162 CHAPTER10. SYMMETRYANDDEGENERACY
Acompletesetofenergyeigenstatesisthen
{
φnm(x,y)=
2
L
sin[
nπx
L
]sin[
mπy
L
], Enm=(n^2 +m^2 )
̄h^2 π^2
2 mL^2
}
(10.62)
TheenergiesEnmaretwo-folddegenerateforn+=m,since
Enm=Emn (10.63)
SincetheHamiltonianH ̃ isinvariantunderreflections ofthex-axisaroundthe
pointx=L/2,andreflectionsofthey-axisaroundy=L/2, wedefinethecorre-
spondingoperators
Rxf(x,y) = f(L−x,y)
Ryf(x,y) = f(x,L−y) (10.64)
ItiseasytoseethatRxandRycommute,
RxRyf(x,y) = RyRxf(x,y)
= f(L−x,L−y))
=⇒[Rx,Ry]= 0 (10.65)
andthattheenergy eigenstatesofeq. (10.62)areeigenstates ofbothRxandRy,
witheigenvalues±1,since
sin
[
nπ(L−x)
L
]
=
{
1 nodd
− 1 neven
×sin[
nπx
L
] (10.66)
However,theHamiltonianH ̃ isalsoinvariantunderaninterchangeofthexandy
coordinates
If(x,y)=f(y,x) (10.67)
andthisoperatordoesnotcommutewithRxandRy:
IRxf(x,y) = f(y,L−x)
RxIf(x,y) = f(L−y,x) (10.68)
TheeigenvaluesofIaredeterminedbythesamereasoningasinthecaseofparity.
SupposeφβisaneigenstateofIwitheigenvalueβ.Then
IIφβ(x,y)=Iφβ(y,x)=φβ(x,y) (10.69)
butalso
IIφβ(x,y)=βIφβ(x,y)=β^2 φβ(x,y) (10.70)