Particle uncertainty principle

[ProPatto16]

Homework Statement

show that |$$\Delta$$E/E| = |$$\Delta$$$$\lambda$$/$$\lambda$$|<<1

Homework Equations

$$\Delta$$E>hbar/2pi$$\Delta$$t

$$\lambda$$=hc/E

The Attempt at a Solution

dunno where to start.

 

[kuruman]

Start by finding an expression for Δλ in terms of E and ΔE.

 

[ProPatto16]

$$\Delta$$$$\lambda$$=($$\Delta$$E/E)$$\lambda$$

...

 

[kuruman]

$$\Delta$$$$\lambda$$=($$\Delta$$E/E)$$\lambda$$

...

If λ = hc/E, what is dλ? Can you find the differential?

 

[ProPatto16]

you mean the derivitive of the LHS? so $$\lambda$$ d$$\lambda$$ ?? that'd just be 1?

 

[kuruman]

Can you find dλ/dE? It is not 1.

 

[ProPatto16]

ah okay just wasnt sure what you were asking.

d$$\lambda$$/dE would be $$\lambda$$d$$\lambda$$ = hc/E dE

which would be 1 = -hc/E² ?

 

[ProPatto16]

no wait. d$$\lambda$$ wouls be a rate of change so then $$\Delta$$$$\lambda$$ = -hc/E² ?

 

[kuruman]

ah okay just wasnt sure what you were asking.

d$$\lambda$$/dE would be $$\lambda$$d$$\lambda$$ = hc/E dE

which would be 1 = -hc/E² ?

How do you figure?
One more time. What is

$$\frac{d}{dE}\left( \frac{hc}{E} \right)$$

 

[kuruman]

no wait. d$$\lambda$$ wouls be a rate of change so then $$\Delta$$$$\lambda$$ = -hc/E² ?

Not quite. Δλ = -(hc/E²) ΔE. Now can you find Δλ/λ ?

 

[ProPatto16]

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?

 

[kuruman]

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.

 

[ProPatto16]

oh so i did do it right.

well then, $$\Delta$$$$\lambda$$/$$\lambda$$ = [(-hc/E²)$$\Delta$$E]/$$\lambda$$

 

[ProPatto16]

oh i think i got it!... that expands to leave ($$\Delta$$E/E²)(-hc/$$\lambda$$) which is $$\Lambda$$E/E so then $$\Delta$$$$\lambda$$/$$\lambda$$=$$\Lambda$$E/E

 

[ProPatto16]

yes i meant to put in deltaE

 

[ProPatto16]

whats the relevance of |$$\Delta$$$$\lambda$$/$$\lambda$$|<<1?

does << mean less than? or something else?

 

[kuruman]

whats the relevance of |$$\Delta$$$$\lambda$$/$$\lambda$$|<<1?

does << mean less than? or something else?

"<" means "less than"; "<<" means "much less than"; "<<<" means "even less than that".

 

[ProPatto16]

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?

 

[kuruman]

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.

 

[ProPatto16]

So it's all good then :)

 

[kuruman]

So it's all good then :)

Unless there is more.

 

[ProPatto16]

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 :)