Calculations Using the Uncertainty Relation

 
One can write down the following expression for the energy and time uncertainties:
 
ΔE · Δt ³  h/4π  =  h/2

and since E = hn:

Dn · Δt ≥  1/4π

Therefore we have the lower border Dn for frequency sharpness of the light impulse having length Δt.

For instance, light impulse having duration of 10 ns may have quite huge bandwidth Dn ³ 1/4p · 10-8s ≈ 10 MHz. And so ultra-short light impulses laying in the femtosecond range (1fs = 10-15s) may have bandwidth of about Dn» 1014 Hz, that almost corresponds to the frequency of visible light, i.e. such light impulse will be seen as white light by us.

For how long time would this fs-impulse be if we send it through a very good monochromator having 10 MHz resolution?
Since the frequency is spectrally narrowed due to this monochromator, one will obtain the time length:               Δt  ≥ 1/4π Dn.≈ 8·10-9s. I.e. the initially very narrow impulse of about 10-15 seconds will be extended on more than 6 orders!


 
Now we would like play with the uncertainty relation in order to measure atomic size: if we talk about electron in hydrogen atom then it will be smeared over Δx = r range. The impulse p will lay in p ≈ Δp ≈ h/4πr range.

The kinetic energy is as follows:

½ mv²  =  p²/2m ≈  Δp²/2m ≈  h²/32π²mr² ,

and potential energy is

Epot  =  - e²/4pe0r

The total energy is 

E  =  h²/32π²mr² − e²/4pe0r

The total energy minimum lies near

dE/dr  = − h²/16π²mr³ + e²/4pe0r²  =  0

®     rmin  =  h²ε0/4πme² ≈  0,13 Å

The minimum of the total energy is as follows:

Emin  =  − me4/2ε0²h² <  0
One can see that it's negative, i.e. electron is bounded.
 

The more accurate calculation gives:

 r  =  a0  = h²ε0/πme²  =  0,52 Å
E(a0)  =  − me4/8ε0h²  =  RH :       Rydberg constant

We can approximately estimate the H-atom size and its ionization energy by using the uncertainty principle.

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