# Heisenberg uncertainity principle

• nouveau_riche
In summary, the wavefunction of a particle and its momentum are related by the De Broglie relation \lambda=h/p where \lambda is the wavelength of the particle and p is the momentum of the particle. The width of the momentum distribution is inversely proportional to the width of the position distribution, subject to the limit set by the Heisenberg Uncertainty Principle. The HUP states that the product of the widths of the measured momentum and position distributions can be no smaller than Planck's constant divided by 4 times pi. This means that while we can measure the momentum and position of a single particle to arbitrary precision, when repeated on a large number of identical particles, the HUP limits the precision of the measured momentum distribution.
nouveau_riche
i am not getting the interconnection between probability wave function of particle with particle's momentum,can anyone help?
i don't want any mathematical equations,any theoretical explanation would suffice

The wavefunction of the particle and its momentum are related by the De Broglie relation
$\lambda=h/p$ where $\lambda$ is the wavelength of the particle and p is the momentum of the particle.

Avijeet said:
The wavefunction of the particle and its momentum are related by the De Broglie relation
$\lambda=h/p$ where $\lambda$ is the wavelength of the particle and p is the momentum of the particle.

i said no mathematical expression,only theoretical answer,suppose you are given with probability wave of a particle(of zero amplitude initially ,then increasing probability which cease down again towards zero amplitude)
how will u relate this uncertainity in position (given by probability wave to that of momentum)?

nouveau_riche said:
i said no mathematical expression,only theoretical answer,suppose you are given with probability wave of a particle(of zero amplitude initially ,then increasing probability which cease down again towards zero amplitude)
how will u relate this uncertainity in position (given by probability wave to that of momentum)?

The width of the momentum distribution (i.e. the uncertainty of the momentum) is inversely proportional to the width of the position distribution (i.e. the uncertainty of the position), subject to the limit that the product of the two uncertainties is never allowed to be smaller than the Planck constant divided by 4 times pi.

Your example is unclearly stated (this is why mathematical descriptions are generally preferred). However, what I think you have done is describe a probability distribution in position space for the particle. As such, the "width" of this distribution is what will have an *inverse* relationship to the width of the momentum distribution, according to the description I gave above. Mathematically, the two distributions are related by a Fourier transform, i.e. the momentum distribution is the Fourier transform of the position distribution, and vice-versa.

SpectraCat said:
The width of the momentum distribution (i.e. the uncertainty of the momentum) is inversely proportional to the width of the position distribution (i.e. the uncertainty of the position), subject to the limit that the product of the two uncertainties is never allowed to be smaller than the Planck constant divided by 4 times pi.

Your example is unclearly stated (this is why mathematical descriptions are generally preferred). However, what I think you have done is describe a probability distribution in position space for the particle. As such, the "width" of this distribution is what will have an *inverse* relationship to the width of the momentum distribution, according to the description I gave above. Mathematically, the two distributions are related by a Fourier transform, i.e. the momentum distribution is the Fourier transform of the position distribution, and vice-versa.

how can you say about the momentum(likely it's velocity) of particle without getting it localized? ?
or do we measure the momentum as the time it takes to go through that probability pattern

nouveau_riche said:
how can you say about the momentum(likely it's velocity) of particle without getting it localized? ?
or do we measure the momentum as the time it takes to go through that probability pattern

I am not talking about measurement at all. I am talking about the intrinsic widths (i.e. uncertainties) of the probability distributions from which the particle is sampled, because it is the relationship between those widths that is limited by the Heisenberg Uncertainty Principle (HUP henceforth).

You are right that to measure the particle's momentum, we need to interact it with a detector, which localizes the particle. So we actually do a position measurement (to arbitrary precision). Then we calculate the momentum, which requires that we know something else about the position of the particle at an earlier time (perhaps we passed it through a narrow slit). Both of those position measurements, and the measurement of the time interval, can be done to arbitrary precision, so we can calculate the momentum to arbitrary precision. From this you can see that in principle, there is no limitation on how precisely we can measure the momentum and position of a single particle.

Where the HUP comes into play is that if you then repeat the same sequence of arbitrarily precise measurements on a large numbers of identically prepared particles (i.e. particles with the same wavefunction, or equivalently particles sampled from the same probability distribution), you will find that your momentum measurements are not all identical, but rather form a probability distribution of possible values for the momentum. The width of this measured momentum distribution for many particles is what is limited by the HUP. In other words, the HUP says that the product of the widths of your measured momentum probability distribution, and the position probability distribution associated with your initial wavefunction, can be no smaller than Planck's constant divided by 4 times pi.

SpectraCat said:
I am not talking about measurement at all. I am talking about the intrinsic widths (i.e. uncertainties) of the probability distributions from which the particle is sampled, because it is the relationship between those widths that is limited by the Heisenberg Uncertainty Principle (HUP henceforth).

You are right that to measure the particle's momentum, we need to interact it with a detector, which localizes the particle. So we actually do a position measurement (to arbitrary precision). Then we calculate the momentum, which requires that we know something else about the position of the particle at an earlier time (perhaps we passed it through a narrow slit). Both of those position measurements, and the measurement of the time interval, can be done to arbitrary precision, so we can calculate the momentum to arbitrary precision. From this you can see that in principle, there is no limitation on how precisely we can measure the momentum and position of a single particle.

Where the HUP comes into play is that if you then repeat the same sequence of arbitrarily precise measurements on a large numbers of identically prepared particles (i.e. particles with the same wavefunction, or equivalently particles sampled from the same probability distribution), you will find that your momentum measurements are not all identical, but rather form a probability distribution of possible values for the momentum. The width of this measured momentum distribution for many particles is what is limited by the HUP. In other words, the HUP says that the product of the widths of your measured momentum probability distribution, and the position probability distribution associated with your initial wavefunction, can be no smaller than Planck's constant divided by 4 times pi.

so what has confinement of wavefunction to do with momentum?

nouveau_riche said:
so what has confinement of wavefunction to do with momentum?

Position and momentum are conjugate variables, mathematically related by a Fourier transform. This means that the widths of their probability distributions are inversely related, as I have said. As to *why* this is the case, that question is harder to answer, particularly without math. In non-relativistic quantum mechanics, it is taken as a postulate .. more specifically, the postulate says that position and momentum along the same coordinate axis are non-commuting operators. This postulate then leads directly to the Heisenberg Uncertainty Principle. I can't explain that any further without going into the math, which you said you weren't interested in.

Spectracat...very clear replies...nice!

## What is the Heisenberg Uncertainty Principle?

The Heisenberg Uncertainty Principle, also known as the Uncertainty Principle, is a fundamental principle in quantum mechanics that states that it is impossible to simultaneously know the precise position and momentum of a subatomic particle.

## Who discovered the Heisenberg Uncertainty Principle?

The Heisenberg Uncertainty Principle was discovered by German physicist Werner Heisenberg in 1927.

## Why is the Heisenberg Uncertainty Principle important?

The Heisenberg Uncertainty Principle is important because it sets a limit to the accuracy with which we can measure certain physical quantities, such as position and momentum. It also plays a crucial role in our understanding of quantum mechanics and the behavior of subatomic particles.

## How does the Heisenberg Uncertainty Principle relate to the concept of wave-particle duality?

The Heisenberg Uncertainty Principle is closely related to the concept of wave-particle duality, which states that particles can exhibit both wave-like and particle-like behavior. The uncertainty in position and momentum described by the principle is a manifestation of this duality.

## Can the Heisenberg Uncertainty Principle be violated?

No, the Heisenberg Uncertainty Principle is a fundamental principle of quantum mechanics and has been repeatedly confirmed by experiments. It is not possible to violate this principle without fundamentally changing our understanding of the behavior of subatomic particles.

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