States, Observables, and Hamiltonians

A symmetry calculation often contains four different statements that use similar symbols: a state is transformed, an observable is transformed, a Hamiltonian is invariant, or a particular state is invariant. These statements are related, but none of them should be silently substituted for another.

The diagnostic table is:

Statement	Formula	Meaning
State transformation	$\lvert\psi\rangle\mapsto S\lvert\psi\rangle$	A new physical state or transformed state representative is being considered.
Observable transformation	$A\mapsto SAS^{-1}$	The observable or apparatus is transformed with the same physical operation.
Hamiltonian symmetry	$SHS^{-1}=H$	The dynamics are invariant under the transformation.
Invariant state	$S\lvert\psi\rangle=e^{i\alpha}\lvert\psi\rangle$	The ray of this particular state is fixed by the transformation.

Here $S$ may be unitary or antiunitary. Most formulas below are written for a unitary $U$ to keep the notation uncluttered; antiunitary transformations require the antilinearity described in Antiunitary Symmetries.

State Transformation

An active state transformation is

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

This statement by itself does not say whether $U$ is a symmetry of the Hamiltonian. It only says how the state changes under the operation.

If $A$ is a fixed observable, then the expectation value in the transformed state is

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

Thus an active state transformation can change measurement probabilities for a fixed apparatus.

Observable Transformation

An observable can also be transformed:

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

If the state and observable are transformed together, the relational expectation value is preserved:

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

This is a covariance statement. It says the same physical relation has been moved together. It is not the same as saying that the fixed observable $A$ has the same expectation value after the state alone is transformed.

For the active/passive convention behind these formulas, see Active and Passive Transformations.

Hamiltonian Symmetry

A transformation is a symmetry of a time-independent Hamiltonian when

$$UHU^\dagger = H.$$

Equivalently,

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

This is a statement about the dynamics, not about a single state. If

$$H\lvert E\rangle = E\lvert E\rangle,$$

then

$$H(U\lvert E\rangle) = E(U\lvert E\rangle).$$

So a Hamiltonian symmetry maps each energy eigenspace into itself. It need not leave each vector in that eigenspace unchanged.

For a strongly continuous one-parameter unitary group

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

Hamiltonian symmetry for all $\alpha$ gives

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

For finite matrices the commutator condition is equivalent to invariance under the whole group. With unbounded generators one must preserve the Hamiltonian domain and use commuting spectral projections; a formal commutator alone need not suffice. This is the starting point for commutators and conservation laws. An antiunitary symmetry can also preserve $H$, but conjugates $i$ and reverses the direction of time evolution.

Invariant State

A particular pure state is invariant under $U$ when its ray is unchanged:

$$U\lvert\psi\rangle = e^{i\alpha} \lvert\psi\rangle.$$

The phase is allowed because pure states are rays. If the equality holds with a phase, all probabilities for that ray are unchanged.

State invariance and Hamiltonian invariance are different conditions; neither implies the other by itself. A Hamiltonian may be rotationally invariant while most individual states are not invariant under every rotation. Instead, rotations may move states around inside a degenerate multiplet.

Example: Harmonic Oscillator Parity

For the one-dimensional harmonic oscillator,

$$H = \frac{P^2}{2m} + \frac{1}{2}m\omega^2X^2.$$

Parity $\Pi$ acts as

$$\Pi X\Pi^{-1}=-X, \qquad \Pi P\Pi^{-1}=-P.$$

Therefore

$$\Pi H\Pi^{-1} = H.$$

Parity is a Hamiltonian symmetry.

The energy eigenstates may be chosen as parity eigenstates:

$$\Pi\lvert n\rangle = (-1)^n\lvert n\rangle.$$

Each energy eigenstate is invariant as a ray because multiplication by $+1$ or $-1$ does not change the ray. But a general superposition such as

$$\lvert\psi\rangle = a\lvert0\rangle+b\lvert1\rangle$$

is transformed into

$$\Pi\lvert\psi\rangle = a\lvert0\rangle-b\lvert1\rangle.$$

For a nonzero vector, this is the same ray if and only if one coefficient vanishes. No choice of relative phase repairs the sign difference when both coefficients are nonzero. The Hamiltonian has parity symmetry; the generic state does not.

Example: Central-Potential Rotations

For a spinless particle in a central potential,

$$H = \frac{\mathbf P^2}{2m} + V(r), \qquad r=\lvert\mathbf R\rvert,$$

rotations are symmetries when the Hamiltonian’s domain and boundary conditions are also rotationally invariant:

$$U(R)HU(R)^\dagger = H.$$

This implies that energy eigenspaces carry representations of the rotation group. For a fixed $\ell$, the states

$$\lvert \alpha,\ell,m\rangle, \qquad m=-\ell,\ldots,\ell,$$

are mixed by rotations. The whole $(2\ell+1)$-dimensional multiplet is invariant as a subspace, but a basis vector with a particular $m$ is not generally invariant under an arbitrary rotation.

The special case $\ell=0$ is different: an $s$ state is rotationally invariant as a ray under ordinary spatial rotations. For $\ell>0$, symmetry usually means organization into multiplets, not invariance of every state.

Example: Spin in a Magnetic Field

Consider a spin-$1/2$ Hamiltonian in a fixed magnetic field along $z$:

$$H = -\gamma B S_z.$$

It is invariant under rotations about the $z$ axis because

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

For $\gamma B\ne0$, it is not invariant under a generic rotation about the $x$ axis, because such a rotation changes $S_z$ into a different spin component. The Hamiltonian has axial symmetry, not full spin-rotation symmetry.

This example is a common source of confusion. If one rotates both the spin and the external magnetic field, the scalar form $-\gamma\mathbf B\cdot\mathbf S$ is covariant. But for the Hamiltonian of a fixed physical setup with $\mathbf B=B\hat z$, only rotations that leave the field direction unchanged are symmetries.

Reading a symmetry statement

Specify the transformed object and the fixed external parameters first. “The $\ell=1$ subspace is invariant” means rotations stay within that three-dimensional space; it does not mean that every vector is fixed. “The magnetic Hamiltonian is covariant” means its form survives when spin and field are rotated together; it does not make every rotation a symmetry at a fixed field.

Exercises

1. Let $\Pi\lvert0\rangle=\lvert0\rangle$ and $\Pi\lvert1\rangle=-\lvert1\rangle$. For which coefficients is $a\lvert0\rangle+b\lvert1\rangle$ invariant as a ray under parity?

Solution

Ray invariance requires

$$a\lvert0\rangle-b\lvert1\rangle = e^{i\alpha} \left( a\lvert0\rangle+b\lvert1\rangle \right).$$

If both $a$ and $b$ are nonzero, this would require $e^{i\alpha}=1$ from the $\lvert0\rangle$ term and $e^{i\alpha}=-1$ from the $\lvert1\rangle$ term, impossible. Thus a nonzero parity-invariant ray in this two-state span must be purely even or purely odd: $b=0$ or $a=0$.

2. A rotationally invariant Hamiltonian has a degenerate $\ell=1$ energy subspace. Does rotational symmetry imply that $\lvert \ell=1,m=0\rangle$ is invariant under every rotation?

Solution

No. Rotational symmetry implies that the $\ell=1$ subspace is mapped into itself and remains at the same energy. A particular $m=0$ basis state is generally mixed with the $m=\pm1$ states by rotations about axes other than $z$. The subspace carries a representation; its individual basis vectors are not all invariant.

3. For $H=-\gamma B S_z$ with $\gamma B\ne0$, which rotations are symmetries if the magnetic field is fixed along $z$?

Solution

Rotations generated by $S_z$ are symmetries because $[H,S_z]=0$. Generic rotations about $x$ or $y$ are not symmetries of the fixed-field Hamiltonian because they rotate $S_z$ into a different component while the external field direction remains fixed. The system has axial symmetry about the field direction.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter 4.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003 reprint of the 1964 edition.