3.9 The time-evolution operator and pictures

Measurement is the one non-unitary step in the theory. The rest of the time, states evolve by postulate 4.

This section packages that evolution into a single operator — and along the way discovers something very practical.

The evolution operator

The Schrödinger equation

\ii\hbar\frac{\dd}{\dd t}\ket{\psi(t)}=\hat H\ket{\psi(t)}\tag{3.9.1}

is first order, so given $\ket{\psi(0)}$ every later instant is uniquely determined. Write that map as an operator:

\ket{\psi(t)}=\hat U(t)\ket{\psi(0)}\tag{3.9.2}

Derivation: the form of U(t)basic~5 min

Substitute $\ket{\psi(t)}=\hat U(t)\ket{\psi(0)}$ into the Schrödinger equation:

\ii\hbar\frac{\dd \hat U}{\dd t}\ket{\psi(0)}=\hat H\hat U(t)\ket{\psi(0)}\tag{3.9.3}

This holds for any initial state, so

\ii\hbar\frac{\dd\hat U}{\dd t}=\hat H\hat U,\qquad \hat U(0)=\hat 1\tag{3.9.4}

If $\hat H$ carries no explicit time dependence, this is the operator version of a scalar differential equation, with solution

\boxed{\ \hat U(t)=\ee^{-\ii\hat Ht/\hbar}\ }\tag{3.9.5}

The exponential of an operator is defined by its power series:

\ee^{\hat A}\equiv\sum_{n=0}^{\infty}\frac{\hat A^n}{n!}\tag{3.9.6}

Check: differentiate term by term, $\frac{\dd}{\dd t}\ee^{-\ii\hat Ht/\hbar}=-\frac{\ii\hat H}{\hbar}\ee^{-\ii\hat Ht/\hbar}$ , and substitute. (This step uses the fact that $\hat H$ commutes with itself, so it can be differentiated like an ordinary number.)

How to actually use it: in the energy basis

An exponentiated operator looks intimidating, but in the eigenbasis of $\hat H$ it is diagonal, and the calculation collapses to one line.

Let $\hat H\ket{n}=E_n\ket{n}$ and expand the initial state as $\ket{\psi(0)}=\sum_n c_n\ket{n}$ . By the “eigenvalues of a function” property from section 3.4, $\ee^{-\ii\hat Ht/\hbar}\ket{n}=\ee^{-\ii E_nt/\hbar}\ket{n}$ , so

\ket{\psi(t)}=\sum_n c_n\,\ee^{-\ii E_nt/\hbar}\ket{n}\tag{3.9.7}

This is exactly the result of section 2.4. Back then the recipe read “expand in stationary states, give each term its own phase factor” — now we see that this is just $\hat U$ written out in the energy basis.

The standard three steps for time-dependent problems: expand in energy eigenstates → multiply each coefficient by $\ee^{-\ii E_nt/\hbar}$ → convert back to the position basis when needed. Every evolving superposition in chapter 2 ran on this recipe.

Unitarity = probability conservation

Proof that U is unitary, and what that meansbasic~4 min

$\hat H$ is Hermitian, so

\hat U^\dagger=\big(\ee^{-\ii\hat Ht/\hbar}\big)^\dagger=\ee^{+\ii\hat H^\dagger t/\hbar}=\ee^{+\ii\hat Ht/\hbar}=\hat U^{-1}\tag{3.9.8}

(conjugation sends $\ii\to-\ii$ , and $\hat H^\dagger=\hat H$ ). That is,

\hat U^\dagger\hat U=\hat 1\tag{3.9.9}

which is the definition of unitary.

Physical consequence one: probability conservation.

\braket{\psi(t)}{\psi(t)}=\bra{\psi(0)}\hat U^\dagger\hat U\ket{\psi(0)}=\braket{\psi(0)}{\psi(0)}=1\tag{3.9.10}

Normalised once, normalised forever. What section 2.2 proved via the continuity equation takes one line here.

Physical consequence two: inner products are preserved.

\braket{\phi(t)}{\psi(t)}=\braket{\phi(0)}{\psi(0)}\tag{3.9.11}

Orthogonal states stay orthogonal forever. Two distinguishable states never evolve into indistinguishable ones — a fact that matters greatly in quantum information (the no-cloning theorem and quantum error correction both rely on it).

Physical consequence three: reversibility. $\hat U(-t)=\hat U(t)^{-1}$ — the evolution can be run backwards. Information is never lost.

Here lies the contrast with measurement: measurement is a projection, irreversible, and information genuinely disappears. The entire tension of the measurement problem sits right here (section 3.8).

When H depends on time

If $\hat H$ carries explicit time dependence, $\hat U(t)=\ee^{-\ii\int\hat H\dd t/\hbar}$ is generally wrong — because $\hat H(t_1)$ and $\hat H(t_2)$ at different times need not commute, and exponentials cannot be merged that way.

The correct form requires the time-ordering operator:

\hat U(t)=\mathcal T\exp\left(-\frac{\ii}{\hbar}\int_0^t\hat H(t')\,\dd t'\right)\tag{3.9.12}

$\mathcal T$ orders the operators later-to-the-left. Expanded, this is the Dyson series — the starting point of scattering theory and time-dependent perturbation theory (section 7.5).

Only when $[\hat H(t_1),\hat H(t_2)]=0$ for all times may $\mathcal T$ be dropped.

The three pictures

Now for the most useful part of the section.

Start with a question

What we ultimately compute is always an expectation value:

\langle A\rangle(t)=\bra{\psi(t)}\hat A\ket{\psi(t)} =\bra{\psi(0)}\hat U^\dagger\hat A\hat U\ket{\psi(0)}\tag{3.9.13}

You may bracket it like this: $\big(\bra{\psi(0)}\hat U^\dagger\big)\hat A\big(\hat U\ket{\psi(0)}\big)$ — the time sits on the states.

Or like this: $\bra{\psi(0)}\big(\hat U^\dagger\hat A\hat U\big)\ket{\psi(0)}$ — the time sits on the operator.

The two bracketings are exactly equivalent. Those are the two pictures.

The picture

The Schrödinger picture (the default)

\ket{\psi_S(t)}=\hat U(t)\ket{\psi(0)},\qquad \hat A_S=\text{constant}\tag{3.9.14}

States move, operators sit still. This is what chapter 2 used throughout.

The Heisenberg picture

\ket{\psi_H}=\text{constant},\qquad \hat A_H(t)=\hat U^\dagger\hat A_S\hat U\tag{3.9.15}

Operators move, states sit still.

\langle A\rangle(t)\ \text{is the same in both pictures}\tag{3.9.16}

because they are merely two bracketings of one and the same expression.

The mathematics

The Heisenberg equation of motion

Differentiate $\hat A_H=\hat U^\dagger\hat A_S\hat U$ (with $\hat A_S$ carrying no explicit $t$ ):

\frac{\dd\hat A_H}{\dd t}=\frac{\ii}{\hbar}\hat U^\dagger(\hat H\hat A_S-\hat A_S\hat H)\hat U\tag{3.9.17}

\boxed{\ \frac{\dd\hat A_H}{\dd t}=\frac{\ii}{\hbar}[\hat H,\hat A_H]\ }\tag{3.9.18}

(add a $\partial\hat A/\partial t$ term when $\hat A_S$ depends explicitly on time).

Compare with classical mechanics:

\frac{\dd A}{\dd t}=\{H,A\}_{\text{Poisson}}\tag{3.9.19}

The forms are exactly parallel; just make $\{\ ,\ \}\to\frac{1}{\ii\hbar}[\ ,\ ]$ . This is the recipe of canonical quantisation.

Conservation laws at a glance

The Heisenberg equation delivers an important conclusion instantly:

[\hat H,\hat A]=0\ \Longrightarrow\ \frac{\dd\hat A_H}{\dd t}=0\tag{3.9.20}

Observables that commute with the Hamiltonian are conserved.

$[\hat H,\hat H]=0$ → energy conservation (when $\hat H$ has no explicit time dependence)
translation invariance → $[\hat H,\hat p]=0$ → momentum conservation
rotation invariance → $[\hat H,\hat L]=0$ → angular-momentum conservation

This is the quantum version of Noether’s theorem, and it is even more direct than the classical one: a symmetry shows up as “some operator commutes with $\hat H$ ”, and the conservation law is read straight off the commutator.

It also answers the question from section 3.8: why do conserved quantities stay put after measurement? Because they commute with $\hat H$ — in the Heisenberg picture they simply do not change with time.

The picture

The interaction picture (the third)

When $\hat H=\hat H_0+\hat V$ , with $\hat H_0$ solvable and $\hat V$ a small perturbation:

\ket{\psi_I(t)}=\ee^{\ii\hat H_0t/\hbar}\ket{\psi_S(t)}\tag{3.9.21}

\hat A_I(t)=\ee^{\ii\hat H_0t/\hbar}\hat A_S\ee^{-\ii\hat H_0t/\hbar}\tag{3.9.22}

Absorb the “easy” evolution generated by $\hat H_0$ into the operators, so the states are driven by $\hat V$ alone:

\ii\hbar\frac{\dd}{\dd t}\ket{\psi_I}=\hat V_I(t)\ket{\psi_I}\tag{3.9.23}

The mathematics

Which one to use

Picture	States	Operators	Best for
Schrödinger	move	still	wave-packet evolution, intuitive pictures
Heisenberg	still	move	conservation laws, field theory, classical comparison
Interaction	move slowly	move fast	perturbation theory, scattering, time-dependent problems

All three coincide at $t=0$ and make identical physical predictions. The choice is purely a matter of which computes most easily.

The interaction picture is the starting point of time-dependent perturbation theory and Fermi’s golden rule (sections 7.5–7.6).

Example: the free particle in the Heisenberg pictureadvanced~5 min

$\hat H=\hat p^2/2m$ . Momentum first:

\frac{\dd\hat p_H}{\dd t}=\frac{\ii}{\hbar}[\hat H,\hat p]=0\tag{3.9.24}

( $\hat p$ commutes with $\hat p^2$ .) So $\hat p_H(t)=\hat p(0)$ is constant — momentum conservation.

Now position:

\frac{\dd\hat x_H}{\dd t}=\frac{\ii}{\hbar}[\hat H,\hat x]=\frac{\ii}{2m\hbar}[\hat p^2,\hat x]\tag{3.9.25}

Using $[\hat p^2,\hat x]=\hat p[\hat p,\hat x]+[\hat p,\hat x]\hat p=-2\ii\hbar\hat p$ :

\frac{\dd\hat x_H}{\dd t}=\frac{\hat p}{m}\tag{3.9.26}

Integrate:

\hat x_H(t)=\hat x(0)+\frac{\hat p(0)}{m}t\tag{3.9.27}

Identical to the classical $x=x_0+vt$ — except every quantity is an operator.

The nice part: wave-packet spreading now takes one line. Using $[\hat x(0),\hat p(0)]=\ii\hbar$ , expand $\sigma_x^2(t)=\langle\hat x_H^2\rangle-\langle\hat x_H\rangle^2$ :

\sigma_x^2(t)=\sigma_x^2(0)+\frac{\sigma_p^2}{m^2}t^2+(\text{cross term})\tag{3.9.28}

For a Gaussian packet with $\langle\Delta x\Delta p+\Delta p\Delta x\rangle=0$ initially, the cross term vanishes; substitute $\sigma_p=\hbar/2\sigma_0$ :

\sigma_x(t)=\sigma_0\sqrt{1+\left(\frac{\hbar t}{2m\sigma_0^2}\right)^2}\tag{3.9.29}

This is exactly the spreading formula of section 2.10, which was obtained there by completing the square inside a Gaussian integral — rather tedious algebra. The Heisenberg picture turns it into a few lines.

Key formulas

Time-evolution operator
$\hat U(t)=\ee^{-\ii\hat Ht/\hbar}$
Only when Ĥ has no explicit time dependence; otherwise time-ordering is needed
Unitarity
$\hat U^\dagger\hat U=\hat 1$
Probability conserved, inner products preserved, evolution reversible
Evolution in the energy basis
$\ket{\psi(t)}=\sum_n c_n\ee^{-\ii E_nt/\hbar}\ket{n}$
The standard three steps for time-dependent problems
Heisenberg equation of motion
$\frac{\dd\hat A_H}{\dd t}=\frac{\ii}{\hbar}[\hat H,\hat A_H]$
The quantum counterpart of the classical Poisson bracket
Conservation criterion
$[\hat H,\hat A]=0\ \Longrightarrow\ \hat A\ \text{conserved}$
The quantum version of Noether's theorem
Interaction picture
$\ii\hbar\frac{\dd}{\dd t}\ket{\psi_I}=\hat V_I(t)\ket{\psi_I}$
The starting point of time-dependent perturbation theory and Fermi's golden rule

Self-check4 questions

What comes next

So far we have assumed we always know which state the system is in.

But what if we don’t? An atom flying out of an oven with a completely random spin direction, say; or a system entangled with its environment, which has no state vector of its own at all.

The next section introduces the tool for these situations — it is also the gateway to chapter 10.

The time-evolution operator and pictures

Recommended first

After this section you should be able to

The evolution operator

Unitarity = probability conservation

The three pictures

The picture

The mathematics

The picture

The mathematics

Key formulas

Self-check4 questions

What comes next

The evolution operator

Unitarity = probability conservation

The three pictures

◑The picture

∑The mathematics

◑The picture

∑The mathematics

∎Key formulas

What comes next

The picture

The mathematics

The picture

The mathematics

Key formulas