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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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    "[...] 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? – noah Jan 17 '22 at 13:30
  • @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? – Ryder Rude Jan 17 '22 at 14:39
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    "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? – BioPhysicist Jan 17 '22 at 14:53
  • @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 – Ryder Rude Jan 17 '22 at 15:06
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    @RyderRude As stated in my answer, in a passive transformation we view the state vector as constant and the operators as changing – BioPhysicist Jan 17 '22 at 15:37
  • I have never heard the "classical mechanics" opinion that passive transformations leave expectation values (etc) unchanged. That would be called a "coordinate transformation" in classical mechanics, not really a transformation at all. The two rigorous terms asked in your question may be https://mathworld.wolfram.com/AlibiTransformation.html and https://mathworld.wolfram.com/AliasTransformation.html – Quantum Mechanic Jan 17 '22 at 15:45
  • @QuantumMechanic Look at page 98 of Shankar's book (if you have it). It's written in the upper half : "This equation has content only if we are talking about the active transformations, for it is true for any o under a passive transformation". This part implies that Shankar's definition of passive transform leaves the state function's unchanged in numerical values. I personally agree with the definition because otherwise, active and passive transformations would be two words for the same thing, wouldn't they? – Ryder Rude Jan 17 '22 at 16:43
  • @QuantumMechanic Alias and Alibi seem to coincide with what I mean by active and passive, yes. – Ryder Rude Jan 17 '22 at 16:46
  • @QuantumMechanic The equation about which the book makes the statement is : $\omega (x,p)=\omega (x',p')$ – Ryder Rude Jan 17 '22 at 16:53
  • @QuantumMechanic There's a more direct definition. There's a heading right above it named "Active transformation". Read under it : "So far, we have viewed the transformation as passive: both (q, p) and (q', p') refer to the same point in phase space described in two different coordinate systems. Under the transformation (q,p) -(q',p'), the numerical values of all dynamical variables are unchanged" – Ryder Rude Jan 17 '22 at 16:57
  • @RyderRude I don't have Shankar's book here so I appreciate the quotes. I think the key word is "view." One can view the same transformation from two perspectives: active and passive. This is like asking how you get to the north pole: either you rotate your position on the earth, or you stand still and the earth rotates toward you; regardless, the physical situation is the same at the end and different from the original physical situation. Of course from this example you can see that one can interpolate between the two transformations by having both you and the earth rotate – Quantum Mechanic Jan 17 '22 at 18:01
  • @QuantumMechanic Noo, don't just focus on the word "view". Read the quote, man : "The numerical values of all dynamical variables remains unchanged". Shankar refers to state functions as dynamical variables. After some googling, I found that the word "passive" transformation is sometimes used for co-ordinate transfornation! It's just that, in the context of QM, it has a different meaning. I tried to connect the two definitions and that confused me – Ryder Rude Jan 18 '22 at 02:51
  • @QuantumMechanic Look at this https://physics.stackexchange.com/questions/51994/active-versus-passive-transformations . Active and passive are being used as alternative terms for alias and alibi . And then there's this : https://math.stackexchange.com/questions/1844714/active-and-passive-transformations-in-linear-algebra . Mathematicians also have a whole another definition – Ryder Rude Jan 18 '22 at 02:55
  • @RyderRude normally dynamical variables refers to things like position and momentum. The links you sent support the opposite of your point. Active transformations we agree on; passive transformations change the coordinates without changing the vector so the vector still moves relative to its surroundings. – Quantum Mechanic Jan 18 '22 at 15:35
  • @RyderRude regardless, the best answer is the final paragraph on your linked question https://physics.stackexchange.com/a/526101/291677: "This should tell you that any field redefinition obtained from a spacetime transformation can be seen in both active and passive interpretations and such vacuous names/interpretations hold no physical or mathematical value. What you should actually care about is how exactly you have defined your new fields and then everything else should follow irrespective of what your mental picture is." – Quantum Mechanic Jan 18 '22 at 15:35
  • @QuantumMechanics OK here's a proof that the QM definition and the definition in the link refer to two different things: The QM definition leaves the basis components of the all the vectors unchanged under a passive transformation. The passive transformation in the link I gave changes the basis components of the vectors. So they are two different definitions of a passive transformation. – Ryder Rude Jan 18 '22 at 16:28
  • @QuantumMechanic In the QM definition of a passive transformation, you leave the vector literally alone. As in, you don't express the same abstract vector in a different basis. You leave even the vector's components untouched. – Ryder Rude Jan 18 '22 at 16:44
  • @RyderRude I think I'll yield on that one. What I picture is that when you leave the vector literally alone but change the dynamical variables, it is as if you've switched to a different basis for the vector. Regardless I think your questions are exactly why it's more precise to use "Schroedinger representation" and "Heisenberg representation" to be precise in QM, and to remember that there are infinitely many "interaction pictures" that mix these up even further – Quantum Mechanic Jan 18 '22 at 18:09
  • @QuantumMechanjc About the distinction between alias and alibi: Take a Hamiltonian field, a co-ordinate system, and a ball. A passive transformation is when you move the co-ordinate system, keeping the ball and the field fixed. An active transformation is when you move the ball, keeping the field and the co-ordinate system fixed. Clearly, both are not merely viewpoints. The states you end up with are very distinct after both transformations. The transformations are equivalent in the link I gave because there's only a field there, and no ball. You need a third party to make the distinction. – Ryder Rude Jan 19 '22 at 04:29

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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  • 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 – Quantum Mechanic Jan 19 '22 at 15:51
  • @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). – Ryder Rude Jan 19 '22 at 16:29
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    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. – Mr. Feynman Nov 23 '22 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.

BioPhysicist
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  • "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. – Ryder Rude Jan 17 '22 at 15:09
  • So I should just take this to mean that a passive transformation in QM is not analogous to one in CM. – Ryder Rude Jan 17 '22 at 15:11
  • @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 (https://en.wikipedia.org/wiki/Rotation_matrix; classical or quantum) says that active and passive both change according to OP's and Shankar's definitions – Quantum Mechanic Jan 17 '22 at 15:47
  • @QuantumMechanic Please read my answer – Ryder Rude Jan 19 '22 at 12:44