^\mathrm T r = \mathbb I$$ and $$\mathrm{det}(r) = +1$$; the set of such matrices is called the special orthogonal group $$\mathrm{SO}(n)$$.

If you would like to remove the requirement that the transformation preserve orientation, then you are talking about the orthogonal group $$\mathrm{O}(n)$$ which preserves inner products. This group includes inversions as well as proper rotations.

More concretely, we can define an orthogonal transformation as a map which takes an orthonormal basis $$\{\hat e_n\}$$ to another orthonormal basis $$\{\hat g_n\}$$, and a rotation as an orthogonal transformation which also preserves the orientation of the basis (i.e. an orthogonal transformation with determinant $$+1$$). It can be shown$$^\ddagger$$ that any orthogonal transformation can be written as a rotation $$r$$ composed with an odd permutation matrix $$i$$; in other words, one can get from any orthonormal basis $$\{\hat e_n\}$$ to any other orthonormal basis $$\{g_n\}$$ by performing a rotation and then, if necessary, an odd permutation of the result.

A symmetric $$n\times n$$ matrix $$M$$ can be diagonalized by an orthogonal transformation which maps an arbitrary basis $$\{\hat e_n\}$$ to an eigenbasis $$\{g_n\}$$ of $$M$$. That is, one finds the transformation $$t$$ which maps $$\hat e_n$$ to $$\hat g_n$$; from there,

$$(t^\mathrm T M t) \hat e_n = (t^\mathrm TM)\hat g_n = t^\mathrm T \lambda_n \hat g_n = \lambda_n t^\mathrm T \hat g_n = \lambda_n \hat e_n$$

as a result, $$t^\mathrm T M t$$ is diagonal in the basis $$\hat e_n$$. As argued above, such a transformation is orthogonal (as it's just a map from one ON basis to another); but we can restrict our attention to special orthogonal transformations by noting that if $$t$$ has determinant $$-1$$, then we can add an inversion $$i$$; the transformation $$i\circ t$$ then also diagonalizes $$M$$ (exercise for the reader), and has determinant $$+1$$.

As a result, for any $$n\times n$$ symmetric matrix $$M$$ there exists a special orthogonal transformation (i.e. a rotation) which diagonalizes it.

Can we always represent it in terms of an infinitesimal generator as a rotation about an axis $$R=e^{\theta K}$$, where $$K^\mathrm T = -K$$?

We can always write a rotation in that form, as per Qmechanic's answer. The proof is not trivial, but for intuition one can note that any rotation can be broken up into $$N$$ smaller rotations, and as $$N\rightarrow \infty$$ the rotations become infinitesimally close to the identity transformation $$R(\theta/N) \approx \mathbb I + \frac{\theta}{N} K$$ for some antisymmetric $$K$$.

However, note that rotations are generically not about any single axis, in the sense that e.g. a rotation in $$d=4$$ dimensions may leave two axes fixed (so specifying a single axis and an angle is not sufficient to define the rotation). Instead, one should think of a rotation as occurring in a plane, see e.g. this wiki article.

Another way to understand this is that a rotation can be defined by writing down an antisymmetric matrix $$K$$ and then letting $$R=e^K$$; this exponential map is surjective, as mentioned above. The antisymmetric matrices constitute a vector space - the Lie algebra corresponding to the Lie group $$\mathrm{SO}(n)$$ - which has dimension $$n(n-1)/2$$. In 3 dimensions, $$3(3-1)/2=3$$ and so the space of antisymmetric matrices is isomorphic to the underlying vector space itself. This means that we can associate each antisymmetric matrix to a vector in $$\mathbb R^3$$, whose magnitude defines an angle and whose direction defines an axis of rotation. But clearly this construction is restricted to the very special case $$n=3$$; in any other dimension, we cannot specify a rotation simply via an axis and an angle, because the former requires a specification of $$n(n-1)/2$$ parameters and the latter provides $$n$$ of them.

As a final note, if we move to complex vector spaces with conjugate-linear inner products, then replacing symmetric $$\leftrightarrow$$ Hermitian and orthogonal $$\leftrightarrow$$ unitary leads to the immediate generalization from $$\mathbb R$$ to $$\mathbb C$$.

$$^\ddagger$$If you're familiar with wedge products, then this can be reformulated as follows: an orthogonal transformation $$t:\hat e_n \mapsto \hat g_n$$ is such that $$\hat g_1 \wedge \ldots \wedge \hat g_n = C \hat e_1 \wedge \ldots \wedge \hat e_n$$ with $$C=\pm 1$$. $$C=+1$$ corresponds to an orientation-preserving orthogonal transformation, i.e. a rotation. From this standpoint, it is obvious that any orthogonal transformation can be expressed as a rotation composed with an inversion; if $$C=-1$$, we can simply add an odd permutation $$\hat g_1 \leftrightarrow \hat g_2$$ to make $$C=+1$$.

• Thank you. I have also read elsewhere about rotation in a plane in 4D, which was one of the underlying doubts. Could you recommend a good book? Does this really go in the direction of Lee groups or is it simply about deeper understanding of Hilbert spaces? Mar 8, 2021 at 14:28
• @Vadim I've updated my answer with some additional insight regarding the axis/plane of rotation issue in higher dimensions. The correspondence between $SO(n)$ and the set of $n\times n$ antisymmetric matrices should be understood as the correspondence between a Lie group and its Lie algebra; for a good reference on the subject, see e.g. these notes by Brian Hall. The facts that orthogonal/unitary transformations simply constitute a shuffling of the basis and that symmetric/Hermitian matrices can be diagonalized by such transformations are crucial[...] Mar 8, 2021 at 15:19
• [...] for understanding Hilbert spaces, and can be found in any text on Linear Algebra. Mar 8, 2021 at 15:20
• I think it is really the knowledge of Lie groups that I lack. Thanks again. Mar 8, 2021 at 15:24