How Do Active and Passive Transformations Differ in Quantum Mechanics?

[etotheipi]

I decided to return to my favourite topic (heavy sarcasm implied...), because somehow this active/passive stuff still trips me up. Let's say we have some operator $A \in L(\mathcal{H}) : \mathcal{H} \rightarrow \mathcal{H}$, and also some unitary transformation $U$ between two sets of basis vectors $|y_i \rangle = U| x_i \rangle$ of the space $\mathcal{H}$. Then it's not too difficult to show that, presuming the bases are orthonormal, $$\begin{align*}

| \phi \rangle &= \sum_i \sum_j |y_i \rangle \langle y_i | x_j \rangle \langle x_j | \phi \rangle \\

\langle y_k| \phi \rangle &= \sum_i \sum_j \langle y_k |y_i \rangle \langle y_i | x_j \rangle \langle x_j | \phi \rangle = \sum_j \langle y_k | x_j \rangle \langle x_j | \phi \rangle \\

&= \sum_j \langle x_j | y_k \rangle ^{\dagger} \langle x_j | \phi \rangle

= \sum_j (\langle x_j | U| x_k \rangle )^{\dagger} \langle x_j | \phi \rangle
\end{align*}$$which tells us that the (column) representations of $|\phi\rangle$ w.r.t. the $\{ |x_i \rangle \}$ and $\{ |y_i \rangle \}$ basis are related by$$\mathcal{M}_y(|\phi \rangle) = \mathcal{M}(U^{\dagger}) \mathcal{M}_x(|\phi \rangle)$$Of course the representation of the operator $A$ must also change, so$$\mathcal{M}_y (A) = \mathcal{M}(U^{\dagger}) \mathcal{M}_x (A) \mathcal{M}(U)$$`That's what I would call a pure "passive transformation", i.e. we only changed the representations of the states and operators, but didn't actually change the states $\in \mathcal{H}$ and operators $\in L(\mathcal{H})$.

Now for the other way of doing things... instead, act the operator $U$ on the states, $$| \phi' \rangle = U | \phi \rangle$$Then, the expectation of $A$ becomes$$\langle \phi' | A | \phi' \rangle = \langle \phi | U^{\dagger} A U | \phi \rangle$$and it's stated that you could interpret this as a new operator $A' = U^{\dagger} A U$ acting on the original states, or the old operator $A$ acting on the transformed states. Confusingly, these two alternatives are also often termed 'active' and 'passive' viewpoints [e.g. as an example, the Schroedinger vs Heisenberg pictures]. To me, it looks like these last two options are both 'active' (i.e. in both cases, we're changing one or the other of the states or the operators, but not our coordinate system).

Is this a case of the 'active'/'passive' terminology being overloaded? Looks to me like a few distinct meanings get lumped under the same umbrella. Clarification would be much appreciated, thanks!

 

[vanhees71]

I don't know, what's the issue with this "active" and "passive" transformations. Usually the meaning is

"active:" You consider an experiment when set up in two different ways. E.g., you can do a double-slit experiment at one place $A$ and the same experiment at another place $B$. Formally that's described by a translation. If the outcomes of the experiment at $B$ are given by applying the corresponding translation operator to the outcomes of the experiment at $A$, one says that this translation is a symmetry. Since the spacetime model of Newtonian or special relativistic physics is translation invariant the physical laws should be translation invariant and that's why we expect that the outcomes of all experiments are "the same" in the above sense wherever we do this experiment.

"passive:" You consider an experiment and describe it in some way and then change the mathematical description to another set of quantities which map one to one to the original ones. If both descriptions are equivalent then there should be a transformation such that the description of the experiment leads to the same physical results and the physics should be invariant under such changes of description. In the above example that means you consider a double-slit experiment at a place $A$ and then describe it once in a reference frame with the origin placed in $A$ and then in a reference frame with the origin placed in $B$. Due to translation invariance of spacetime the description using both reference frames must give the same physical outcome.

In quantum theory a symmetry is expressed in terms of unitary operators (or antiunitary ones, but that is only necessary when you consider time-reversal symmetry, so let's stick to unitary transformations). Indeed quantum theory is invariant under arbitrary unitary transformations, i.e., if you transform all vectors like $|\psi \rangle \rightarrow |\psi' \rangle=\hat{U} |\psi \rangle$ and all operators representing observables by $\hat{A} \rightarrow \hat{A}' = \hat{U} \hat{A} \hat{U}^{\dagger}$ with a unitary operator $\hat{U}$, obeying $\hat{U}^{-1} = \hat{U}^{\dagger}$, all physically observable quantities like probabilities for the outcome of measurements, expectation values of observables, the possible values of operators (the spectra of the operators representing observables), etc. do not changed under such a unitary transformation.

A unitary transformation then is called a symmetry if the time evolution is invariant, i.e., if $\hat{H}'=\hat{U} \hat{H} \hat{U}^{\dagger}=\hat{H}$. This is of course equivalent to $[\hat{H},\hat{U}]=0$.