In Griffiths' Intro. to QM 3rd, Sec. 6.2, transforming an observable $Q$ by the translation operator $T$ is found to be $$ Q' = T^\dagger Q\ T $$ the same for the parity operator $\Pi$ instead of $T$ we have $Q' = \Pi^\dagger Q\ \Pi$.

But in other texts, e.g., Tannoudji, QM, 2nd ed, Vol. I, Complements of chapter VI, Complement B$_{VI}$, 5. Rotation of observables, and also in other questions here and here the transformation on the observable $A$ by a unitary transformation $U$ should be $$ A' = UA\ U^\dagger $$ where $U$, as I understand, should be an active transformation, as $T$ above and I expected that the two equations should be the same. But it seems that the two definitions are not equivalent, or is there any mistake?


Griffiths Definition:

The transformed operator $\hat Q'$ is defined to be the operator that gives the same expectation value in the untranslated state $\psi$ as does the operator $\hat Q$ in the translated state $\psi'$ $$ \langle\psi'|\hat Q|\psi'\rangle = \langle \psi | \hat Q' |\psi \rangle $$ There are two ways to calculate the effect of a translation on an expectation value. One could actually shift the wave function over some distance (this is called an active transformation) or one could leave the wave function where it was and shift the origin of our coordinate system by the same amount in the opposite direction (a passive transformation). The operator $\hat Q'$ is the operator in this shifted coordinate system.

Using Eq. 6.1, $$ \langle\psi|T^\dagger\hat Q\ \hat T|\psi\rangle = \langle \psi | \hat Q' |\psi \rangle $$

Tannoudji Definition:

Let us assume the system to be in the eigenstate $|u_n\rangle$ of $A$: the device for measuring $A$ in this system will give the result $a_n$ without fail. But just before performing the measurement, we apply a rotation $\scr R$ to the physical system and, simultaneously, to the measurement device; their relative positions are unchanged. Consequently, if the observable $A$ which we are considering describes a physical quantity attached only to the system which we have rotated (that is, independent of other systems or devices which we have not rotated), then, in its new position, the measurement device will still give the same result $a_n$ without fail. Now, after rotation, the device, by definition, measures $A'$, and the system is in the state: $$ |u_n'\rangle = R|u_n\rangle $$ We must therefore have: $$ A|u_n\rangle = a_n|u_n\rangle \implies A'|u_n'\rangle = a_n|u_n'\rangle $$ that is: $$ R^\dagger A' R |u_n\rangle = a_n|u_n\rangle $$

Note that $\scr R$ is the rotation the physical 3-dimensional space and $R$ is the its representative operator in the Hilbert space.


1 Answer 1


There are two physically different ideas with different mathematical properties when defining the (active) action of a symmetry on observables in quantum physics.

Assume that, according to the Wigner theorem, $U$ is an either unitary or anti unitary transformation of state vectors $\psi$ corresponding to an active action on the states of a quantum system.

If $A$ is an observable, we have the dual action, $$A \to S_U(A) := U^{-1}A U$$ and the inverse dual action $$A \to S^*_U(A) := UAU^{-1}\:.$$

The former has the meaning of an action on the physical measurement instruments such that the effect on the outcomes on the unchanged state is the same as the outcomes of the changed states on the unchanged observables. I.e. instead of translating the system along $x$, I translate the instruments along $-x$.

The latter has the meaning of an action on the measurement instruments which cancels the action of the symmetry on the system as far as the outcomes of measurements are concerned.

The proofs of these facts are trivial from the basic QM formalism (see the final Note).

There is a fundamental mathematical difference when discussing the action of a symmetry group $G$ represented by a unitary (or projective unitary) representation on the state vectors $$G\ni g \mapsto U_g\:.$$ As usual, (up to phases) $$U_gU_h =U_{g\circ h}\:, \quad U_e = I$$ where $\circ$ is the product in $G$ and $e$ is the identity element. I henceforth use the shorthand $S_g := S_{U_g}$ and similarly for $S^*$.

The inverse dual action defines a proper representation of $G$: $$S^*_g S^*_h = S^*_{g\circ h}\:,$$ whereas the dual action defines a left representation $$S_g S_h = S_{h\circ g}\:.$$ The use of one or another action is matter of convenience and depends on the physical interpretation. In QFT the natural action of the group of isometries of the spacetime on field observables is usually implemented through $S^*$.


If $$A = \int_{\sigma(A)} \lambda dP^{(A)}(\lambda)$$ is the spectral decomposition of the selfadjoint operator $A$ and $U$ is a unitary or antiunitary operator, then $$UAU^{-1} = \int_{\sigma(A)} \lambda dUP^{(A)}(\lambda)U^{-1}\:.$$ In other words, the spectral measure $P^{(UAU^{-1})}(E)$ of $UAU^{-1}$ is just $UP^{(A)}(E)U^{-1}$.

Hence, the probability that the outcome of $A$ stays in $E\subset \mathbb{R}$ when the state is represented by the unit vector $\psi$ is $$||P^{(A)}(E)U \psi||^2 = ||U^{-1}P^{(A)}(E)U \psi||^2 = ||P^{(U^{-1}AU)}(E)||^2 = ||P^{(S_U(A))}(E) \psi||^2\:,$$ giving rise to the said interpretation of $S_U(A)$: acting on $A$ with $S_U$ and leaving fixed the state is equivalent to acting on $\psi$ with $U$ and leaving $A$ unchanged.

In particular, in particular regarding expectation values, $$\langle\psi| S_U(A) \psi \rangle = \langle U\psi| A \:U\psi \rangle$$

Similarly, $$||P^{(S^*_{U}(A))}(E)U \psi||^2 = ||U^{-1}UP^{(A)}(E)U^{-1}U \psi||^2 = ||P^{(A)}(E) \psi||^2\:,$$ giving rise to the said interpretation of $S^*_U(A)$: the action on $A$ with $S_U^*$ cancels the action of $U$ on $\psi$.

In particular, in particular regarding expectation values, $$\langle U\psi| S^*_U(A) U\psi \rangle = \langle\psi| A \psi \rangle$$


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