I know the mathematical form of HUC and its statement i.e.It is impossible to determine simultaneously. , the exact position and momentum of an electron. i don't know why this principle exists. Moreover, i read that this all is not due to technique inadequancies, but due to the nature of electron. What all is this? I tried to understand derivation for the mathematical form of HUC (just to get the idea why such thing exist), but it used advanced mathematics (I'm in class 11). Please give me the idea about why this principle exists. Why it is so that we are unable to measure velocity and position at the same time to high accuracy. (No usage of earlier established principles and maths please, in the answer.)

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    $\begingroup$ When you'll be ready to digest all the math, you'll see. Before that, just take our word for it. $\endgroup$ – Ivan Neretin Aug 8 '19 at 12:22
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    $\begingroup$ The usual/standard explanation is that the act of measuring something (at an atomic level ) is such that it inevitably perturbs what is being measured. However, you should really ask the physics StackExchange community for a proper non-mathematical explanation as to what the current thinking on this is. $\endgroup$ – porphyrin Aug 8 '19 at 12:30
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    $\begingroup$ You might want to take look at this: physics.stackexchange.com/questions/287126/… , noting that it's not due to the nature of the electron specifically, but rather due to the wave nature of matter that reveals itself with any particles that are sufficiently small. $\endgroup$ – theorist Aug 8 '19 at 12:57
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    $\begingroup$ @porphyrin It's probably useful to note that the current understanding is that the uncertainty principle is independent of the observer effect. $\endgroup$ – theorist Aug 8 '19 at 13:00
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    $\begingroup$ Unfortunately, you can look this in one of two ways. (1) The real life analogy is all around you because it's how the universe works, or (2) there is no real life analogy because your everyday intuition is totally at odds with how quantum mechanics works. $\endgroup$ – Zhe Aug 8 '19 at 13:32

I can provide some physical intuition, but it's still going to require math and previous work to make it happen (aside on this at the end).

Let's consider a particle in just one dimension. There are two quantities of interest: (1) the position of the particle along this coordinate, (2) the scalar momentum (related to the velocity) along this coordinate.

I'm going to represent this particle as a plot on a graph. The horizontal axis will measure position. The vertical axis represents some notion of how likely it is to find the particle at this position (formally, it would be squared and normalized). Finally, we also stipulate that momentum of the particle is going to be the frequency of what I draw. This may seem strange when we're talking about an arbitrary plot, but it's actually completely reasonable from the perspective of representing functions from as Fourier series. You're just going to have to assume that this is true because a formal analysis does require math (see other answers).

Now, let's make this more concrete with some plots.

Here's a standard sine function. It has a very well defined frequency, so we know the momentum exactly. Where is the particle? Well, the sine wave stretches out to infinity in both directions, so we have no idea where it "is."


Here's a different graph. Where is the particle? It's very close to $x=0$. What is it's momentum? It's kind of hard to pick out single frequency here, so you cannot easily establish what the momentum is.


And that's basically it. Identifying frequency (momentum) and position for a plot like this are traded-off against one another. You'll probably end up in the intermediate case where you know some information about both. This is wave packet.

wave packet

This particle is roughly centered around $x = 0$, and we can roughly say what its momentum is.

The uncertainty principle simply formalizes this idea except using the formal mathematical structure of quantum mechanics.

Now, for the aside I promised. Your question was entirely reasonable until you got to the last line, and I quote: "No usage of earlier established principles and maths please."

This is impossible. Science does exist in a vacuum. It is based on the past observations and structure built by previous generations of scientists. If we were to start from scratch, would we arrive at the same science that we have today? Possibly, possibly not. But the understanding we are working with is built upon foundations that we are stuck with, for better or for worse.

You also can't really do this without math. Part of science is to describe relationships, and mathematics allows us to do so unambiguously. Words suck at conveying ideas because words mean different things.

Lastly, it's fine to try to build intuition on a concept by relating it to the macroscopic world. However, you should not assume that this is possible. It may be philosophically satisfying that objects behave in similar ways regardless of size, but that's not the way the world is.

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There is a simple way to understand this if you understand the relationship between frequency and wavelength in a vibrating string

But you can't avoid at least some mathematics.

They key to understanding where the position momentum uncertainly relationship comes from is to think about the relationship between the frequencies present in a musical note and the time over which the note is sampled. This can be understood if you know about Fourier transforms but that involves mathematics. But the basic principles can be understood without that detail.

A vibrating string can produce a nearly pure note. Just one dominant frequency is present. If you look at a long time period while the string is producing that note and analyse the frequencies present you will get a very narrow set of frequencies. But, if you only sample the vibration for a very short period of time, you will see other frequencies. To put it another way, if you look at an infinite pure sine wave, it will have a single frequency: the curve is smooth and there are no sharp edges when the vibration is plotted on a chart. But, if you take a small sample (over a short period of time) both the start and end of the chart will show sharp edges where the sample starts and stops. If you now translate that into a chart of the frequencies present, there will be many other frequencies present.

Real musical instruments also show some of this behaviour. A plucked string will emit a wider range of frequencies than a smoothly vibrated one.

The link to Heisenberg is this. If you have infinite time to observe the frequency you can get a pure tone with little uncertainly about the frequencies present. But if you only have a short time to observe the vibration you inevitably have other frequencies and the "certainty" about the frequencies present is much larger. This is very clear in the mathematics of Fourier transforms (which describe the relationship between the worlds of vibrations over time and the frequencies present).

The key takeaway is that you cannot simultaneously have a very precise frequency and a very small localisation in time when observing the vibration. The more time you have, the tighter your mix of frequencies will be: the less time, the wider the range of frequencies you will see.

It isn't a perfect analogy to position and momentum uncertainly in the quantum world but it is a similar type of tradeoff. And it isn't a practical limitation which we could bet with better instruments or techniques: it is baked in to the very physical definitions of the things we observe (in music: the relationship between observed vibrations over time and the frequencies present in the mix).

The position and momentum of particles are two points of view of the same thing just as a vibration over time is one view of a musical note and the frequencies present is another. But the process of translating from one point of view to the other (mathematically a Fourier transform) cannot give us precision in both frequency and time simultaneously. For quantum particles it is the same: we cannot know at the same time how fast something is moving and where it is.

The mathematics would give a more precise picture, but the physical analogy provides some intuition as to why this is so.

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  • $\begingroup$ This is a nice explanation but surely there is more in it than just a Fourier transform between conjugate variables? If you observe for a long (effectively infinite) time then just one energy will be measured but if for a short time many waves, i.e. a localised wavepacket is observed and hence many energies exist; why? Where do these different energies come from and are there just because we observe for a different length of time? $\endgroup$ – porphyrin Aug 9 '19 at 14:20
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    $\begingroup$ No, not really. The Heisenberg Uncertainty Principle is just a result of the the variables being related by a Fourier transform. See e.g. en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle $\endgroup$ – Ian Bush Aug 9 '19 at 15:08
  • $\begingroup$ @porphyrin It isn't a perfect analogy, but the mathematics is very similar. And the analogy does give a non-mathematical picture that people familiar with the basics of sound will understand. Ian Bush has it right. $\endgroup$ – matt_black Aug 9 '19 at 18:09

As you can see from figures in Zhe's answer that it is possible by adding waves to make a wavepacket. The characteristic of a wave is that superpositions are possible and these lead to interference/diffraction phenomena. By analysing the wavepacket's spread in wavelength and position, the expression $\Delta k.\Delta x \ge 1$ is found where $k=2\pi/\lambda$. The exact value of the product$\Delta k.\Delta x $ will depend on the shape of the wavepacket but $1$ is the minimum value. This expression is also found by Fourier analysis between what are called conjugate pairs, such as position $x$ and its reciprocal $k$. Time and frequency are another such pair. The wavepacket at $x$ with width characterised by $\Delta x$ and that of $\Delta k$ (width of wavevector) are related by Fourier Transforms as described in other answers.

However, the relationship $\Delta k.\Delta x \ge 1$ is not an inherent property of quantum mechanics but is a property of Fourier Transforms and well known in classical physics. An example is that it is not possible to form a train of electromagnetic waves for which it is possible to measure, at the same time, the position and wavelength with infinite accuracy.

Quantum mechanics enters when a particle is given wavelike properties as satisfied by the de Broglie relation. A material particle (such as an electron or a molecule) of energy $E$ and momentum $\pmb p$ is now associated with a wave of angular frequency $\omega=2\pi\nu$ and wavevector $\pmb k$ i.e. $E=\hbar\omega; \pmb p=\hbar \pmb k$. The fact that the Schroedinger equation is linear and homogeneous means that for particles a superposition principle applies which gives them wavelike properties.

(Aside: The Schroedinger equation describes what happens to a particle under the influence of a potential (such as in an harmonic oscillator) and allows the calculation of the energies and wavefunctions. The Schroedinger equation has no derivation, it is assumed to be correct, as is the de Broglie relationship, and is justified by experience as it has not yet been found to fail to describe experiments.

The small value of Planck's constant makes the limitations of the uncertainty principle totally negligible for anything macroscopic, i.e. greater than approx micron size. )

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It's simply a consequence of wave mechanics and operators. Any observable, i.e., any observable which can be measured in a physical experiment should be associated with a quantum-mechanical operator (like + or - operators, just more complicated). Furthermore, if these operators are intrinsically non-communicative, or if the order in which they are applied matters (like the - sign; 5-3 $\neq$ 3-5), then they can't be known simultaneously. This is important because your measurements effect the quantum mechanical state. If you're measuring 1 then 2 and you get a different answer than if you measured 2 then 1, then you can't get anything meaningful out of that measurement because you don't really know whats what. It gets more complicated with eigenvalues and eigenstates so hopefully we can leave it at that.

So from quantum mechanics we know that the operators for position and momentum are $\hat{x} = x$ and $\hat{p_x} = -i\hbar\frac{\partial}{\partial x}$. If these two operators did commute, then the difference between their resultants would be zero.

However, we can test these operators via a 'trial' function to see if they commute:

$$\hat{p_x} = -i\hbar\frac{\partial}{\partial x} ; \hat{x} = x ; f = f(x)$$

$$\hat{x}\cdot\hat{p_x} \cdot f = \hat{x}(\hat{p_x}\cdot f) = x(-i\hbar\frac{\partial f}{\partial x}) = -i\hbar x \frac{\partial f}{\partial x}$$

$$\hat{p_x} \cdot \hat{x} \cdot f = \hat{p_x} (\hat{x} \cdot f) = -i\hbar\frac{\partial}{\partial x}xf = -i\hbar x \frac{\partial f}{\partial x} - i\hbar f$$

Since those results differ, the operators are non-commutable, and thus you can't know momentum and position precisely and simultaneously.

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    $\begingroup$ "It's simply a consequence of wave mechanics and operators." But those would not mean a thing if they did not just happen to describe our universe correctly. So the answer to the question "why?" is more like "We don't know but that is what we observe". $\endgroup$ – Karsten Theis Aug 8 '19 at 20:25
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    $\begingroup$ Yes you can ask enough why questions of every domain of science until you bottom out at the answer which is "well, that's just how it is" $\endgroup$ – Michael Green Aug 8 '19 at 20:55
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    $\begingroup$ This is all fine, but it basically concludes that you need to understand some fairly complex maths to get the answer. And the OP wanted an answer that didn't do that. $\endgroup$ – matt_black Aug 9 '19 at 18:13

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