Chapter 9: Q1P (page 343)

A hydrogen atom is placed in a (time-dependent) electric $field E = E (t) \overset{⏜}{k .} calculate all four matrix elements H_{ij}^{,} of the perturbation \overset{⏜}{H^{,}} = eEz$ between the ground state (n = 1 ) the (quadruply degenerate) first excited states (n = 2 ) . Also show ${thatH}_{ii}^{,} = 0$ for all five states. Note: There is only one integral to be done here, if you exploit oddness with respect to z; only one of the n = 2 states is “accessible” from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration—assuming transitions to higher excited states can be ignored.

Short Answer

Expert verified

$H_{100, 210}^{,} = - 0.7449 e E a .$

Step by step solution

Concept of time varying electric field

Consider a hydrogen atom placed in a time-varying electric field:

$E = E (t) \overset{⏜}{z}$

Where E(t) is some arbitrary function of time. The spatial wave functions formula is given by:

$ψ_{n / m} = R_{nl} Y_{l}^{m}$

Calculating all for matrix elements and showing that Hii,=0

$ψ_{n / m} = R_{n |y| m} . From Tables 4.3 and 4.7 :$

$ψ_{100} = \frac{1}{\sqrt{{πa}^{3}}} e^{- r / a}; ψ_{200} = \frac{1}{\sqrt{8^{'} {πa}^{3}}} (1 - \frac{r}{2 a}) e^{- r / 2 a} ψ_{210} = \frac{1}{\sqrt{32 {πa}^{3}}} \frac{r}{a} e^{- r / 2 a} \cos θ; ψ_{21 \pm 1} = \pm \frac{1}{\sqrt{64 {πa}^{3}}} \frac{r}{a} e^{- r / 2 a} \sin θ e^{\pm i ϕ}$

$But r cosθ = z and r {sinθe}^{\pm iϕ} = r \sin (cosϕ \pm i sinϕ) = r sinθ cosϕ \pm ir sinθsinϕ . so {|ψ|}^{2 is} an even function of z in all cases, and hence \int z {|ψ|}^{2} dxdydz, {soH}_{ii}^{,} = 0 moreover, ψ_{100} is even in z, and so are ψ_{200}, ψ_{211}, and ψ_{21 - 1}, so H_{ij}^{,} = 0 for all except H_{100, 210}^{,} = eE \frac{1}{\sqrt{{πa}^{3}}} \frac{1}{\sqrt{32 {πa}^{3}}} \frac{1}{a} \int e^{- r / a} e^{- r / 2 a} z^{2} d^{3} r = \frac{eE}{4 \sqrt{2 {πa}^{4}}} \int e^{- 3 r / 2 a} r^{2} \cos^{2} {θr}^{2} sinθdrdθ . = \frac{eE}{4 \sqrt{2 {πa}^{4}}} \int_{0}^{\infty} r^{4} e^{- 3 r / 2 a} dr \int_{0}^{π} \cos^{2} θsinθdθ \int_{0}^{2 π} dϕ = \frac{eE}{4 \sqrt{2 {πa}^{4}}} 4! {(\frac{2 a}{3})}^{5} \frac{2}{3} 2 π = (\frac{2^{8}}{3^{5} \sqrt{2}}) or 0 : 7449 eEa .$

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Short Answer

Step by step solution

Concept of time varying electric field

Calculating all for matrix elements and showing that Hii,=0

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