Lecture 2: Angular momentum and rotation

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1 Lecture : Angular momentum and rotation Angular momentum of a composite system Let J and J be two angular momentum operators. One might imagine them to be: the orbital angular momentum and spin of a particle; the angular momentum of two spin-less particles; the spin of two particles in bound state. and a-priori we do not know the relative orientation of the two vectors. And indeed, this orientation can only within a small range of quantized values. Classical angular-momentum addition works by simply adding components. In quantum mechanics, we cannot know the three components simultaneously, but we can know the magnitude-squared, J (the Casimir operator) and one component; J z being the conventional choice. The total angular momentum operator and z-component have the eigenvalues: J j, m = j ( j + ) j, m J j, m = j ( j + ) j, m J z j, m = m j, m J z j, m = m j, m () though let s drop the from now on. Consider combined operators: J = (J x + J x ) + (J y + J y ) + (J z + J z ) J z = J z + J z acting on ψ( j, j, m, m ) = j, m j, m ψ is an eigenfunction of J z with eigenvalue m = m + m but it is not generally an eigenfunction of J. However, linear combinations of ψ can produce eigenfunctions Ψ( j, m) of J with eigenvalues j( j + ), with j j < j < j + j. Ψ( j, m) = j m = j j m = j C( j, j, j, m, m ) ψ( j, j, m, m ) () The coefficients C( j, j, j, m, m ) are the Clebsch-Gordan coefficients. They give the probability (=C ) that we measure, from a combined system of j, m j, m, a combined total angular momentum of j( j + ). Clebsch and Gordan were two 9th century mathematicians who identified these coefficients in the development of Lie algebra.

2 Let s derive these coefficients by first noting that the angular momentum operators obey commutation relation: [J i, J j ] = ɛ i jk i J k () where ɛ i jk = + (xyz) (zyx) 0 (otherwise) [J, J i ] = 0 (i = x, y, z) (4) And define raising / lowering operators which, from eq., satisfies Using this commutation relation: J ± = J x ± ij y (5) [J z, J ± ] = ±J ± (6) J z J j, m = (J J z J ) j, m = J (J z ) j, m = (m ) J j, m Similarly m + is the eigenvalue of J z applied to the raised state: (J + j, m ) As the eigenvalue of the raised state drops out so simply, we know there can only be a constant term in front of the raised eigenstate. We write, J ± j, m = C ± ( j, m) j, m ± (7) with the boundary condition that C ± is required to be 0 for J j, j and J + j, +j. To derive these constants we note that J + is complex-conjugate of J and from bra-ket algebra (noting J + J +) reverse eq. 8 j, m J j, m + = C ( j, m + ) j, m j, m = C ( j, m + ) = j, m J + j, m + (8) j, m + J + j, m = C + ( j, m) j, m + j, m + = C + ( j, m) so the lowering coefficient of the m + state is the same as the raising coefficient of the m state: C is real. Thus applying both operators successively: where the double operator can be broken down to: from where we can easily identify the eigenvalues. C ( j, m + ) = C + ( j, m) = C (9) J J + j, m = C j, m (0) J J + = J x + J y + i(j x J y J y J x ) = J x + J y + J z J z + i[j x, J y ] = J J z + J z = J J z (J z ) () C = j( j + ) m(m ) C ( j, m) = j( j + ) m(m ) () C + ( j, m) = C +( j, m) = C ( j, m + ) = j( j + ) m(m + )

3 An example application and calculation Consider two particles j, m and j, m forming a combined state j, m where j = and j = /. Evidently, j can be / or /. Two key points: Coefficients should normalise to When raising (lowering) a state to an m higher (lower) than J ( J), state = 0 / states: Evidently the max. (/) and min. (-/) m states are: Using definition: Now operating on the combined state, =,,, =,, C ± ( j, m) = j( j + ) m(m ± ) J, =, J, =, 0 J, 0 =, J, = 0 In analogous way: J, = J,,, =, 0, +,,, =, 0, +,,, =, 0, +,, / states:, = a,, + b, 0, with a + b =. Now apply J + J +, = 0 = a,, + b,, a + b = 0 a = and b =

4 Our Clebsch-Gordan coefficients for j = and j = j m m m / / / / / / + / / And for a j = and j = combination j 0 m 0 0 m m / / + / / In the exam, Clebsch-Gordan coefficient tables are provided, though you should still be able to derive them if asked. They are also recording in most of the recommended texts and in the PDG: > Reviews, Tables, Plots > Mathematical Tools 4 Rotations Consider a small rotation ɛ, about the y-axis: x = R y (ɛ) x x cos ɛ y = z sin ɛ sin ɛ x y cos ɛ z And its inverse: x = R y (ɛ) x x cos ɛ y = z sin ɛ sin ɛ x y cos ɛ z Impose invariance: ψ ( x ) = ψ( x) = ψ(r y (ɛ) x ) 4

5 Without losing generality take a specific point x = a and find a relation between ψ ( a) and ψ( a): ψ ( a) = ψ(r y (ɛ) a) = ψ(a x cos ɛ a z sin ɛ, a y, a z cos ɛ + a x sin ɛ) = ψ(a x ɛa z, a y, a z + ɛa x ) as ɛ 0 ψ( a) ψ( a) ψ( a) + ɛa x ɛa z z +... Taylor expansion x [ ( )] ψ( a) + ɛ a x z a z x [ i ] [ ɛj y ψ( a) as : J y = i z } {{ } x x ] z U r (ɛ) Conservation Invariance (Ehrenfest theorem) Consider the time variation of U r : [ ] d d dt φ(t) U r ϕ(t) = dt φ(t) U r ϕ(t) + φ(t) du [ ] r d ϕ(t) + φ(t) U dt r t ϕ(t) du = φ(t) r ϕ(t) + i }{{} dt φ(t) HU r U r H } {{ } ϕ(t) 0 by assumption commutes with Hamiltonian Hence, U r is invariant if J y (the tricky part of U r ) is independent of time and commutes with the Hamiltonian. Commutation with the Hamiltonian means the eigenvalues of U r are also energy eigenvalues, i.e. the system is defined by its angular momentum, specifically the total angular momentum plus one component. Angular momentum is conserved by the wavefunction be invariant under rotation. Rotation is a Lie group and so any rotation can be expressed as the successive application of the infinitesimal rotation: or for some, general D rotation: U r ( θ) = e i θ ˆ J. U r (β) = lim ( i β n n J y ) n = e i βj y () In the language of operators, the angular momentum operator J y is said to be the generator of rotations about the y axis. Euler angles Generic rotation can be parameterised by three Euler angles defining three successive rotations. Most usefully :. rotation by angle α about the original z-axis.. rotation by angle β about the intermediate y-axis.. rotation by angle γ about the final z-axis. 5

6 Thus, a general rotation of a wavefunction can be represented by D-matrices: D j m,m(αβγ) = j, m e i/ αj z e i/ βj y e i/ γj z j, m = e i/ (αm+γm) d j m,m(β) (4) where the change of the z-projection (m m ) only occurs during the rotation around ŷ. The two rotations around the original ẑ leave the quantum numbers unchanged. Rotation matrix: d j m,m (β) Although the y-projection is unchanged, the z-direction has changed so the quantum number m is not the same it is now projected onto a new z -axis. A state j, m transforms under a rotation β about the y-axis into a linear combination on the ( j + ) j, m states. e (i/ )βj y j, m = dm m(β) j j, m (5) Which we apply to the final state j, m to examine the rotation from m to m projection. m d j m m (β) = j, m e (i/ )βj y j, m (6) So, lets calculate the matrix the specific case of j =. There are projection (, 0, ) so there are elements of d j m m in this case. I calculate a couple here, the rest can be done as an exercise.. From inspecting the expansion of e iβj y (with = ): e iβj y. Look seperately for solutions of J n+ y = iβj y! β J y + i! β J y + 4! β4 J 4 y +... j, m and J n y j, m. Recall the raising/lowering operators J ± = J x ±ij y and the Clebsch-Gordon coefficients C jm ± = j( j + ) m(m ± ) J + J = J x + ij y J x + ij y J y = i [J + J ] so, J y, = i [ C+ 0 C, 0 ] with C = = i, 0 6

7 4. Operate again with J y J y, = i J y, 0 = (C0 +, C 0, ) = (,, ) as C0 + = C 0 = 5. And again: J y, = J y(,, ) = ( i [J + J ], i [J + J ], ) = i 4 (, 0 +, 0 ) = i, 0 = J y, 6. Note the cyclical pattern and conclude: J n+ y, = i, 0 J n y, = (,, ) 7. Which then leads to, for each specific j, m state:, J n+ y, = 0, J n+ y, = 0, 0 J n+ y, = i, J n y, =, J n y, =, 0 J n y, = 0 8. With these relations the d j coefficients are calculated as: m m d =, e iβj y, =, + ( iβj y ) +! ( iβj y) + i! ( iβj y ) + 4! ( iβj y) , = β! + i β ! = ( + (! β + 4! β4...)) = ( + cos β) d =, e iβj y, = ( (! β + 4! β4...)),, = 0 of course = ( cos β) 7

8 Example: e e + µ µ + The incoming left-handed electron annihilates with the right-handed positron. In the electromagnetic process, a γ is exchanged with the outgoing µ-pair. Final state particles must have opposite helicity in conserving j. A = A(e L e+ R µ L µ+ R ) d, = ( + cos θ) A = A(e L e+ R µ R µ+ L ) d, = ( cos θ) dσ d cos θ = A + A + cos θ The two amplitudes should be of equal intensity due to parity conservation with an electromagnetic interaction. See the following reproduction of e + e µ + µ data from the TASSO experiment at DESY (c. 980). BTW: the asymmetry is evidence for the off-shell influence of the parity-violating Z 0 interfering with the γ exchange. 8

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