Generators, Unitary Evolution, and Stone's Picture
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You should be comfortable with the definition of the adjoint operator \(A^\dagger\) and the concept of a self-adjoint operator (\(A^\dagger = A\)). Familiarity with the Taylor series expansion for functions is also assumed.
Why this matters
This lecture bridges the gap between static linear algebra and dynamic physical systems.
In classical and quantum mechanics, the evolution of a state vector over time is governed by a continuous transformation. Understanding how an infinitesimal change (the generator) dictates the finite evolution (the unitary operator) is the key to interpreting conservation laws, symmetries, and the probabilistic structure of quantum theory. We move from the algebraic properties of matrices to their role as time-evolution operators.
The Idea in Plain Language
Imagine a rigid object rotating in space. If you know the direction and speed of the rotation axis (the generator), you can predict the object's orientation at any future time. In linear algebra, the "generator" is a matrix \(A\). If \(A\) is anti-Hermitian (\(A^\dagger = -A\); for a real matrix, antisymmetric), it acts like a rotation generator. The resulting transformation \(U(t)\) is unitary, meaning it preserves lengths and angles. This is crucial because in physics, preserving the "length" of a state vector often corresponds to preserving the total probability or energy norm. The generator tells us *how* the system changes; the unitary operator tells us *where* it ends up.
Generators and the Matrix Exponential
Consider a linear system described by the differential equation \(\dot{x} = Ax\), where \(x(t)\) is a vector in a Hilbert space \(\mathcal{H}\) and \(A\) is a bounded linear operator. The solution to this initial-value problem, with \(x(0) = x_0\), is given by the matrix exponential. As established in the cited derivation, the series \(\exp(tA) = \sum_{n=0}^{\infty}\frac{t^{n}A^{n}}{n!}\) converges for all \(t \in \mathbb{R}\).
The operator \(U(t) = \exp(tA)\) forms a one-parameter group. This means \(U(t+s) = U(t)U(s)\) and \(U(0) = I\). The generator \(A\) is recovered by differentiating \(U(t)\) at \(t=0\): \(A = \left. \frac{d}{dt} U(t) \right|_{t=0}\). This relationship is fundamental: the infinitesimal behavior of the evolution determines the global evolution.
Unitarity and Inner-Product Preservation
Not all generators produce unitary evolution. For \(U(t)\) to be unitary for all \(t\), it must preserve the inner product. That is, for any two vectors \(x, y \in \mathcal{H}\), we require \(\langle U(t)x, U(t)y\rangle = \langle x, y\rangle\). This condition ensures that the "distance" between states and the "length" of states remain constant over time.
From the cited derivation, \(U\) is unitary if and only if \(U^\dagger U = I\). For the one-parameter group generated by \(A\), this condition translates to a constraint on \(A\) itself. Specifically, \(U(t)\) is unitary for all \(t\) if and only if the generator is skew-Hermitian (or anti-Hermitian), meaning \(A^\dagger = -A\). This is a critical result: the generator of a unitary flow must be anti-Hermitian.
Stone's Picture and Self-Adjoint Observables
In quantum mechanics, physical observables are represented by self-adjoint operators (\(H^\dagger = H\)). However, the time-evolution operator is unitary. How do these connect? We use the convention that the generator of time evolution is \(A = -iH\), where \(H\) is the Hamiltonian (a self-adjoint operator).
If \(H^\dagger = H\), then \(A^\dagger = (-iH)^\dagger = iH^\dagger = iH = -(-iH) = -A\). Thus, \(A\) is anti-Hermitian, satisfying the condition for unitary evolution. The time-evolution operator becomes \(U(t) = \exp(-iHt)\). This is the essence of Stone's theorem in its finite-dimensional form: every strongly continuous one-parameter unitary group is generated by a self-adjoint operator. The "picture" is that self-adjoint operators (observables) generate unitary transformations (evolutions).
The derivations
Matrix Exponential and One-Parameter Flows
Let \(A\) be a bounded linear operator on a finite-dimensional (or, more generally, a Hilbert) space \(V\). The operator series \(\exp(tA)=\sum_{n=0}^{\infty}\frac{t^{n}A^{n}}{n!}\) converges in operator norm for every \(t\in\mathbb{R}\) and defines the unique differentiable operator-valued function \(U(t)\) satisfying \(\dot U=AU\), \(U(0)=I\). Consequently \(x(t)=U(t)x_{0}\) is the unique solution of the initial-value problem \(\dot x=Ax\), \(x(0)=x_{0}\), the family \(\{U(t)\}_{t\in\mathbb{R}}\) is a one-parameter group, and \(U(t)\) is unitary for all \(t\) if and only if the generator is skew-adjoint, \(A^{\dagger}=-A\) — equivalently \(A=-\tfrac{i}{\hbar}H\) with \(H\) self-adjoint, in which case \(H\) is the physical generator of the flow.
Unitary Operators as Inner-Product Isometries
Let \(U\) be a bounded linear operator on a complex Hilbert space \(\mathcal{H}\) with a well-defined adjoint \(U^\dagger\), and take the inner product antilinear in its first slot, linear in its second. Then \(U\) preserves the inner product, \(\langle Ux, Uy\rangle = \langle x, y\rangle\) for all \(x,y\in\mathcal{H}\), if and only if \(U^\dagger U = I\). On a finite-dimensional space this is equivalent to full unitarity \(U^\dagger U = U U^\dagger = I\), and every eigenvalue \(\lambda\) of \(U\) then satisfies \(\lvert\lambda\rvert = 1\), i.e. \(\lambda = e^{i\theta}\) for some real \(\theta\).
A common pitfall is confusing the sign convention in the exponent. If you define the generator as \(A = iH\) instead of \(A = -iH\), the direction of time evolution reverses. While mathematically valid, it contradicts the standard physical convention where \(H\) is the energy operator and \(t\) increases forward. Always verify that \(A^\dagger = -A\) holds for your specific definition of \(A\). C
Worked example
Problem. Consider a two-level quantum system with Hamiltonian \(H = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). Calculate the time-evolution operator \(U(t)\) for \(t = \pi/2\) and verify that it is unitary.
- Identify the generator: \(A = -iH = -i \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}\).
- Compute \(A^2\): \(A^2 = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} = \begin{pmatrix} (-i)(-i) & 0 \\ 0 & (-i)(-i) \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I\).
- Use the series expansion for \(\exp(tA)\). Since \(A^2 = -I\), the even and odd terms separate: \(\exp(tA) = \sum_{k=0}^{\infty} \frac{t^{2k} A^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{t^{2k+1} A^{2k+1}}{(2k+1)!}\). Note \(A^{2k} = (-1)^k I\) and \(A^{2k+1} = (-1)^k A\).
- Simplify: \(\exp(tA) = \cos(t)\,I + \sin(t)\,A = \cos(t)\,I - i\sin(t)\,H\).
- Substitute \(t = \pi/2\): \(\cos(\pi/2) = 0\), \(\sin(\pi/2) = 1\). Thus \(U(\pi/2) = A = -iH = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}\).
- Verify unitarity: \(U^\dagger = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\), and \(U^\dagger U = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} = \begin{pmatrix} (i)(-i) & 0 \\ 0 & (i)(-i) \end{pmatrix} = I\).
Units check. The Hamiltonian \(H\) has units of energy (Joules). The term \(Ht\) must be dimensionless for the exponential to be defined. Since \(\hbar\) is typically absorbed into the definition of \(H\) in natural units or explicitly included as \(H/\hbar\), the argument of the sine and cosine functions is dimensionless, ensuring the result is a dimensionless unitary matrix.
Three quick tests
- If \(A\) is a real symmetric matrix, is \(U(t) = \exp(tA)\) unitary?
Answer
No. For real symmetric \(A\), \(A^\dagger = A\). The condition for unitarity is \(A^\dagger = -A\). Since \(A \neq -A\) (unless \(A=0\)), \(U(t)\) is not unitary; it is a real symmetric matrix that generally does not preserve the Euclidean norm. - What is the generator \(A\) if the time-evolution operator is \(U(t) = \exp(i \omega t \sigma_z)\), where \(\sigma_z\) is the Pauli Z matrix?
Answer
Comparing with \(U(t) = \exp(tA)\), we identify \(A = i \omega \sigma_z\). Note that \(\sigma_z\) is Hermitian, so \(A\) is anti-Hermitian, consistent with unitary evolution. - Why must the eigenvalues of a unitary operator \(U\) lie on the unit circle in the complex plane?
Answer
If \(Uv = \lambda v\), then \(\langle Uv, Uv\rangle = \langle v, v\rangle\). By unitarity, \(\langle Uv, Uv\rangle = \langle v, v\rangle\). Also, \(\langle Uv, Uv\rangle = \langle \lambda v, \lambda v\rangle = \lambda \bar{\lambda} \langle v, v\rangle = |\lambda|^2 \langle v, v\rangle\). Thus \(|\lambda|^2 = 1\), so \(|\lambda| = 1\).
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-088 · D-083 · Threads: symmetry, waves, matter