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Tuffi

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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