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{Uncertainty Principal} Uncertainty in position and wave-vector

A Wave packet is described by

[tex]\Psi[/tex](x)= sin [tex]\frac{kx}{x}[/tex]

Make reasonable estimates of the uncertainty in position and wavevector and show that this function obeys the uncertainty priciple

[tex]\Delta[/tex]x[tex]\Delta[/tex]k>1

solutionish...

sin [tex]\frac{kx}{x}[/tex] = sin ([tex]\pm[/tex][tex]\Delta[/tex]x[tex]\Delta[/tex]k) = 0

hence [tex]\Delta[/tex]x must = some interger of [tex]\pi[/tex]...?

this is about where i get lost don't know if i'm going in the right direction!?
 

Answers and Replies

18
0
isn't [tex]\sin{\frac{kx}{x}}=\sin{k}[/tex]? do you mean [tex]{\frac{\sin{kx}}{x}}[/tex] instead?
 
yeah ta!
 

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