Mackey’s theory on the construction of the basic observables in the quantum theory from projective unitary representations of the Galilean group

Galilean group element: (a, u, s, g), (&, v, t, h) where a, u € ®3,(s,p) € R x 50(3). Let G be the Galilean group. Then

G = V®SH,H = Rx 50(3),V = R3 x R3

Buy (a, u, s, g), we mean (a, u)o(s,g). Its composition law is ((a,u)o(s,g))o((b, v)o(t,h)) = ((s,g)[(b, u)] + (a,u), (s,g)0(t, h)) where

= (s,g)o(b, v)o(s,p)-1 = (s,p)o(b, v)o(-s,s-1) (s,0)o(b,u)o(-s,g-1)(t, r) = (s,g)o(b, v).(t-s, g^r)

= (s, g)o(t—s, g~1r+v(t—s)+b)

= (i, r + gv(t - s) + gb) = (g(b - vs),gv)(t, r)

Thus, we get

Hamiltonian density of the electromagnetic field in curved space-time in terms of position and momentum fields

t = Ky/^F^F^ 7tp = dC/dAp.o =

  • 4 A' y^gFOp H = itpAp$ - C = A'v^(4F°^p,0 - F^) = Av^(4Fo'Mm.o - FppFpp) Now,
  • 7TP = 4K y/^ggOa gpli Fa/3

= 4K^[g00gpmF0m + g(>"‘9I>() Fm() + gOmgpkFmk)

= 4Ky/^[(g00gpm - g0mgp0)F0m + gOmgpkFmk]

We need to solve for Ap.o in terms of Ap>m and Tvm. Note that tt° = 0. This is one of the constraints to be incorporated using the Dirac bracket formalism. We write the above equation as

s a ta- /—-[/ 00 sm „0m s0 77 . sk rp 1

% =4Ky/-g{(g g -g g )tOm--g g

We define the 3 x 3 matrix

sm ^00 ^sm 0m sO

7 =((7 )),7 =9 9 ~9 9

and then write the above as

tts/4A^ = ysmFOm + gOmgskFmk

Now writing ((7sm)) as the inverse of the matrix ((7sm)), we get

FOs = ysm[7rm/4K^g - gOrgkmFrk]

Now we define the electric and magnetic fields as

Es = FOs,Bs = -e(skm)Fkm

Then, we can write

Es = ysm m/4Kx/^ + gOrgkmc(rkl)Bl] or equivalently,

Trs/4A'v^ = ysmEm - g^g^mkl^Bt

We then have

H = ttsAs,0 -£ = KJ=-ghsmEm - - F^F^]

We note that

TrpApgd3x = f TrsAStod3x =

14K^gF0sASt0d3x = 4K [ V-9F0s(ASt0 - j40,s + A0s)d3x

= 4K I FOsFOsy/=gd3x + 4K f ^F0sA0,sd3x

= 4K I FOsFOs^d3x + 4K I(F0sA0^)tSd3x-4K f (F0s^)tSA0d3x

= 4K I F0sF0s^d3x in the absence of charges since in the absence of charges, the Maxwell equations read

(FOs^),s = 0 and also we have noted that the spatial volume integral of a three divergence is zero by Gauss’ theorem. This results in

H = 4KFOsFOs^- KF^F^^

K^(4F°sF0s - F^F^ = K^g(2FOsFOs - FrsFrs) Noting that

FOs = Es,Frs = -e{rsm)Bm

FOs = (g°°gsm - gOmgOs)FOm + gokgsmFkm

= YmEm+gokgsm^kml)Bl

Also

Frs = (grmgs0 - grOgsm)FmO + grmgskFmk

= -8rmsEm - grmgske(jnkp)Bp

where

arms _ rm sO _ rO sm

i ¿7 eZ J J

Thus,

H = Ky/^(2FOsFOs - FrsFrs)

= 2K^YmEm+gokgsme(kml)Bl)Es

-K^-g^rsm^E, + grl gske(lkp)Bp)Bm

Exercise: [1] Verify that in the case of flat Minkowskian space-time, this expression reduces to the familiar special relativistic formula

H = (1/2)(E2 + B2)

provided that we take K = —1/4. Note that in flat space-time, •ysm = — 6sm = gsm,g°° = lg0s = Q

(2) Denote

Ho = (1/2)(E2 + B2)

Then show that if the metric is

= n/iu T hpljl(x}

ie represented as a small perturbation of flat space-time, we can then write

H = Ho + Ck(rs,x)ErEs + C?(rs, x)BrBs + Cs(rs,x)ErBs

where Cj(rs,x) are linear combinations of h^^x). Now we are in a position to express the Hamiltonian in terms of the canonical position and momentum fields. Note that the Bs are functionals of the position fields Ar(x) while the equation

Trs/4ZysmEm - gOmgske(mkl)Bi

Es = ■ysm^-n8 / + gOmgsk e{mkl)Bi)

Exercise: Using the above equations, express the Hamiltonian density in terms of the position and momentum fields As,tts,s = 1,2,3 in a curved space-time upto linear orders in the metric perturbations.

Coulomb scattering

The wave operator Q+ does not exist. In fact, writing

Ho = P2/2m, V(Q) = Ze2/Q

we find that

|| V(Q)exp(-itH0)f II dt Jo

is infinite. In fact, we have

exp(itHo)Q .exp(—itHo) = exp(it.ad(P2)/2m)(Q) = Q + Pt/rn

and on the other hand,

exp(—irnQ2/2t)P.exp(imQ2/2t) = exp(—im.ad(Q2)/2t)(P)

= P+mQ/t = (m/t)(Q+Pt/rn)

Thus,

exp(itH0)V(Q).exp(-itH0) = V(Q + Pt/m) = V((t/m)(F + mQ/t))

= ZtV{Pt/m)Z/

where

Zt = Zt(Q) = exp(—imQ2/2t)

Thus,

|| V(Q).exp(—itH0)f ||=|| V(Pt/m).Z/f ||

<|| V(Pt/m) < Q >_”|| . <||< Q >n / ||

where

=(1 + Q2)1/2

Assuming that

J(l + Q2)nf(Q)2d3Q

It follows that the wave operator Q+ will exist provided that

t ->|| V(Pt/m) < Q >-”||

is integrable on [0,oo). Now for any q > 2, we have

->ll V(Pt/m) < Q >-”||<|| V{Qt/m) ||, . ||< Q >-"||9

We have

ll-nll9= (/ (i + Q2rnq/2d3QV

and for this integral to be finite, we require that 2 — nq/2 < — 1 or equivalently, nq > 6. On the other hand, for t —>|| V{Qt/m) ||9 to be integrable over [0, oo), we can derive the condition:

(I |V(Qt/m)|’d3Q)1/<’ = (m3/3t3)1/’( J |V(Q)|’d3Q)1/9

and this is integrable if t~3'q is integrable over (<5, oo). This will happen provided that '3/q > 1 or equivalently, q < 3 and also V € L''(R3). Thus for the existence of the wave operator acting on a function f(Q), we require three conditions:

Scattering matrix in quantum mechanics

Q+ = I + i U(-t)V.U°(t)dt Jo tr+ = I-i / u°(-t)VU{t)dt Jo

V_ = I-i U(-t)VU°(t)dt

= I-i U(t)VU°(-t)dt

Jo

a*_ = i + i [ u°(t)vu(-t)dt = i + i [ u°(-t)vu(t)dt JO J — oo

q; - Q*_ = - f U°(-t)VU(t)dt

J — oo

r = s -1 = = (n; -

= - [ U°(-t)VU(t)il_dt = - [

JR JR

where we have used

C/(t)Q_ = Q_[/°(t)

Thus,

-R= [ U°(-t)V(I -i [ U(s)VU°(—s)ds)U°(t)dt

JR Jo

= f U°(-t)VU°(t)dt - i [ U0{-t)VU(s)VU0(t - s)dtds

JR JRxR+

= I exp(it(X — /x)')E0(dX')VE0(dp.)dt Jw

—i / Eo(dX)Vexp(—isH)VEo^dp)exp(i{X — p.)t + iXs)dtds

JR3xR+

= 2tt [ 8(X - p)E0(dX)VE0(dp)

Jv2

-2% [ 8(X - p)E0(dX)V.(H - X)-1 .VE0(dp)

2

Equivalently,

-7?/2tt= [ E'0(X)V E'0(X)dX — [ E'0(X)V.(H - A)-1 ,V.E'0(X)dX Ju ./r

This gives us explicitly the form of the scattering matrix at energy A: 5(A) - I = 7?(A) = -2tt(^(A)(V - V.(H - A)-1V)E^(A)

It is clear that the matrix element < f | S( A) |p > exists for all A if and only if f, g belong to the absolutely continuous spectrum of Ho, for that would imply the existence of the derivatives E'0(X)f = dEo(X)f /dX and E'0(X)g = dEo(X)g/dX.

 
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