Question 11.30: In this problem we will recover the results Section 11.2.1 d...

(a)

E =-\nabla \varphi-\frac{\partial A }{\partial t}=-\frac{ E _{0}}{\omega} \cos ( k \cdot r -\omega t)(-\omega)= E _{0} \cos ( k \cdot r -\omega t) .

B = \nabla \times A =\frac{1}{\omega} \sin ( k \cdot r -\omega t) \nabla \times E _{ 0 }-\frac{1}{\omega} E _{ 0 } \times \nabla (\sin ( k \cdot r -\omega t)) .

=0-\frac{1}{\omega} E _{0} \times[\cos ( k \cdot r -\omega t) k ]=\frac{ k \times E _{0}}{\omega} \cos ( k \cdot r -\omega t)=\frac{1}{\omega} k \times E .

(note that E_0 is a constant, so its derivatives are zero).

(b) From Eq. 4.188

H=\frac{1}{2 m}( p -q A )^{2}+q \varphi             (4.188).

H=\frac{1}{2 m}( p -q A ) \cdot( p -q A )+q \varphi+V=\frac{p^{2}}{2 m}+V-\frac{q}{2 m}( p \cdot A + A \cdot p )+\frac{q^{2}}{2 m} A^{2} .

=H^{0}-\frac{q}{2 m}( p \cdot A + A \cdot p )+\frac{q^{2}}{2 m} A^{2} .

Now, acting on a test function f(r),

( p \cdot A + A \cdot p ) f=-i \hbar[ \nabla ( A f)+ A \cdot(\nabla f)]=-i \hbar[( \nabla \cdot A ) f+ A \cdot( \nabla f)+ A \cdot( \nabla f)] ,

But

\nabla \cdot A =\frac{1}{\omega} \nabla \cdot\left[ E _{0} \sin ( k \cdot r -\omega t)\right]=\left.\frac{1}{\omega}[( \nabla \cdot E ) 0) \sin ( k \cdot r -\omega t)+ E _{ 0 } \cdot( \nabla \sin ( k \cdot r -\omega t))\right] .

=\frac{1}{\omega}\left[0+ E _{0} \cdot \cos ( k \cdot r -\omega t) k \right]=0 .

\left(\text { since } E _{0} \cdot k =0 .\right)   So (dropping the test function)

( p \cdot A + A \cdot p )=2 A \cdot p .

Ignoring the A² term, then,

H^{\prime}=\frac{e}{m} A \cdot p =\frac{e}{m \omega} \sin ( k \cdot r -\omega t) E _{0} \cdot p =\frac{e}{2 \operatorname{im\omega}} e^{i k \cdot r } E _{0} \cdot p e^{-i \omega t}-\frac{e}{2 i m \omega} e^{-i k \cdot r } E _{0} \cdot p e^{i \omega t} .

(c) Referring to Eq. 11.36, we have

H_{b a}^{\prime}=\frac{V_{b a}}{2} e^{-i \omega t}, \quad H_{a b}^{\prime}=\frac{V_{a b}}{2} e^{i \omega t}                     (11.36).

V_{b a}=\frac{e}{i m \omega} E_{0}\left\langle b\left|p_{z}\right| a\right\rangle .

But (using Eqs. 2.52 and 3.65)

[x, \hat{p}]=i \hbar                    (2.52).

[\hat{A} \hat{B}, \hat{C}]=\hat{A}[\hat{B}, \hat{C}]+[\hat{A}, \hat{C}] \hat{B}                    (3.65).

\left[H^{0}, z\right]=\frac{1}{2 m}\left[p^{2}, z\right]=\frac{1}{2 m}\left[p_{z}^{2}, z\right]=\frac{1}{2 m}\left(p_{z}\left[p_{z}, z\right]+\left[p_{z}, z\right] p_{z}\right)=\frac{-i \hbar}{m} p_{z} .

so (referring to Eqs. 11.18 and 11.40)

\omega_{0} \equiv \frac{E_{b}-E_{a}}{\hbar}                 (11.18).

H_{b a}^{\prime}=-\wp E_{0} \cos (\omega t), \quad \text { where } \wp \equiv q\left\langle\psi_{b}|z| \psi_{a}\right\rangle               (11.40).

V_{b a}=\frac{e}{i m \omega} E_{0} \frac{i m}{\hbar}\left\langle b\left|\left[H^{0}, z\right]\right| a\right\rangle=\frac{e E_{0}}{\hbar \omega}\left\langle b\left|\left(H^{0} z-z H^{0}\right)\right| a\right\rangle=\frac{e E_{0}}{\hbar \omega}\left(E_{b}-E_{a}\right)\langle b|z| a\rangle .

=\frac{e E_{0}}{\hbar \omega} \hbar \omega_{0}\langle b|z| a\rangle=-\frac{\omega_{0}}{\omega} E_{0} \wp .

This differs from Eq. 11.41, which did not include the ratio of the !’s, but it does not affect our conclusions in Sections 11.2.3 or 11.3, because we only considered transitions at resonance, \omega=\omega_{0} (the probability of a transition at other driving frequencies being negligible).

V_{b a}=-\wp E_{0}                    (11.41).