Heisenberg's Uncertainty Principle common misconceptions

In summary: This is the so-called " uncertainty principle ". It's a fundamental characteristic of quantum objects, and has nothing to do with the effects of measurement.The Nottingham professor is trying to avoid saying that the uncertainty principle is about the effects of measurement.
  • #36
I wrote:
I don't agree that the physical state must be expressed in space-time.
You wrote:
If that claim is on topic, then disputing it, which is what I've been doing, is also on topic. There are two ways to dispute this claim:

(1) If we're talking about the position vs. the momentum basis, then any state can of course be expressed in either basis (or in any of an infinite number of other possible bases). If your claim just means "we can use the momentum basis instead of the position basis", then of course that's true. But there is no state that can only be expressed in the momentum basis, not the position basis. And in defending your claim, you have been appearing to defend the latter claim; for example, your very next sentence in post #15 was:
(I wrote)
For example that would be impossible if the state is defined by its energy-momentum.
This appears to be saying that an energy/momentum eigenstate does not have a position representation, which is false. That's why I objected.
If you look at #13 you'll see that the context was one in which an observer had made a choice of basis. You cannot choose two incompatible bases at the same time. That is what I was saying. (The topic is the HUP after all.)
 
Last edited:
Physics news on Phys.org
  • #37
phinds said:
Then you have been taught incorrectly since that point of view says that the HUP is a measurement problem, which is is not. It is a fundamental characteristic of quantum objects and has nothing to do with the effects of measurement.

This seems very unlikely since Susskind certainly knows better. Are you sure you understood him correctly? Can you give a citation with an exact quote?
HUP is a theory that just imposes limit on the fact that a pair of physical parameters associated with a specific object cannot be measured simultaneously accurately... that's it... if you have further queries please don't hesitate to ask.
 
Last edited by a moderator:
  • #38
Marco Masi said:
It is unfortunate that so many physicists continue to believe in the old version. It was excusable at the times of Heisenberg, no longer today.
So what is the latest version of the HUP, in mathematical form?
 
  • #39
This thread has been off topic since post #18.
 
  • #40
ΔxΔp<=h/2π
 
  • #41
Agni101 said:
HUP is a theory that just imposes limit on the fact that a pair of physical parameters associated with a specific object cannot be measured simultaneously accurately... that's it... if u hv further queries pls don't hesitate to ask.
The uncertainty principle is not about limits from measurements.
You got the direction of the inequality wrong in your last post, and there is a missing factor of 2.
 
  • Like
Likes vanhees71
  • #42
Agni101 said:
HUP is a theory that just imposes limit on the fact that a pair of physical parameters associated with a specific object cannot be measured simultaneously accurately... that's it... if u hv further queries pls don't hesitate to ask.
No. It is exactly what I said it is.

EDIT: Oops. I didn't notice that mfb had already beat me to it.
 
  • #43
Agni101 said:
ΔxΔp<=h/2π
but not DpDx>=h/2pi-greek ?
 
  • #44
Karolus said:
So what is the latest version of the HUP, in mathematical form?
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$
For position and momentum components thus it's
$$\Delta x_j \Delta p_k \geq \frac{\hbar}{2} \delta_{jk}.$$
 
  • #45
mfb said:
The uncertainty principle is not about limits from measurements.
You got the direction of the inequality wrong in your last post, and there is a missing factor of 2.
yes it will be ΔxΔp≥h/4π...when sleepy even known eqns get wrong. Thanks for rectifying.
 
  • #46
mfb said:
The uncertainty principle is not about limits from measurements.
You got the direction of the inequality wrong in your last post, and there is a missing factor of 2.
and it is factor of 1/2 not 2
 
  • #47
That is a missing factor of 2 at the other side. In general, "something is wrong by a factor of 2" doesn't specify the direction.
 
  • #48
vanhees71 said:
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$
For position and momentum components thus it's
$$\Delta x_j \Delta p_k \geq \frac{\hbar}{2} \delta_{jk}.$$
vanhees71 said:
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$
For position and momentum components thus it's
$$\Delta x_j \Delta p_k \geq \frac{\hbar}{2} \delta_{jk}.$$

it seems identical to the one formulated by Heisenberg ... or not?
 
  • #49
PeroK said:
Essentially, yes. Note that you cannot say you will get different measurements, just that you no longer have the "certainty" about the momentum, say. This is because position and momentum are incompatible. The HUP tells you theoretically why position and momentum are incompatible: their operators do not commute. The Susskind experiment, in a way, tells you practically why they are incompatible. The interesting question, I think, is the extent to which the two are related.
You can say that it's diffraction if you adopt the wave-particle duality. Modern QM, however, tends to avoid the wave-particle duality and prefers a purely probabilistic QM explanation for why the particle behaves as it does.

There was a thread on here recently about why the wave-particle duality is now a historical footnote to modern QM.

Thanks for all your help with this. I was thinking some more about HUP and I couldn't resolve this scenario i dreamed up:

1) If a laser is rated at a specific wavelength and you make the aperture the laser light escapes through smaller and smaller then the uncertainty in the momentum goes up and the dot that hits the wall/screen spreads out. When you see this experiment done, why doesn't the laser ever change colour since isn't momentum is linked to wavelength? If the aperture is really small then the spot on the screen spreads out but since it has hit the screen and you see it, does this not make its position very well known? And since momentum must be very uncertain could it not be uncertain enough to change its colour?
 
  • #50
Jimmy87 said:
Thanks for all your help with this. I was thinking some more about HUP and I couldn't resolve this scenario i dreamed up:

1) If a laser is rated at a specific wavelength and you make the aperture the laser light escapes through smaller and smaller then the uncertainty in the momentum goes up and the dot that hits the wall/screen spreads out. When you see this experiment done, why doesn't the laser ever change colour since isn't momentum is linked to wavelength? If the aperture is really small then the spot on the screen spreads out but since it has hit the screen and you see it, does this not make its position very well known? And since momentum must be very uncertain could it not be uncertain enough to change its colour?

Let me give you part of the answer. We are talking here about momentum in the transverse direction, not total momentum. For an EM wave, diffraction is about a change in direction, not about a change in wavelength or energy.
 
  • Like
Likes vanhees71 and Jimmy87
  • #51
PeroK said:
Let me give you part of the answer. We are talking here about momentum in the transverse direction, not total momentum. For an EM wave, diffraction is about a change in direction, not about a change in wavelength or energy.

So total momentum is conserved then, just the x component changes? And to change the wavelength the total momentum has to change? Is it not misleading when textbooks say "lasers produce photons in a momentum eigenstate because they have a precise wavelength" - surely when it diffracts and its momentum spreads out it is no longer in a momentum eigenstate/definite wavelength but you said it will be because total momentum/wavelength is conserved?
 
  • #52
Jimmy87 said:
So total momentum is conserved then, just the x component changes? And to change the wavelength the total momentum has to change? Is it not misleading when textbooks say "lasers produce photons in a momentum eigenstate because they have a precise wavelength" - surely when it diffracts and its momentum spreads out it is no longer in a momentum eigenstate/definite wavelength but you said it will be because total momentum/wavelength is conserved?
In this case you have light with a well defined wavelength, hence a well defined energy and momentum, in the sense of magnitude of momentum. If the light reflects off a surface, or diffracts, then the direction of its momentum changes, but not the magnitude.
 
  • Like
Likes Jimmy87
  • #53
Noether's Theorem and quantum mechanics are related by the notion from Hamiltonian mechanics that every dynamical variable can be interpreted as an infinitesimal generator of some canonical transformation, or the quantum mechnical notion that every Hermitian operator generates a unitary transformation.

The Heisenberg principle is true of any variable with a continuous spectrum and the infinitesimal generator of translations in that variable, just because these variables always have a nonzero commutator in every possible state. Position and momentum, angle and angular momentum, charge and phase, these are all conjugates in classical mechanics. The charge operator generates infinitesimal rotations in the phase of charged-particle wavefunctions, not changes in potential.

Noether's Theorem states that when translations of a certain variable are a symmetry, the infinitesimal generator of those translations is conserved. So translations in x, translations in angle, and translations in phase give conservation of momentum, angular momentum, and charge. But these generators obey the HUP with their conjugate variables.

It is important to understand that conjugate pairs like position and momentum relate as generators of translations. If these translations leave the system unchanged, i.e. you have a symmetry generated by the conjugate momentum, then the conjugate momentum must be conserved.

In quantum theory the conjugate pairs are not independent. This is because on a Hilbert space the generators of translations and the coordinate they act on relate like the derivative and the coordinate, which don't commute.

So it really comes down to the geometry of the phase space of a system. In classical physics the phase space is just a normal manifold with the usual geometric structure. But in quantum theory the construction using translations on the Hilbert space result in a non-commutative geometry.
There are other conjugate variables, other than position and momentum. Here you can read about the Heisenberg Uncertainty Principle.

https://brilliant.org/wiki/heisenberg-uncertainty-principle
 
  • Like
Likes vanhees71

Similar threads

Replies
1
Views
935
  • Quantum Physics
Replies
25
Views
676
Replies
26
Views
4K
  • Quantum Physics
Replies
15
Views
994
  • Quantum Physics
Replies
12
Views
2K
  • Quantum Interpretations and Foundations
Replies
23
Views
4K
  • Quantum Physics
Replies
11
Views
2K
  • Quantum Physics
3
Replies
71
Views
7K
Replies
18
Views
2K
  • Quantum Physics
Replies
3
Views
2K
Back
Top