# On the interpretation of Heisenberg's principle

This question (hoping it's not completely irrelevant) is about the interpretation of Heisenberg's principle and a ficticious opposite to the relation.

$$\sigma_x \sigma_p \geq \frac{\hbar}{2}$$

I'm currently studying the basics of Quantum Mechanics (following D.J. Griffiths' Introduction to QM) and trying to incorporate to my 'natural' understanding of things Heisenberg's principle. Doing the maths it seems very valid to me and I think I can grasp its interpretation. Anyways I've been wondering this "what if" question:

Question: What would be the (fundamental) consequences on the Heisenberg principle if it'd be "reversed"? Does it make sense to ask this?

$$\sigma_x \sigma_p \leq \frac{\hbar}{2}$$

Not only because I find it interesting but also because discussing it might be a way to invalidate this "option" if we had no math (it'd also help with intuition).

Thanks in advance.

• Note there are states for which $\sigma_x\sigma_p=\hbar/2$ so at least the border case of the proposed inequality does make sense. – ZeroTheHero Oct 2 '20 at 23:26
• Are you asking for the consequences of newly proposed HUP in a fictitious universe? Or are you asking how do we rule out this inequality in our universe? – Superfast Jellyfish Oct 3 '20 at 4:48
• @SuperfastJellyfish I'm asking for (fictitious) consequences of this hypothetical HUP to counterexample why this should never be the case (except for $\sigma_x \sigma_p = \hbar /2$, as ZeroTheHero correctly states) – holahola Oct 3 '20 at 4:55
• I understand why $8\ge 3$, but what would be the consequences if this inequality were reversed? – WillO Oct 3 '20 at 13:30
• @WillO that's exactly the kind of arguments I expect. If we elaborate, it's clear that if the inequality is reversed the math must work in a different way or it simply doesn't make sense. Anyways, for example in a mid or high school class context it might be easier to explain and extrapolate consequences of assuming $8 \leq 3$ (in this case) and inspect the lack of sense rather than actually proving that it doesn't make any sense with actual maths. Am I explaining myself? It's like making conclusions about natural (and hypothetical) phenomena bypassing the use of maths. – holahola Oct 7 '20 at 12:38

## 6 Answers

First thing to note is that the available phase space for the new inequality is much smaller than the older one. This can be visualised in the following way.

Here, the blue region is the phase space available due to the original HUP and the red region corresponding to the new UP is the complement of it.

The old HUP imposed a lower bound on possible momentum measurements given a known spatial wavefunction. Whereas the new UP provides an upper bound on it.

This means that if the spread in position, $$\sigma_x=1$$, then the spread in the momentum must be less than $$\hbar/2$$. This is indicated by the green line in the image. As ZeroTheHero pointed out, this would imply arbitrary linear combinations of states no longer are valid. This means our entire quantum mechanics formalism would no longer be valid because the new UP is inconsistent with it.

A peculiar consequence is that the minimum is now $$0$$. The new UP in completeness is $$0\le\sigma_x\sigma_p\le\frac{\hbar}{2}$$

Some of the physical consequences (in no particular order) of such an imposition are:

1. In the double slit experiment, the interference pattern will be truncated
2. Classical trajectories must become more and more common as we look closer and closer to the atomic scale.
3. Since electrons are bound to atoms and are small, they must move extremely slowly!
4. There are no physical restrictions on the speed of point particles and they can be at rest. This implies there is no consistent physical reason within this model to disallow temperature $$T=0K$$

It does not make sense to propose your "alternative HUP" without changing everything else about quantum mechanics, too. The uncertainty principle is not some sort of axiom of quantum mechanics, it is a direct consequence of the Robertson-Schrödinger relation, which holds generally for all observables on all Hilbert spaces. We cannot talk about changing the uncertainty principle without changing the entire theory.

Therefore talking about specific consequences is impossible, too, since assuming that the rest of quantum mechanics works as usual is inconsistent. "Changing the uncertainty principle" is not a proposal for a different theory, it is just ill-defined.

• I upvoted this (and left a comment on the OP to the same effect), but to be fair, the OP did not explicitly rule out abandoning QM entirely, in which case we can certainly fulfill the proposed inequality (and more) by retreating to classical mechanics, where $\sigma_x\sigma_p=0$. – WillO Oct 3 '20 at 15:18
• @WillO Sure, the OP did not rule that out, but they asked if it makes sense to ask for consequences of reversing the uncertainty principle. I say that it doesn't - you can propose an entire alternative theory where this reverse principle might hold, but "What happens if the reverse the UP?" on its own is simply not a well-defined question. In order to make it well-defined OP would have to supply the rest of the alternative theory, too. – ACuriousMind Oct 3 '20 at 16:21
• I agree completely. – WillO Oct 3 '20 at 16:24

Imagine an infinite well, so that $$\sigma_x$$ is clearly bounded. Then you would have a maximum on $$\sigma_p$$, which mean you could not construct arbitrary solutions as linear combos of the eigensolutions.

For instance an initial state $$(\psi_1(x)+\psi_n(x))/\sqrt{2}$$ would have a maximum value of $$n$$ else $$\sigma_p$$ would be “too large”.

Hence some linear combinations are disallowed, which violate the linearity of the time-dependent Schrodinger equation.

There are some problems with the "reversed" Heisenberg principle you are mentioning. It just does not makes sense in the frame of standard QM.

Hand wavy physical argument:

First of all, since the product $$\sigma_x \sigma_p$$ is interpreted as an area in phase-space (or configuration-space) it would include the case where this area is zero, which (for $$\sigma_x = \sigma_p = 0$$, which is not excluded by your "reversed" inequality.) refers to the classical point of view where you can pin-point the configuration in phase-space. So it should be possible to create a quantum state with certain momentum and certain position.

Standard QM formalism:

Furthermore since the uncertainty relation follows directly of the inner product space properties of the Hilbert space, namely the Caushy-Schwarz inequality, there is just no way to combine your "reversed" version with standard quantum physics. So therefore there is no way to describe the motion of quantum particles obeying your "reversed" inequality in a sensible way.

• Does suggesting that “it should be possible to create a quantum state with certain momentum and certain position, which is not achievable by any experiment” not assume the inequality goes in the “standard” $\ge$ direction? – ZeroTheHero Oct 3 '20 at 0:13
• I just gave a physical argument why it does not make sense to reverse the inequaity. The main argument is that the formalism simply does not allow such a change, which is also mentioned in the answer of ACuriousMind. I'll edit for clairity. – AlmostClueless Oct 3 '20 at 12:01

For a “natural” understanding of the uncertainty principal, consider that each subatomic “particle” is associated with a wave packet of finite size. (The magnitude of the wave disturbance at each point determines the probability that the particle will interact with something else at that point.) To “locate” the particle with some degree of accuracy requires that the wave packet be small. But Fourier analysis tells us that the frequency (or energy) of a small packet cannot be accurately determined. At the subatomic scale, there is always a trade-off.

Heisenberg principle comes from the Fourier theory. You can't reverse it without dropping half of the mathematical apparatus of QM.

But if you have a better QM... :)