- #1
rgoerke
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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
[tex] H = H_0 - F D_z [/tex]
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 [tex] F^2 [/tex], and is in general given by
[tex] E_0^2 =-\sum_n \frac{|<n|(-F D_z)|0>|^2}{E_n - E_0}[/tex]
What I don't understand: I have seen it claimed in a couple of sources that the second order correction to energy is
[tex] E_0^2 = -\frac{1}{2}\alpha F^2 [/tex]
Where [tex]\alpha[/tex] 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 [tex]D_z = \alpha F[/tex], or equivalently
[tex] \alpha = \frac{<F|D_z|F>}{F} [/tex]
where [tex] |F> [/tex] 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
[tex] <F| H_0 - F D_z |F> = E_0 - \frac{1}{2}\alpha F^2[/tex]
by setting [tex] |F> = |0> + F|F'> [/tex], where [tex] |0> [/tex] is the ground state of [tex] H_0 [/tex] and I can get to an expression like
[tex] \frac{<F|D_z|F>}{F} = <F'|H_0|F'> + \frac{1}{2}\alpha [/tex]
which seems to suggest
[tex] F^2 <F'|H_0|F'> = \frac{1}{2}\alpha F^2= -E_0^2[/tex]
I'm not sure if this is true, or if I've made a mistake. Is it generally true, that if [tex]\psi_0^1[/tex] is the first order correction to the wave function, that we have
[tex] <\psi_0^1|H_0|\psi_0^1> = -E_0^2 [/tex] ?
Any advice or ideas would be much appreciated.
Thanks!
[tex] H = H_0 - F D_z [/tex]
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 [tex] F^2 [/tex], and is in general given by
[tex] E_0^2 =-\sum_n \frac{|<n|(-F D_z)|0>|^2}{E_n - E_0}[/tex]
What I don't understand: I have seen it claimed in a couple of sources that the second order correction to energy is
[tex] E_0^2 = -\frac{1}{2}\alpha F^2 [/tex]
Where [tex]\alpha[/tex] 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 [tex]D_z = \alpha F[/tex], or equivalently
[tex] \alpha = \frac{<F|D_z|F>}{F} [/tex]
where [tex] |F> [/tex] 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
[tex] <F| H_0 - F D_z |F> = E_0 - \frac{1}{2}\alpha F^2[/tex]
by setting [tex] |F> = |0> + F|F'> [/tex], where [tex] |0> [/tex] is the ground state of [tex] H_0 [/tex] and I can get to an expression like
[tex] \frac{<F|D_z|F>}{F} = <F'|H_0|F'> + \frac{1}{2}\alpha [/tex]
which seems to suggest
[tex] F^2 <F'|H_0|F'> = \frac{1}{2}\alpha F^2= -E_0^2[/tex]
I'm not sure if this is true, or if I've made a mistake. Is it generally true, that if [tex]\psi_0^1[/tex] is the first order correction to the wave function, that we have
[tex] <\psi_0^1|H_0|\psi_0^1> = -E_0^2 [/tex] ?
Any advice or ideas would be much appreciated.
Thanks!