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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

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    $\begingroup$ "[...] passive transformations are supposed to be $U^\dagger A U$ applied to all operators $A$, leaving $|\psi\rangle$ unchanged." Can you give a source for that? $\endgroup$
    – noah
    Jan 17, 2022 at 13:30
  • $\begingroup$ @noah It's in Shankar's Principles of Quantum Mechanics, chapter on symmetries. page 284. Anyway, do you also agree that this is just an alternative way of doing an active transformation? $\endgroup$
    – Ryder Rude
    Jan 17, 2022 at 14:39
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    $\begingroup$ "IMO a passive transformation should be something that transforms |ψ⟩, while also applying some suitable changes to the operators to leave the expectation values unchanged". So you're just asking to be convinced why your preference isn't the same as the standard definitions? $\endgroup$ Jan 17, 2022 at 14:53
  • $\begingroup$ @BioPhysicist I didn't pull that opinion out of my ass. That's what passive transformation means in classical mechanics. Same state represented by different numbers. I don't see why the word "passive" should imply anything else $\endgroup$
    – Ryder Rude
    Jan 17, 2022 at 15:06
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    $\begingroup$ @RyderRude As stated in my answer, in a passive transformation we view the state vector as constant and the operators as changing $\endgroup$ Jan 17, 2022 at 15:37

2 Answers 2

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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.

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  • $\begingroup$ This seems like a good way to organize things. I'll be picky and say Field 1 can be a quasiprobability distribution in phase space, not just 1 or 0, and you're limiting things to only scalar Hamiltonians, but the ideas should generalize with no problems $\endgroup$ Jan 19, 2022 at 15:51
  • $\begingroup$ @QuantumMechanic Yeah, I'm trying to generalise the idea to include probability fields in field 1. That should make this closer to QM. I haven't understood quasiprobability distribution though. So far, I'm trying to make Field 1 a usual probability distribution. Or maybe a wave-function having complex values on the phase space points. (they exist in classical mechanics too. See KvN mechanics). $\endgroup$
    – Ryder Rude
    Jan 19, 2022 at 16:29
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    $\begingroup$ I agree. In literature "passive" is used with two different meanings, that are the real passive and the type 2 you mentioned, mostly in QM books. $\endgroup$ Nov 23, 2022 at 16:20
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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.

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  • $\begingroup$ "Note that expectation values should not depend on active vs passive distinction, as these are values we can actually measure." In classical mechanics, there is measurable difference between active and passive. They are not merely interpretations. $\omega (x,p)$ does not change its numerical value under a passive transformation. In an active transformation, it generally does. $\endgroup$
    – Ryder Rude
    Jan 17, 2022 at 15:09
  • $\begingroup$ So I should just take this to mean that a passive transformation in QM is not analogous to one in CM. $\endgroup$
    – Ryder Rude
    Jan 17, 2022 at 15:11
  • $\begingroup$ @RyderRude I am interested to know where classical mechanics has passive transformations that are just equal to coordinate transformations. I personally distinguish between the two. For example, wikipedia talking about rotations (en.wikipedia.org/wiki/Rotation_matrix; classical or quantum) says that active and passive both change according to OP's and Shankar's definitions $\endgroup$ Jan 17, 2022 at 15:47
  • $\begingroup$ @QuantumMechanic Please read my answer $\endgroup$
    – Ryder Rude
    Jan 19, 2022 at 12:44

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