Math Infinitum

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Intuition

So far the state has moved and the observables have stood still. The opposite bookkeeping works just as well: freeze the state at its initial value and let every observable move instead. This is the Heisenberg picture, and its equations of motion look like the classical ones — the rate of change of an observable is its commutator with the Hamiltonian.

Walking past a statue, or standing still while the statue is carried past you: the view is the same. The Schrödinger picture moves the state past fixed observables, the Heisenberg picture moves the observables past a fixed state.

The averages of $X$ and $Y$ against $\omega t$ for the state $\ket{+}$ under $\hat{H}=\tfrac{\hbar\omega}{2}Z$ . In the Heisenberg picture the operator itself turns, $X_{H}(t)=X\cos\omega t-Y\sin\omega t$ , and the averages in the fixed state follow.

Moving operators instead of states

In the Heisenberg picture the state is fixed at $\ket{\psi(0)}$ and each observable is replaced by $\hat{A}_{H}(t)=\hat{U}^{\dagger}\hat{A}\hat{U}$ . Every expectation value comes out the same as in the Schrödinger picture, and the observables obey Heisenberg's equation of motion.

\hat{A}_{H}(t)=\hat{U}(t)^{\dagger}\hat{A}\,\hat{U}(t),\qquad \frac{d\hat{A}_{H}}{dt}=\frac{i}{\hbar}\bigl[\hat{H},\hat{A}_{H}\bigr]

The two pictures

Averages agree: $\bra{\psi(0)}\hat{A}_{H}(t)\ket{\psi(0)}=\bra{\psi(t)}\hat{A}\ket{\psi(t)}$ .

Heisenberg’s equation of motion

\frac{d}{dt}\bigl(\hat{U}^{\dagger}\hat{A}\hat{U}\bigr)=\frac{i}{\hbar}\bigl[\hat{H},\hat{U}^{\dagger}\hat{A}\hat{U}\bigr]

Differentiate the product with the product rule and use the Schrödinger equation for $\hat{U}$ and for its adjoint. Because $\hat{H}$ commutes with $\hat{U}$ , the two terms combine into a commutator with .

Proof steps

\frac{d}{dt}\bigl(\hat{U}^{\dagger}\hat{A}\hat{U}\bigr)=\dot{\hat{U}}^{\dagger}\hat{A}\hat{U}+\hat{U}^{\dagger}\hat{A}\dot{\hat{U}}

The product rule, for an $\hat{A}$ with no explicit time dependence.

\dot{\hat{U}}=-\tfrac{i}{\hbar}\hat{H}\hat{U},\qquad \dot{\hat{U}}^{\dagger}=\tfrac{i}{\hbar}\hat{U}^{\dagger}\hat{H}

The Schrödinger equation for $\hat{U}$ and its adjoint.

=\tfrac{i}{\hbar}\bigl(\hat{U}^{\dagger}\hat{H}\hat{A}\hat{U}-\hat{U}^{\dagger}\hat{A}\hat{H}\hat{U}\bigr)

Substitute.

\hat{U}^{\dagger}\hat{H}=\hat{H}\hat{U}^{\dagger},\qquad \hat{H}\hat{U}=\hat{U}\hat{H}

$\hat{U}$ is a function of $\hat{H}$ , so they commute.

=\tfrac{i}{\hbar}\bigl(\hat{H}\hat{A}_{H}-\hat{A}_{H}\hat{H}\bigr)=\tfrac{i}{\hbar}[\hat{H},\hat{A}_{H}]

Move $\hat{H}$ to the outside of each term.

Applications

Practice

The Operator Moves

In the Heisenberg picture the state stays at its initial value and each observable is transformed by the evolution operator on both sides.

\hat{A}_{H}(t)=\hat{U}^{\dagger}\hat{A}\hat{U}

Try it

In the Heisenberg picture, what depends on time?