Introduction to Quantum Mechanics · Problem 11.30

Quantum Dynamics — David J. Griffiths

Physics — Quantum Mechanics · electromagnetic-wave-perturbation-theory

Problem

In this problem we will recover the results Section 11.2.1 directly from the Hamiltonian for a charged particle in an electromagnetic field (Equation 4.188). An electromagnetic wave can be described by the potentials A=E0ωsin⁡(k⋅r−ωt),φ=0\mathbf{A} = \frac{\mathbf{E}_0}{\omega}\sin(\mathbf{k}\cdot\mathbf{r} - \omega t), \qquad \varphi = 0 where in order to satisfy Maxwell's equations, the wave must be transverse (E0⋅k=0\mathbf{E}_0 \cdot \mathbf{k} = 0) and of course travel at the speed of light (ω=c∣k∣\omega = c|\mathbf{k}|). (a) Find the electric and magnetic fields for this plane wave. (b) The Hamiltonian may be written as H0+H′H^0 + H' where H0H^0 is the Hamiltonian in the absence of the electromagnetic wave and H′H' is the perturbation. Show that the perturbation is given by H^′(t)=e2imωeik⋅rE0⋅p^ e−iωt−e2imωe−ik⋅rE0⋅p^ eiωt,(11.127)\hat{H}'(t) = \frac{e}{2im\omega}e^{i\mathbf{k}\cdot\mathbf{r}}\mathbf{E}_0\cdot\hat{\mathbf{p}}\,e^{-i\omega t} - \frac{e}{2im\omega}e^{-i\mathbf{k}\cdot\mathbf{r}}\mathbf{E}_0\cdot\hat{\mathbf{p}}\,e^{i\omega t}, \tag{11.127} plus a term proportional to E02E_0^2 that we will ignore. *Note:* the first term corresponds to absorption and the second to emission. (c) In the dipole approximation we set eik⋅r≈1e^{i\mathbf{k}\cdot\mathbf{r}} \approx 1. With the electromagnetic wave polarized along the zz direction, show that the matrix element for absorption is then Vba=−ω0ω℘E0.V_{ba} = -\frac{\omega_0}{\omega}\wp E_0. Compare Equation 11.41. They're not exactly the same; would the difference effect our calculations in Section 11.2.3 or 11.3? Why or why not? *Hint:* To turn the matrix element of p\mathbf{p} into a matrix element of r\mathbf{r}, you need to prove the following identity: im[H^0,r^]=ℏp^im\left[\hat{H}^0, \hat{r}\right] = \hbar\hat{\mathbf{p}}.

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