Infinitesimal Transformations

An infinitesimal transformation is the first-order approximation to a continuous unitary transformation near the identity. If

$$U(\alpha)=e^{-i\alpha G/\hbar},$$

then for a small parameter increment $\delta\alpha$,

$$U(\delta\alpha) = I-\frac{i}{\hbar}\delta\alpha\,G +O(\delta\alpha^2).$$

This first-order term is where commutators enter quantum symmetry. A generator acts on states directly, while it acts on observables through commutators.

Expanding a unitary group

The algebra below is exact to the stated order for bounded operators. For unbounded generators it applies on common invariant domains with enough differentiability; a state in $D(G)$ has an $o(\delta\alpha)$ remainder, while the $O(\delta\alpha^2)$ estimate needs $D(G^2)$.

Let $G$ be the self-adjoint generator of a one-parameter unitary group. For a small real parameter $\delta\alpha$,

$$U = I-\frac{i}{\hbar}\delta\alpha\,G +O(\delta\alpha^2), \qquad U^\dagger = I+\frac{i}{\hbar}\delta\alpha\,G +O(\delta\alpha^2).$$

The opposite signs in $U$ and $U^\dagger$ are the source of the commutator in operator transformations.

The parameter $\delta\alpha$ may have dimensions. For a translation, $\delta\alpha=\delta a$ has dimensions of length. For a rotation, $\delta\alpha=\delta\theta$ is dimensionless. In all cases $\delta\alpha\,G/\hbar$ must be dimensionless.

Infinitesimal State Transformation

With the active convention used in this volume, states transform as

$$\lvert\psi\rangle \mapsto \lvert\psi'\rangle = U\lvert\psi\rangle.$$

Using the first-order expansion,

$$\lvert\psi'\rangle = \lvert\psi\rangle -\frac{i}{\hbar}\delta\alpha\,G\lvert\psi\rangle +O(\delta\alpha^2).$$

Thus

$$\delta\lvert\psi\rangle \equiv \lvert\psi'\rangle-\lvert\psi\rangle = -\frac{i}{\hbar}\delta\alpha\,G\lvert\psi\rangle.$$

The vector $-iG|\psi\rangle/\hbar$, rather than the operator $G$ itself, is the tangent to this curve at $|\psi\rangle$. All displayed $\delta$ rules below keep only first-order terms.

Operator Transformation with the System

If the observable or apparatus is transformed together with the physical system, the active operator transformation is

$$A \mapsto A' = UAU^\dagger.$$

Substitute the first-order expansions:

$$\begin{aligned} A' &= \left(I-\frac{i}{\hbar}\delta\alpha\,G\right) A \left(I+\frac{i}{\hbar}\delta\alpha\,G\right) +O(\delta\alpha^2) \\ &= A -\frac{i}{\hbar}\delta\alpha\,GA +\frac{i}{\hbar}\delta\alpha\,AG +O(\delta\alpha^2). \end{aligned}$$

Therefore

$$\delta A = A'-A = -\frac{i}{\hbar}\delta\alpha\,[G,A].$$

Equivalently,

$$\delta A = \frac{i}{\hbar}\delta\alpha\,[A,G].$$

This is the central local rule: generators act on operators by taking commutators.

Operator Transformation with the State Only

Many calculations instead transform the state while keeping the measured operator fixed. Then

$$\langle A\rangle_{\psi'} = \langle U\psi|A|U\psi\rangle = \langle\psi|U^\dagger A U|\psi\rangle.$$

The effective operator acting in the original state is

$$A_{\rm eff} = U^\dagger A U.$$

To first order,

$$\delta A_{\rm eff} = \frac{i}{\hbar}\delta\alpha\,[G,A].$$

This sign is the opposite of the active transformation $A\mapsto UAU^\dagger$. Both formulas are correct; they answer different questions. For the detailed convention distinction, see Active and Passive Transformations.

Translation Example

For translations on the line,

$$T(\delta a) = I-\frac{i}{\hbar}\delta a\,P +O(\delta a^2).$$

The active wavefunction transformation is

$$(T(\delta a)\psi)(x) = \psi(x-\delta a).$$

Expanding the right-hand side gives

$$\psi(x-\delta a) = \psi(x) -\delta a\,\frac{d\psi}{dx} +O(\delta a^2).$$

Comparison with the unitary expansion gives the position-space generator

$$P=-i\hbar\frac{d}{dx},$$

under the usual domain assumptions.

For the position operator $X$, the state-only or Heisenberg-style transformed operator is

$$T^\dagger(\delta a)XT(\delta a) = X+\frac{i}{\hbar}\delta a\,[P,X] +O(\delta a^2).$$

Using $[X,P]=i\hbar$, this becomes

$$T^\dagger(\delta a)XT(\delta a) = X+\delta a +O(\delta a^2).$$

This shifts the expected position to the right. For translations on the full line the relation is actually exact: $T^\dagger(a)XT(a)=X+aI$, because the first commutator is a multiple of $I$ and all higher nested commutators vanish.

Rotation Example

For a rotation by a small angle $\delta\theta$ about a unit vector $\hat{\mathbf n}$,

$$U(\hat{\mathbf n},\delta\theta) = I-\frac{i}{\hbar}\delta\theta\, \hat{\mathbf n}\cdot\mathbf J +O(\delta\theta^2).$$

For any operator $A$, the active operator change is

$$\delta A = -\frac{i}{\hbar}\delta\theta\, [\hat{\mathbf n}\cdot\mathbf J,A].$$

When $A$ is a vector operator, this commutator encodes the ordinary infinitesimal rotation of vector components. The angular-momentum-specific operator tests are collected in Commutators with Angular Momentum. For example, the angular momentum algebra itself,

$$[J_i,J_j] = i\hbar\sum_k\epsilon_{ijk}J_k,$$

says that angular momentum components rotate into one another under rotations.

Hamiltonian Invariance Test

Infinitesimal transformations give a local test for symmetry of a Hamiltonian. Suppose

$$U(\delta\alpha) = I-\frac{i}{\hbar}\delta\alpha\,G +O(\delta\alpha^2).$$

The Hamiltonian is invariant under the active transformation if

$$U(\delta\alpha)HU^\dagger(\delta\alpha) = H +O(\delta\alpha^2).$$

Using the operator formula,

$$\delta H = -\frac{i}{\hbar}\delta\alpha\,[G,H].$$

Therefore infinitesimal invariance requires

$$[G,H]=0.$$

For finite matrices this local condition implies invariance under every finite transformation in the group. For unbounded operators, strong commutation (commuting spectral projections) is a sufficient precise condition; a formal commutator alone is not. The conservation-law interpretation is developed in Commutators and Conservation Laws.

Finite Transformations Need More Than First Order

The infinitesimal rule is local. To recover a finite transformation, one must integrate or exponentiate it:

$$A(\alpha) = U(\alpha)AU^\dagger(\alpha).$$

Differentiating with respect to $\alpha$ gives

$$\frac{dA(\alpha)}{d\alpha} = -\frac{i}{\hbar} [G,A(\alpha)]$$

when $G$ is independent of $\alpha$. This differential equation has the formal solution

$$A(\alpha) = e^{-i\alpha G/\hbar} A e^{i\alpha G/\hbar}.$$

If generators at different parameter values do not commute, as in a time-dependent Hamiltonian or a path through a nonabelian group, finite transformations require ordered products rather than a single ordinary exponential.

Exercises

1. Starting from $A'=UAU^\dagger$ and $U=I-i\delta\alpha\,G/\hbar+O(\delta\alpha^2)$, derive the first-order change in $A$.

Solution

Use

$$U^\dagger = I+\frac{i}{\hbar}\delta\alpha\,G +O(\delta\alpha^2).$$

Then

$$\begin{aligned} A' &= \left(I-\frac{i}{\hbar}\delta\alpha\,G\right) A \left(I+\frac{i}{\hbar}\delta\alpha\,G\right) +O(\delta\alpha^2) \\ &= A -\frac{i}{\hbar}\delta\alpha\,GA +\frac{i}{\hbar}\delta\alpha\,AG +O(\delta\alpha^2). \end{aligned}$$

Therefore

$$\delta A = A'-A = -\frac{i}{\hbar}\delta\alpha\,[G,A].$$

2. Use $[X,P]=i\hbar$ to show that $T^\dagger(\delta a)XT(\delta a)=X+\delta a+O(\delta a^2)$.

Solution

For the state-only or Heisenberg-style transformed operator,

$$T^\dagger(\delta a)XT(\delta a) = X+\frac{i}{\hbar}\delta a\,[P,X] +O(\delta a^2).$$

Since $[X,P]=i\hbar$, one has $[P,X]=-i\hbar$. Hence

$$\frac{i}{\hbar}\delta a\,[P,X] = \frac{i}{\hbar}\delta a(-i\hbar) = \delta a.$$

Thus

$$T^\dagger(\delta a)XT(\delta a) = X+\delta a+O(\delta a^2).$$

3. Let $U(\delta\alpha)=I-i\delta\alpha\,G/\hbar+O(\delta\alpha^2)$. If $UHU^\dagger=H+O(\delta\alpha^2)$, show that $[G,H]=0$.

Solution

The first-order change in the Hamiltonian under the active transformation is

$$\delta H = -\frac{i}{\hbar}\delta\alpha\,[G,H].$$

The condition $UHU^\dagger=H+O(\delta\alpha^2)$ says that the first-order term vanishes for arbitrary small $\delta\alpha$. Therefore

$$[G,H]=0.$$

4. Suppose $A$ commutes with $G$. What happens to $A$ under the infinitesimal active transformation generated by $G$?

Solution

The active first-order change is

$$\delta A = -\frac{i}{\hbar}\delta\alpha\,[G,A].$$

If $[G,A]=0$, then $\delta A=0$ to first order. For a connected one-parameter group with appropriate domain control, $A$ is invariant under the full finite transformation as well.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.