ANGULAR MOMENTUM. We have seen that when the system is rotated through an angle about an axis α the unitary operator producing the change is

Transcription

1 ANGULAR MOMENTUM We have seen that when the system is rotated through an angle about an ais α the unitary operator producing the change is U i J e Where J is a pseudo vector obeying the commutation relations J,J iє J i j ijk k J,J i 0 J, iє We note that the rotation rotates the apparatus i j ijk k any vector i J, 0 any scalar We can now use the commutation relations to determine the eigenvalues and eigenstates of J. We note that J,J i 0 assures us that there eists a complete set of simultaneous eigenstates of J and any one component of J. We choose J. Then there is a complete set of eigenstates J,J' To find them we define Then J J ij y

2 J,J J,J ij y J,J i J,J y ij i ij ij J J ij J y y y Then JJ J,J' JJ JJ JJ J,J' JJ' J J,J' J' 1J J,J' Hence J J,J ' Is an eigenstate of J with eigenvalues J ' 1 We know that J ' J Hence the series of states generated by J must terminate at J and ' min J. ' ma J J,J 0 ' min J,J'min JJ J,J'min 0 But Hence y y JJ J ij J ij J J i J,J J J i ij J J J y y J,J JJ J J J' J' J,J J,J 0 'min 'min 'min min 'min 'min

3 J J J' 0 'min min imilarly J J,J' ma 0 J,J JJ J,J' 0 'ma ma JJ J J J J,J' J J J,J' J J' J J,J' J,J' ma ma ma 'ma ma ma J J J' 0 'ma ma But J' J' n n 0,1,, ma ma J J' J' J J' n J' n min min min min J' min nj' min n n n n1 n J' min n1 J' min J' ma n Let n j Then j j J j j i, J' min J' ma where

4 1 3 j 0,,1,, We normalie the states as follows: J J,J ' Has length J,J 'J J J,J ' J J J ij J ij J J i J,J y y y y J J iij J J J 1/ "length" J J ' J ' Thus if J,J is normalied we find the normalied state produced by J is where J = m. Hence 1 1 J J,J ' J J,J 1/ 1/ ' J J ' J ' jj1mm1 1 J,m1 J 1/ J,m jj1mm1 Let s compare the magnitude of the components of J in the y plane to J. j, m J J jm j, m J J jm j j 1 m y

5 Hence J J y j 1 m m 1/ This is least when m = j. Then 1/ J y j j1 j 1 1/ J j j Hence for large j (classical case) there is very little uncertainty in J, but for the quantum case of j = ½, for eample the uncertainty is greater than the magnitude! Orbital and pin Angular Momentum The orbital angular momentum classically is L r p We define this to be the QM orbital angular momentum. We have L,Lyp p,p p y y imilarly Also y p,p p y,p p p p,p p,p p p y y yp,p,p p y yp,p,p p y i yp i p i L y L,L i L ; L,L i L y y

6 L,L i L,L i Ly L L,L i L,L i y L,L i 0 Hence Now consider L L,L L,L L L L,L L,L L i i y i y i y y L L,L L,L L i y i L L LL L L LL L L LL i1j j i1j j y ij j ij j y i3j j i3j j L L L L L y 13 LL 31 y L L L L L L L y y 31 y L L L L L L L 13 y y 31 LL LL LL LL LL LL y y y y LL LL LL LL LL LL y y y y L,ε p, k j ijk i j j L,L i ε L i j ijk k L,L i 0 ε p,, p ε i i ε iε ijk i j j i j j ijk i i ikj kji i imilarly L,iε i j ijk k

7 Also L,p iε p i j ijk k L, i εkjip, k j 1 3 ε p, kji k j 1 3 ε p, ε p, p, kji k j j kji k j j j j j j ε i 0 kji k j ince kji is antisymmetric and k j is symmetric. imilarly L,p i 0 and L, i p 0 From our previous results J,L i j iεijklk J i,l 0 However J,L j J1 J J 3,L j J J,L J,L J J J,L J,L J J J,L J,L J 1 1 j 1 j 1 j j 3 3 j 3 j 3 Jiε L iε LJJiε L iε LJ Jiε L iε LJ 1 1jk k 1jk k 1 jk k jk k 3 3 jk k 3jk k 3 iε JL LJiε JL LJiε JL LJ 1jk 1 k k 1 jk k k 3jk 3 k iε JL L J ijk i k k i ince ijk is antisymmetric in i and k this will only be 0 if (J i L k + L k J i ) is symmetric in i and k. This is only true if J = L. Hence, in general J,Lj 0 k 3

8 Net we consider what L actually does. We know that a translation through a distance Produces a change given by the unitary operator ap i U a e If we consider a sequence of such translations we get ince everything commutes this is n n1 1 anp a1p i i Ua Ua Ua e e n p i a j Ap i j1 Ue e Where A is the total displacement. Thus if we move along any closed loop A = 0, U I the identity. Net consider moving the system in a circle centered at the origin and lying in a plane with unit normal n. If we move through an angle d we translate by da nˆ d Then da p n p dn p d n L d ˆ ˆ ˆ Hence ˆn Ld id L i i U d e e Thus dnl d1l L i i i U e e e

9 since everything commutes. Note that this operator has only shifted the apparatus without rotating it The difference between the made up by the spin, apparatus effect of J and L is which rotates the or J L We now find the eigenstates and eigenvalues of L just as before. ince the commutation relations are the same as for J (ecept for the ) we find L,L i ε L i j ijk k L,L i 0 L 1 L min, Lma In the case of J we had l = 0,1/,1,. Consider rotating about the ais. Then

10 U i J e Applying this to the state L,m we get i J i m e L,m e L,m If = we must get 1. Hence im e 1 m 0,1,,3, Thus we must have l = 0,1,,3, so that only integer values of l are allowed. pin Angular Momentum We define the spin angular momentum as J L Consider the commutator, i j J, i j L, i jiεijkk εijki k 0 imilarly,p 0 i j Thus has nothing to do with the position or momentum of the object. It is entirely internal. However, since does generate a rotation it will have the same commutation relations between its components and magnitude as J or L, i ε i j ijk k, i 0

11 Thus we have a complete set of states which are simultaneous eigenstates of and one component of. Taking the component to be we have the states,' Most systems will have a definite value of. Hence we will deal with eigenstates of but not necessarily. uch a state is then given by Then am,m m and,m' a' m m,m,m As usual applying a rotation in this fashion (rotate object internal, but don t change its location) is represented by the operator If the rotation is about the ais we have Then U e i i U ˆ e m i i ' e,m e,m Now if = the values of m will be. If = they will be,., etc. Thus for the state 1 1,

12 we find Thus to get back to the original state we need i 1 1 ' e, 4 In other words, it takes revolutions to return to the original position! Just another indication that involves internal motions, not eternal ones. In this case we have two basis states pin ½ Particles ,,, The operator is then represented by the following matrices where ,,, Now as before we define i y Then i y From earlier we have

13 1 1, ,,, ,,, 1/ 1/ Hence 1 1, Hence y 0 i y i i

14 y i 0 y Hence Thus y 0 i i Note that I i0 i 1 0 y I i 0 i I Now suppose we measure in a new system inclined at an angle relative to. There are only two possible results: 1/. We want the probability of each. uppose the new ais lies in the direction (sin cos, sin sin, cos) ' ˆ ˆ ˆ ˆ sin cos sinsiny cos We are measuring the quantity.

15 be the state where a measurement of ' Let will be certain to give / and be the state where the measurement will be certain to give /. Then we need a b a b Once we have the a and b we can predict the probability of getting either /. The probability of /is a if we start in, etc. We have a Epand in terms of and. Then We have ' ˆ ' ˆ sin cos sinsin cos y Then i 1 0 ' ˆ sin cos sin sin cos 1 0 i i cos sin e i sin e cos i cos sin e i sin e cos

16 i cossin e a 1cos i i sin e cos sin e We must also have 1 sin cos 1cos 1 sin 1cos 1cos sin 1 cos 1 sin sin sin cos sin cos sin cos cos \ Hence sin cos sin e i sin cos e i cos e sin a cos i Thus the probability of getting / when you start in the state is

17 cos Note that the epectation value of in the state +> is ẑ' we write ẑ' cos sin cos sinsin y sin cos sinsin cos i Hence ẑ' cos This is eactly what we would epect classically. Note that we could also calculate it as ẑ' probof endingin But as epected. prob of ending in cos sin cos sin cos 1 cos cos sin cos 1 pin 1 Particles In this case we have 3 basis states: = - 1,0,1. Then

18 ince we know that is Hermeitan we have * * * Also, since we have Finally, since changes by only 1 we have Thus we need only matri elements Thus * 1/ / *1 1/ 1 1/

19 imilarly we find 0 i y i 0 i i