130_notes.dvi

look at them=− 1 recursion relationswitham=bm= 0 and solve the equations.

(s−κ)a 0 +Zαb 0 = 0

−Zαa 0 + (s+κ)b 0 = 0

(

(s−κ) Zα

−Zα (s+κ)

)(

a 0

b 0

)

= 0

s^2 −κ^2 +Z^2 α^2 = 0

s^2 =κ^2 −Z^2 α^2

s=±

√

κ^2 −Z^2 α^2

Note that whileκis a non-zero integer,Z^2 α^2 is a small non-integer number. We need to take the
positive root in order to keep the state normalized.

s= +

√

κ^2 −Z^2 α^2

As usual, theseries must terminate at somem=nrfor the state to normalizable. This can be
seen approximately by assuming either thea’s or theb’s are small and noting that the series is that
of a positive exponential.

Assume the series forFandGterminate at the samenr. We can then take the equations in the
coefficients and setanr+1=bnr+1= 0 to get relationships betweenanrandbnr.

anr=−

√

k 2

k 1

bnr

bnr=−

√

k 1

k 2

anr

These are the same equation, which is consistent with our assumption.

The final step is to use this result in therecursion relations form=nr− 1 to find acondition
onEwhich must be satisfied for the series to terminate. Note that this choice ofmconnectsanr
andbnrto the rest of the series giving nontrivial conditions onE. We already have the information
from the next step in the recursion which givesanr+1=bnr+1= 0.

−am+ (s+m+ 1−κ)am+1−

√

k 2

k 1

bm+Zαbm+1= 0

−

√

k 1

k 2

am−Zαam+1−bm+ (s+m+ 1 +κ)bm+1= 0

−anr− 1 + (s+nr−κ)anr−

√

k 2

k 1

bnr− 1 +Zαbnr= 0

−

√

k 1

k 2

anr− 1 −Zαanr−bnr− 1 + (s+nr+κ)bnr= 0

−

√

k 1

k 2

anr− 1 + (s+nr−κ)

√

k 1

k 2

anr−bnr− 1 +Zα

√

k 1

k 2

bnr= 0

−

√

k 1

k 2

anr− 1 −Zαanr−bnr− 1 + (s+nr+κ)bnr= 0