## uncertainty principle, relating the uncertainty in position to the uncertainty

Prove the uncertainty principle, relating the uncertainty in position (A=x) to the uncertainty in energy ($$B=p^2/(2m + V)$$):

$$\sigma x\sigma H \geq \hbar/2m |<P>|$$

For stationary states this doesn't tell you much -- why not??

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 Quote by ttiger654 Prove the uncertainty principle, relating the uncertainty in position (A=x) to the uncertainty in energy ($$B=p^2/(2m + V)$$): $$\sigma x\sigma H \geq \hbar/2m | |$$ For stationary states this doesn't tell you much -- why not??

solution-
[x,p2/2m+V]=1/2m[x, p2]+[x,V];

[x, p2]= xp2 − p2x = xp2 − pxp + pxp − p2x = [x, p]p + p[x, p].

using the equation [x,p]=ih{this is known as canonocal commutation relation}

[x, p2]= ihp + pih = 2ihp. and And [x, V ] = 0,
so [x,p2/2m+ V]=1/2m(2ihp) = ihp/m

The generalized uncertainty principle says, in this case,

σ2xσ2H≥{(1/2i)(ih/m)<p>}^2={h/2m<p>}^2⇒ σxσH ≥h/2m|<p>|. QED

For stationary states σH = 0 and p = 0, so it just says 0 ≥ 0.

## uncertainty principle, relating the uncertainty in position to the uncertainty

for reference u can use {griffiths_d.j._introduction_to_quantum_mechanics__2ed.}