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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?

glS
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slimmerikko
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3 Answers3

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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.

MonteNero
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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.
DaftWullie
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The deeper reason lies in the fact that angular momentum is the generator of rotations. To understand this conceptually, we can think of objects with angular momentum as being capable of rotation, much like objects with linear momentum are capable of translation (therefore momentum is the generator of translations).

Another fact is that if $A$ is a hermitian operator, then $e^{\mathrm{i}A}$ is an unitary operator. When applied to rotations, the exponent should be propotional to the rotation angle $\theta$. The rotation operator $R(\mathbf{\hat{n}}, \theta)$ should be equal to $e^{\mathrm{i}K\theta}$. Based on the previous statement, $K$ should be the angular momentum operator $J$, or more generally, $\mathbf{J} \cdot \mathbf{\hat{n}}$, which is the generator of rotations. If you find the above reasoning difficult to understand, there is a good alternative way to look at it! In the exponent $\mathrm{i}K\theta$, $K$ is treated as a constant with respect to $\theta$, meaning it is independent of the rotation angle $\theta$. The key idea is that $K$ is conserved under rotation. In this sense, $K$ corresponds to the angular momentum $\mathbf{J} \cdot \mathbf{\hat{n}}$ along a specific direction, which is conserved in rotations around that axis.

The complete expression is： $$ R(\mathbf{\hat{n}}, \theta) = \exp \left( -\frac{\mathrm{i} \mathbf{J} \cdot \mathbf{\hat{n}} \theta } { \hbar } \right) $$ where the factor of $\hbar$ in the denominator is introduced to ensure that the exponent is dimensionless.

In the case of a two-dimensional space, where the vector is two-component, the angular momentum operator $J$ becomes the spin operator $S$. The spin operators $S$ can be represented using the Pauli matrices. $$ \mathbf{J} = \mathbf{S} = \frac{\hbar}{2}\boldsymbol{\sigma} $$

Thus, the rotation operator in 2D can be written as: $$ R(\mathbf{\hat{n}}, \theta) = \exp \left( -\frac{\mathrm{i} \boldsymbol{\sigma} \cdot \mathbf{\hat{n}} \theta } { 2 } \right) $$

If you want to learn more about the details of generators, you can refer to J. J. Sakurai's Modern Quantum Mechanics.

Xiaomin Cui
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