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Some websites and textbooks refer to $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ as the correct formula for the uncertainty principle whereas other sources use the formula $$\Delta x \Delta p \geq \hbar.$$ Question: Which one is correct and why?

The latter is used in the textbook "Physics II for Dummies" (German edition) for several examples and the author also derives that formula so I assume that this is not a typing error.

This is the mentioned derivation:

$\sin \theta = \frac{\lambda}{\Delta y}$

assuming $\theta$ is small:

$\tan \theta = \frac{\lambda}{\Delta y}$

de Broglie equation:

$\lambda = \frac{h}{p_x}$

$\Rightarrow \tan \theta \approx \frac{h}{p_x \cdot \Delta y}$

but also:

$\tan \theta = \frac{\Delta p_y}{p_x}$

equalize $\tan \theta$:

$\frac{h}{p_x \cdot \Delta y} \approx \frac{\Delta p_y}{p_x}$

$\Rightarrow \frac{h}{\Delta y} \approx \Delta p_y \Rightarrow \Delta p_y \Delta y \approx h$

$\Rightarrow \Delta p_y \Delta y \geq \frac{h}{2 \pi}$

$\Rightarrow \Delta p \Delta x \geq \frac{h}{2 \pi}$

So: Which one is correct and why?

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  • $\begingroup$ You should supply the derivation you mentioned. $\endgroup$ – David H Jun 30 '13 at 11:24
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    $\begingroup$ It really is not important though h_bar/2 is what comes from solutions of various QM systems en.wikipedia.org/wiki/Uncertainty_principle . The importance lies in the "order of h_bar" $\endgroup$ – anna v Jun 30 '13 at 11:27
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    $\begingroup$ The derivation above is not a derivation of the uncertainty principle. The steps where he introduces approximations contribute to additional uncertainty. A more general analysis will of course allow for sharper bounds on uncertainties. $\endgroup$ – David H Jun 30 '13 at 12:02
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    $\begingroup$ The factor 1/2 needs to be included for the delta terms to represent standard deviations. This becomes apparent when doing a rigorous derivation of the uncertainty principle. $\endgroup$ – Johannes Jun 30 '13 at 15:36
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    $\begingroup$ Saying the 2 is not important is like saying kinetic energy is something like $mv^2$! Sure it works for contrived cases, but it's generally false, it will get you in trouble if you ever try to do quantitative physics, and moreover any derivation of the formula is just as easy to do correctly as incorrectly. $\endgroup$ – user10851 Jul 1 '13 at 3:11
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The strongest limit without loss of generality is $$ \Delta p\Delta x \ge \frac12 \hbar, $$ this is always true. Whilst $\Delta p\Delta x \ge \hbar$ might often be true, it is not always true.

The $\frac12$ is often omitted, because, as mentioned in the comments, often only the magnitude of the right-hand-side is important, and not its precise value. Also, it might be omitted for brevity/simplicity.

A further reason is historical: Heisenberg's original statement of his uncertainty principle was a rough estimate that omitted $\frac12$. Only later was his estimate refined with a formal calculation and the $\frac12$ added.

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Suppose $A$ and $B$ two observables (hermitian operators).

Take $$A' = A - \langle A\rangle ,\qquad B' = B - \langle B\rangle $$

Then $$V(A) = \langle A'^2\rangle , V(B) = \langle B'^2\rangle $$ where $V$ is for variance.

Suppose that $[A,B] = iC$, where $C$ is an hermitian operator. Then, you have also $[A',B'] = iC$.

The Cauchy-Schwartz inequality gives :

$$\langle A'^2\rangle \langle B'^2\rangle \ge |\langle A'B'\rangle |^2$$

Writing $$A'B' = (\frac{A'B' + B'A'}{2}) + (\frac{A'B' - B'A'}{2}) = R+i \frac{C}{2}$$ (where $R = \frac{A'B' + B'A'}{2}$). This gives: $$\langle A'B'\rangle = \langle R\rangle + i \frac{\langle C\rangle }{2}$$

$R$ and $C$ are hermitian operators, so $\langle R\rangle$ and $\langle C\rangle $ are real quantities.

So $$|\langle A'B'\rangle |^2 = |\langle R\rangle |^2 + \large \frac{|\langle C\rangle |^2}{4}$$

Finally : $$\langle A'^2\rangle \langle B'^2\rangle \ge |\langle R\rangle |^2 + \frac{|\langle C\rangle |^2}{4}$$

That is : $$V(A) V(B) \ge |\langle R\rangle |^2 + \frac{|\langle C\rangle |^2}{4}$$

By definition of the standard deviation ($(\Delta X)^2 = V(X)$), you have : $$(\Delta A)^2 (\Delta B)^2 \ge |\langle R\rangle |^2 + \frac{|\langle C\rangle |^2}{4}$$

So : $$(\Delta A) (\Delta B) \ge \sqrt {|\langle R\rangle |^2 + \frac{|\langle C\rangle |^2}{4}}$$

So : $$(\Delta A) (\Delta B) \ge \frac{|\langle C\rangle |}{2}$$

By choosing $A=X, B=P, C= \hbar ~~Id$, we get :

$$(\Delta X) (\Delta P) \ge \frac{\hbar}{2}$$

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Some websites and textbooks refer to $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ as the correct formula for the uncertainty principle whereas other sources use the formula $$\Delta x \Delta p \geq \hbar.$$ Question: Which one is correct and why?

Both are correct. The first is is the maximum uncertainty for 4-dimensional case. while the latter is the uncertainty for 2-dimensions.

I just recently realized, that Planck constant might be $\frac{2h}{\pi}=(\frac{e^1}{2c})^4$, where $(\frac{e^1}{2c})$ presents the single dimension.

As $$4\Delta x \Delta p = \frac{4\hbar}{2}=\frac{2h}{\pi}=(\frac{e^1}{2c})^4$$

So, though both are correct, the most important thing to understand is, where the uncertainty comes. It comes from the fact, that you cant measure movement without time. And as everything is just movement of light, (velocity) and you can never measure it correctly in 4-dimensions. The fourth dimension will always be missing, so you allways get;

$\Delta x_1 \Delta p+\Delta x_2 \Delta p+\Delta x_3 \Delta p+\Delta x_4 = 3\Delta x \Delta p$
Though it truly is $4\Delta x \Delta p$, we just can't never see it so.

Please note, that even if my interpretation for Planck's constant $$h=\frac{\pi}{2}(\frac{e^1}{2c})^4=(\frac{\pi*2.71828}{32*299792458})^4=6.6358^{-34}$$

is not correct, the 4-dimensional interpretation for Heisenberg's uncertainty principle remains valid.

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    $\begingroup$ Hi @JokelaTurbine, what I find totally nonsense is your sentence: "It comes from the fact, that you cant measure movement without time. And as everything is just movement of light, you So you can never measure it correctly." $\endgroup$ – pppqqq Oct 8 '16 at 11:24
  • $\begingroup$ @rhetoricalphysicist I've read MTW. physics.stackexchange.com/questions/241606/… But it turned out to be just a nice math-book. $\endgroup$ – Jokela Oct 8 '16 at 11:36
  • $\begingroup$ @pppqqq Thanks for the answer. Now I understand what you mean. And notice that I used bad-language. "movement" should be "velocity". -my foult sorry. $\endgroup$ – Jokela Oct 8 '16 at 11:38

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