# 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® _{S}H,H* = Rx 50(3),

*V*= R

^{3}x R

^{3}

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~ ^{1}r+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^ 7t ^{p} = dC/dAp.o =*

- 4 A'
*y^gF*A'v^(4F°^^{Op }H = it^{p}A_{p}$ - C =_{p},_{0}-*F^)*= Av^(4F^{o}'M_{m}.o -*F*Now,_{pp}F^{pp}) - 7T
^{P}=*4K y/^gg*^{Oa}g^{pli}F_{a/}3

*= 4K^[g ^{00}g^{pm}F_{0m} + g^{(>}"‘9^{I>()} F_{m()} + g^{Om}g^{pk}F_{mk})*

*= 4Ky/^[(g ^{00}g^{pm} - g^{0m}g^{p0})F_{0m} + g^{Om}g^{pk}F_{mk}]*

We need to solve for *A _{p}.o* in terms of

*A*

_{p}_{>m}and

*Tv*Note that tt° = 0. This is one of the constraints to be incorporated using the Dirac bracket formalism. We write the above equation as

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

*% =4Ky/-g{(g g -g g )t _{Om}--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

tt^{s}/4A^ = *y ^{sm}F_{Om} + g^{Om}g^{sk}F_{mk}*

Now writing ((7s_{m})) as the inverse of the matrix ((7^{sm})), we get

*F _{Os} = y_{sm}[7r^{m}/4K^g - g^{Or}g^{km}F_{rk}]*

Now we define the electric and magnetic fields as

*E _{s} = F_{Os},B_{s} = -e(skm)F_{km}*

Then, we can write

*E _{s} = y_{sm}
^{m}/4K_{x}/^ + g^{Or}g^{km}c(rkl)B_{l}] *or equivalently,

Tr^{s}/4A'v^ = *y ^{sm}E_{m} - g^g^mkl^Bt*

*We* then have

*H = tt*^{s}A_{s},_{0} -£ = KJ=-gh^{sm}E_{m} - - F^F^]

We note that

*Tr ^{p}A_{p}gd^{3}x = f Tr^{s}A_{St}od^{3}x =*

*14K^gF ^{0s}A_{St0}d^{3}x = 4K [ V-9F^{0s}(A_{St0} - *j4

_{0},

_{s}+

*A*

_{0s})d^{3}x*= 4K I F ^{Os}F_{Os}y/=gd^{3}x + 4K f ^F^{0s}A_{0},_{s}d^{3}x*

*= 4K I F ^{Os}F_{Os}^d^{3}x + 4K I(F^{0s}A_{0}^)_{tS}d^{3}x-4K f (F^{0s}^)_{tS}A_{0}d^{3}x*

*= 4K I F ^{0s}F_{0s}^d^{3}x *in the absence of charges since in the absence of charges, the Maxwell equations read

(F^{Os}^),_{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 = 4KF ^{Os}F_{Os}^- KF^F^^*

*K^(4F° ^{s}F_{0s} - F^F^ = K^g(2F^{Os}F_{Os} - F_{rs}F^{rs}) *Noting that

F_{Os} = *E _{s},F_{rs} = -e{rsm)B_{m}*

*F ^{Os} = (g°°g^{sm} - g^{Om}g^{Os})F_{Om} + g^{ok}g^{sm}F_{km}*

*= Y ^{m}E_{m}+g^{ok}g^{sm}^kml)B_{l}*

Also

*F ^{rs} = (g^{rm}g^{s0} - g^{rO}g^{sm})F_{mO} + g^{rm}g^{sk}F_{mk}*

*= -8 ^{rms}E_{m} - g^{rm}g^{sk}e(jnkp)B_{p}*

where

*arms _ rm sO _ rO sm*

i ¿7 eZ *J J*

Thus,

*H = Ky/^(2F ^{Os}F_{Os} - F_{rs}F^{rs})*

*= 2K^Y ^{m}E_{m}+g^{ok}g^{sm}e(kml)B_{l})E_{s}*

*-K^-g^rsm^E, + g ^{rl} g^{sk}e(lkp)B_{p})B_{m}*

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

*H =* (1/2)(E^{2} + B^{2})

provided that we take *K* = —1/4. Note that in flat space-time, *•y ^{sm} = — 6_{sm} = g^{sm},g°° _{= lg}0s _{= Q}*

(2) Denote

*H _{o} =* (1/2)(E

^{2}+

*B*

^{2})Then show that if the metric is

* ^{=} n/iu* T

*h*

_{pljl}(x}ie represented as a small perturbation of flat space-time, we can then write

*H = Ho + C _{k}(rs,x)E_{r}E_{s} + C?(rs, x)B_{r}B_{s} + Cs(rs,x)E_{r}B_{s}*

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 *B _{s}* are functionals of the position fields

*A*while the equation

_{r}(x)Tr^{s}/4Z^{sm}E_{m} - g^{Om}g^{sk}e(mkl)Bi

*E _{s} = ■ysm^-n^{8} / + g^{Om}g^{sk} e{mkl)Bi)*

Exercise: Using the above equations, express the Hamiltonian density in terms of the position and momentum fields A_{s},tt^{s},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

*H _{o} = P^{2}/2m, V(Q) = Ze^{2}/Q*

we find that

*|| V(Q)exp(-itH _{0})f* II

*dt Jo*

is infinite. In fact, we have

*exp(itHo)Q .exp(—itHo) = exp(it.ad(P ^{2})/2m)(Q) = Q + Pt/rn*

and on the other hand,

*exp(—irnQ ^{2}/2t)P.exp(imQ^{2}/2t) = exp(—im.ad(Q^{2})/2t)(P)*

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

Thus,

*exp(itH _{0})V(Q).exp(-itH_{0}) = V(Q + Pt/m)* = V((t/m)(F +

*mQ/t))*

*= Z _{t}V{Pt/m)Z/*

where

*Z _{t} = Z_{t}(Q) = exp(—imQ^{2}/2t)*

Thus,

*|| V(Q).exp(—itH _{0})f ||=|| V(Pt/m).Z/f ||*

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

where

=(1 + Q

^{2})^{1}/^{2}

Assuming that

*J(l + Q ^{2})^{n}f(Q)^{2}d^{3}Q*

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^{-n}ll_{9}= (/ (i + *Q ^{2}r^{nq/2}d^{3}QV*

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)|’d^{3}Q)^{1}/^{<}’ = (m^{3}/3t^{3})^{1}/’( *J* |V(Q)|’d^{3}Q)^{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''(R

^{3}). 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 [ U ^{0}{-t)VU(s)VU^{0}(t - s)dtds*

JR JRxR_{+}

*= I exp(it(X — /x)')E _{0}(dX')VE_{0}(dp.)dt Jw*

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

JR^{3}xR_{+}

= 2tt *[ 8(X - p)E _{0}(dX)VE_{0}(dp)*

*Jv ^{2}*

-2% *[ 8(X - p)E _{0}(dX)V.(H - X)-^{1} .VE_{0}(dp)*

*J« ^{2}*

Equivalently,

-7?/2tt= *[ E' _{0}(X)V E'_{0}(X)dX — [ E'_{0}(X)V.(H* - A)

^{-1}

*,V.E'*./r

_{0}(X)dX JuThis gives us explicitly the form of the scattering matrix at energy A: 5(A) - *I* = 7?(A) = -2tt(^(A)(V - *V.(H -* A)-^{1}V)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.