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 an energy eigenspace into another state with the same energy. It need not leave each vector in that eigenspace unchanged.

For a continuous family

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

Hamiltonian symmetry for all $\alpha$ gives

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

That is the starting point for conservation laws and good quantum numbers.

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.

This condition is stronger than Hamiltonian symmetry for a particular state. 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

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

This is generally not the same ray unless one coefficient vanishes or the relative phase condition is special. 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:

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

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

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

It is not invariant under a 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.

How to Diagnose a Claim

When reading or writing a symmetry statement, ask:

1. What object is being transformed: state, observable, Hamiltonian, external field, or basis?
2. Is the transformation active or passive?
3. Is the statement about one state, one operator, a subspace, or the full Hamiltonian?
4. Is equality exact, equality up to a phase, or equality only after transforming parameters?
5. Does the conclusion require nondegeneracy, simultaneous diagonalization, or a choice of basis inside a degenerate subspace?

These questions prevent the common mistake of proving a true statement about one object and then applying it to another.

Common Mistakes

• Saying “the state has the symmetry” when only the Hamiltonian has the symmetry.
• Treating a multiplet subspace as if every basis vector in it were invariant.
• Forgetting that invariant pure states are invariant only up to a global phase.
• Confusing covariance of a transformed apparatus with invariance of a fixed observable.
• Ignoring external fields when deciding whether a Hamiltonian is symmetric.
• Assuming that a symmetry of $H$ fixes a unique energy eigenstate in a degenerate eigenspace.

Cross-Links

• Quantum Symmetries
• Active and Passive Transformations
• Unitary Symmetries
• Antiunitary Symmetries
• Symmetry Groups and Representations
• Symmetry Constraints on Hamiltonians
• Commutators and Conservation Laws
• Parity
• Central Potentials and Rotational Symmetry
• Spin in Magnetic Fields
• Expectation Values

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.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003.

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

Solution

Rotations generated by $S_z$ are symmetries because $[H,S_z]=0$. 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.