Particle uncertainty principle

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Homework Statement



show that |[tex]\Delta[/tex]E/E| = |[tex]\Delta[/tex][tex]\lambda[/tex]/[tex]\lambda[/tex]|<<1

Homework Equations



[tex]\Delta[/tex]E>hbar/2pi[tex]\Delta[/tex]t

[tex]\lambda[/tex]=hc/E

The Attempt at a Solution



dunno where to start.
 

Answers and Replies

  • #2
kuruman
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Start by finding an expression for Δλ in terms of E and ΔE.
 
  • #3
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[tex]\Delta[/tex][tex]\lambda[/tex]=([tex]\Delta[/tex]E/E)[tex]\lambda[/tex]

...
 
  • #4
kuruman
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[tex]\Delta[/tex][tex]\lambda[/tex]=([tex]\Delta[/tex]E/E)[tex]\lambda[/tex]

...
If λ = hc/E, what is dλ? Can you find the differential?
 
  • #5
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you mean the derivitive of the LHS? so [tex]\lambda[/tex] d[tex]\lambda[/tex] ?? that'd just be 1?
 
  • #6
kuruman
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Can you find dλ/dE? It is not 1.
 
  • #7
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ah okay just wasnt sure what you were asking.

d[tex]\lambda[/tex]/dE would be [tex]\lambda[/tex]d[tex]\lambda[/tex] = hc/E dE

which would be 1 = -hc/E2 ?
 
  • #8
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no wait. d[tex]\lambda[/tex] wouls be a rate of change so then [tex]\Delta[/tex][tex]\lambda[/tex] = -hc/E2 ?
 
  • #9
kuruman
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ah okay just wasnt sure what you were asking.

d[tex]\lambda[/tex]/dE would be [tex]\lambda[/tex]d[tex]\lambda[/tex] = hc/E dE

which would be 1 = -hc/E2 ?
How do you figure?
One more time. What is

[tex]\frac{d}{dE}\left( \frac{hc}{E} \right)[/tex]
 
  • #10
kuruman
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no wait. d[tex]\lambda[/tex] wouls be a rate of change so then [tex]\Delta[/tex][tex]\lambda[/tex] = -hc/E2 ?
Not quite. Δλ = -(hc/E2) ΔE. Now can you find Δλ/λ ?
 
  • #11
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oh gee. i feel so dumb -.- this should be so easy.

the notation i use for this differential would be:

hc/E dE
= hcE-1 dE
= (-1)hcE-2
= (-hc)/E2

how else would i derive that with respect to e :/... and i just left h and c as is since theyre constant?
 
  • #12
kuruman
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oh gee. i feel so dumb -.- this should be so easy.

the notation i use for this differential would be:

hc/E dE
= hcE-1 dE
This is correct except you need E2 in the denominator.
= (-1)hcE-2
= (-hc)/E2
This is not. You just can't drop the dE.
 
  • #13
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oh so i did do it right.

well then, [tex]\Delta[/tex][tex]\lambda[/tex]/[tex]\lambda[/tex] = [(-hc/E2)[tex]\Delta[/tex]E]/[tex]\lambda[/tex]
 
  • #14
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oh i think i got it!!.... that expands to leave ([tex]\Delta[/tex]E/E2)(-hc/[tex]\lambda[/tex]) which is [tex]\Lambda[/tex]E/E so then [tex]\Delta[/tex][tex]\lambda[/tex]/[tex]\lambda[/tex]=[tex]\Lambda[/tex]E/E
 
  • #15
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yes i meant to put in deltaE
 
  • #16
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whats the relevance of |[tex]\Delta[/tex][tex]\lambda[/tex]/[tex]\lambda[/tex]|<<1?

does << mean less than? or something else?
 
  • #17
kuruman
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whats the relevance of |[tex]\Delta[/tex][tex]\lambda[/tex]/[tex]\lambda[/tex]|<<1?

does << mean less than? or something else?
"<" means "less than"; "<<" means "much less than"; "<<<" means "even less than that".
 
  • #18
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well in this particular question deltaE = 4.1neV and E=2.58eV so deltaE/E would be much much much much less than one. which shows that the equality would only hold if deltalambda/lambda was equaly small. would that be sufficient reason?
 
  • #19
kuruman
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You have shown that Δλ/λ = ΔΕ/Ε. This equality holds no matter what. If ΔΕ/Ε << 1, then Δλ/λ << 1 no matter what.
 
  • #20
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So it's all good then :)
 
  • #22
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Nah the other parts to the question I could do. I found all the answers just wasn't sure how to actually show that proof. So thanks :)
 

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