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Almost all uncertainties (for example the position-momentum uncertainty or time-energy uncertainty) are greater than ${\hbar}/{2} $. But what is the derivation of this uncertainty by Heisenberg? Is there any sort of intuitive explanation behind the magnitude of uncertainty? I know why uncertainty happens but I do not why the value. It'd be great if somebody could provide a simple explanation.

$$ \mathrm{uncertainty} \geq \frac{\hbar}{2} $$

Qmechanic
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rahulgarg12342
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2 Answers2

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The general form of the uncertainty principle for two operators $A$ and $B$ is given by

$$\Delta A\Delta B \ge\frac12|\langle[A,B]\rangle|.$$

For the uncertainty between momentum and position, we have $[P,X]=-i\hbar$, which leads us to

$$\Delta X\Delta P\ge\frac{\hbar}{2}.$$

As one can see, the factor $1/2$ follows from the derivation of the general result, which can be found for example in chapter 3.5 of these excellent lecture notes, while the appearance of the reduced Planck constant is due to the commutation relation between the operators.

Frederic Brünner
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Hiesenbergs uncertainty principle is based on measuing two things simultaneously, for example: $m_1$ and $m_2$, and to do that in quantum mechanics you have to apply corresponding hermitian operators: $\hat{M_1}$ and $\hat{M_2}$ to state $|\Psi\rangle$: $$ \tag{1.2} \begin{split} 1) \quad \hat{M_1}\hat{M_2}|\Psi\rangle\qquad\qquad 2)\quad\hat{M_2}\hat{M_1}|\Psi\rangle \\ \qquad m_2\hat{M_1}|\Psi\rangle \qquad\qquad\qquad m_1\hat{M_2}|\Psi\rangle \\ \qquad m_2m_1|\Psi\rangle \qquad\qquad\qquad m_1m_2|\Psi\rangle \end{split} $$ $$ m_1 m_2 = m_2 m_1 $$ So it should be obvious that $\hat{M_1}\hat{M_2}=\hat{M_2}\hat{M_1}$ or $\hat{M_1}\hat{M_2} - \hat{M_2}\hat{M_1} = 0$ which is a commutator of $\hat{M_1}$ and $\hat{M_2}$: $[\hat{M_1},\hat{M_2}] = 0$

So if we will do same with position and momentum we will find out that commutator: $[\hat{x}, \hat{p}]$ isn't zero but equals to $i\hbar$, so this is where the $\hbar$ comes from, the equation that describes uncertainty is: $$ \Delta A\Delta B \geq \frac{|\langle[A,B]\rangle|}{2} $$ So after replacing $A$, $B$ you will get: $$ \Delta x \Delta p_x \geq \frac{\hbar}{2} $$