Particle uncertainty principle

In summary, the homework statement is that deltaE/E = delta\lambda/\lambda. If λ = hc/E, then dλ is found to be hcE-1 dE.
  • #1
ProPatto16
326
0

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

...
 
  • #4
ProPatto16 said:
[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
you mean the derivitive of the LHS? so [tex]\lambda[/tex] d[tex]\lambda[/tex] ?? that'd just be 1?
 
  • #6
Can you find dλ/dE? It is not 1.
 
  • #7
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
no wait. d[tex]\lambda[/tex] wouls be a rate of change so then [tex]\Delta[/tex][tex]\lambda[/tex] = -hc/E2 ?
 
  • #9
ProPatto16 said:
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
ProPatto16 said:
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
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
ProPatto16 said:
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
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
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
yes i meant to put in deltaE
 
  • #16
whats the relevance of |[tex]\Delta[/tex][tex]\lambda[/tex]/[tex]\lambda[/tex]|<<1?

does << mean less than? or something else?
 
  • #17
ProPatto16 said:
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
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
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
So it's all good then :)
 
  • #21
ProPatto16 said:
So it's all good then :)
Unless there is more.
 
  • #22
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 :)
 

1. What is the particle uncertainty principle?

The particle uncertainty principle, also known as the Heisenberg uncertainty principle, is a fundamental principle of quantum mechanics that states that it is impossible to simultaneously know the exact position and momentum of a particle. This principle arises from the wave-particle duality of quantum particles, which means that they can exhibit both wave-like and particle-like behavior.

2. How does the particle uncertainty principle affect our understanding of particles?

The particle uncertainty principle has a significant impact on our understanding of particles because it sets a limit on the precision with which we can measure their properties. This means that there will always be some degree of uncertainty in our knowledge of a particle's position or momentum, and this uncertainty increases as the precision of our measurements increases.

3. Is the particle uncertainty principle a limitation of our measurement tools?

No, the particle uncertainty principle is not a limitation of our measurement tools. It is a fundamental property of quantum particles that cannot be overcome. This means that even with perfect measurement tools, we would still encounter this uncertainty in our measurements.

4. How does the particle uncertainty principle relate to the observer effect?

The particle uncertainty principle is closely related to the observer effect, which is the idea that the act of observing a system can affect its behavior. In the case of the particle uncertainty principle, the act of measuring a particle's position or momentum inevitably disturbs the particle, making it impossible to know both properties with certainty.

5. Can the particle uncertainty principle be applied to macroscopic objects?

While the particle uncertainty principle is a fundamental property of quantum particles, it is not typically applicable to macroscopic objects. This is because the effects of quantum mechanics are generally only observable at the subatomic level. However, some experiments have shown that the principle can also apply to larger objects, such as molecules, under certain conditions.

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