I'm self-studying Quantum Computation from Nielsen and Chuang's book. In section 4.2 they discuss that for any unit vector $\hat n$, the rotation operator $R_{\hat n}(\theta) = \exp(-i\theta\hat n \cdot \vec\sigma/2)$ rotates the Bloch vector about the $\hat n$ axis by an angle $\theta$.

While I can work through the calculations to prove that this is the case, I'm having trouble understanding why, on a deeper level, rotations are given by exponentials of Pauli matrices. I understand that rotations are always of the form $\exp(iK\theta)$ for some Hermitian matrix $K$, but I don't understand why we should specifically take $K = \hat n \cdot \vec \sigma/2$ to get a rotation around the $\hat n$ axis in the Bloch sphere.

Does anyone have an intuitive explanation for this?


2 Answers 2


There are a number of points here, and I'm not exactly sure which are most relevant to what you're trying to get at, but let me try...

  • Any unitary is a rotation. The important property here is that unitaries maintain the inner product: if we have two initial states $|\psi\rangle$ and $|\phi\rangle$, they have a particular inner product $\langle\psi|\phi\rangle$. If we apply a unitary to them, $|\psi\rangle\rightarrow U|\psi\rangle$, then the inner product is $$ \langle\psi|\phi\rangle\rightarrow \langle\psi|U^\dagger U|\phi\rangle=\langle\psi|\phi\rangle. $$ It does not change.
  • The unitary also has particular axes which are unchanged by the rotation (the eigenvectors of $U$).
  • Any unitary can be written as $e^{-iHt}$ for a hermitian matrix $H$. You don't have to write it in this way, but usually the physical description of a system is its Hamiltonian, $H$, and its evolution in time comes from the Schrödinger equation $$ \frac{d|\psi\rangle}{dt}=-iH|\psi\rangle, $$ so if $H$ is constant in time, the solution is $U=e^{-iHt}$. Thus, it is natural to make this connection between the physical system and the way we choose to write the operators. (Also note that the eigenvectors of $H$ and of $U$ are the same, modulo some degeneracy issues, so $H$ fixes the axes of the rotation.)
  • For a qubit, the Pauli matrices + identity are a basis, meaning any matrix can be written as a linear combination of them. For a Hermitian matrix $H$, the linear factors are always real. When converting to a unitary, the identity in $H$ just converts to an irrelevant global phase.
  • $\begingroup$ Thank you for your answer. As I stated in another comment, the part that I'm not getting is why exponentiating $\hat n \cdot \vec\sigma$ gives a rotation around the $\hat n$ axis on the Bloch sphere. Or in matrix language: why is $\hat n$ the Bloch vector of the eigenstates of $\exp(-i\theta \hat n \cdot \vec\sigma/2)$? $\endgroup$ Jul 17, 2023 at 20:24
  • 1
    $\begingroup$ It's kind of by definition of the matrix exponential. If I have a matrix $A$ with eigenvalues $\lambda$ and eigenvectors $|\lambda\rangle\langle\lambda|$ then $e^A=\sum_{\lambda}e^{\lambda}|\lambda\rangle\langle\lambda|$. So, the unitary has the same eigenvectors as the Hamiltonian. Moreover, $(\vec{n}\cdot\vec{\sigma})^2=I$, so $\vec{n}\cdot\vec{\sigma}$ has eigenvalues $\pm 1$. Thus, the projectors onto the eigenvectors are $|\lambda\rangle\langle\lambda|=\frac12(I\pm \vec{n}\cdot\vec{\sigma})$, so there is a very clear connection between the eigenvectors and the Bloch vector. $\endgroup$
    – DaftWullie
    Jul 18, 2023 at 7:02

Pauli matrices are rotation generators that are essentially the "axes" around which we rotate. In 3D space, we would typically rotate around the $x, y$ or $z$ axis. In quantum mechanics, we're working in the Hilbert space of states, which is a complex vector space, and our "axes" are the Pauli matrices. Each Pauli matrix corresponds to rotations around a different axis in the Bloch sphere. Since Paulis are rotation generators (they generate SU(2)) the operation of exponentiation of their linear combinations is actually a way of creating members of SU(2).

Therefore, the operator $n \cdot \sigma$ is essentially a linear combination of the Pauli matrices, which can be considered a vector in the space of Hermitian operators. Therefore $n \cdot \sigma$ points in a certain direction in this space. The value of each component of this vector determines the amount of rotation in each axis of the Bloch sphere.

  • $\begingroup$ Thank you for your answer. I'm still struggling to understand specifically why exponentiating $\hat n \cdot \vec \sigma$ gives a rotation around the $\hat n$ axis, but maybe this requires more knowlegde of the connection between SU(2) and SO(3). $\endgroup$ Jul 16, 2023 at 21:29
  • $\begingroup$ Look into Lie group for deeper understanding en.m.wikipedia.org/wiki/Special_unitary_group $\endgroup$
    – MonteNero
    Jul 16, 2023 at 23:40

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