Active and Passive Transformations

An active transformation changes the physical state, system, or operator being considered while the description frame is held fixed. A passive transformation changes the coordinates, basis, or reference frame used to describe the same physical object. The same unitary matrix can appear in both stories, but it usually appears with an inverse or adjoint in one of them.

This volume uses active transformations by default unless a page explicitly says otherwise. That convention fixes signs in translation, rotation, and generator formulas.

The Basic Distinction

Viewpoint	What changes?	What stays fixed?	Typical formula
Active	the physical state or system	coordinate axes or basis labels	$\lvert\psi\rangle\mapsto U\lvert\psi\rangle$
Passive	the coordinate system or basis	the abstract state or physical situation	components transform with $U^\dagger$ or $U^{-1}$

The distinction is not about whether a matrix is present. Both viewpoints may use unitary matrices. The distinction is what the matrix is being used to do.

For a unitary $U$, an active state transformation is

$$\lvert\psi\rangle \mapsto \lvert\psi_{\rm act}\rangle = U\lvert\psi\rangle.$$

If the observable $A$ is kept fixed, the new expectation value is

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

If the observable is transformed along with the state, the transformed operator is

$$A_{\rm act} = UAU^\dagger,$$

and the physical relation is represented consistently:

$$\langle U\psi|UAU^\dagger|U\psi\rangle = \langle\psi|A|\psi\rangle.$$

That last equality is not saying the state did not change. It says that if one moves the whole physical setup, including the state and the apparatus represented by $A$, the relational prediction is unchanged.

Passive Basis Change

In a passive basis change, the abstract vector is not changed. Only its coordinates change.

Let a new orthonormal basis be related to the old one by

$$\lvert e_i'\rangle = U\lvert e_i\rangle.$$

The coordinate components of the same state are then

$$c_i' = \langle e_i'|\psi\rangle = \langle e_i|U^\dagger|\psi\rangle.$$

In column-vector language,

$$c' = U^\dagger c.$$

The matrix of the same abstract operator changes as

$$A' = U^\dagger A U.$$

This is why a passive change often looks like the inverse of an active transformation. The physical state did not move; the coordinate grid used to describe it moved.

For the general linear-algebra convention, see Change of Basis. For the quantum state convention, see Change of Basis.

Why Both Descriptions Are Used

Both descriptions are useful because they answer different questions.

Use the active description when asking:

• What state results after applying a translation, rotation, parity operation, or time evolution?
• Is a Hamiltonian invariant under a physical transformation?
• What generator produces this continuous transformation?
• Which quantum numbers are conserved by a symmetry?

Use the passive description when asking:

• What are the components of the same state in a new basis?
• How does an operator matrix look in another representation?
• How do spherical components or spin components change when axes are relabeled?
• Why do two references write inverse rotation matrices for the same physics?

The danger is switching descriptions mid-calculation. A sign error in a generator is often an active/passive error in disguise.

Example: Spatial Translation

With the active convention, translating a wavefunction to the right by $a$ gives

$$(T(a)\psi)(x) = \psi(x-a),$$

where

$$T(a) = \exp\left( -\frac{iaP}{\hbar} \right).$$

The argument is $x-a$ because the translated wavepacket value at the old coordinate $x$ came from the old wavepacket value at $x-a$. This is the convention used in Translations and Momentum.

A passive relabeling of coordinates has the opposite flavor. If the same physical point is described by a new coordinate

$$x' = x-a,$$

then the same physical wavefunction is represented by

$$\psi_{\rm pass}'(x') = \psi(x'+a).$$

The two formulas differ because the first moves the state through a fixed coordinate system, while the second changes the coordinate labels for the same state.

Example: Spin Rotation Versus Axis Rotation

For an active spin-$1/2$ rotation by angle $\theta$ about $\hat{\mathbf n}$,

$$\lvert\psi\rangle \mapsto U(\hat{\mathbf n},\theta)\lvert\psi\rangle,$$

with

$$U(\hat{\mathbf n},\theta) = \exp\left( -\frac{i\theta}{2} \hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

If a spin is actively rotated while the measurement axis remains $z$, the probability for spin up along $z$ becomes

$$\left| \langle +_z| U(\hat{\mathbf n},\theta) |\psi\rangle \right|^2.$$

By contrast, if the physical spin state is unchanged and the spin basis is rotated, the new basis vector is

$$\lvert +_{z'}\rangle = U(\hat{\mathbf n},\theta)\lvert +_z\rangle,$$

and the component of the same state in the rotated basis is

$$\langle +_{z'}|\psi\rangle = \langle +_z| U^\dagger(\hat{\mathbf n},\theta) |\psi\rangle.$$

The adjoint appears because the bra belongs to the rotated basis. This is the same inverse relation that appears in ordinary coordinate changes.

Generator Sign Conventions

For an active continuous unitary transformation,

$$U(\alpha) = \exp\left( -\frac{i\alpha G}{\hbar} \right),$$

the infinitesimal state change is

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

For an active transformation of an operator along with the system,

$$A \mapsto UAU^\dagger,$$

so to first order

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

For the passive or Heisenberg-style convention

$$A \mapsto U^\dagger A U,$$

the sign is instead

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

Both conventions are legitimate. Mixing them silently is not. When comparing two books, identify whether their $U$ moves states, moves axes, changes basis, or evolves operators.

Symmetry Tests Use Active Transformations

When this volume asks whether a transformation is a symmetry of a Hamiltonian, it uses the active condition

$$SHS^{-1} = H.$$

For a unitary symmetry,

$$UHU^\dagger = H.$$

This is a statement about the physical Hamiltonian being invariant under a physical transformation. It is not merely the statement that the matrix of $H$ can be rewritten in another basis.

A passive change of basis can make the entries of a Hamiltonian matrix look different while all predictions remain the same. A true symmetry is stronger: it says the transformed Hamiltonian is the same operator under the stated transformation.

Common Mistakes

• Inferring a physical symmetry from a passive basis change.
• Translating a wavefunction to the right with $\psi(x+a)$ while using the active convention elsewhere.
• Comparing rotation formulas from different books without checking whether axes or vectors are being rotated.
• Forgetting that bras in a changed basis bring in $U^\dagger$.
• Treating $A\mapsto UAU^\dagger$ and $A\mapsto U^\dagger A U$ as interchangeable notation rather than different conventions.
• Changing both the state and the basis and then interpreting the result as if only one had changed.

Cross-Links

• Quantum Symmetries
• States, Observables, and Hamiltonians
• Unitary Symmetries
• One-Parameter Unitary Groups
• Generators
• Infinitesimal Transformations
• Translations and Momentum
• Rotations Preview
• Rotations in Three Dimensions
• Angular Momentum Algebra
• Spin Rotations
• Operator Conventions
• Angular Momentum Conventions
• Quantum Change of Basis
• Linear-Algebra Change of Basis

References

• 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.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. With the active translation convention $(T(a)\psi)(x)=\psi(x-a)$, show that a narrow wavepacket centered near $x_0$ moves to a wavepacket centered near $x_0+a$.

Solution

The translated wavefunction has large magnitude where $\psi(x-a)$ has large magnitude. If the original wavefunction is centered near $x_0$, then $\psi(x-a)$ is largest when $x-a\approx x_0$, or $x\approx x_0+a$. Thus the active transformation moves the packet to the right by $a$.

2. Suppose $\lvert e_i'\rangle=U\lvert e_i\rangle$. Why do the coordinates of the same state transform as $c'=U^\dagger c$ rather than $c'=Uc$?

Solution

Coordinates are overlaps with basis bras. Since

$$\lvert e_i'\rangle = U\lvert e_i\rangle,$$

the corresponding bra is

$$\langle e_i'| = \langle e_i|U^\dagger.$$

Therefore

$$c_i' = \langle e_i'|\psi\rangle = \langle e_i|U^\dagger|\psi\rangle.$$

In matrix form this is $c'=U^\dagger c$. The adjoint appears because this is a passive coordinate change, not an active transformation of the state.