Galilean boost symmetry 1 D. E. Soper 2 University of Oregon 4 April 2011

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1 Gallean boost symmetry 1 D. E. Soper 2 Unversty of Oregon 4 Aprl 2011 I offer here some background for Chapter 4 of J. J. Sakura, Modern Quantum Mechancs. Ths chapter dscusses symmetres other than translatons and rotatons, ncludng party, translatons n a lattce, and tme reversal. It omts boost nvarance, so I dscuss boost nvarance here. 1 Introducton Non-relatvstc classcal mechancs s nvarant under a change of reference frame to one that s movng wth a fxed velocty wth respect to the reference frame orgnally consdered. To be more precse, suppose that we descrbe a system n the lab frame usng coordnates (t, x, y, z). Then we swtch to a movng frame usng coordnates (t, x, y, z ). The transformaton s t = t, x = x + v t. Note that f we were dong relatvstc physcs, we would have t t, but lfe s smpler wth non-relatvstc physcs. For understandng quantum mechancs, t s smpler to thnk of leavng the coordnates alone but gvng the system a boost by velocty v: we smply add v to the velocty of every partcle. 2 Boost operator We are thus led to consder the untary operator U( v) that boosts the system. The operator U( v) does nothng to spns. For ths reason, we consder partcles wthout spns. We consder a system made of partcles 1 through N wth masses m 1 through m N. We begn wth a momentum space descrpton, n whch the bass vectors are p1, p 2,..., p N. The operator U( v) adds v to the velocty of every partcle. Thus for a partcle wth mass m, U( v) adds m v to the momentum of the partcle: U( v) p1, p 2,..., p N = p1 + m 1 v, p 2 + m 2 v,..., p N + m N v. (1) 1 Copyrght, 2011, D. E. Soper 2 soper@uoregon.edu 1

2 Now let s see what ths does n a poston space descrpton. We have U( v) ( x1, x 2,..., x N = (2π) 3N/2 d p 1 d p N exp ) p x U( v) p1, p 2,..., p N ( = (2π) 3N/2 d p 1 d p N exp ) p x p1 + m 1 v, p 2 + m 2 v,..., p N + m N v = (2π) 3N/2 d k 1 d k N ( exp ) ( k1 k m v) x, k 2,..., (2) k N ( = exp ) m v x (2π) 3N/2 d k 1 d k N ( exp ) k1 k x, k 2,..., k N ( = exp m v x ) x1, x 2,..., x N. That s U( v) x1, x 2,..., x N = exp (M v R ) x1, x 2,..., x N (3) where M s the total mass of the system, M = m (4) and R s the operator that gves the poston of the center of mass of the system R = 1 m x. (5) M We learn two thngs from ths. Frst, the boost operator s very smple n the poston representaton. Second, the generator of boosts s M R. That s rather smlar to the generator of space translatons beng P. 2

3 3 Transformaton of operators These results tell us how the operators x and p transform under boosts. (We could call these x,op and p,op, but we don t do that here because we are not n danger of confusng the operators wth ther egenvalues.) We have U( v) 1 x U( v) = x, U( v) 1 p U( v) = p + m v. One operator of some mportance n quantum mechancs s the total momentum, P = p. (7) (6) It transforms accordng to U( v) 1 P U( v) = U( v) 1 p U( v) = ( p + m v). (8) Thus U( v) 1 P U( v) = P + M v. (9) Let us take the hamltonan to be H = p 2 + V ( x 1,..., x N ). (10) 2m Wth ths form, we can see what the boosted hamltonan s That s U( v) 1 HU( v) = = + ( p + m v) 2 2m + V ( x 1,..., x N ) p 2 2m + V ( x 1,..., x N ) p v m v 2. (11) U( v) 1 HU( v) = H + P v M v 2. (12) Thus the hamltonan s not boost nvarant. Rather, the hamltonan and the total momentum operator transform n a smple way that s ndependent of the detals of the potental. 3

4 4 The nternal hamltonan Defne H nt. = H 1 2M P 2. (13) Here nt. denotes nternal, M s the total mass of the system and P s the total momentum of the system. The nternal hamltonan represents the energy n a reference frame n whch the total momentum of the system vanshes. That s, f the system s n an egenstate of P, then we can boost t wth velocty v = P /M and thus make the egenvalue of P vansh. Then H nt. s the hamltonan for the sutably boosted system. Of course, the system n general wll not be n an egenstate of P and v s a numercal vector whle P s an operator, so n general we cannot boost to make P = 0. However, Eq. (13) s a relaton among operators that can defne what we mean by H nt.. How does H nt. transform under boosts? We note that U( v) 1 1 2M P 2 U( v) = 1 2M P 2 + P v M v 2. (14) Thus H nt. s boost nvarant U( v) 1 H nt. U( v) = H nt.. (15) 5 Translaton nvarant potental For the remander of these notes, we wll take V wll be a functon only of the absolute values of dfferences of the coordnates, x x j. Thus t s nvarant under space-translatons and rotatons. However, ths nvarance s not ndcated explctly n the notaton. Wth a translaton nvarant potental, we have [ P, V ] = 0. (16) Then [ P, H] = 0. (17) That means that we can choose P and H to both be dagonal. From the defnton (13), we have H = H nt M P 2. (18) 4

5 Choosng P to be dagonal makes the second term just the knetc energy for a partcle wth momentum P and mass M. We can say that H nt. gves the nternal energy of the system and then the full energy equals the nternal energy plus the knetc energy from the moton of the center of mass. If we want the energy levels of a system, t s really H nt. n whch we are nterested. The knetc energy from the center of mass moton s trval and unnterestng. To construct H nt., we smply need the momenta We note that the operators k are boost nvarant, k = p m M P. (19) U( v) 1 k U( v) = p + m v m M P m M M v = k. (20) Furthermore, k = p when P = 0. Thus f the states happen to be egenstates of P, then k = p f we boost the system to make the egenvalue of P vansh. Note that, usng p = P and m = M, we have k = 0. (21) A smple calculaton gves k 2 2m = p 2 2m P 2 2M. (22) Wth ths, we have H nt. = k 2 2m + V ( x 1,..., x N ). (23) 6 Internal coordnates and momenta We have already defned nternal momenta k, k = p m M P. (24) 5

6 wth k = 0. (25) Then the complete system can be descrbed usng the k and the total momentum P. Smlarly, we can defne nternal coordnates wth r = x R. (26) m M r = 0. (27) Then the complete system can be descrbed usng the r and the poston of the center of mass, R. A smple calculaton shows that R and P are conjugate operators n the sense that ther commutaton relaton s [R k, P l ] = δ kl. (28) A smple calculaton also shows that k commutes wth R, [R k, k l ] = 0. (29) (Indeed, M R s the generator of boosts, so ths says that k s boost nvarant.) A smple calculaton also shows that r commutes wth P, [r k, P l ] = 0. (30) (Indeed, P s the generator of translatons, so ths says that r s translaton nvarant, whch s evdent because t s the dfference of two postons.) Thus we have one set of operators, R and P, that tell about the overall moton of the system and another set of operators, the k and r, that tell about the nternal moton. The nternal moton operators commute wth R and P, so we can separate the dynamcs of the quantum system nto two ndependent parts, descrbed by commutng sets of operators. Snce we consder that the potental depends only on dfferences of the postons x, t can be wrtten as a functon of the r. V ( x 1,..., x N ) = V ( r 1,..., r N ). (31) 6

7 Then the hamltonan H nt. depends only on the nternal operators. Unfortunately, the commutaton relatons between the r and the k are a lttle complcated snce the operators k are not ndependent because of Eq. (25) and nether are the operators r, because of Eq. (27). We have [r k, k l j] = δ kl (δ j m j M ). (32) Suppose that you want to solve for the nternal moton of three partcles. Then you need to reduce the three poston vectors to just two, wth two conjugate momentum vectors. There are more than one ways to do ths, but we wll not pursue them here. What we want to notce s that for any number of partcles the nternal moton factors from moton of the center of mass. Takng P to be dagonal, we can use a wave functon e P R ψ( r 1,, r N ), (33) where here P s the egenvalue of the operator P. To fnd ψ( r 1,, r N ), we use H nt.. 7 The nternal hamltonan for two partcles Wth just two partcles, ths constructon s pretty smple. Wth our assumpton about V, the potental s a functon of just x 1 x 2, so that Then H = p 2 1 2m 1 + p 2 2 2m 2 + V ( x 1 x 2 ). (34) H nt. = k 2 1 2m 1 + k 2 2 2m 2 + V ( x 1 x 2 ). (35) Snce k 1 + k 2 = 0, we can use just one momentum operator k wth k 1 = k and k 2 = k. Then [ 1 H nt. = + 1 ] 1 m 1 m 2 2 k 2 + V ( x 1 x 2 ). (36) Defne the reduced mass µ by 1 m m 2 = 1 µ. (37) 7

8 Then H nt. = k 2 2µ + V ( x 1 x 2 ). (38) For space coordnates, we can use R and r = r 1 r 2 : R = 1 M (m 1 x 1 + m 2 x 2 ). r = x 1 x 2. (39) Then We know already that H nt. = k 2 + V ( r ). (40) 2µ [R k, P l ] = δ kl, [R k, k l ] = 0, [r k, P l ] = 0. (41) A smple computaton gves Thus n a coordnate representaton, [r k, k l ] = δ kl. (42) k l = r l. (43) In a coordnate representaton, we look for a wave functon e P R ψ( r), (44) where here P s the egenvalue of the operator P. We can dagonalze the operator P because t commutes wth H nt. and wth P 2 /(2M). Then ψ( r) s the nternal wave functon. The Schrödnger equaton for ψ( r) s the equaton that we are used to, except that the reduced mass µ appears n the knetc energy. 8

9 8 Angular momentum The angular momentum operator for our partcles s J = x p. (45) Ths transforms under translatons and boosts accordng to e P a Je P a = ( x + a) p, U( v) 1 JU( v) = x ( p + m v). (46) That s, e P a Je P a = J + a P, U( v) 1 JU( v) = J + R M v. (47) We can make up an nternal angular momentum that s nvarant under translatons and boosts as S = r k. (48) Snce both r and k are nvarant under translatons and boosts, so s S. Let us see how S s related to J. We have S = ( x ( R) p m M ) P = x p m M x P R p + m R M P (49) = J R P R P + R P = J R P. That s where J = L + S, (50) L = R P. (51) 9

10 Thus the total angular momentum J s composed of the orbtal angular momentum L assocated wth the moton of the center of mass plus the nternal angular momentum S. The orbtal angular momentum transforms under translatons and boosts as an ordnary angular momentum, e P a Le P a = L + a P, U( v) 1 LU( v) = L + R M v. (52) The nternal angular momentum s nvarant under translatons and boosts, e P a Se P a = 0, U( v) 1 SU( v) = 0. (53) It s easy to check that L has the usual angular momentum commutaton relatons wth tself, [L k, L l ] = ɛ kln L n. (54) Also, L commutes wth S, snce both R and P commute wth S. Then but so [L k, S l ] = 0, (55) [J k, J l ] = [L k + S k, L l + S l ] = ɛ kln L n + [S k, S l ] (56) [J k, J l ] = ɛ kln J n, (57) [S k, S l ] = ɛ kln (J n L n ). (58) That s, the nternal angular momentum has the usual angular momentum commutaton relatons wth tself, [S k, S l ] = ɛ kln S n. (59) To summarze, we have J = L + S where L and S have the usual angular momentum commutaton relatons wth themselves and L commutes wth S. These are the same relatons as we had for the orbtal angular momentum and the spn angular momentum of a sngle partcle. We smplfed our calculatons by consderng a bound system made of many spn 0 partcles. We could have used spn 1/2 partcles. Then the 10

11 operator we are callng S here would be made of the nternal orbtal angular momenta of the consttuent partcles, r k, plus the sum of ther spn angular momenta, s. For example, a smple model for a proton s that t s made of three spn 1/2 quarks combned, wth some orbtal angular momentum, to make a system wth total angular momentum 1/2. 9 Moton of the system as a whole We have consdered a system of partcles that nteract wth each other va a potental that depends only on absolute values of the separatons between pars of partcles. The most mportant case s that there s a contrbuton e e j x x j for each par of partcles. The analyss that we have appled works for any state, but t s most useful for bound states. Thus, let s consder a bound state here. We can make the system more complcated by supposng that there s an addtonal force that acts on our system. The smplest case s that we have a bound state, an atom, say, and we place t n an external electrc feld. Then we have an addtonal potental V = e Φ( x ). (60) Assumng that our potental s slowly varyng, we can approxmate ths by V Q Φ( R), (61) where Q = e s the total charge of the atom. Then we have H = H nt M P 2 + Q Φ( R). (62) Here H nt. s the same hamltonan for the nternal moton that we had before. We can magne solvng for the nternal moton n the form H nt. Ψ = Ent. Ψ. (63) 11

12 Then the full (tme ndependent) Schrödnger equaton s [ ] 1 Ψ P 2M 2 + Q Φ( R) = (E Ent.) Ψ. (64) Our atom moves n the potental Φ( R). A more exact approxmaton to the nteracton of the atom wth the external feld s V Q Φ( R) e r E( R), (65) where E = Φ. (66) Ths next approxmaton s especally mportant f the lowest order approxmaton s zero because the atom s electrcally neutral, Q = 0. Now we have an nteracton between the external feld and the electrc dpole moment of the atom. Ths couples the nternal and external motons. Perhaps we can come back to ths when we deal wth approxmaton methods n the next quarter. 10 Group multplcaton law There s a pecularty wth symmetry operatons nvolvng the boost operator. Suppose that we multply together U = e + P a U( v) 1 e P a U( v). (67) Ths corresponds to the followng sequence of coordnate transformatons: x x + v t a boost x + v t + a a translaton x + a an nverse boost x an nverse translaton. (68) That s, the product of these transformatons leaves the coordnates alone. If our quantum operators obeyed the group multplcaton laws, we would have U = 1. However, we know how P transforms under boosts, so U( v) 1 e P a U( v) = e ( P +M v) a. (69) 12

13 Thus U = e M v a. (70) We do not get the unt operator. But what we get s a number wth absolute value 1, a phase factor. Ths phase factor s appled to every vector n the space of quantum mechancal states. It then cancels whenever we calculate a probablty φ ψ 2. We see that the quantum theory provdes a representaton of the Galle group up to a phase. Ths phase doesn t matter. When we dealt wth rotatons and translatons by themselves, we could have had somethng smlar, but there we got a true representaton wth no phase. At least that s what we got f you ddn t have spn 1/2 partcles. Wth a spn 1/2 partcle, a rotaton through 2π, whch s the same as not rotatng at all, gves a phase factor 1. 13

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