Is action truly an observable in physics?

[JustinLevy]

I ran across one of Christoph Schiller's physics posts, and this one got me scratching my head:

http://www.motionmountain.net/wiki/index.php/Teach_clearly!#On_action_as_an_observable
"On action as an observable
Numerous physicists finish their university studies without knowing that action is a physical observable. Students need to learn this. Action is the integral of the Lagrangian over time. It is a physical observable: action measures how much is happening in a system over a lapse of time. If you falsely believe that action is not an observable, explore the issue and convince yourself - especially if you give lectures."

Is there any sense in which the viewpoint he is pushing is true?
Since momentum and position have an uncertainty relation, it seems like the naive means of trying to build an observable from an integral containing the observables of position and momentum would fail. Or in the path integral formulation of quantum mechanics, it seems clear to me that one cannot uniquely specify the action between an initial and final state, because the whole point is that multiple paths are taken.

It's always interesting to learn another viewpoint or interpretation, but this just appears to be plain wrong. So please let me know if I am missing something here.

 

[zenith8]

I ran across one of Christoph Schiller's physics posts, and this one got me scratching my head:

http://www.motionmountain.net/wiki/index.php/Teach_clearly!#On_action_as_an_observable
"On action as an observable
Numerous physicists finish their university studies without knowing that action is a physical observable. Students need to learn this. Action is the integral of the Lagrangian over time. It is a physical observable: action measures how much is happening in a system over a lapse of time. If you falsely believe that action is not an observable, explore the issue and convince yourself - especially if you give lectures."

Is there any sense in which the viewpoint he is pushing is true?
Since momentum and position have an uncertainty relation, it seems like the naive means of trying to build an observable from an integral containing the observables of position and momentum would fail. Or in the path integral formulation of quantum mechanics, it seems clear to me that one cannot uniquely specify the action between an initial and final state, because the whole point is that multiple paths are taken.

It's always interesting to learn another viewpoint or interpretation, but this just appears to be plain wrong. So please let me know if I am missing something here.

In the de Broglie-Bohm formulation of quantum mechanics, the action S is related to the phase of the Schroedinger wave function. I say 'related to' rather than 'equal to' since the S that appears in the relevant `quantum Hamilton-Jacobi' equation is a function of the endpoints of the path whereas the action has fixed endpoints (something like the difference between a definite or an indefinite integral). The phase of the wave function has observable consequences (Berry etc.) though don't push me for details.

In terms of path integrals, in de Broglie-Bohm theory you can reduce the sum over an infinite number of paths to a single path (corresponding to the one the electron actually follows). You can therefore uniquely define the action between the initial and final state. The only catch is that you have to include the `quantum potential' in the definition of the Lagrangian that you integrate along the path, as well as the classical potential - but this is hardly surprising since that is what makes quantum trajectories different from classical ones.

I explained all this in https://www.physicsforums.com/showpost.php?p=2677122&postcount=23" if it helps.

 

[tom.stoer]

Formally the action is a self-adjoint operator and can therefore be used as an observable.

What else is meant here by the term "observable"?

 

[Maaneli]

In terms of path integrals, in de Broglie-Bohm theory you can reduce the sum over an infinite number of paths to a single path (corresponding to the one the electron actually follows). You can therefore uniquely define the action between the initial and final state. The only catch is that you have to include the `quantum potential' in the definition of the Lagrangian that you integrate along the path, as well as the classical potential - but this is hardly surprising since that is what makes quantum trajectories different from classical ones.

I explained all this in https://www.physicsforums.com/showpost.php?p=2677122&postcount=23" if it helps.

What's beautiful about this is that, in order to determine the initial quantum potential, all you need to know is the initial |psi|^2 distribution, and then you can calculate the general wavefunction for all future times from that single path integral.

 

[Demystifier]

I think the guy is talking about the observable in CLASSICAL (not quantum) mechanics. (Which does not mean that what he says makes much sense.)

 

[Maaneli]

Formally the action is a self-adjoint operator and can therefore be used as an observable.

What else is meant here by the term "observable"?

It's worth adding that the sense in which action is an "observable" is that it is deduced from position measurements of an ensemble of particles, and not a single particle. Merely calling action a physical "observable" obscures this fact.

 

[Maaneli]

BTW, Zenith, I have a small gripe to pick with this statement in your linked post:

<< Note that this elevates the de Broglie-Bohm theory from being an 'interpretation' to a mathematical reformulation of quantum mechanics equivalent in status to Feynman's. >>

The standard formulation of pilot-wave theory (S.E. plus G.E.) is already sufficient to make it a full-fledged alternative *theory* of QM, and not merely a mathematical reformulation (let alone just an interpretation) of QM. So, it's status is rather different from Feynman's path integral formulation, since the latter is really just a mathematical formulation of standard QM (it has no ontology, requires the usual measurement postulates, and has no apparent possibility of leading to predictions which differ from standard QM in the Schroedinger or Heisenberg pictures).

 

[zenith8]

BTW, Zenith, I have a small gripe to pick with this statement in your linked post:

<< Note that this elevates the de Broglie-Bohm theory from being an 'interpretation' to a mathematical reformulation of quantum mechanics equivalent in status to Feynman's. >>

The standard formulation of pilot-wave theory (S.E. plus G.E.) is already sufficient to make it a full-fledged alternative *theory* of QM, and not merely a mathematical reformulation (let alone just an interpretation) of QM. So, it's status is rather different from Feynman's path integral formulation, since the latter is really just a mathematical formulation of standard QM (it has no ontology, requires the usual measurement postulates, and has no apparent possibility of leading to predictions which differ from standard QM in the Schroedinger or Heisenberg pictures).

Absolutely fair enough - I agree. Mind you, I did write that post quite a while ago (and I think I borrowed the phrase from something I read anyway..).

 

[Maaneli]

Mind you, I did write that post quite a while ago (and I think I borrowed the phrase from something I read anyway..).

Ya, it's from Mike Towler's deBB lecture course.

 

[zenith8]

Ya, it's from Mike Towler's deBB lecture course.

That'll be it..

 

[Demystifier]

In terms of path integrals, in de Broglie-Bohm theory you can reduce the sum over an infinite number of paths to a single path (corresponding to the one the electron actually follows). You can therefore uniquely define the action between the initial and final state. The only catch is that you have to include the `quantum potential' in the definition of the Lagrangian that you integrate along the path, as well as the classical potential - but this is hardly surprising since that is what makes quantum trajectories different from classical ones.

I explained all this in https://www.physicsforums.com/showpost.php?p=2677122&postcount=23" if it helps.

Is there a generalization of this idea to particles (not necessarily relativistic) with spin?

 

[tom.stoer]

I think you can use the Pauli-equation for the wave function but still use the same guiding equation for the particles. That means spin is a property of the wave function, not of the particles in the Broglie-Bohm theory

 

[Maaneli]

I think you can use the Pauli-equation for the wave function but still use the same guiding equation for the particles. That means spin is a property of the wave function, not of the particles in the Broglie-Bohm theory

I think one can also use the generalized quantum potential which includes a spin-dependent contribution.

 

[tom.stoer]

I think this is the same. The quantum potential is caculated via the Pauli equation and has therefore non-vanishing spin dependent terms. Forrmally the guiding equation is not affected.

 

[Maaneli]

Zenith, another thing. You wrote:

"So to conclude, Feynman's paths are mathematical tools for computing the evolution of , while (if it is the case that particles actually exist) one among the de Broglie-Bohm paths is the actual motion of the particle as deduced from the equations of QM, which exists in addition to the wave field."

This is not all - The deBB particle path (velocity) between two points in spacetime is also the amplitude-weighted mean of all the Feynman paths (velocities) between those two points in spacetime. You can see this by plugging in the Feynman path-integral into the guiding equation and computing the first derivative. This point was originally made in Antony Valentini's PhD thesis. IMO, this strongly suggests that the deBB particle path (velocity) is some kind of average, rather than an exact property of the particle.

 

[Maaneli]

I think this is the same. The quantum potential is caculated via the Pauli equation and has therefore non-vanishing spin dependent terms. Forrmally the guiding equation is not affected.

I'm referring to some thing else - the guiding equation obtained from the NR limit of the Dirac current yields an additional spin-current component to the usual NR guiding equation. If one takes the time derivative of this modified guiding equation and obtains the resulting quantum Hamilton-Jacobi equation, one finds that this quantum Hamilton-Jacobi equation also has an additional quantum potential due to the spin-current term. This is in addition to the terms arising from the spinor properties of the wavefunction.

 

[tom.stoer]

I have to aks again: The guiding equation for the n-th particle reads

[tex]\frac{dq_n}{dt} = \frac{1}{m_n}\nabla_n S[/tex]

with S derived from the wave function. Using the Pauli equation, S itself changes due to spin, but the form of the equation itself is not affected.

Are you saying that using the Dirac equation the guiding equation itself is changed, i.e. that the r.h.s has a more complicated form? If yes, how does it look like?

 

[Maaneli]

I have to aks again: The guiding equation for the n-th particle reads

[tex]\frac{dq_n}{dt} = \frac{1}{m_n}\nabla_n S[/tex]

with S derived from the wave function. Using the Pauli equation, S itself changes due to spin, but the form of the equation itself is not affected.

Are you saying that using the Dirac equation the guiding equation itself is changed, i.e. that the r.h.s has a more complicated form? If yes, how does it look like?

Yes, that's what I'm saying. See equations 2.8 - 2.10 and 4.1 - 4.4 of this paper:

Implications of Lorentz covariance for the guidance equation in two-slit quantum interference
Peter Holland, Chris Philippidis
http://arxiv.org/abs/quant-ph/0302076

 

[tom.stoer]

But you can absorb the log term in an appropriate S'.

That's what I am saying: the form of S does change, but the guiding equation always contains the gradiant of an appropriate S.

 

[Maaneli]

But you can absorb the log term in an appropriate S'.

That's what I am saying: the form of S does change, but the guiding equation always contains the gradiant of an appropriate S.

Of course one can arbitrarily define things, but the point is that the physics and the trajectories are different.

Also, these additional terms are not a consequence of the spinor properties of the Pauli wavefunction (they persist even in the Schroedinger case).

 

[JustinLevy]

Thanks for all the responses, but this thread seems to have gotten off topic.

Formally the action is a self-adjoint operator and can therefore be used as an observable.

What else is meant here by the term "observable"?

In the quote from the first post, Schiller calls it a "physical observable". So I assume he means it can be measured and has direct physical meaning. But that seems impossible to me.

The action being a functional, it takes in a path of the fields through configuration space and spits out a number. In quantum mechanics, we cannot measure this path / which path is taken, so how can the action possibly be observable?

I think the guy is talking about the observable in CLASSICAL (not quantum) mechanics. (Which does not mean that what he says makes much sense.)

Alright, let's consider this classically.

If I changed the action equation describing a system by a constant:
[tex]S \rightarrow S + C_0[/tex]
the equations of motion yielded from the stationay paths are not affected. So at most, only the change in action is measurable.

However, if I scaled the Lagrangian by a constant
[tex]\mathcal{L} \rightarrow C_0 \mathcal{L}[/tex]
the equations of motion yielded from the stationay paths are not affected either. So the magnitude of the action cannot be measured either.

There are an infinite number of functionals the action could be and still yield the same motion for a classical system. So if someone said: the action the system went through between time t1 and time t2 is 10 Joule-sec, this statement tells you nothing physical.

So it seems we can't just measure the action of a classical system. At least classically though, we could precisely measure the path everything takes. If one _defined_ the action in terms of the state variables, one could then essentially _define_ the action of each path. But the value of the action along a path itself is not measureable. It would be like claiming classically we can measure the components of the electrodynamic vector potential. We can't, as that would be claiming we can measure the gauge the universe chose. But classically we can measure the scalar [itex]\mathbf{E}\cdot\mathbf{B}[/itex] for the electric or magnetic fields, which is gauge independent. Similarly it seems classically we can measure the paths objects take, but not the action.

 

[tom.stoer]

In the quote from the first post, Schiller calls it a "physical observable". So I assume he means it can be measured and has direct physical meaning. But that seems impossible to me.

The action being a functional, it takes in a path of the fields through configuration space and spits out a number. In quantum mechanics, we cannot measure this path / which path is taken, so how can the action possibly be observable?

It depoends what you mean by "observable". An observable in QM is nothing else but a self-adjoint operator; so L is an observable. Of course it need not make sense to calculate S via a time integral, but formally that doesn't matter.

"Observable" does not mean that you can measure it in practice; the theory only tell's you that you can do it in principle.

 

[JustinLevy]

It depoends what you mean by "observable".

Okay, so I guess some of this can be reduced to a terminology issue. Thanks for clarifying that.

"Observable" does not mean that you can measure it in practice; the theory only tell's you that you can do it in principle.

I still don't understand this though. It seems impossible even in principle, for two reaons:
1) since in the path integral formulation any path can be (and is) taken, it is impossible even in principle to specify the action, unless all paths had the same action.

2) The action tells you about something of the system at many points in past as well as present. Since different paths have different actions, if this was measureable, it would mean you could "collapse" the history of possible paths to what you measured. Your measurement would have to affect the past!

So it seems to me, that while one may give a particular operator the name: Action. It seems that such a measurement is not possible even "in principle". So defining an observable as above, is not sufficient to mean it can be measured in principle. For example, could one construct a "global phase observable" using the above definition? And that clearly is not measureable even in principle.

 

[tom.stoer]

Let me ask a question: is the scattering matrix an observable?

 

[JustinLevy]

Well, we can construct it as a self adjoint operator that operates on a state to give us a new state. So in that sense, sure. If we did a measurement, it would collapse the state into one of the possible eigenstates of the operator.


The "action" though, cannot possibly be an operator on hilbert space. It doesn't take a state and give us a new state. It is a functional. It requires a whole path through hilbert space, and spits out a number. So it can't even have eigenstates.

Maybe the previous arguments against the action being measurable even in principle should have instead made me reconsider if the action was even an operator on hilbert space at all. So the more I think about it, I'm beginning to doubt that one can consider action as a self adjoint operator. Is there some different sense in which we can consider it an operator on hilbert space?

 

[tom.stoer]

Let my explain why I asked this question: the S matrix in QFT is to be understood as an operator; so we are not only interested in its matrix elements [tex]S_{fi} = <f|S|i>[/tex] but in the operator S itself.

Now the S-matrix is not an observable in the formal sense as S is a unitary operator (not a self-adjoint operator).

The formal definition of S in the interaction picture is

[tex]S \propto T e^{-i\int dt H_{int}(t)}[/tex]

where the integral runs over minus infinity to plus infinity and T means the time ordered product.

This similar to the action in the part integral where one uses

[tex]e^{i\int dt L(t)} = e^{iS}[/tex]

I think that it should be possible to promote this expression to an unitary operator as well which means one can interpret S as observable in the formal sense.

 

[JustinLevy]

Hmm... you are right, the scattering operator S is a unitary operator, not a self-adjoint operator. Now I'm confused. Why are we able to measure the matrix elements?

Obviously we can see what goes in and what comes out. Can we only measure the magnitude of the matrix element, and therefore not the phase, and therefore what we call a "measurement" of a particular in -> out matrix element is actually a self-adjoint operator made from the unitary S-matrix?


I need to think more about your Action example. <f|i> can be related to a path integral, with contains a reference to the action ... but the action is a functional assigning a number to each path, which the integral then sums. This doesn't sound like an operator on a hilbert space at all. An operator should operate on a single state, mapping to new states. The action, being a functional, takes in an entire path mapping to a number... this seems incredibly distinct to me.

I have a feeling I'm misunderstanding your point quite severely. Thank you for having the patience to discuss, and answer so many of my questions. It is very much appreciated, and I (hopefully) am learning to think about these things more precisely.

---
EDIT:
Still thinking, but is the following correct at least? (Getting back to the openning post)

There isn't an "action observable" in the sense that given an initial and final state it tells you how much "action"/"change" occurred. I assume no, because there are many paths connecting any two states, and the answer is different for each path.

Along your suggestions, that however doesn't mean the Lagrangian cannot be interpretted as an operator -- an infinitesimal generator of time evolution in some sense? I'm still confused on how to think about this point succinctly. Instead of e^iS, what if I just took this action operator S, and stuck it between two states to get a matrix element. What does it tell me?

 

[tom.stoer]

In order to understand S as an observable in the formal sense you must get rid of paths. An observable exists as an operator in a Hilbert space where no path does exist.

All what I wanted to do is to start with a construction of something like an operator-valued action S in a Hilbert space. The scattering matrix detour is because of two reasons: Firstly it should explain why the time integral may not be a problem for the action, either. There are examplex where physically meaningful operators can be constructed which are not defined for one single instant of time but for a time interval. Secondly it should explain that perhaps we do really need a self-adjoint operator but that perhaps a unitary operatoir is sufficient as we can reconstruct enough information from it.

Once you have constructed S living in a Hilbert space you can start to find out how it is related to a classical action along a specific path. One idea is to sandwich S between energy eigenstates which relates the matrix elements to something like the transition induced by the interaction <m|V|n>. Another idea is to sandwich S between peaked states related by a classical trajectory. So one could e.g. introduce Gaussian wave pakets for t=0 and t=T located at the classical positions for t=0 and t=T respectively and check what this S tells you.

To be honest: I don't know anything about the result; I simlyl want to indicate that mathematically it may work, wereas physically the interpretation and the relation to the classical S is less clear.

(I am not referring to dBB here as in dBB the action play a different role)