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Classically, we obtain the equations of motion by finding a path which has an action that is stationary with respect to small changes in the path. That is the path for which:

$\delta S =0$

Scaling the action by a constant should therefore do nothing. However, some books seem to consider the overall sign of the action important (since if we changed the sign we could find a path with arbitrarily negative action). Does a scaling factor or change in sign actually matter?

For quantum mechanics we have:

$K(x,y;T) = \langle x;T|y;0 \rangle = \int_{x(0)=x}^{x(T)=y} e^{i S /\hbar } Dx $

Now it looks like scaling the action $S$ will cause changes in how paths will interfere, but the overall sign of the action looks like it still won't matter. So now we can somehow measure the absolute scale of the action?

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up vote 7 down vote accepted

Well, you have basically answered you question yourself. Rescaling of the action is the same thing as rescaling of the Planck's constant. Obviously this can't have any effect classically. But on the quantum level $\hbar$ measures the non-commutativity of observables and in the extreme limit $\hbar \to 0$ you recover classical mechanics.

As for the sign, this doesn't matter either classically or quantum mechanically. We are not interested in only minimizing the action but in all extrema. Reverting the sign only means that we swap the meaning of maximum and minimum but the solutions won't change at all.

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It's a quibble unrelated to the Question, but the limit $\hbar\rightarrow 0$ is delicate enough that one doesn't recover classical mechanics without other considerations. There is, for example, measurement incompatibility for all finite $\hbar$, so there is a discontinuity in the violation of Bell inequalities between $\hbar$ however so small and $\hbar=0$. Useful answer, nonetheless. – Peter Morgan Mar 20 '11 at 15:09
@Peter: No problem, good to point it out. I know it's delicate but I didn't want to get into gory details of quantization and deformations of Poisson algebra. But at least on the intuitive level it's true and in hand-wavy way it's considered quite often when talking about classical limit (e.g. when deriving Hamilton-Jacobi equation, WKB approximation, etc.). – Marek Mar 20 '11 at 15:32

For classical mechanics, the magnitude of a scaling factor does not matter, but the sign may depend on how you formulate your action principle. The least action principle can be taken a bit too literal by some, but as you note it is whether the path is stationary that is the more rigorous definition. For short enough time differences, in classical mechanics for particles the action is always a minimum (well, or a maximum depending on sign choice). See "When action is not least" by Taylor and Gray. However even for short times, the "path" of fields in classical mechanics are saddle points in the action. So at some point one needs to give up taking the "least" in least action literally.

For Quantum mechanics, the scaling will cause an effect. This in one of the root reasons why classical actions that lead to the same equations of motion, can actually lead to different quantum theories. So this effect is indeed measurable in principle. The sign however is still not measurable, as the sign choice is just a convention.

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Thanks for taking the time to explain why some people care about the sign. As for the quantum part, while it sounds reasonable, can someone explain mathematically why the sign of the quantum case isn't measurable? – John Mar 20 '11 at 11:49
@John: look at the path integral. Changing the sign is the same thing as complex conjugation of integrand and therefore also the whole integral. But quantum mechanics is a projective theory (remember that states correspond to rays, not vectors) and so this global change of phase can't change physics. – Marek Mar 20 '11 at 12:12

Your argument is absolute correct; a change of sign/scale factor won't change the solutions to the 'classical' equation $\delta S = 0.$ However, observables do change. If you've seen some field theory, you know an observable F is calculated like $$\langle F \rangle = \frac{\int\mathrm{d}\phi F[\phi] e^{iS[\phi]/\hbar}}{\int\mathrm{d}\phi e^{iS[\phi]/\hbar}};$$ changing the scale of S does change what you measure. It is indeed equivalent to a change of scale in $\hbar,$ so practically any observable will somehow change its value!

Furthermore, when authors say that the sign does matter, it helps to think about the link between quantum and classical field theories. For example, if you 'Wick rotate' the free QFT, you get $$S[\phi] = \int\mathrm{d}^Dx \frac{1}{2}(\nabla\phi(x))^2 + \frac{1}{2}m^2\phi(x)^2, $$ which is by construction positive everywhere: $S[\phi] \geq 0.$ In that case, expressions like $$\langle F \rangle = \frac{\int\mathrm{d}\phi F[\phi] e^{-\beta S[\phi]}}{\int\mathrm{d}\phi e^{-\beta S[\phi]}}$$ make sense, since the integrals converge. If you change the sign of $S$, the integrals suddenly diverge horrendously - you suddenly don't have a nice field theory anymore. This is a very general property: large field values and large deviations should be 'punished' by being exponentially supressed; if you change the sign, large values and deviations are being exponentially favoured.

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Euclideanized action is more like energy than standard action (which you apparently acknowledge by writing $\beta$ there) and therefore, like energy, one needs to have it bounded from below. But this is an unrelated issue to the complex phase change, in my opinion. – Marek Mar 20 '11 at 12:17
Surely one Wick rotates in the direction needed to make the integrals converge, not in an absolute direction. – Peter Morgan Mar 20 '11 at 15:01
@Marek: I definitely agree with you that it has very little to do with the phase change. But since John posted his question in very general terms (just look at the tags), I thought it might be useful to include an example of an action where a sign change does matter. – Gerben Mar 20 '11 at 17:51

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