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The Uncertainty Principle states that the more you know about a particles position, the less you know about its velocity and vice-versa, and there is an equality $\sigma_x\sigma_p \geq \frac{h}{4\pi}$ which states this in a more precise way. I was wondering if this inequality could be altered and generalized to relate different derivatives of position, such as position and acceleration, velocity and acceleration, velocity and jerk, or really any two (or possibly even more than two) derivatives of position. I do not have a physics background.

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    $\begingroup$ What would be a time derivative of the position in quantum mechanics? $\endgroup$
    – GiorgioP
    Jul 3 at 5:59
  • $\begingroup$ As I said, I do not have a physics background. I had assumed it would have been velocity, but based on your question, I'm assuming that the time derivative of position doesn't make much sense in quantum mechanics. $\endgroup$ Jul 3 at 6:13
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It is possible to "generalize" the Uncertainty Relations (UR), in the sense that it is possible to obtain other UR's, from the algebra of position and momentum, although only for observables that depend on them.

The starting point is the generalized form of the UR, valid for the two generic symmetric operators $A$ and $B$: $$ \sigma_A \sigma_B \geq \left| \frac{1}{2i} \langle[A,B] \rangle \right|, $$ where $\sigma_A$ and $\sigma_B$ are the standard deviations of the distributions of the values of the two observables in a given quantum state, and $\langle[A,B] \rangle $ is the expectation value of the commutator $AB-BA$ in the same state.

Therefore, all we need is the commutator of $A$ and $B$. This can be easily obtained from the basic commutator of positions and momenta $[x_i,p_j]=i\hbar \delta_{ij}$ for any observable which could be written as an analytical function of them (one has to use some basic commutator identities).

Notice that to obtain the quantum UR for classical quantities, one has to provide a reasonable expression of these quantities as operators in terms of position and momentum. For example, a velocity operator can be naturally defined as ${\bf p}/m$. For higher-order derivatives, the problem is to find a reasonable definition in terms of position and momentum.

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The Heisenberg Uncertainty Principle is one that has been successful to describe observations at the particle level, but can now be derived from the basic theory of quantum mechanics. (here is the history of how it was developed by Heisenberg) The one you quote is about position and momentum being constrained when measured together. It depends on the mathematics of quantum mechanics.

There are more possible uncertainty relations, but they are derived from the mathematical theory, so not arbitrary generalizations , as you suggest.

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  • $\begingroup$ Your last sentence confuses me. Deriving new uncertainty relations from the existing mathematical theory is along the lines of what I was suggesting, but perhaps not stated clearly. $\endgroup$ Jul 3 at 6:10
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    $\begingroup$ they are different variables, for example energyand time, hyperphysics.phy-astr.gsu.edu/hbase/uncer.html . Your question seems to imply the delta(x) to be one of the variablse $\endgroup$
    – anna v
    Jul 3 at 7:30
  • $\begingroup$ Ahhh. Thank you for your original answer and your clarification. $\endgroup$ Jul 3 at 7:52
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    $\begingroup$ uncertainties come with coupling two variables, but yes it deals with momentum and space. As the other answer says, you can try to use the dxdp HUP to try and see if there are quantum mechanical operators that would give what you are explicitly asking, based on the dpdx basic uncertainty. Mathematics is powerful. $\endgroup$
    – anna v
    Jul 3 at 8:44
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    $\begingroup$ @NullSimplex mass in special relativity is used for particle physics, and yes it is "messier" . It is the "length" of the energy momentum four vector.See hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html $\endgroup$
    – anna v
    Jul 4 at 4:21
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As Anna is saying, the Heisenberg uncertainty principle is a fundamental property of nature. It is not that this relation was defined using mathematical concepts only, but was derived rigorously using results and postulates of quantum mechanics such as the Robertson-Schrodinger uncertainty relation $$\sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2\mathrm{i}}\langle[\hat{A},\hat{B} ]\rangle \right)^2 $$

and the commutator relation $$[\hat{x},\hat{p}] = i\hbar$$ the concept that a quantum state can be expressed as the linear combination $$| \psi \rangle = \sum_n c_n |n\rangle$$ and the expectation value of quantities, for example energy $$ \langle E\rangle = \langle \psi | H | \psi \rangle$$

along with other quantum mechanical concepts and postulates. If you could determine the uncertainty relations you want, these would be mathematical relations with no physical meaning.

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