# Conservation of uncertainty

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?

• For the OP: what is your understanding of a closed system and of uncertainty? – DanielC Jan 30 '18 at 15:25
• More importantly, what do you mean by "total" uncertainty? Are you hoping to add all the uncertainties you can think of? – Mike Jan 30 '18 at 15:27
• What about the classic Gaussian wave packet and wave-packet spreading? Its uncertainty product grows rapidly. – JEB Jan 30 '18 at 15:34
• ? Are you kidding? Have you looked at a free Gaussian wavepacket in the first month of your QM class? – Cosmas Zachos Jan 30 '18 at 15:35

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.

• 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. – S. McGrew Jan 30 '18 at 18:42
• 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. – S. McGrew Jan 30 '18 at 18:52
• @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 $\psi$. – J. Murray Feb 1 '18 at 1:44
• @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. – Lio Elbammalf Feb 1 '18 at 9:08
• 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". – JEB Feb 2 '18 at 14:22

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.

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.