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Math Help - Heisenberg's Uncertainty Principle

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Heisenberg's Uncertainty Principle

    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
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Chris L T521 View Post
    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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