Heisenberg's Uncertainty Principle

**Chris L T521** · July 18th 2008, 01:54 PM

Last night in physics, we were diving further and further into Quantum Mechanics, and my professor gave us a simple derivation of Heisenberg's Uncertainty Principle. However, there is something I don't get. Maybe you guys can help me out. I get all the stuff at the beginning, but its pretty much the last bit where I'm lost...

$\sin(\vartheta)=\frac{\lambda}{W}$
$\tan(\vartheta)=\frac{\Delta p_y}{p_x}$

Due to Taylor Approximations, for small values of $\vartheta , \ \sin(\vartheta)\approx\tan(\vartheta)\approx\varth eta$

Thus, $\frac{\lambda}{W}=\frac{\Delta p_y}{p_x}$

From deBroiglie's Equation, we discover that $p_x=\frac{h}{\lambda}$ .

Therefore, $\frac{\lambda}{W}=\frac{\Delta p_y}{\frac{h}{\lambda}}\implies \frac{\lambda}{W}=\frac{\Delta p_y \lambda}{h}$ .

Using the idea that $\sin(\vartheta)\approx\tan(\vartheta)\approx\varth eta$ , We can see that $\frac{\lambda}{W}=\frac{\lambda}{\Delta y}\implies W=\Delta y$ .

Thus, $\frac{1}{\Delta y}=\frac{\Delta p_y}{h}\implies \Delta p\Delta y=h{\color{red}\geqslant\frac{\hbar}{2}}$

I understand numerically that $h\geqslant\frac{\hbar}{2}\implies 1\geqslant\frac{1}{4\pi}$ , but why is it that we say in the Uncertainty Principle that $h\geqslant\frac{\hbar}{2}$ ???

I'd appreciate any input.

--Chris

**CaptainBlack** · July 18th 2008, 02:12 PM

Originally Posted by Chris L T521

Thus, $\frac{1}{\Delta y}=\frac{\Delta p_y}{h}\implies \Delta p\Delta y=h{\color{red}\geqslant\frac{\hbar}{2}}$

I understand numerically that $h\geqslant\frac{\hbar}{2}\implies 1\geqslant\frac{1}{4\pi}$ , but why is it that we say in the Uncertainty Principle that $h\geqslant\frac{\hbar}{2}$ ???

I'd appreciate any input.

--Chris

The derivation shows that:

$\Delta p\Delta y=h$

and the next bit is just a statement that this is compliant with the uncertainty principle which reqires that:

$\Delta p\Delta y \ge \frac{\hbar}{2}$

RonL

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