QuantumPhysics.dvi

The translation equationT†XT=X+aIthen allows us to compute the commutator

[X,P] =i ̄hI (5.33)

The position eigenstates become labeled by the continuous eigenvalues of the position oper-

ator, and including the appropriate continuum normalization factorofN, we have

|n〉

√

N/(2πL) =|x;X〉 an+x 0 =x

X|x;X〉=x|x;X〉 (5.34)

We have included the labelX in the kets|x;X〉to make it clear that the basis of states

corresponds to eigenstates of the operatorX. Since the operatorP is manifestly diagonal

in the same basis whereT is diagonal, we may interchangeably use the notations|km;T〉=

|km;P〉. The relation between momentum and position eigenstates becomes,

P|km;P〉= ̄hkm|km;P〉 |km;P〉=

∫+πL

−πL

dx eixkm|x;X〉 (5.35)

The momentum eigenvaluespm= ̄hkmremain finite and discrete during this limit, since

km=

φm

a

=

2 πm

Na

=

m

L

m∈Z (5.36)

The completness relations are now,

I=

∑

m∈Z

|km;P〉〈km;P|=

∫+πL

−πL

dx|x;X〉〈x;X| (5.37)

from which it follows that the inner product relations between|x;X〉states are

〈x;X|x′;X〉=δ(x−x′) (5.38)

We conclude by producing the energy levels for this problem. Sinceφm=akm, all the phases

tend to 0 asa→0, and all energy levels would be then degenerate. To make things more

interesting, we can choose a more interesting limit, where

A 0 = 2A 2 =

h ̄^2

Ma^2

(5.39)

for a fixed parameterM. As a result, the limiting energy levels are given by

Em=

(pm)^2

2 M

(5.40)

which is the standard kinetic energy formula for a free particle on a circle, and the parameter

Mshould be thought of as the mass.