ecture 6: Angulr Momentum II Ph851 Fll 9
The Angulr Momentum Opertor The ngulr momentum opertor is defined s: r r R r P It is vector opertor: r r e r e r e According to the definition of the crossproduct the components re given : YP ZP XP ZP XP YP
Commuttion Reltions The commuttion reltions re given : [ [ [ These re not definitions the re just consequence of [XP]ih ] ih ] ih ] ih An three opertors which oe these reltions re considered s generlied ngulr momentum opertors Compct nottion: [ j k ] ihε jkl Summtion over l is implied l ε jkl evi Cevit tensor if n two indices re the sme 1 cclic permuttions of (or 13) -1 cclic permuttions of (or 31)
Simultneous Eigensttes In the HW will see tht: Where: [ ] [ ] [ ] This mens tht simultneous eigensttes of nd eist et: Wh nd nd not or? We wnt to find the llowed vlues of nd.
Algeric solution to ngulr momentum eigenvlue prolem In nlog to wht we did for the Hrmonic Oscilltor we now define rising nd lowering opertors: Anlog is not ect: i i ( ) 1 A X ip 1 A ( X ip) ets consider the ction of first: recll our pproch ( ) ( )( A A A ε ε 1 A ε ) for SHO: ( ) i [ ] ih ih [ ] ih ih ( ) ih h i ( i ) h( i ) ( ) ( h)
dder Argument ( ) ( h) Conclusion: if is n eigenstte of with eigenvlue then there is lso n eigenstte h with eigenvlue h Thus we cn s: C Similrl we cn redil show tht: So tht we must hve lso: This llows us to s: h ( ) ( h) C h 1. If eists then h eists or C. If eists then -h eists or C -
Estlishing Some ower Bounds Strting from: It follows tht Which mkes sense ecuse is n oservle Therefore we hve: Result: ZP YP P Z Y P ZP Y P P Z YP
The Trick We note tht: This mens tht: B definition we hve: Comining the two gives us: ( ) ( ) ( ) ( )
The conclusion Conclusion: 1. There is min for ever llowed. There is m for ever llowed min ( ) m ( ) We re just sing tht is ounded from ove nd elow Thus we must hve: ( m ) So tht no sttes with higher cn eist Which requires: C ( ) m Similrl C min ( )
Just ittle Further to Go C C m ( ) min ( ) If we hve: m ( ) Then we must hve: m ( ) But: ( i )( i ) i i h i[ ]
Strting to see the light t the end of the tunnel h This mens: m ( ) ( h ) m ( m h ) m m m hm m ( ) hm ( )
Not quite done et m ( ) m ( ) hm ( ) Similrl we cn show tht: min ( ) Which together with: Gives us: ( i )( i ) i i i[ ] h min ( ) hmin ( )
Keep moving forwrd ( ( h) ) ) m ( m ( ( h) ) ) min ( min From we get: m ( ) hm ( ) min ( ) hmin ( ) min ( ) hmin ( ) m ( ) hm ( ) The solution is: min 1 ( h ± h 4 m 4h ) m 1 ( h ± ( ) min h m min h m min m min < m This is perfect squre -
dder Termintion Requirement MAIN POINT SO FAR: For given min () nd m () re the onl vlues from which the lowering/rising opertors do not crete new sttes. So we cn uild ldder strting t min nd cting with to crete new stte t min h etc min min h min mh The onl w the ldder will terminte is if we hve m For some integer N min Nh With min - m this leds to: m N h m Nh This mens tht the mimum -component of ngulr momentum must lws e hlfinteger or whole integer times h
Quntition of ngulr momentum ets replce m with the smol l ln/. The llowed eigenvlues of re then: mh ; m l l 1 l... l From the condition: m ( ) m ( ( h) ) We see tht the corresponding eigenvlue of is: ( hl h) hl ( 1) h l l Conclusion: We cn relel the sttes s: l m Then we hve: l m h l( l 1) l m l m hm l m
The FINA SIDE Now we hve: ( ) C h C h ikewise: C h h l(l 1) hm ± lm h l(l 1) m(m ±1) lm ±1