A quick uncertainty question

1. Jan 31, 2012

StephenD420

If an atom transitions from an initially excited state emitting a photon E=hf and going to the ground state, and since ΔxΔk=1 show that ΔEΔt = hbar.

So I know Ephoton=hf=hbar*sqrt(k/m) = hbar * sqrt(1/Δx*m)
and since E=pc and p=hbar*k=hbar/Δx

but I am not sure what to do from here???
Any help would be greatly appreciated.
Thank you.
Stephen

2. Feb 1, 2012

wukunlin

i'll give you a few clues:

$$k = \frac{2 \pi}{\lambda}$$
$$c = f \lambda$$

3. Feb 1, 2012

StephenD420

ok so...
E = pc=hbar*kc=hbar*2pi/λ*fλ=hf=Ephoton
if E= hf=pc=hbar*sqrt(k/m)
hbar^2/(4pi^2) = p^2*c^2*m/k
and since ΔxΔy=1
so h=p^2*c^2*m*Δx=E*p^2*Δx
so would this lead to h=ΔE*p^2*Δx
or ΔE*Δt=h
and the uncertainty principle for RMS values be ΔEΔt=h/4pi

I am not sure what to do to get the answer from here.
Any help would be appreciated.
Thanks.
Stephen

4. Feb 4, 2012

5. Feb 4, 2012

wukunlin

I think you are just over complicating the problem, don't use E=pc (is that even true?). Just use the identities I gave you and rearrange ΔxΔk=1 into ΔEΔt = hbar.

6. Feb 4, 2012

StephenD420

How??? I am not seeing it. Would you show me the first couple of steps....

Thanks.
Stephen

7. Feb 4, 2012

wukunlin

since $k = \frac{2 \pi }{\lambda}$ and $c = f \lambda$
then $k = \frac{2 \pi f}{c}$

all of those are constants except for f, therefore

$$Δk = \frac{2 \pi}{c} Δf$$

8. Feb 4, 2012

StephenD420

ok so
k=2*pi/λ and c=f*λ and k=2*pi*f/c
so Δk= 2*pi*Δf/c
since ΔxΔk=1
1=2*pi/c * ΔxΔf
h=2*pi/c *Δx*h*Δf
hbar=ΔEΔt

and since the RMS value would be the average so 1/2*the uncertainty value so h/4*pi = ΔEΔt

is this it????

Thanks for the help.
Stephen

9. Feb 4, 2012

wukunlin

i don't see what RMS has to do with this question? hbar is your uncertainty (or the product of two uncertainties...)

10. Feb 4, 2012

StephenD420

ok am I right on the preceding work???

11. Feb 4, 2012

wukunlin

looks good, just make sure you understand every step

12. Feb 5, 2012

StephenD420

13. Feb 5, 2012

wukunlin

also stated on the site that it is an approximation for extreme relativistic velocities

14. Feb 5, 2012

StephenD420

which is needed for a photon which is why Ephoton=pc

15. Feb 5, 2012

wukunlin

ah right, sorry

16. Feb 5, 2012

StephenD420

no problem at all and thanks so much for the help....I am really just repeating what my prof said as I am trying to understand all this..

Thanks again.
Stephen