Understanding the quadratic Stark effect

[rgoerke]

This is what I understand right now: The stark effect is when we perturb a system with hamiltonian H_0 by applying a constant electric field, so that

H = H_0 - F D_z

where F is the field, aligned in the z direction, and D_z is the z-component of the induced dipole. The first order correction to the ground state energy in perturbation theory is 0 because of symmetry, but the second order correction should be propositional to F^2, and is in general given by

E_0^2 =-\sum_n \frac{|<n|(-F D_z)|0>|^2}{E_n - E_0}

What I don't understand: I have seen it claimed in a couple of sources that the second order correction to energy is

E_0^2 = -\frac{1}{2}\alpha F^2

Where \alpha is the polarizability. I have not problem with making this definition, since it has the correct F-dependence, but it is claimed that this definition is consistent with the classical definition of polarizability: that to first order D_z = \alpha F, or equivalently

\alpha = \frac{<F|D_z|F>}{F}

where |F> is the ground state of the full perturbed hamiltonian.

I'm having trouble convincing myself that these definitions are consistent. I have tried expanding the left hand side of

<F| H_0 - F D_z |F> = E_0 - \frac{1}{2}\alpha F^2

by setting |F> = |0> + F|F'>, where |0> is the ground state of H_0 and I can get to an expression like

\frac{<F|D_z|F>}{F} = <F'|H_0|F'> + \frac{1}{2}\alpha

which seems to suggest

F^2 <F'|H_0|F'> = \frac{1}{2}\alpha F^2= -E_0^2

I'm not sure if this is true, or if I've made a mistake. Is it generally true, that if \psi_0^1 is the first order correction to the wave function, that we have

<\psi_0^1|H_0|\psi_0^1> = -E_0^2 ?

Any advice or ideas would be much appreciated.

Thanks!

 

[fzero]

by setting |F> = |0> + F|F'>, where |0> is the ground state of H_0

If you want to compute the energy of a state to 2nd order, you must go to 2nd order in the expansion of the state. See, e.g. http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Time_Ind_PT.htm

 

[rgoerke]

If you want to compute the energy of a state to 2nd order, you must go to 2nd order in the expansion of the state. See, e.g. http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Time_Ind_PT.htm

Right, but if I include the second order (|F> = |0> + F|F'> + F^2|F''>) term I get

\frac{<F|D_z|F>}{F} = <F'|H_0|F'> + F^2<F''|H_0|F''> + \frac{1}{2}\alpha

but the claim is only that

\frac{<F|D_z|F>}{F} = \alpha

to first order, so the F^2<F''|H_0|F''> should not contribute.

 

[fzero]

It is possible to show that

\langle \psi_0^1|H_0|\psi_0^1\rangle = -E_0^2 + E_0^0 \langle \psi_0^1|\psi_0^1\rangle

by projecting \langle\psi_0^1| onto the linear terms and then using the 2nd order result. The value of \langle \psi_0^1|\psi_0^1\rangle is convention dependent and generally not 1.

In Sakurai's conventions where \langle \psi_n^0 | \psi\rangle =1,

<br /> \langle \psi_0^1|\psi_0^1\rangle=-\sum_n \frac{|\langle \psi^0_n | D_z | \psi^0_0 \rangle |^2}{(E_n - E_0)^2}.<br />

The main thing to conclude from this is that \langle \psi_0^1|\psi_0^1\rangle itself isn't well defined and we should press further in perturbation theory to relate this to expressions using the unperturbed states.


To get back to your equivalency check, I think you're making things overcomplicated. The point is that the energy shift is equal to -\alpha F^2/2 (this is the expression that corresponds to the classical result), where


<br /> \alpha = 2 \sum_n \frac{|\langle \psi^0_n | D_z | \psi^0_0 \rangle |^2}{E_n - E_0}<br />

is independent of the external field. If you do textbook perturbation theory you can obtain this result.

 

[rgoerke]

It is possible to show that

\langle \psi_0^1|H_0|\psi_0^1\rangle = -E_0^2 + E_0^0 \langle \psi_0^1|\psi_0^1\rangle

by projecting \langle\psi_0^1| onto the linear terms and then using the 2nd order result. The value of \langle \psi_0^1|\psi_0^1\rangle is convention dependent and generally not 1.

In Sakurai's conventions where \langle \psi_n^0 | \psi\rangle =1,

<br /> \langle \psi_0^1|\psi_0^1\rangle=-\sum_n \frac{|\langle \psi^0_n | D_z | \psi^0_0 \rangle |^2}{(E_n - E_0)^2}.<br />

The main thing to conclude from this is that \langle \psi_0^1|\psi_0^1\rangle itself isn't well defined and we should press further in perturbation theory to relate this to expressions using the unperturbed states.

Thanks for your help, I will think about this.

To get back to your equivalency check, I think you're making things overcomplicated. The point is that the energy shift is equal to -\alpha F^2/2 (this is the expression that corresponds to the classical result), where


<br /> \alpha = 2 \sum_n \frac{|\langle \psi^0_n | D_z | \psi^0_0 \rangle |^2}{E_n - E_0}<br />

is independent of the external field. If you do textbook perturbation theory you can obtain this result.

Again, thanks for your help. I'm finishing my undergrad and only recently have been introduced to perturbation theory. I have no doubt I'm making things overcomplicated, but it's still not clear to me how E_0^2 = -\alpha F^2/2 is equivalent to the classical definition D_z = \alpha F. I have been working with the definition of \alpha you quoted in my project, and I understand how to derive it given E_0^2 as above, I'm just not clear on connecting it to the classical definition.

Thanks

 

[fzero]

Again, thanks for your help. I'm finishing my undergrad and only recently have been introduced to perturbation theory. I have no doubt I'm making things overcomplicated, but it's still not clear to me how E_0^2 = -\alpha F^2/2 is equivalent to the classical definition D_z = \alpha F. I have been working with the definition of \alpha you quoted in my project, and I understand how to derive it given E_0^2 as above, I'm just not clear on connecting it to the classical definition.

Thanks

The energy of a induced dipole in an external field is (there is a factor of 1/2 with respect to the analogous formula for a permanent dipole)

U = \frac{1}{2} \vec{D}\cdot\vec{E} ,

so if \vec{D} = \alpha \vec{E}, then

U = - \frac{1}{2} \alpha |\vec{E}|^2.

 

[rgoerke]

Of course, that just comes from classical E&M. Thanks a lot, I understand now.