Lecture 26: Angular Momentum II

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1 ecture 6: Angulr Momentum II Ph851 Fll 9

2 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

3 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)

4 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.

5 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)

6 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 -

7 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

8 The Trick We note tht: This mens tht: B definition we hve: Comining the two gives us: ( ) ( ) ( ) ( )

9 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 ( )

10 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[ ]

11 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 ( )

12 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 ( )

13 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 -

14 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

15 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

16 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

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