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In QM, active transformations are $|\psi (a)\rangle=U(a)|\psi \rangle$. And passive transformations are supposed to be $U^{\dagger} A U$ applied to all operators $A$, leaving $|\psi \rangle $unchanged. My question is, isn't a passive transformation just a re-labeling of states, without actually changing any state function?

e.g. In classical mechanics, a passive transformation re-labels every point, and leaves the values of the state functions at each point unchanged. The state functions have to undergo a change in functional form (in terms of the new co-ordinates) to accomodate for this.

But, $U^{\dagger} A U$ changes the expectation values of the operators. So isn't it an active transformation?

IMO a passive transformation should be something that transforms $|\psi \rangle$, while also applying some suitable changes to the operators to leave the expectation values unchanged

BioPhysicist
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Ryder Rude
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2 Answers2

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I got an idea. Take (on classical phase space):

  1. A co-ordinate system

  2. Field1 - A probability distribution describing the state.

  3. Field 2- A Hamiltonian scalar field under which the object moves

A Passive transformation moves the co-ordinate system and leaves Field1 and Field2 unmoved.

An Active transformation type-1 moves Field 1, and leaves the co-ordinate system and Field 2 unmoved.

An Active transformation type-2 moves Field 2, and leaves the co-ordinate system and Field 1 unmoved

Type-1 and Type-2 are clearly two viewpoints of the same thing. Passive transformation is distinct from both.

Now, on to Quantum Mechanics : What we have named active and passive transformations in QM are actually analogous to Active Type-1 and Type-2 transformations respectively.

The analogue of a passive transform in QM would just be a change of basis, of both the state vector and the operators.

Ryder Rude
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The difference is in what is changing over time: the state vector or the operators.

In the active transformation, the state vectors change via unitary transformation as you have given:

$$|\psi (t)\rangle=U(t)|\psi(0)\rangle$$

while the operators $A$ are constant.

With a passive transformation the state vector $|\psi\rangle$ is constant, but the operators change over time.

$$A(t)=U^{\dagger}(t) A U(t)$$

So the active/passive distinction cares about the state vectors in terms of "active is changing" and "passive is not changing", not the operators or expectation values. Note that expectation values should not depend on active vs passive distinction, as these are values we can actually measure.

BioPhysicist
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