What is the force on a neutral atom of polarizability α that is placed a short distance d from an ideal mirror

Question

Hint: Assume a classical oscillator model of mass m.

Answer 1

IIRC, the London force between neutral atoms goes like one over r^7 or maybe r^6.  But if the dipole is close enough to the mirror that it oscillates in perfect phase with its image, it drops off slower maybe.  I'll see if I can find a derivation.  And in this case, you aren't actually inducing a dipole in the image--the image just IS a dipole automatically.

So if we treat it as an oscillator of mass m and spring constant k:

ground state energy = h-bar omega

= 1/2 h-bar sqrt (k/m)

<PE> = 1/2 k <x^2> = 1/2 GSE = 1/4 h-bar sqrt (k/m)

<x^2> = 1/2 h-bar / sqrt (km)

Can we express the spring constant in terms of something more real?

p = alpha E = qx

F = kx = qE = q^2 x / alpha

k = q^2 / alpha

Plug that back in:

<x^2> = h-bar sqrt (alpha/m) / 2q

The rms dipole moment is:

<p^2> = q^2<x^2>

= q h-bar sqrt (alpha/m) / 2

If we assume, then, that the atom is perfectly in phase with its image, then you just get the 1 / r^4 dipole/dipole force.

<F> = 6 k <p^2> / r^4, where k is now Coulomb's constant, not the made-up k I imagined before.

= 3 k q h-bar sqrt (alpha/m) / r^4

= 3/16 k q h-bar sqrt (alpha/m) / d^4

(because r is just 2d).

The answer has a couple parameters that aren't in the question.  The effective "mass" and "charge" that will be drawn out in an actual atom isn't obvious unless it's hydrogen (although Ze and Z*electronmass may be close enough).  The force is repulsive if the dipole is in phase with its image (head-to-head and tail-to-tail as if light traveled instantaneously.  If it were out of phase, the force could be attractive or non-existant.

So multiply the force by cos (2d omega/ c)

= cos (2dq / (c sqrt (m alpha) ) )