In this lecture we will go through the method of coupling of angular momentum.

Transcription

1 Lecture 3 : Title : Coupling of angular momentum Page-0 In this lecture we will go through the method of coupling of angular momentum. We will start with the need for this coupling and then develop the method for evaluating coupled state in terms of uncoupled states. First we will take up the case of coupling of two generalized angular momenta and then utilize this to take the case of three angular momenta.

2 Page- In previous lecture, we have seen that the electron-electron repulsion term is responsible for the splitting of energy levels. The reason for this is the one electron quantum numbers are not the good quantum numbers for the total Hamiltonian. e Because, H does not commute with r i, but Hcommutes with L and S where ij L= i ; S = Si. It is equivalent to saying that H is invariant under a rotation of i i spatial and spin coordinates. Proof: [ H, L] = e k e k i< j k rij i< j r r ij ij where, = ir k k k Allowing the operators to act on arbitrary function ψ k ψ = kψ + ψ k r ij rij rij k = k + k r ij rij r ij = k irk k r ij r ij So [ H, L] = e k e k r r [, ] = 0 [ ] i< j k ij i< j k = e k e k + ie rk k i< j k rij i< j k r ij i< j k r ij ri rj ri rj Now, r k k = rk δ 3 ik + rk δ 3 jk k r ij k rij rij r r r r =+ = 0 So H L Similarly H, S = 0 i j i j 3 3 rij rij ij

3 Page- The angular momentum for an atomic electron can be visualized in terms of a vector model where the angular momentum vector is seen as precessing about a direction in space. Let us consider the set of states having a common value of the quantum number j. For each of these states the length of the orbital angular momentum vector, = j( j+ ) in units of. The z-component of this vector is m in units of. For example if j = then there are five projections as shown in the figure. Z ( + ) In this model, we can understand that, for two electron system, the electron-electron repulsion is responsible to couple l and l to form the resultant vector L.Similarly, s and s to form the resultant vector S L S l s l s

4 Page-3 Generalized angular momentum operator J is defined as a vector operator with Hermitian components J x, J yand J z satisfying J J = ij The components are defined as J = + ( J x ij y) +, J0 Jz And the inverse relations are Jx J J = ( ) and J = ( J + J ) + y = and J = ( J + x ij y) + The commutation relations between them are [ J J ] J + = +, [ J, J 0 ] J =, and [ J+, J ] = J0 0, Since J, J and J are Hermitian, the total angular momentum operator x y z J = J + J + J = J J + J J J x y z Using the Dirac notation, the eigen functions are represented as jmwhere, m is the projection of J, we have ( ) J jm = j j+ Jm Jz jm = m jm j m jm = δ δ jj mm Similarly we have J+ jm, = [ j( j+ ) mm ( + ) ] jm, + And J jm, = [ j( j+ ) mm ( ) ] jm,

5 Page-4 Coupling of Angular Momenta Two angular momentum operators J and J operate on two different spaces j, m and j, m such that, ( ) J jm = j j+ jm Jz jm = m jm J jm = j( j + ) jm Jz jm = m jm Let us define, J = J + J Jx = Jx + J x y y y Jz = Jz + Jz ( ) With understanding J = J + J ( ) And J = J + ij t x y J = J + J ( Jx Jx) i( Jy Jy) ( Jx ijy) ( Jx ijy) = = t t t Now the question is whether J will be an angular momentum? For this it should follow all the commutation relation for generalized angular momentum. J, J = J + J, J + J x y x x y y = Jx, J y + Jx, J y + Jx, J y + Jx, J y ( for different spaces) = Jx, J y + Jx, J y = ij + ij = i J + ij = ij So, J, J = ij [ ] z z z z z x y z

6 Page-5 The other relations will also follow. So we can conclude that J J = ij And this relation is sufficient to identify J as an angular momentum operator. So the orthonormal Eigen function corresponding to this operator is: j, j, j, m j, m The meaning of this notation is that, the coupled jm should arise from two uncoupled Eigen function jm and jm. So, Where, ( ) J jm = j j+ j j jm J jm = m j j jm z 3 j = 0,,,,... and =,,,..., ( + ) m j j j j J values Now we have to setup the relationships between coupled Eigen function jm and uncoupled Eigen functions jm and jm. Similarly, coupled Eigen values j, m with uncoupled Eigen values j, j and m, m. Let us construct a product wavefunction, j j m m j m j m

7 Page-6 As we know that, the products and their linear combinations are also the Eigen functions of J J and J J., z, z ( ) ( ) ( ) ( ) J j j m m = J j m j m = j j + j m j m = j j + j j m m J j j m m = J j m j m = j j + j m j m = j j + j j m m J j j m m = J j m j m = m j m j m = m j j m m J = = = z z z j jmm Jz jm jm m jm jm m j jmm Now, let us apply J z on the uncoupled Eigen functions: ( ) Jz j jmm = Jz + Jz j jmm = Jz jm jm + Jz jm jm = ( m+ m) jm jm = m jm jm If m= m+ m, then j jmm is an Eigen function of J z. J = J + J = J + J + JJ Since ( ) j jmm will not be the Eigen function of So, we have to find a common Eigen function of J. J and J z. We construct a linear combination of the uncoupled Eigen function, such that j j jm = j j m m j j jm j j m m m, m Eigen functionof Numericvalueknownas Eigen functionof J and J Vector AdditionCoefficients J z and J or, Clebsch GordonCoefficients We have to find out the relationships between the quantum numbers representing the coupled and uncoupled Eigen functions.

8 Page-7 First applying J z on both sides of the equation, we get: ( ) m j j jm = m + m j j m m j j jm j j m m m, m It suggests that the value of j jmm j j jm = 0 if m m+ m We have to find the possible values of j We know that j m j and j m j Since m= m+ m We have j m m j and j m m j Let m assume its maximum value j, m its maximum value j and m its maximum value j then j j j j and j j j j So j j j j+ j and j j j j+ j Combining these two we get j j j j+ j For example if, j =, m =, and j =, m =, 0,, then the possible values of j =, m=+, and j =, m=+, +,, because m= m + m 3 = =, =+ = = + 0 =, =+ + 0=+ 3 = + =+, =+ + =+

9 Page-8 It is also important to put the condition that j + j + j = n, where n is an integer. This follows that j + j j j j + j 0. This is known as triangular conditions. j+ j + j Considering all these we have, j = j+ j, j+ j, j+ j,..., j j And m= j, j,..., j Coupling of three angular momenta Let take three angular momenta J, J and J 3 such that J = J+ J + J3 There are three ways to couple these. As an example, j =, j =, j3 = Case -, First couple j = and j =, 0 This gives j, j j =, Now we couple j with j 3 = This gives, ( ), 3, 3 = j j j j j j j, { 0 },, { }, 0, { },, { },

10 Page-9 Case- First couple j = and j 3 =, This gives j, j 3 j 3 = 3, Now we couple j3with j =,, 0,, This gives j j, j3( j3), j j3, j = 3,, 3,, Case -3 First couple j = and j 3 =, This gives j, j3 j3 = 3, Now we couple j3 with j =

11 Page-0,, 0,, This gives j j, j3( j3), j j3, j = 3,, 3,, Note: When three angular momentum operators are coupled to form the total angular momentum, the possible values of the total angular momentum will be the same, independent of the method of coupling scheme. However, the coupled wavefunctions (discussed later) will depend on the coupling scheme.

12 Page- Recap In this lecture, we have learnt that the two angular momenta get coupled to form new angular momentum. The reason for coupling of angular momentum is to form a new basis set that will be the eigenfunction of the total Hamiltonian. In case of three angular momenta, the coupled angular momentum does not depend on the coupling scheme. However, the coupled wavefunction arising from the uncoupled state does depend on the coupling scheme. We will understand this in the next lectures.