Factor 2 in Heisenberg Uncertainty Principle: Which formula is correct?

Question

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?

Answer 1

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.

Answer 2

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 :

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

Answer 3

Heisenberg proposed $$\Delta x \Delta p \geq \hbar,$$ but he only gave a heuristic argument and did not use a precise definition of uncertainty. Kennard (1927) defined uncertainty precisely, to coincide with the definition of standard deviation in statistics. $$\langle \Delta K \rangle^2 = \langle K^2 \rangle - \langle K \rangle^2 .$$ He then proved $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ Kennard's inequality is generally known as the Heisenberg uncertainty principle. This is the form which will be used in text books, whereas Heisenberg's original proposal is likely to appear only in popular accounts.

Answer 4

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.