I was reading about how the Planck's Constant can be measured with LEDs, which made me think about this question.

The Heisenberg Uncertainty Principle states that: $$\Delta x \Delta p \ge {\hbar\over2}$$ a.k.a, there is always some uncertainty while measuring things.
This is a fundamental law. Nothing escapes it (as far as I know).

Now, when we are attempting to measure the Planck's Constant, wouldn't there be some uncertainty? This should mean that I can never be able to measure the Planck's Constant to full accuracy, ever. But if the amount of uncertainty depends on $\hbar$ (which is uncertain), doesn't this lead to uncertainty of uncertainty?

Note: I'm not a physicist. Just a physics enthusiast. Try not to get too technical, please.

  • $\begingroup$ I think like many people you have a slight misunderstanding of what the uncertainty principle actually means (not that I or anyone really understands it). Also I would like to see the experiment for measuring Planck's length because it is exponentially shorter than even the wavelength of gamma rays. $\endgroup$ – Jaywalker Mar 22 '16 at 10:48
  • $\begingroup$ @Jaywalker I never said Planck Length. I was talking about the Constant. Anyway, I don't think it's possible at all to measure the Planck Length. $\endgroup$ – Udit Dey Mar 22 '16 at 10:59
  • $\begingroup$ The Planck Constant is a factor in the uncertainty principle yes, but the Planck's constant is an actual constant with a definite value. $\endgroup$ – Jaywalker Mar 22 '16 at 11:00
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    $\begingroup$ A reminder that comments are not to be used for answering the question. $\endgroup$ – David Z Mar 22 '16 at 13:06
  • $\begingroup$ Related: physics.stackexchange.com/q/24068/2451 , physics.stackexchange.com/q/114133/2451 and links therein. $\endgroup$ – Qmechanic Mar 22 '16 at 15:33

In your post you've intertwined two distinct notions:

"The Heisenberg Uncertainty Principle states that . . . there is always some uncertainty while measuring things."

The uncertainty while measuring things is called measurement error and it is due to the experimental apparatus.

The uncertainty described by Heisenberg says that there is a fundamental limit, expressed in terms of $\hbar$, to the precision with which certain pairs of physical properties of a particle, known as complementary variables, such as position $x$ and momentum $p$, can be known simultaneously.

Now, when we are attempting to measure the Planck's Constant, wouldn't there be some uncertainty? This should mean that I can never be able to measure the Planck's Constant to full accuracy, ever.

There will be uncertainty related to the experimental technique used. However, the uncertainty in the knowledge of ℏ does not change the Uncertainty principle and its consequences.

But if the amount of uncertainty depends on ℏ (which is uncertain), doesn't this lead to uncertainty of uncertainty?

The amount of uncertainty of the measurement does not depend on ℏ's uncertainty, quite the opposite the precision of the measurement determines the significant digits of the result.

If two complementary physical properties can describe perfectly the state of a quantum object. Then, to me $\frac{\hbar}{2}$ is some sort of nature's machine epsilon that sets a limit to the amount of information we may know about the state of the quantum object.

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The objects on the l.h.s. of the position-momentum uncertainty relation $$ \Delta x \Delta p \geq \frac{\hbar}{2}$$ are standard deviations of quantum mechanical operators, defined for any operator $A$ by $$ \Delta A:=\sigma_A=\sqrt{\langle A^2 \rangle - \langle A\rangle^2}$$ where $\langle \dot{}\rangle$ denotes taking the expectation value with respect to a fixed state. Also, the quantum mechanical uncertainty relation has a priori nothing to do with imprecision of measurements, in particular, it does not refer to any limitation of the measurement apparatus, see e.g. this question. It refers to an intrinsic property of quantum states not having simultaneously well-defined classical values for all possible observables.

$\hbar$ is a constant, not a quantum mechanical operator, so this uncertainty relation means nothing for it. It just doesn't apply. All "uncertainty" on $\hbar$ is of the ordinary experimental kind, which has never stopped us from using fixed "true" values for constants in our theoretical considerations. Of course, for any computations that are sensitive to very small changes in $\hbar$ you'll need to take that experimental uncertainty into account, but that has nothing to do with quantum mechanics or the uncertainty principle, it's just how using experimentally measured values always works.

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    $\begingroup$ +1. You can be certain about uncertainty, but errors are mostly guesswork and statistics. $\endgroup$ – Joe Bloggs Mar 22 '16 at 11:52
  • $\begingroup$ I'm no physicist but do you happen to be missing a square root? $\endgroup$ – user541686 Mar 23 '16 at 9:08
  • $\begingroup$ @Mehrdad: Indeed, thanks! 19 upvotes and you're the first to notice that... $\endgroup$ – ACuriousMind Mar 23 '16 at 10:52
  • $\begingroup$ :) I would also prefer $\sqrt{\left\langle \left(A - \left\langle A \right\rangle\right)^2 \right\rangle}$ if that's the actual definition in physics, since it's also more intuitive. $\endgroup$ – user541686 Mar 23 '16 at 19:59

First of all, the uncertainity principle can be rigorously derived from the framework of quantum mechanics as given by Von Neumann's postulates. This uncertainity is different from the measurement error caused by experimental apparatus. The error due to experimental apparatus can be curtailed by performing different experimental procedures instead of a single one. e.g you can use the spectra of blackbody radiation, photoelectric effect, the Stern-Gerlach experiment and many others in order to get a bound on the experimental error in the determination of Planck constant. But the Heisenberg uncertainity is a characteristic of nature, and cannot be removed away by experimental accuracy.

In general, all physical constants/quantities are measured experimentally upto some error. That is a characteristic of not just the Planck constant but of all physically measureable quantities. That doesn't mean that the theory is invalidated.

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I am not satisfied of the published replies, so I will try my own, as a metrologist (expert in measurement units, but not in theoretical physics). The question clearly is referring to the experimental frame while all the answers are referring only to the theoretical frame, so they do not talk nor understand with each other.

Here we are dealing with two physical quantities, position (SI unit m) and momentum (SI unit kg m/s) and a quantity that is a fundamental constant of physics, the Planck one (unit J s = kg m^2/s).

The equation between quantities does not depend on the units in which they might be measured, and in it the Planck constant has a univocal exact value. However, this numerical value is unknown, as any ‘true’ value of a quantity in nature. One basic reason is that the measured value depends on the measurement units used. It is fundamental to understand that, the difference between value and numerical value, a need I think valid for both sides, the theorists and the practitioners, otherwise they would not understand the system of units.

The equation is obviously a model that is supposed to be presently satisfactory. The notion of ‘exact model’ does not exist, just because it is a model of the nature, not handled to us by ‘Nature’. The model in question model was not disproved so far, so it is considered acceptable –or, colloquially, exact if you wish, without any addition to it. But nobody can say if changes should be necessary in future. All models can be found imperfect.

Totally another issue is what happens if you want, in each particular case, verify if your experimental situation is such that, having a measure (numerical value) of a Δx and a measure (numerical value) of a Δp, both affected by (experimental) uncertainty, you want to verify if the equation holds. In this case you have to use on the right the best current numerical value of the Planck constant (today 6.626 059 57(29) ·10^−34 J s, where the two digits in italics are uncertain by the standard deviation indicated in parentheses). You can compute the mean value of the expression on the left and associate to it the resulting uncertainty computed with the rule of propagation of uncertainty, and compare with the expression on the right. The sign > holds if the result is compatible with it. This means that the uncertainty interval on the left does not overlap the one on the right, remaining higher in values; or, it overlaps the latter on the lower side but remaining not lower in value than the mean value of the constant. This is a verification of the theory: it would invalidate it is the results are found sound and contradict the theory. However, a single experiment is not deemed sufficient, just due to the experimental uncertainty that can comprise systematic errors. The quantistic uncertainty reported in one of the answers does not enter at all in the above.

Therefore, in my opinion, there was a good reason for asking the question, but not all the answers are correct because they did not reply to the question.

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