1. Jan 12, 2015

Matt atkinson

1. The problem statement, all variables and given/known data
A Stark effect experiment is performed on the rubidium D1 line $(5p \ {}^2P_{1/2} → 5s \ {}^2S_{1/2})$ at 780.023 nm. Given that the polarizabilities of the $5p \ {}^2P_{1/2}$ and $5s \ {}^2S_{1/2}$ levels are $6.86 × 10^{−16}$ and $2.78 × 10^{−16} cm−1 m^2 V^{−2}$ , respectively, deduce the field strength that would have to be applied to shift the wavelength by 0.001 nm, stating
whether the wavelength shift would be positive or negative.

2. Relevant equations

3. The attempt at a solution
So i know that basically; $\vec{p}=\alpha \vec{E}$
and so $\Delta \epsilon=-\frac{1}{2}\alpha E^2$ by integrating from $E=0$ to $E=E$.
I know that both levels shift by an amount proportional to their polarization but, im just completely stuck on how to go about solving it.
I drew a diagram and understand the process i think, just when i go about solving it I end up with stupid answers.

2. Jan 12, 2015

Staff: Mentor

Please show what you did then, so we can see what went wrong.

3. Jan 13, 2015

Matt atkinson

Okay, here's what I was trying to do;
So both levels the ${}^2P_{1/2}$ and the ${}^2S_{1/2}$ would be shifted by an applied electric field.
If that shift was equal to;
$\Delta \epsilon_1=-\frac{1}{2}\alpha_1E^2$
and
$\Delta \epsilon_2=-\frac{1}{2}\alpha_2E^2$
for the P and S state respectively.
The photon in the transistion given off then has a shift of $0.001nm$ so I did
$\frac{hc}{\lambda }=\frac{hc}{(780.023nm)}+(\Delta \epsilon_1 +\Delta \epsilon_2)=\frac{hc}{(780.023nm)\pm (0.001nm)}$
This is as far as i got, I was wondering if I can use a $+$ for the shift because the polarizabilities are positive, but im not sure.

4. Jan 13, 2015

Staff: Mentor

Why do you add the two energy shifts?

You can plug in the formulas and solve for E. One sign for the shift will give a solution, one will not.

5. Jan 13, 2015

Matt atkinson

I added the two shifts because they would shift the total energy also, or do you mean why is it a plus? I suppose it should be a $\pm$ right?

6. Jan 13, 2015

Staff: Mentor

Your photon comes from the transition between the two states. If you know the energies E1 and E2 (let's say E1>E2), what is the energy of the transition? What changes if you increase E1 by some value X and E2 by a different value Y?

7. Jan 13, 2015

Matt atkinson

Oh right, so as before $\Delta E=E_2 - E_1$ then after shift $\Delta E'=(E_2 + Y) - (E_1 +X)$.

So in the case above id have;
$\frac{hc}{780.023nm}+\Delta\epsilon_2 - \Delta\epsilon_1=\frac{hc}{(780.023nm)\pm(0.001nm)}$
and so
$\frac{hc}{780.023nm}-\frac{1}{2}\alpha_2E^2+ \frac{1}{2}\alpha_1E^2=\frac{hc}{(780.023nm)\pm(0.001nm)}$
and then solve for E, and like you mentioned previously one either the plus or minus will give a solution.

Which would have to be the positive shift because its energy needs to be greater so the solution isn't complex.
Thankyou!

I got;

$E=\sqrt(2(\frac{hc}{(780.023nm)\pm(0.001nm)}-\frac{hc}{780.023nm})*(\alpha_1-\alpha_2)^{-1})$

This would only work for a "-" wavelength shift uhm.

Last edited: Jan 13, 2015
8. Jan 13, 2015

Staff: Mentor

Check the signs. the higher energy level gets a larger shift downwards.

9. Jan 13, 2015

Matt atkinson

$\alpha_1$ is for the higher energy level with a larger polarization factor, which means with my equation now, if i swapped any signs wouldnt it mean that the answer would be complex?

10. Jan 13, 2015

Staff: Mentor

No, it means you have to reconsider the sign of the wavelength shift after you fixed the energy shift.

11. Jan 14, 2015

Matt atkinson

Oh i see, to the signs of the alpha would be the other way round, which would make it a positive wavelength shift.