The Heisenberg Picture (Preview)

So far time dependence has lived entirely in the state: $|\psi(t)\rangle = U(t)|\psi(0)\rangle$ while operators stayed fixed. This is the Schrödinger picture. There is an equivalent description in which the operators carry the time dependence and the state stands still — the Heisenberg picture. Both predict identical measurable results.

Two ways to package the same physics

Every prediction in quantum mechanics is an expectation value (or, more generally, a matrix element):

\langle \hat A\rangle(t) = \langle\psi(t)|\,\hat A\,|\psi(t)\rangle = \langle\psi(0)|\,U^\dagger(t)\,\hat A\,U(t)\,|\psi(0)\rangle.

There are two equally valid ways to read the right-hand side, just by choosing where to group the $U$ factors:

Schrödinger picture: keep $U(t)$ with the state. The state evolves, $|\psi(t)\rangle = U(t)|\psi(0)\rangle$ , and the operator $\hat A$ is fixed.
Heisenberg picture: attach $U(t)$ to the operator. The state is frozen at $|\psi(0)\rangle$ , and the operator evolves as $\hat A_H(t) = U^\dagger(t)\,\hat A_S\,U(t).$

Because the grouping is just bookkeeping on the same expression, the two pictures give the same $\langle \hat A\rangle(t)$ for every observable, state, and time.

The Heisenberg equation of motion

Differentiating $\hat A_H(t) = U^\dagger \hat A_S U$ and using $i\hbar\,\dot U = \hat H U$ yields the Heisenberg equation of motion:

\frac{d}{dt}\hat A_H(t) = \frac{i}{\hbar}\,[\hat H, \hat A_H(t)] + \left(\frac{\partial \hat A_S}{\partial t}\right)_H.

Notice the structure: it is identical to Ehrenfest's theorem from the previous lesson, but now it is an equation for the operator rather than for its expectation value. Taking the expectation value of both sides in the (fixed) state recovers Ehrenfest exactly. The Heisenberg picture is, in a sense, Ehrenfest's theorem promoted from averages to operators.

What carries over and what does not

Eigenvalues are picture-independent: $\hat A_H$ and $\hat A_S$ are related by a unitary similarity transformation, so they share the same spectrum. Measurable outcomes are unchanged.
Commutators evolve covariantly: equal-time commutators keep their Schrödinger form, e.g. $[\hat x_H(t), \hat p_H(t)] = i\hbar$ , because conjugating both factors by the same unitary preserves the commutator.
The Hamiltonian (if time-independent) is the same in both pictures, since it commutes with $U(t) = e^{-i\hat H t/\hbar}$ : $\hat H_H = U^\dagger \hat H U = \hat H$ .

Why have two pictures at all

The Schrödinger picture is intuitive for state preparation and for visualising wavefunctions spreading or qubits rotating on the Bloch sphere. The Heisenberg picture shines when you care about the dynamics of observables themselves — it makes conservation laws transparent (a conserved operator is literally constant in time), it is the natural language for quantum field theory, and it exposes the close analogy between quantum commutators and classical Poisson brackets.

The takeaway

The Schrödinger and Heisenberg pictures are two equivalent bookkeeping choices for the same unitary evolution $U(t) = e^{-i\hat H t/\hbar}$ : either the state moves and operators stay fixed, or operators move via $\hat A_H = U^\dagger \hat A_S U$ and the state stays fixed. They agree on every observable prediction, and the Heisenberg equation of motion is the operator version of Ehrenfest's theorem.