Uncertainty in Energy-Time Relation for Atomic Transitions

[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

 

[wukunlin]

i'll give you a few clues:

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

 

[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

 

[StephenD420]

bump...please any help

 

[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.

 

[StephenD420]

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

Thanks.
Stephen

 

[wukunlin]

okay, how about:

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

 

[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

 

[wukunlin]

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

 

[StephenD420]

ok am I right on the preceding work?

 

[wukunlin]

looks good, just make sure you understand every step

 

[StephenD420]

Great thanks so much for the help!

And yes E=pc look at this website to confirm (hyperphysics)
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/debrog2.html#c1

 

[wukunlin]

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

 

[StephenD420]

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

 

[wukunlin]

ah right, sorry

 

[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