# Heisenberg Uncertainty Principle: Which formula is correct?

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 $\Delta x \Delta p \geq \hbar$ .

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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You should supply the derivation you mentioned. –  David H Jun 30 '13 at 11:24
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" –  anna v Jun 30 '13 at 11:27
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. –  David H Jun 30 '13 at 12:02
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. –  Johannes Jun 30 '13 at 15:36
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. –  Chris White Jul 1 '13 at 3:11

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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Thank you for your rigorous answer! –  curiousGeorge119 Feb 7 '14 at 20:46