2B24 Atomic and Molecular Physics 4Lecture18 proof

Proof of equation (40) from Lecture 18 J is the total angular momentum vector, and V is a vector operator.

The commutation relations for the components of J and V are [Jx , Vx ] = 0, [Jx , Vy ] = i¯ hVz , [Jx , Vz ] = −i¯ hVy [Jy , Vx ] = −i¯ hVz , [Jy , Vy ] = 0, [Jy , Vz ] = i¯ hVx [Jz , Vx ] = −i¯ hVy , [Jz , Vy ] = −i¯ hVx , [Jz , Vz ] = 0

(1) (2) (3)

Also, the commutation relations for the individualt components of J are: [Jx , Jy ] = i¯ hJz , [Jy , Jz ] = i¯ hJx [Jz , Jx ] = i¯ hJy

(4)

Using these we can show that: J × V + V × J = 2i¯ hV

(5)

and also that h

h

J 2, J 2, V

ii

= 2¯ h2 J 2 V + V J 2 − 4¯ h2 (V · J) J h

(6)

ii

h

Now the matrix element hψ| J 2 , J 2 , V |ψi of the left hand side of equation 6 for |ψi = |lsjmj i is equal to zero, so we must have:

hlsjmj | J 2 V + V J 2 |lsjmj i = 2hlsjmj | (V · J) J|lsjmj i

(7)

and so: j(j + 1)¯ h2 hlsjmj |V |lsjmj i = hlsjmj | (V · J) J|lsjmj i

(8)

Now recall the energy shift from the lecture: ∆ES =

Z

ψ∗

µB B zˆ · Sψdτ h ¯

where we have put B = Bz. This energy is: µB ∆ES = Bhψ|Sz |ψi h ¯ 1

(9)

(10)

So we can use equation 8 by setting V = S and taking the z− component, which gives; 1 hψ|(S · J)Jz |ψi j(j + 1)¯ h2 mj = hψ|(S · J)|ψi j(j + 1)¯ h

hψ|Sz |ψi =

(11)

We can now write L = J − S, so that L2 = J 2 + S 2 − 2S · J. And: 1 hψ|J 2 + S 2 − L2 |ψi 2 1 2 = h ¯ (j(j + 1) + s(s + 1) − l(l + 1)) 2

hψ|S · J|ψi =

(12)

which by inserting into the above gives: "

j(j + 1) + s(s + 1) − l(l + 1) hψ|Sz |ψi = mj h ¯ 2j(j + 1)

#

(13)

And so the energy shift is: "

j(j + 1) + s(s + 1) − l(l + 1) ∆ES = mj µB B 2j(j + 1) as given in the lectures (equation 40).

2

#

(14)