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The uncertainty principle is usually given as an example: $$\Delta Q\Delta P \geq \hbar.$$ The product of $\Delta Q$ and $\Delta P$ is a measure of uncertainty, and amounts to an integral of probability density over phase space.

Is it correct that in a closed system the total uncertainty cannot change?

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  • $\begingroup$ For the OP: what is your understanding of a closed system and of uncertainty? $\endgroup$
    – DanielC
    Jan 30, 2018 at 15:25
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    $\begingroup$ More importantly, what do you mean by "total" uncertainty? Are you hoping to add all the uncertainties you can think of? $\endgroup$
    – Mike
    Jan 30, 2018 at 15:27
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    $\begingroup$ What about the classic Gaussian wave packet and wave-packet spreading? Its uncertainty product grows rapidly. $\endgroup$
    – JEB
    Jan 30, 2018 at 15:34
  • $\begingroup$ ? Are you kidding? Have you looked at a free Gaussian wavepacket in the first month of your QM class? $\endgroup$
    – Cosmas Zachos
    Jan 30, 2018 at 15:35

3 Answers 3

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As referenced, but not explained, in the comments an emitted wave is an example of increased uncertainty.

These are commonly referred to as free Gaussian wavepackets, a wave spreading out as it travels. You can treat this as your traditional wave in a square well where the width of the well changes with time. The area of confinement, $\Delta x$, is therefore expanding but, with no reason to expect the uncertainty on the momentum to be decreasing their product, $\Delta x(t) \Delta p$ will increase with time.

The link provided explains this in math, if you truly want to learn about quantum mechanics this is a language you should become comfortable in, but hopefully this answers your question without steeping you in the mechanics you've yet to learn.

Edit: To clarify, if you can prove that, for some given closed system uncertainty is not conserved then the conservation of uncertainty as a universal law has been disproved. The expanding wave is such an example and, as such, shows that any law stating that uncertainty must be conserved is false.

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  • $\begingroup$ I understand that the confinement in both momentum space and configuration space can increase. However, when that happens the probability density decreases. Clearly the integral of the probability density over phase space cannot increase beyond 1, nor can it decrease below 1. $\endgroup$
    – S. McGrew
    Jan 30, 2018 at 18:42
  • $\begingroup$ So I guess my question boils down to the definition of "uncertainty". Δx(t)Δp is a vague notion unless it is written as the probability density integrated over x and p. $\endgroup$
    – S. McGrew
    Jan 30, 2018 at 18:52
  • $\begingroup$ @S.McGrew In this context, the definition of the uncertainty of an observable $A$ in the state $\psi$ is $\Delta A = \sqrt{\langle A^2\rangle - \langle A \rangle^2}$, where $\langle A\rangle=\langle\psi|A|\psi\rangle$ is the expectation value of $A$ in the state $\psi$. $\endgroup$
    – J. Murray
    Feb 1, 2018 at 1:44
  • $\begingroup$ @S.McGrew If you define your closed system to include the size of the expanded wave then you'll get both the initial state and final - ie a closed system with increasing uncertainty. As soon as we can find an example for which a rule, like conservation of uncertainty, doesn't work then we can count it as disproved unless you want to start introducing caveats. $\endgroup$
    – Lio Elbammalf
    Feb 1, 2018 at 9:08
  • $\begingroup$ Allusion: "an expression designed to call something to mind without mentioning it explicitly; an indirect or passing reference." So title should be "Referenced, but not Explained". $\endgroup$
    – JEB
    Feb 2, 2018 at 14:22
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The answer to my question appears to be within the Wigner formulation of quantum mechanics, hidden behind ambiguities in the meaning of "uncertainty". The integral of the Wigner quasiprobability density over all phase space is constant. That integral is what I was looking for. It appears that extending this formulation to multiparticle systems and to curved space is not trivial.

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It depends on the situation mostly. For the momentum-position uncertainty $\Delta p \Delta x\ge\frac{\hbar}{2}$ it is only conserved for systems who are in an eigenstate, that is in one of the fundamental functions that you can use to decompose the wavefunction. One can deduce that under eigenstates the energy-time uncertainty is 'undefined' as $\Delta E=0$ and $\Delta t$ is diverging to infinity, because energy can't change, simply by showing only one energy can exist in an eigenstate.

In non-eigenstates, such as the case with the free Gaussian wave packet, $\Delta p\Delta x$ are not conserved as noted by others, it linearly grows. However, several eigenstates exist, thus you have probability over energies so $\Delta E$ is now non zero. It does however take some rigorous math.

Because the energy in that example is directly proportional to the square of the momentum, and that distribution does not change, one can show that $<E>,<E^2>,\Delta E$ also do not change in the case of gaussian wavepacket. This is exactly what we get in classical mechanics as well, if you do not put energy in the system, there is no reason that it should. If $\Delta t$ is the time it takes to change said energy, the energy-time uncertainty relation is meaningless, as $\Delta t$ would have to be very long, and up to this point I assumed potentials that do not change in time.

With time dependent potentials, like an electromagnetic wave, it may cause change in the energy distribution, and thus $\Delta E$, but in accordance to the time-energy uncertainty. $\Delta E$ Can't change too much too fast, there is a limit to it, and that is the energy-time uncertainty.

Some other relations like the angular momentum also have similar relations.

In summary, the relations:

$$\Delta p\Delta x\ge\frac{\hbar}{2}$$ $$\Delta E\Delta t\ge\frac{\hbar}{2}$$

only speak about the minimal quantity the uncertainty can assume. In a closed system one might expect that only $\Delta E\Delta t$ to not change, while other quantities may change over time.

For the more rigorous stuff, showing : https://www.colorado.edu/physics/phys2170/phys2170_fa06/downloads/Gaussian.pdf

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  • $\begingroup$ The definition of "uncertainty" is still not clear from the answers that have been given. For example, the wave function of an electron is spread out over the entire universe - even if it is "confined" in a box. The probability density outside the box drops off exponentially with distance from the walls of the box, but never vanishes completely. So what is its Δx? $\endgroup$
    – S. McGrew
    Jan 30, 2018 at 21:42
  • $\begingroup$ What would be wrong with equating "uncertainty" to the "volume" of phase space occupied by the particle, defined by the integral of dxdp over all of phase space where each element dxdp is weighted by the probability that the particle is in that element? $\endgroup$
    – S. McGrew
    Jan 30, 2018 at 21:48
  • $\begingroup$ The quantity of delta of some operator, be it x or p, E, t etc. is something that comes out of statistics. When one realizes that integrating the inner product of the wavefunction by itself, it should be the probability. Average quantity of an operator is just operating on that inner product (specifically the ket, as per dirac's notation. operating on the bra would be by the operator's conjugate) $\endgroup$
    – Dark BabyIon
    Jan 30, 2018 at 23:16
  • $\begingroup$ so average of x, I just have to calculate <ψ|x|ψ> and I get the average value of x. It is where the particle would be on average. For momentum similarly <ψ|p|ψ>. For energy I may want to know <ψ|p^2|ψ> because it is directly proportional to the kinetic energy for a particle, let's through <ψ|x^2|ψ> too in there. This is where it gets relevant: the square of the Delta of some operator is actually the difference between <ψ|O^2|ψ> and <ψ|O|ψ>^2. What it measures is not the difference of the limits of the operator, but in a way, how much it varies. $\endgroup$
    – Dark BabyIon
    Jan 30, 2018 at 23:21
  • $\begingroup$ In probability and statistics theory it is called Variance, and it is useful quantity because (informally) it tells you how spread out the function is. (trivia: that has a direct relation with the width of a gaussian packet) for the position-momentum uncertainty it is basically stating that one can't know where a particle is and where it is going with certainty, you need to sacrifice accuracy of one for another. With energy-time uncertainty, you can know how much time you need to wait for a certain system to gain or lose energy (loosely speaking) $\endgroup$
    – Dark BabyIon
    Jan 30, 2018 at 23:26

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