# Difference between active and passive transformations.

1. Oct 7, 2012

### thepopasmurf

I'm taking a quantum field theory course and the topic of active vs passive transformations came up. I have previously taken a physics course and active/passive transformations were never explicitly discussed.

What is the difference between the two?

In particular I'm trying to follow the following argument:

Consider a scalar field f(x) which transforms under a Lorentz transformation x->L(x).

It transforms as

f(x) -> f'(x) = f(L_inverse(x))

Why is it L_inverse(x) instead of L(x)?

Thanks.

2. Oct 7, 2012

### HallsofIvy

Staff Emeritus
That's not a mathematics question. I am going to move this to "quantum physics".

3. Oct 8, 2012

### haushofer

A scalar function f(x) transforms as

f'(x') = f(x)

where x'= Lx. Applying the inverse of L on both arguments of this relation gives your relation. The difference between passive and active is whether you apply the transfo on the point on the manifold, or the coordinate chart describing it.

4. Oct 8, 2012

### haushofer

For some extra fun, this topic could be moved to thr GR-forum :P

5. Oct 11, 2012

### Ilja

Assume you have a field A_i(x) which is itself not gauge-invariant but interacts in the usual gauge-invariant way with some other field \psi.

Now you can apply gauge transformations in the usual way. But there are two possible meanings of them: First, they simply describe another choice of the basis for \psi. To obtain the same formula for the interaction, we also have to change the definition of A_i(x). This is what is named passive gauge transformation.

But there is also another possibility. We replace A_i(x( by another field, without changing anything in the coordinates. The field A'_i(x) is, then, another, physically different field. If we are only able to observe this difference via its interaction with \psi(x), and, moreover, cannot even compare the phases of \psi(x) in different points, we may be unable to observe this difference. But if we have other possibilities to observe, given the assumption of non-gauge-invariance ot the theory it would be possible to distinguish these two field configurations.

The Bohm-Aharonov effect may be used to illustrate this. Let's cover some torus-like part outside the solenoid with two charts. Now, we can use arbitrary passive gauge transformations on above charts. But this will not change the size of the Bohm-Aharonov effect - it is only a change in the description of the same fields.

But if we consider, instead, an active one, say, only on one of the charts, and only on the part which does not intersect with the other chart, so that the particle going through obtains an additional phase factor. On this chart, this active transformation would be described by the same formula as a gauge transformation. Given that the gauge transformation would be constant in the part where the first chart interacts with the second one, we have no problem to combine the modified field on one chart with the unmodified one on the other. But the field as a whole is now
really a different one, because it gives another size for the Bohm-Aharonov effect.

Let's compare with the first, passive situation. Again, we change nothing on one chart, and use the same transformation, which seems nontrivial only inside the other chart, on the second chart. But we can see that now the formulas of how to transform the field \psi(x) from one chart to the other has to be changed - once we have changed the coordinates on one chart but not on the other. For at least one of the two intersections between the two charts this transformation has to be nontrivial. As the result, the correct passive transformation - as a simple change of coordinates - does not change anything in physics, so, also not the Bohm-Aharonov effect.

So, we have to be careful to distinguish a modification of the field itself - as a physical field - from a gauge transformation caused by a change of coordinates.

6. Oct 11, 2012

### ianhoolihan

What are the chances you are doing Partt III thepopasmurf? If so, I posted a similar question (https://www.physicsforums.com/showthread.php?t=643070) which follows from yours. Anyway, see the notes of Tong that I linked in that post --- it's a few pages before, and about a temperature field. If I rotate my field actively, by sending $x\to x'$, and if I want to write the new field in terms of the old one, I've got to undo this with the inverse.