Angular momentum operators in spherical coordinates

Angular momentum operators in spherical coordinates PhL 1.23.08 From Saxon page 262 we have p = -i and so L = r x p = -i r x But in spherical coordinates we know that ( see T.K page) = r + (1/r) + (1/rS) = ipr + ipθ + ipφ Therefore (note" in MM ponderings, cross product is form invariant under rotations! ) iL = r x = x + x (1/S) = - (1/S) From this we could say that iL = iL = - (1/S) iLr = 0 which seems to suggest that = r + (1/r) iL - iL These seem deceptively simple, but I think they are correct. [ supported by the L2 calculation later on in this doc.] These are dimensionless operators. Consider the picture below, but imagine the unit vector triad translated to the origin. A function f(r,θ,φ) has to change in the θ direction to "have some angular momentum about the axis" because you can see that sort of implies "rotation" about the axis (moved to the origin), put right hand thumb in . Similarly, your function f(r,θ,φ) has to change in the φ direction to "have some angular momentum about the axis" because you can see that sort of implies "rotation" about the axis, put right hand thumb in . You can see that the sign is negative because at the equator, points in the - direction which is the normal direction associated with the φ rotation direction shown. At this equator we expect to get iL = - because the situation is exactly analogous to the iL = situation. As we move toward the north pole, I guess you need a huge amount of amplitude to generate your iL because is moving into the φ rotation plane, so to speak. As for iLr = 0, you would need some ψ Euler angle to get some action for angular momentum about this axis, thinking Goldstein. But we don't have any ψ action here! I guess we could sort of add it and then say iLr = ∂ψ. We usually deal with the Cartesian components, so repeat the above, L = -i [ - (1/S)] Then ( see another T.K. page for the various spherical coordinate dot products ) Lx = - i [ - (1/S)] = -i [ (-S) - (CC )(1/S)] = i [ S + cot C ] Ly = - i [ - (1/S)] = -i [ (C) - (CS )(1/S)] = i [ - C + cot S ] Lz = - i [ - (1/S)] = -i [ (0) - (-S )(1/S)] = -i [] The raising and lower operators are: L = Lx i Ly = i [ S + cot C ] i { i [ - C + cot S ]} = i [ S + cot C ] { [ C cot S ]} = { iS C} + { i cot C ∓ cot S } = { iS C} + cot { i C ∓ S } = { iS C} + i cot { C iS } = { C iS} + i cot { C iS } = ei [ + i cot ] // agrees with "multipole expannsions.doc Next, a calculation which takes a lot of lines to carry out, at least by this method: L2 = Lx2 + Ly2 + Lz2 = { i [ S + cot C ]}2 + { i [ - C + cot S ]}2 + { -i []}2 = [ S + cot C ]2 [ - C + cot S ]2 []2 So now have to be careful about what the derivatives act on, so write it out as is: L2 = [ S + cot C ]2 + [ - C + cot S ]2 + []2 = [ S2 2 + cot2 C22 + S cot C + cot C S ] f(,) + [ C2 2+ cot2 S22 C cot S cot S C ] + 2 where each differential operator acts on everything to its right (imagine some f(,) sitting there). If we ignore the four cross terms for the moment, we get = 2 + cot2 2 + 2 = 2 + (1/sin)2 2 Now come the cross terms 1,2,3,4: 1 S cot C = (S C ) [cot f() ] = (S C ) [ csc2 + cot ] f So get 1 = (S C ) [ csc2 + cot ] 2 cot C S = (cot ) C [ S f] = (cot ) C [C +S ]f So get 2 = (C [C +S ] ) (cot ) = (C2 + S C )(cot ) 3 C cot S = (S C ) [ csc2 + cot ] 4 cot S C = (cot) S [ C f] = (cot) S [ -S + C] = S [ -S + C](cot) = [ S2 SC] (cot) So combine these four terms to get 1+2+3+4 = (S C ) [ csc2 + cot ] + (C2 + S C )(cot ) (S C ) [ csc2 + cot ] + (S2 SC) (cot) = { (C2 + S C ) + (S2 SC)} (cot) = (cot) So our total result is now L2 = 2 + (1/sin)2 2 + (cot) = 2 + cot + (1/sin)2 2 You can write the first two terms as (1/S) [ S ] so we get L2 = 2 cot (1/sin)2 2 L2 = (1/S) [ S ] (1/sin)2 2 // agrees with multipole Summary of all results: Lx = i [ S + cot C ] L = ei [ + i cot ] Ly = i [ C cot S ] L2 = [ 2 + cot + (1/sin)2 2] Lz = -i L2 = [ (1/S) [ S ] + (1/sin)2 2 ] Exercise: Compute L2 entirely in spherical coordinates. -L2 = -L L = + iL iL = ( - (1/S)) ( - (1/S)) = ( - (1/S)) - (1/S) ( - (1/S)) Now we have to account for derivatives of the unit vectors, which we know are these So we have many terms in the above to think about. T1 = () = ()+ () =0 + () = 2 T2 = - ( (1/S)) = - ( ) (1/S) - ((1/S)) = + (1/S) - ((1/S)) = 0 + 0 = 0 T3 = - (1/S) () = - (1/S) ()- (1/S) () = - (1/S)[- Cθ - Sθ ] + 0 = - (1/S)[- Cθ ] = cotθ T4 = (1/S)[ (1/S)] = (1/S)( )(1/S)+ (1/S) (1/S) = (1/S)( Cθ )(1/S)+ (1/S)2 = 0 + (1/S)22 = (1/S)22 So, adding the three non-vanishing terms, we get -L2 = 2 + cotθ + (1/S)22 and this, happily, agrees with the first form shown above. And this seems a little easier, if you happen to know all those unit vector derivatives. Note added 8.29.08. How do we know that [ L2 , a ] = 0 , where a = constant vector ? Our usual answer is that a transforms as a rotational scalar, and L2 commutates with any such scalar. But let's just look a bit into the details. Let's start off computing the commutator in a primed spherical coordinate system in which a' = a ' . That is, we just choose our primed coordinate system to line up in this way since we are free to choose it in any way we want. So we have ' = ' : [ L'2 , a'' ] = [ L'2 , (a ')' ] = [ L'2 , a ∂r' ] = 0 where the last equality comes from the fact that L'2 is a differential operator only in θ' and φ' and does not involve r' at all. For example, [∂θ',∂r'] = 0. So, in this primed frame we have proven our result. Now let's go to some unprimed frame according to r = R r' where R is a 3x3 rotation matrix. Then we know that r' = R-1r which says r'i = R-1ij rj and dr'i = R-1ij drj and (dr'i/drj) = R-1ij So i = ∂i = d/dri = by the chain rule acting on some function f = R-1ji d/dr'j by what was just shown above = Rij'j since R-1= RT , rotations are real orthogonal so we have shown that = R' which is the same as r = R r' In this bit of detail, we have just proven that the gradient vector transforms in the same manner as the position vector, in case there was any doubt about it. Now, in this new unprimed frame we have a = R a' . So let's start with our primed frame result and work from there: 0 = [ L'2 , a'' ] = [ L2, (R-1a) ( R-1) ] = [ L2, a ] QED In the first step we use the fact that L2 is the same in all rotationally related frames. We can make this proof a little more specific by using rotation operators R = exp(-iγnL) which are exponentiated differential operators in θ and φ, with rotation angle γ, and we then start again with 0 = [ L'2 , a'' ] and apply R R-1 around both sides to get 0 = [ RL'2 R-1 , R(a'') R-1] where R R-1 around something is how we move an operator from the primed to unprimed frame. Thus we have RL'2 R-1 = L2 // R does this to any scalar operator like L'2 R(a'') R-1 = a' R' R-1 = a' R-1 // R does this to any vector operator like ' = (R a') = a and therefore we have 0 = [ RL'2 R-1 , R(a'') R-1] = [L2, a ] and we are finally done. Now that we are dead sure that [L2, a ] = 0, we can examine it in spherical coordinates. We know that a = ar∂r + aθ(1/r)∂θ + aφ(1/rsinθ)∂φ False Statement: "Since a is an arbitrary vector, each term above must be zero. " When I made this statement, I had in mind just pulling a out of the commutator and writing [L2, a ] = Σiai [L2, i] sum i is over the spherical components, (not true!) But of course we are thinking of a = ar + aφ + aθ . But we know that L2 is going to act on these unit vectors, so you cannot just pull a out in this way! Had this been a true statement, we would then have concluded that [L2, i] = 0 for each spherical component = r + (1/r) + (1/rS) => 0 = [L2, r] = [L2, (1/r)] = [L2, (1/rS)] and the 1/r slips though L2 so we would conclude that [L2, Lφ ] = [L2, Lθ ] = 0 // not true! This looks appealing thinking of the Cartesian world, but these equations are not true in spherical coordinates! You can show this by direct calculation using L2 given as angles. Action of L2 on unit vectors: -L2 = [ 2 + cot + (1/sin)2 2] = () + cot() + (1/sin)2() = () + cot () + (1/sin)2( Sθ ) = – + cot () + (1/sin)2 Sθ [ - Cθ - Sθ ] = – + cot () - (1/sin)2 Sθ Cθ - (1/sin)2 Sθ Sθ = – + cot () - cot - = - 2 What does this means that L2 = 2 = l(l+1) , saying that somehow is an eigenfunction of L2 with angular momentum l = 1. Well, is a rotation group vector which transforms by a 3x3 matrix which is equivalent to a 3x3 representation of the rotation group and this is the l = 1 representation. So pretty nice how that came out. Let's try another unit vector: -L2 = [ 2 + cot + (1/sin)2 2] = () + cot() + (1/sin)2 ( ) = ( – ) + cot( – ) + (1/sin)2 ( Cθ ) = ( – ) + cot( – ) + (1/sin)2 Cθ [- Cθ - Sθ ] = – + cot( – ) – cot2θ - cotθ = – (1 + cot2θ) - 2 cotθ = – csc2θ - 2 cotθ So we find this result: L2 = csc2θ + 2cotθ I have no idea how to interpret this, but as expected, it is non-zero.