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I was wondering about the dimensions of the Planck constant ($h$) and the dimensions of the action, which are obviously same. Then a train of thought led me to conclude that uncertainty principle can be stated as "minimum action required to make an observation is $\hbar/2$".

Also, principle of least action is the governing law for classical dynamics and uncertainty principle is a consequence of commutation relations which are stated as postulate in theory of quantum mechanics.

Now, my question is that can this postulate of quantum mechanics be interpreted/proved as setting the limit for action i.e. $|S| \ge kh$ or the vice-versa i.e. $|S| \ge kh \implies \Delta{x}\Delta{p} \ge \hbar/2$, where $k$ is some positive real number.

I have not seen this been interpreted like this. So this could be a trivial interpretation. If this is the case please refer me to the relevant reference. If this is not the case then this could have consequences for interpretation of action, uncertainty principle and QM in general, so please help prove or disprove this relation.

These are some of the observations that an answer may address:

  1. Principle of least action leads to Feynman's path integral formulation of quantum mechanics.
  2. Principle of least action leads to classical Hamiltonian mechanics which can be expressed through Poisson brackets notation which directly corresponds to commutators in QM.
  3. Classically, for SHM kinetic energy and potential energy are the same over a time period so casually it may seem that the action taken over a time period is 0 but considered more carefully the Lagrangian is a function of q and $\dot{q}$ which follow uncertainty principle so the integral can be non-zero.

Edit

Looking at the lack of responses, I have to ask - is this a question worth investigating further? If yes, can someone point to some authority in foundations of quantum physics? I am a amateur physicist at the present, working alone and do not know how to proceed.

prateek
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  • I doubt you can find a relation like this. The absolute value of the action is not a terribly meaningfully quantity. I can always add a constant $V_0$ to my potential energy and this will shift my action by $-V_0(t_1-t_0)$, without changing the path of least action. This means I can make the action as small as I like without changing any of the physics. – By Symmetry Feb 18 '21 at 15:45
  • I have never seen the uncertainty principle expresses in this way but I do not see anything wrong wit your interpretation. Whether or not it is trivial I will leave for others to address – Lewis Miller Feb 18 '21 at 15:47
  • In addition the action is bounded by $|S| \le (t_1-t_0)\mathrm{max}(|L|)$, so I can make my action very small by simply considering a very short time interval. What your saying would then imply that there is no classical limit over sufficiently short times. However I don't think this is true. The energy-time uncertainly relation cannot be treated the same way as the position momentum relation due to the lack of a time operator in QM and I have never seen a treatment of the classical limit that depended in an obvious way on the timescales involved. – By Symmetry Feb 18 '21 at 15:49
  • @BySymmetry " there is no classical limit over sufficiently short times" is that not generally the case when we take infinitesimal time? Also, I am not asking to treat energy-time uncertainty principle same as position-momentum one. But generalise the uncertainty principle in this way to mean that action is itself limited by planck constant. But your first comment poses serious problems to the interpretation, maybe it can be fixed by fixing the potential's zero point. – prateek Feb 18 '21 at 16:07
  • @prateek It is certainly not obvious to me. If I have a reasonably localized wave-packet with some average position an momentum, so that I can approximate it with a classical particle, then a short time later I will have a very similar wave-packet that I can presumably still approximate with a classical particle and who's average position an momentum will, by Ehrenfest's theorem, have evolved according to Newton's laws. This question of what happens at short times boils down to how you interpret the energy time uncertainty relation – By Symmetry Feb 18 '21 at 16:09
  • Related: https://physics.stackexchange.com/q/287514/2451 , https://physics.stackexchange.com/q/28957/2451 and links therein. – Qmechanic Feb 18 '21 at 16:10

1 Answers1

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Bohr-Sommerfeld quantization is based on a relationship like the one you are discussing \begin{equation} \oint \vec{p} \cdot d \vec{q} = n h, n \in \mathbf{N} \end{equation} where the integral is over a closed path in phase space, $n$ is an integer, and $h$ is Planck's constant. The left had side has units of action.

The modern point of view is that this relationship arises as a consequence of the WKB approximation (up to a small correction where $n\rightarrow n+1/2$). However, note that this approach only applies to bound systems.

In general there is no bound on the action in quantum mechanics. For example, in the path integral approach, there is a contribution from all possible paths weighted by $e^{i S/\hbar}$, where $S$ is the action of the path, and there is no bound on the action.

Andrew
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  • I am not meaning to redefine uncertainty principle only to interpret a bit differently. Also, about action, I am saying if we treat absolute value of action as a quantum observable then does it follow a minimum bound? In path integral approach, there is no bound on action but it is nowhere explicitly mentioned/determined that the quantum of action can approach arbitrarily close to 0. – prateek Feb 21 '21 at 18:01
  • @prateek I don't know exactly what you are proposing, but I don't see a way for it to work. Let's say we want $\langle S \rangle > k \hbar$, where $S$ is the "action operator", for some $k$. This can't be true in general. Think about an attractive square well potential. Consider a particle with a large momentum, so $p^2/2m > V$, then the Lagrangian is positive, and can be arbitrarily small. Then consider bound states, which by definition have $V<p^2/2m$, and so have negative Lagrangian. We can set up the potential so the highest lying bound state is negative and as close to zero as we want. – Andrew Feb 21 '21 at 18:31
  • I calculated $\langle |S| \rangle i.e. \int \langle |L| \rangle dt $, the expected value of absolute value of Lagrangian in the ground state of harmonic oscillator comes out to be $\hbar \omega \sqrt{2/\pi e}$ which is smaller than the ground state energy $\hbar\omega/2$. I am suggesting something along these lines. – prateek Feb 22 '21 at 10:08
  • @prateek I encourage you to explore it and see where it goes, but at the moment I don't see any reason that $\langle |S|\rangle$ should have a lower bound. – Andrew Feb 22 '21 at 13:23
  • for example, in 1d quantum mechanics, put a free particle into a Gaussian wavepacket with momentum $p$. Then the expected value of the action should be roughly $p^2 T/2m$ where $T$ is the observation time. For fixed $T$, you can take $p$ to be as small as you want and therefore make the action as small as you want. – Andrew Feb 22 '21 at 13:33
  • the quantity $p^2T/2m$ has to be greater than $\hbar/2$ because of energy-time uncertainty. And for a free particle, Lagrangain and Hamiltonian are same, so this should hold for both. – prateek Feb 22 '21 at 13:45
  • @prateek The $\Delta E \Delta T$ uncertainty relationship is more subtle than the other ones and doesn't directly apply here. For example, a free particle can be in a delta function state in momentum (zero uncertainty in energy), and since momentum is conserved for a free particle this is a stationary state. So in that case, $\Delta E = 0$, and $T$ can be arbitrarily long or short. – Andrew Feb 22 '21 at 14:52
  • actually sorry, I realized the $\Delta E \Delta T$ uncertainty relationship is relevant. For $\Delta E=0$, we know the energy exactly, so the state is a stationary state, so the lifetime is infinite. However this doesn't prevent us from taking $T$, the time over which we integrate the action, to be as long or as small as we want. Furthermore, the actual value of the energy $E$ can be as large or as small as we want. So we can easily arrange an average action value of $\ll \hbar$ by computing the action for a free particle over an interval $T$ with energy $E$ such that $ET \ll \hbar$. – Andrew Feb 22 '21 at 18:12