Is there any sense in which we can consider Action a physical observable? What would experiments measuring it even look like? I am interested in answers both in classical and quantum mechanics.

I ran across a physics textbook called "Motion Mountain" today, with volumes covering a broad swath of physics, written over the past decade by a dedicated German physicist with support from some foundations for physics outreach. So it appears to be a serious endeavor, but it's approach to many things is non-standard and often just sounds wrong to me. In discussing his approach to some topics the author says:

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.


Further on the author also discusses measurements of this physical observable, saying

No single experiment yields [...] action values smaller than hbar

So I think he does mean it literally that action is physically measurable and is furthermore quantized. But in what sense, if any, can we discuss action as an observable?

My current thoughts:
Not useful in answering the question in general, but hopefully explains where my confusion is coming from.

I will admit, as chastised in the quote, I did not learn this in university and actually it just sounds wrong to me. The textbook covers classical and quantum mechanics, and I really don't see how this idea fits in either.

Classical physics
The system evolves in a clear path, so I guess we could try to measure all the terms in the Lagrangian and integrate them along the path. However, multiple Lagrangians can describe the same evolution classically. The trivial example being scaling by a constant. Or consider the Lagrangian from electrodynamics which includes a term proportional to the vector potential, which is itself not directly measurable. So if Action was actually a "physical observable", one could determine the "correct" Lagrangian, which to me sounds like nonsense. Maybe I'm reading into the phrasing too much, but I cannot figure out how to interpret it in a manner that is actually both useful and correct.

Quantum mechanics
At least here, the constant scaling issue from classical physics goes away. However the way to use the Lagrangian in quantum mechanics is to sum over all the paths. Furthermore, the issue of the vector-potential remains. So I don't see how one could claim there is a definite action, let alone a measurable one. Alternatively we could approach this by asking if the Action can be viewed as a self-adjoint operator on Hilbert space ... but the Action is a functional of a specific path, it is not an operator that acts on a state in Hilbert space and gives you a new state. So at first blush it doesn't even appear to be in the same class of mathematical objects as observables.

Ultimately the comments that experiments have measured action and show it is quantized, make it sound like this is just routine and basic stuff I should have already learned. In what sense, if any, can we discuss action as a physical observable? What would experiments measuring it even look like?

  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/9686/2451, physics.stackexchange.com/q/41138/2451 and links therein. $\endgroup$
    – Qmechanic
    Jul 23, 2017 at 3:08
  • $\begingroup$ Classically, if energy is measurable at all times then integral of energy over time is measurable. For a particle is integral of 1/2mv^2. I believe we should check the mathematical definitions in physics of action not deduce them only from a rant about teaching philosophy that pre-supposes we are familiar with the definition. Definition of action: en.m.wikipedia.org/wiki/Action_(physics) Stationary Action Principle: en.m.wikipedia.org/wiki/Stationary_Action_Principle , should be in the question $\endgroup$
    – Al Brown
    Jul 28, 2021 at 22:03
  • $\begingroup$ Also relevant... quora.com/Is-Motion-Mountain-Physics-a-good-reference-book $\endgroup$
    – Andrew
    Jul 29, 2021 at 2:53

3 Answers 3


The book propagates a myth.

Experiments measure angular momentum, not action - even though these have the same units. One finds empirically that angular momentum in any particular (unit length) direction appears in multiples of $\hbar/2$, due to the fact that its components generate the compact Lie group SO(3), or its double cover U(2).

That Planck's constant $\hbar$ is called the ''quantum of action'' is solely due to historical reasons. It does not imply that the action is quantized or that its minimal value is $\hbar$. Early quantum theories such as Bohr-Sommerfeld approximation used quantized action but this was an approximation to the more general quantization of Dirac, etc... One must additionally take care not to confuse the word action in "action-angle coordinates" with the action of variational calculus.

Indeed, the action of a system defined by a Lagrangian is a well-defined observable only in the very general and abstract sense of quantum mechanics, where every self-adjoint operator on a Hilbert space is called an observable, regardless of whether or not we have a way to measure it. The action of a system along a fixed dynamically-allowed path depends on an assumed initial time and final time, and it goes to zero as these times approach each other - this holds even when it is an operator. Hence its eigenvalues are continuous in time and must go to zero when the time interval tends to zero. This is incompatible with a spectrum consisting of integral or half integral multiples of $\hbar$.

  • 3
    $\begingroup$ The value of the action between two time slices aka Hamilton's function is an observable on the phase space. $\endgroup$ Dec 5, 2020 at 8:09
  • $\begingroup$ @Prof.Legolasov: How can it be observed? $\endgroup$ Dec 6, 2020 at 15:15
  • 2
    $\begingroup$ it is a function on the phase space, therefore it can be observed by observing the coordinate and momentum and plugging them into the model-specific formula for the Hamilton-Jacobi function. In QM it becomes an operator that acts on the Hilbert space. $\endgroup$ Dec 12, 2020 at 7:18
  • 1
    $\begingroup$ Arnold, can you answer the post below? It suggests that there is an error in your argument. $\endgroup$
    – user85598
    Jan 26, 2021 at 6:28
  • 1
    $\begingroup$ @Christian: By the way, the answer of Motion Mountain does not exhibit an error in my arguments but proves his contrary position by arguments from authority, which carry little substantial weight. $\endgroup$ Jan 26, 2021 at 21:16

In the language of the OP, action is a functional, since it is an integral of the Lagrangian... but over an arbitrary path. In other words, it is an abstract mathematical object, which has no counterpart in the real world.

This functional is then minimized in respect to all possible trajectories. In quantum mechanical terms the action along the optimal trajectory corresponds to the phase of a wave function, which is measurable (although defined up to a constant), e.g., in the experiments on Aharonov-Bohm effect, and any other interference experimente. The fact was recognized long before the advent of the path integrals - Landau&Livshitz derive quasiclassical approximation is an eiconal expansion of the phase of the wave function, which they openly call action.

  • $\begingroup$ I believe youre supposed to take the actual path. It cannot be calculated if the path isnt known. Secondly, recall the principal of least action when calculating: en.wikipedia.org/wiki/Principle_of_least_action Classically, if energy is measurable at all times then integral of energy over time is measurable. For a particle is integral of 1/2mv^2. $\endgroup$
    – Al Brown
    Jul 28, 2021 at 22:05
  • 1
    $\begingroup$ @AlBrown Not sure what you disagree with, since you say yourself that action is known only for a specified path. It is like a function: $f(.)$ is an abstract object that cannot be measured, but for any given point $x$ the value of function at this point, $f(x)$ is a number that could be measurable. $\endgroup$ Jul 29, 2021 at 12:47
  • $\begingroup$ That makes sense. I see $\endgroup$
    – Al Brown
    Aug 1, 2021 at 22:28

One can measure an on-shell action by counting cycles and noting the phase within the cycle.


$∂τ$ is a proper time step of the system consisting of observable simultaneous components. By observing the components one can find repeated patterns. Basically all systems cycle.

$τ=∫∂τ$ is the constant information of the system or Hamilton's principal function.

$$0=\frac{dτ}{dt}=\frac{∂τ}{∂x}\;\frac{∂x}{∂t}+\frac{∂τ}{dt}\\ W=\frac{∂τ}{dt}=-\frac{∂τ}{∂x}\;\frac{∂x}{∂t}=-H\\ H=pẋ=mẋ²$$

Note, $p = mẋ$ is by observation, i.e. the physical content. Then it is assigned to $\frac{∂τ}{∂x}$ by definition, leading to $H=mẋ²$. $∂τ=mẋ∂x$ then says that $ẋ$ and $∂x$ contribute independently to a time step $∂τ$.

Splitting off some non-observable part of the system and associating it to the location of the observed part $H(ẋ,x)=T(ẋ)+V(x)$ half-half, makes $T(ẋ)=mẋ²/2$. Half-half is the usual but not mandatory choice for $V$.

$W=-H$ stays constant. It is the cycle's information divided by the period time. $I=∮(∂τ/∂t)dt=∮(-H)t=Wt$. One cannot minimize $∫Wdt$ because it increases monotonously, counting until the system ceases to exist.

$L(x)=mẋ²+W=mẋ-H$ oscillates and returns to 0 in a cycle. $J=∫Ldt$ returns to the same value after one or many cycles. Minimizing this produces conditions to attribute observables to the same system time step (the equations of motion).

$$0 = \frac{δJ}{δx} = \frac{1}{δx}∫\left(δx\frac{∂L}{∂x}+δẋ\frac{∂L}{∂ẋ}\right)dt = \frac{1}{δx}∫δx\left(\frac{∂L}{∂x}-\frac{d}{dt}\;\frac{∂L}{∂ẋ}\right)dt \\ \frac{∂L}{∂x} = \frac{d}{dt}\;\frac{∂L}{∂ẋ} \\ F=ṗ$$

One can measure an action satisfying the equations of motion by counting cycles and noting the phase within the cycle.

Measuring our every-day time is also via measuring action.

Comparing system changes to our time unit motivates energy $W$. $W=\frac{∂τ}{dt}$ is system time divided by our time, which is frequency times a factor to keep the units consistent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.