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.

Active states and passive coordinates

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.

Choosing a viewpoint

Use an active transformation to predict how a prepared state changes relative to a fixed detector. Use a passive transformation to express the same state and detector in new coordinates. In either case, the probability is the overlap between the state and the measurement state; changing coordinates alone cannot change that overlap.

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 Momentum Operator as Generator.

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 one-parameter unitary group with a fixed self-adjoint generator,

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

the state change to first order in a small parameter $\alpha$ 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

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

For unbounded generators these expansions apply to vectors in the necessary operator domains. Identify whether $U$ moves states, moves axes, changes basis, or evolves operators before choosing a sign.

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 Hamiltonian symmetry additionally says the transformed Hamiltonian is the same operator under the stated transformation. For antiunitary $S$, the condition $SHS^{-1}=H$ reverses time evolution rather than commuting with it; see Antiunitary Symmetries. The basis changes above are time-independent: a moving basis introduces an additional term in the Schrödinger Hamiltonian.

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.

3. In the $S_z$ basis, rotate $|+_z\rangle=(1,0)^T$ about $y$ through $\theta$. Compare the actively rotated column with the components of the unchanged state in the rotated basis.

Solution

Since $-i\sigma_y=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$,

$$U_y(\theta)=\begin{pmatrix} \cos(\theta/2)&-\sin(\theta/2)\\ \sin(\theta/2)&\cos(\theta/2) \end{pmatrix}.$$

The active column is $(\cos(\theta/2),\sin(\theta/2))^T$. The passive column is $U_y^\dagger(\theta)(1,0)^T=(\cos(\theta/2),-\sin(\theta/2))^T$. The opposite signs reflect different questions, not different physical laws. For a passive change, transform the observable matrix too: $A'=U_y^\dagger A U_y$, so $c'^\dagger A'c'=c^\dagger A c$.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013. doi:10.1007/978-1-4614-7116-5.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter 3.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.