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Why is Heisenberg's uncertainty principle not an experimental error since it is the error created by photons striking on elementary particles?

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marked as duplicate by ACuriousMind, 0celo7, John Rennie, Qmechanic Sep 4 '15 at 22:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Please read the link Qmechanic editted in. The uncertainty principle is far more general than the specific experimental situation you mentioned. – Michael Brown Mar 16 '13 at 13:38
Possible duplicates: and links therein. – Qmechanic Mar 16 '13 at 18:32
I voted this up because there are so many "good" undergraduate modern physics books which include this "thought experiment." Common misconceptions are worth answering. – levitopher Mar 17 '13 at 3:49
I think this question should be tagged an FAQ. I've tried to give a clear and precise answer which will hopefully settle this question once and for all, on physics.SE. – Siva Mar 17 '13 at 8:59
I don't think any of the answers to this question are ideal. The reason is that they all dismiss "photons striking on elementary particles" as completely irrelevant to Heisenberg's uncertainty principle. But it's not. Because the position and momentum can't simultaneously exist, you can't simultaneously measure them, which means any experimental attempt to do so which can be described classically will be foiled by some consideration such as photons striking on elementary particles. – Peter Shor Mar 17 '13 at 13:15
up vote 13 down vote accepted

it is the error created by photons striking on elementary particles

It's not. Heisenberg's uncertainty principle actually has nothing to do with any particular experiment, or any particular interaction. It's a purely mathematical statement about waves.

Its true meaning is explained in detail on the Wikipedia page, but the gist is that if you have a wave, you can express it as a function of position, $\psi(x)$, or of momentum, $\phi(p)$. These two functions are Fourier transforms of each other. You can then calculate the variance of each function, $\sigma_x^2$ and $\sigma_p^2$ respectively, using formulas given on Wikipedia, and you will find that these two quantities obey the relationship

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

Since $\sigma$ is a measure of how tightly concentrated the wave is around one particular point, this tells you that a wave which is tightly concentrated in position must be fairly spread out in momentum, and vice versa. (For the proper definitions of "concentrated" and "spread out," of course.)

The only way this connects to measurement is that, if you make a series of position measurements on objects with the same quantum state, the variance of those measurements will tend to the variance of the wavefunction. And similarly for momentum. So with a large number of measurements of both position and momentum, if you compute their variances, you'll find that they have to satisfy that inequality. In a sense, it's a statement about the particle's state before it gets hit with a photon (or something else), not some effect of the photon hitting it.

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The statement: "in quantum mechanics a particle is represented by a wavefunction and the position an momentum are found by operating on the wave function solution of the dynamic problem." is missing somewhere, an important semantic confusion for newbies. – anna v Mar 17 '13 at 5:00

The uncertainty principle is a mathematical statement about the dispersion of pairs of observables. $A$, $B$ It can be written as

$$\Delta A \Delta B\geq \alpha\langle[A,B] \rangle $$

for a proper constant $\alpha$. So, if the pair of observables do not commute, you can't have arbitrary small dispersion for both of them

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I understand what you're saying, but for someone who hasn't taken quantum mechanics, this isn't going to make any sense at all. – Nick Mar 17 '13 at 2:38
@Nick that's okay. The point of this being an expert-level site is that we should assume people asking questions understand some basic things about the subjects they're asking about. – David Z Mar 17 '13 at 5:14

Due to the wave-particle duality the wavelength of matter wave at quantum levels becomes more pronounced and hence when another wave like photon or even matter-wave like electron is used to see the elementary particles which itself are matter-waves, there is a redistribution of energy.

Even if photons are avoided like in Atomic force microscopy or Scanning tunneling microscopy we don't see the actual elementary particles in the atoms but just the deflection of the cantilever due to vanderWaals forces, capillary forces, chemical bonding, electrostatic forces, magnetic forces of the atoms according to Hook's law.

Hence if we use wave to measure momentum(particle property) then we cannot be sure of its position(wave property) and if we don't use wave and measure its position(particle property) then we cannot be sure of its momentum(wave property).

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Often Heisenberg is presented as though the error arises from the measuring device bumping or changing the particle. That happens but the limitation on measurement is more fundamental. There is no simultaneous reality of position and momentum. Elementary particles are not classical objects for which we can know position and momentum to arbitrary certainty.

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The most important reason why Heisenberg’s uncertainty principle, in measurements of mechanically conjugate quantities, is not an experimental error, lies in the fact that this “error” is determined by a fundamental constant of nature, $h = 6.63\times 10^-{34}$Js, Planck’s constant, which we have no access in order to control it; to make it smaller or larger. There are several conjugate quantities in mechanics. Some of them are: position, x, and momentum, p, of a particle, energy, E, and time, t, angular momentum, L, and its anglular of orientation, ${\bf \phi}$.

For example let us imagine we want to measure the position and momentum of a particle so that we know exactly where it is and exactly how fast is moving. There is an inherent limitation in the accuracy with which we can do this, and it is determined by the set of inequalities:

$\Delta x\Delta p_x\ge \hbar$

$\Delta y\Delta p_y\ge \hbar$

$\Delta z\Delta p_z\ge \hbar$

where $\Delta Q$ stands for the uncertainty in the measurement of the physical quantity $Q$.

These tell us that the more accurately we measure the position of the particle in one direction, the less accurately we can then measure its momentum in that direction. These uncertainties do not reflect limitations of the apparatus we use, it is far more subtle than this. We do have the liberty to decide on the accuracy in measuring the position of the particle, in the x-direction say, and we know the limitations of our device for that. However, once we have measured the position of the particle with the “desired” accuracy, which has some uncertainty $\Delta x$, the measurement of the momentum of the particle in the x-direction is at the mercy of Planck’s constant:

$\Delta p_x\ge \frac{\hbar}{\Delta x}$.

There is no way we can pin point the value of $p_x$. We can only make compromises between knowing more of one quantity and less of the other. We can never know both of them exactly even with the use of perfect apparatus. This bizarre situation is an expression of the fact that, the behaviour of particles in the quantum world is completely determined by laws that are not those of Newtonian mechanics. Our knowledge about the behaviour of matter in the quantum world is formed by wave functions which obey Schrodinger’s equation and the inherent laws of probability. Whether we like it or not it is completely irrelevant, this is how it is, and this is what nature tells us. We may chose to be in perpetual denial, but there is nothing we can do about it.

It is nature’s choice of doing things, not ours

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"several conjugate quantities in mechanics" . Should insert: "In quantum mechanics these quantities turn become operators that act on the QM wavefunction to give the probability of finding a value for them". or some such. to stress that the QM language differs from the classical uncertainty. – anna v Mar 17 '13 at 4:52

Heisenberg's uncertainty principle has nothing to do with measurement. It tells you about the fuzziness in the very notion of observable properties of a quantum state. It tells you about the fuzziness you'd see even if you had perfect measurement and good statistics (I've made that precise below). You couldn't do better than the Heisenberg limit because the observables are not defined any more precisely. (If nothing else, I hope I've driven that point home.)

The standard mathematical proof (which you can see in any intro quantum textbook) will show that for any two observables $A$ and $B$ you try to define, for a state $|\psi\rangle$ $$ \Delta A \; \Delta B \geq \frac{\hbar}{2} \langle \psi| [A, B] |\psi \rangle$$ Note: The constant factor ($\frac{1}{2}$ here) varies in different derivations, depending on how exactly you define $\Delta A$ and $\Delta B$, but the essence is the same.

If you set out to measure observable properties of a quantum state, there are further sources of "error".

  1. Say we had no measurement errors. If we could clone arbitrarily many copies of a quantum state, then we could run the measurement multiple times (once on each cloned copy, since measuring would collapse the wavefunction). Then we could do statistics on all the results, to make a measurement which saturates the Heisenberg limit. But, linearity of quantum mechanics prevents you from cloning states. So even if you had no measurement errors, you couldn't saturate the Heisenberg limit.
  2. What if we didn't have a perfect measuring setup: like your example of the probe photon affecting the system? Then the best we can do is definitely worse than the Heisenberg limit and that can also be calculated. The calculation is slightly involved and I'll leave a reference.

Ref: Some of these things are spelled out more concretely at

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I'm surprised that none of the other answers (from last year) mention interaction-free measurement. The way we know that the uncertainty principle is not a side effect of interactions during measurement is that it applies even when there is no interaction (and even when the measuring device is nowhere near any part of the system).

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