Physics 232 Gauge ivariace of the magetic susceptibilty Peter Youg (Dated: Jauary 16, 2006) I. INTRODUCTION We have see i class that the followig additioal terms appear i the Hamiltoia o addig a magetic field: H = H 1 + H 2 (1) where H 1 = H 2 = e (p A + A p), (2) c e2 c 2 A2, (3) where A meas A(x), i.e. the value of the vector potetial at the positio of the electro. Note that, except i the Coulomb gauge where A = 0, A(x) does ot commute with p ad so we eed to keep the order p A + A p i H 1. For simplicity of otatio, we cosider here just sigle electro, but the discussio is easily geeralized to may electros simply by icorporatig a sum over the electros. We cosider here a time-idepedet magetic field. For the case of a uiform field we chose, i class, the followig gauge A = 1 2 x H, (4) where H is the applied field. This is a particularly coveiet gauge because we showed i class that, i it, H 1 = e h c L H, (5) which I call the paramagetic cotributio, ad which vaishes for closed shell atoms where the total orbital agular mometum L is zero. I this gauge, the remaiig piece, H 2, which I call the diamagetic cotributio, is small ad egative.
2 I a gauge trasformatio we make the replacemet A A = A + χ, (6) where χ(x) is ay fuctio of x. This trasformatio clearrly leaves the value of the field H uchaged but gives rise to the followig additioal terms i the Hamiltoia: H gauge = e (p χ + χ p) + e2 ( c c 2 2 χ A + ( χ) 2). (7) However, the fial aswer to ay calculatio should be gauge ivariat, i.e. idepedet of these extra terms ivolvig χ. I this hadout we shall show explicitly that the magetic susceptibility is gauge ivariat. The susceptibility requires the chage i the free eergy up to secod order i the field, or equivaletly, up to secod order i A. The free eergy is give by F = k B T log Z = k B T log e βe. (8) Hece if we ca show that the eergy levels are ipededet of the gauge fuctio χ up to secod order, we will have show that the susceptibility is gauge ivariat. II. THE SUSCEPTIBILITY Let us deote a eergy eigestate of the uperturbed Hamiltoia (i.e. i the absece of a field) by ad its eergy by E. To determie the chage i a eergy level k up to secod order i the field we eed to treat H 2 to first order i perturbatio theory (because it is already secod order i the field), ad H 1 to secod order (because it is oly first order i the field). Hece, if we take some gauge, the chage i E k o addig the field is give by E k = e k A p + p A k + e2 c c 2 k A2 k + e2 (c) 2 k A p + p A 2 E k E. (9) Now suppose we do a gauge trasformatio. The additioal terms i Eq. (7) are the added to
3 the Hamiltoia, but they should ot chage E k. I other words we should fid that 0 = E gauge k = (10) 1 k χ p + p χ k c (11) + e2 k χ A k (12) mc2 + e2 c 2 k ( χ)2 k (13) + e2 k χ p + p χ A p + p A k 4m 2 c 2 E k E (14) + e2 4m 2 c 2 + e2 4m 2 c 2 k A p + p A χ p + p χ k E k E (15) k χ p + p χ 2 E k E. (16) It seems a formidable task to show that these terms add up to zero. However, we will fid that it is ot as bad as it seems. Furthermore, o the way we will derive a useful result, a special case of which leads to the importat f-sum rule. III. AN IMPORTANT RESULT We see that may of the terms i E gauge k i Eq. (10) ivolve χ p + p χ. We ow derive a alterative expressio for it, which will tur out to be useful. We take the uperturbed Hamiltoia to be H = p2 + V (x), (17) so there are o velocity depedet forces i the absece of the magetic field. The commutator of χ(x) with H is give by [χ, H] = 1 [χ, p2 ] = 1 [χp2 p(χ p) + p (χp) p 2 χ] = 1 ([χ, p] p + p [χ, p]). (18) Sice [χ, p] = i h χ we have which ca be writte as [χ, H] = i h ( χ p + p χ) (19) χ p + p χ = 2i [χ, H]. (20) m h
4 Next we take matrix elemets of this equatio betwee states k ad : χ p + p χ k = 2i χh Hχ k. (21) m h Sice ad k are exact eigestates of H with eigevalues E ad E k, we ca write this last expressio as χ p + p χ k = 2i m h (E E k ) χ k, (22) which will be useful i the remaiig sectios. IV. A SPECIAL CASE: THE f-sum RULE. To simplify thigs ad to make cotact with a well-kow result i quatum mechaics, i this sectio we take the special case χ(x) = x α, (23) where α deotes oe of the cartesia coordiates. It follows that χ p = p χ = p α. Hece Eq. (22) ca be writte as (E k E ) x α k = i h m p α k. (24) Multiplyig by k x α gives (E k E ) x α k 2 = i h m k x α p α k, (25) ad iterchagig with k gives (E k E ) x α k 2 = i h m k p α x α k. (26) Addig Eqs. (25) ad (26), summig over, ad dividig by 2, gives (E E k ) x α k 2 = i h ( k x α p α k k p α x α k ) = i h k x αp α p α x α k = h2, (27) where we used the completeess of the eigestates = 1. Our result is therefore (E E k ) x α k 2 = h2 (28)
5 which is kow as the f-sum rule. It plays a importat rule i atomic ad solid state physics. The key assumptio made i derivig it is the absece of velocity-depedet forces. It is of iterest to verify that the f-sum rule works for the simple harmoic oscillator. For that model we have E k = (k + 1/2) hω, k x k + 1 = h/ω k + 1, ad k x k 1 = h/ω k, so the left had side of Eq. (28) is equal to hω h h (k + 1) + ( hω) ω ω k = h2 (29) as required. If oe multiplies Eq. (24) by k p α rather tha by k x α ad the follows the similar steps to those above, oe fids a sum rule for the momets of the mometum p α k 2 E E k = m 2. (30) (The term with = k ca be omitted because k p k = 0 accordig to Eq. (24).) Icidetally, if we had ot take the special case χ(x) = x α but had kept a geeral form for χ(x), ad used similar reasoig, we would have obtaied the followig geeralized f-sum rule : (E E k ) χ(x) k 2 = h2 k ( χ)2 k. (31) V. GAUGE INVARIANCE OF THE SUSCEPTIBILITY We ow show that a gauge trasformatio gives o chage i the susceptibility, i.e. the sum of the terms for E gauge k i Eq. (22) is zero. First of all cosider Eq. (11). Accordig to Eq. (22) with = k Eq. (11) is equal to zero. Next, substitutig Eq. (22) ito Eq. (14) (with ad k iterchaged) gives e 2 2im 4m 2 c 2 h Similarly Eq. (15) becomes Combiig Eqs. (32) ad (33) gives k χ A p + p A k = ie2 k χ(a p + p A) k. (32) 2 hmc2 ie2 k (A p + p A)χ k. (33) 2 hmc2 ie 2 k χ(a p + p A) (A p + p A)χ k = ie2 2 hmc2 k [χ, p] A + A[χ, p] k 2 hmc2 = e2 k χ A k, (34) mc2
6 which cacels Eq. (12). Hece we have foud that Eqs. (12), (14) ad (15) sum to zero. Next substitute Eq. (22) for the left had factor i Eq. (16). This gives ie 2 k χ( χ p + p χ) k. (35) 2 hmc2 Alteratively we could have substituted for the right had factor i Eq. (16) obtaiig ie2 k ( χ p + p χ)χ k. (36) 2 hmc2 Takig the average of Eqs. (35) ad (36), Eq. (16) ca be writte as ie 2 k χ( χ p + p χ) ( χ p + p χ)χ k = ie2 4 hmc2 k [χ, p] χ + χ[χ, p] k 4 hmc2 = e2 c 2 k ( χ)2 k, (37) which cacels Eq. (13). Hece we have show that Eqs. (13) ad (16) sum to zero. Altogether we have show i this sectio that Eqs. (11) (16) sum to zero. Hece the susceptibility does ot deped o choice of gauge as expected o geeral grouds.