- #1
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!
| \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!
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