Calculation of the dipole polarizability of the hydrogen atom

[Tuffi]

A hydrogen atom placed in an electrical field results in a changed energy level and a changed eigenfunction, compared to the free atom. To examine this effect, we choose a homogenous electrical field of the force F, whose field lines run along parallel to the z-axis. The Schrödinger equation is in atomic units:

Ĥ(F)|ψ(F)> = (-1/2 Nabla^2 - 1/r+ Frcosθ) |ψ(F)>
=(Ĥ_0+ Frcosθ)|ψ(F)> = ε(F) |ψ(F)>.
Here, Ĥ_0 is the Hamilton operator for the hydrogen atom in absence of the field. We want to calculate the upper limit for the ground state energy ε(F) in the field through the variational principle. As a testfunction we use:
|ϕ ̃> =c_1 |1s> + c_2 |2p_z>,
Where |1s> and |2p_z> are the exact, normalized eigenfunctions for the free hydrogen atom:
|1s> = √π e^(-r)
|2p_z> = √(32π) re^(-r⁄2) cosθ.
Why are the functions |2p_x> and |2p_y> not used in ϕ ̃>?
The actual variational problem consists in minimizing the expectancy value <ϕ ̃|Ĥ|ϕ ̃> according to the side condition <ϕ ̃|ϕ ̃> =1. Formulate the extremely initially with the method of the Lagrange multipliers. Show that this problem is equivalent to finding the eigenvalues and eigenvectors of a (2x2)-Hamilton-matrix. (What is figured out for the matrix elements <1s| Ĥ(F) |1s>, <1s| Ĥ(F) |2p_z>, <2p_z | Ĥ(F) |1s>, and <2p_z | Ĥ(F) |2p_z> is not yet of interest at this point. These matrix elements can be abbreviated as H_11, H_12, H_21, and H_22 at this point, yet.)
Now determine the optimal values for the coefficients c_1 and c_2 and the upper limit for the energy ε(F) of the ground state in the field. (Hereto you must certainly calculate the matrix elements.)
To determine the electrical dipole polarizability α, ε(F) is developed from part c) in a Taylor series like (1 + x)^2 ≈1+ x⁄2…:
ε(F) ≈ ε(F=0)- 1/2 αF^2+ ….
Determine the value of α. How could you diminish the deviation?

Hints:
〖 Ĥ〗_0 |1s> = -1/2|1s> 〖 Ĥ〗_0 |2p_z> = -1/8|2p_z>
∫_0^∞▒〖r^n e^(-αr) dr= n!/α^(n + 1) 〗




Ĥ(F)|ψ(F)> = (-1/2 Nabla^2 - 1/r+ Frcosθ) |ψ(F)>
=(Ĥ_0+ Frcosθ)|ψ(F)> = ε(F) |ψ(F)>

|ϕ ̃> =c_1 |1s> + c_2 |2p_z>

|1s> = √π e^(-r)
|2p_z> = √(32π) re^(-r⁄2) cosθ

ε(F) ≈ ε(F=0)- 1/2 αF^2+ …

〖 Ĥ〗_0 |1s> = -1/2|1s> 〖 Ĥ〗_0 |2p_z> = -1/8|2p_z>
∫_0^∞▒〖r^n e^(-αr) dr= n!/α^(n + 1) 〗

To a) In my opinion these functions are perpendicular to the field so that a dipole polarizability is not possible.

To b) I guess my function is
|ϕ ̃> =c_1 |1s> + c_2 |2p_z>,
which is
|ϕ ̃> =c_1 | √π e^(-r)> + c_2 |√32π re^(-r⁄2) cosθ>.
The side condition is obviously
<ϕ ̃|ϕ ̃> =1.
Now my problem is how to set these two equations in a manner of Lagrange multipliers. Do I have to square the first equation to get equivalence to the second equation? What is my x and what is my y; r and θ? For my side condition I get F_λ=0=0. It does not make any sense to me. I would appreciate it if somebody could help me out of my confusion.

Thank you in advance.



Tuffi

 

[Nino MMP]

To b) The Lagrangian for the variational problem is L(c_1, c_2, λ) = <ϕ ̃|Ĥ|ϕ ̃> - λ[<ϕ ̃|ϕ ̃> - 1]. The equation to solve is ∂L/∂c_1 = 0, ∂L/∂c_2 = 0, ∂L/∂λ = 0.These three equations will form a system of equations for c_1, c_2 and λ. Once the optimal values for c_1 and c_2 are found, the upper limit for the energy ε(F) can be calculated by plugging in these values into the Lagrangian. For the matrix elements, the expectation value of the Hamiltonian with respect to the test function must be computed. This involves computing integrals of the form <ϕ ̃|Ĥ|ϕ ̃> = ∫ ϕ ̃*(r, θ) (Ĥ_0 + Frcosθ)ϕ ̃(r, θ) drdθ. The integrals can be evaluated using standard methods such as integration by parts or series expansion. To c) To determine the electrical dipole polarizability α, we need to expand ε(F) in a Taylor series. This can be done by first expanding the testfunction in powers of F, |ϕ ̃> = c_1 |1s> + c_2 |2p_z> + c_3 |1s>F + c_4 |2p_z>F + ... and then plugging this into the Hamiltonian and solving for the coefficients c_i. The Taylor series for ε(F) can then be written as ε(F) = ε(F=0) + c_3 <1s|Ĥ|1s>F + c_4 <2p_z|Ĥ|2p_z>F + ... From this, we can find the dipole polar