Problem Set 8 Solutions

Transcription

1 Physics 57 Proble Set 8 Solutions Proble The decays in question will be given by soe Hadronic atric eleent: Γ i V f where i is the initial state, V is an interaction ter, f is the final state. The strong interaction preserves isospin, so V ust be an isospin singlet. We apply the Wigner-Eckart theore to copute the required ratios: i V f i f i V f where i V f is an isospin singlet which cancels upon taking ratios of two atrix eleents which differ by an isospin rotation. Thus, we can copute ratios of certain rates by taking ratios of Clebsch-Gordan coefficients. Consider the first ratio: We have: Γ(ρ π π ) Γ(ρ π + π ), ) (,,,, Thus,, (,, ) the nuerator vanishes, so Now consider the second ratio Finally, the third ratio is: Γ(K + K + π ) Γ(K + K π + ) Γ( + p π ) Γ( + n π + ) Γ(ρ π π ) Γ(ρ π + π ) 3 3, (,, (,, (,, (, ), ) /3, /3 ), ) /3, /3 In reality, isopin syetry is broken by a nuber of effects, including electroagnetic interactions (e.g. the proton the neutron have different charges), the weak interaction, the light-quark asses. However, for decays ediated by the strong force, isospin is a good syetry, should be approxiately conserved. For other decays, isospin is not conserved. For instance, isospin conservation would suggest that all the pions are stable with equal asses, since they for an isospin triplet all lighter particles are isospin singlets (apart fro bare quarks). However, the π ± are in fact 5 MeV heavier than the π, all pions are unstable. The π has a lifetie of s decays electroagnetically, whereas the π ± have equal lifeties of 8 s decay via the weak interaction.

2 Physics 57 Proble () The atrix eleents D (R) are defined by, D(R), where D(R) is the operator associated with the spatial rotation R. Now consider the direct product state ;,,. D(R) rotates each factor equally: Therefore, D(R) ; [D(R), ] [D(R), ] ; D(R) ;, D(R),, D(R), D (R) D (R) so () D, D(R), ; D(R) ; ; ; ; D(R) ; ; ;,, ; ; D (R) D (R) ; ; () Since d () (θ)d (R(θ)), where R(θ) is a rotation about the y-axis by an angle θ, we have () d (θ), ; ; d (θ) d (θ) ; ; We are free to choose any, which satisfy +. Thus, to find the spin-3/ atrices, we take /. We find: 3/ d 3/,3/ (θ) 3, 3, ;, / d, / (θ) d /,/ (θ), ;, 3, 3 d, (θ) d /,/ (θ) ( + cos θ) cos θ cos 3 θ 3/ d 3/,/ (θ) 3, 3, ;, d /, (θ) d /, / (θ), ;, 3, 3 +, 3, ;, d /, (θ) d /,/ (θ), ;, 3, d / 3, (θ) d /, / (θ)+ d / 3, (θ) d /,/ (θ) 3 (+cos θ) sin θ sin θ cos θ 3 3 sin θ cos θ

3 Physics 57 3/ 3 d /,/ (θ),, ;, d /, (θ) d /, / (θ), ;, 3, 3 +,, ;, d /, (θ) d /,/ (θ), ;, 3, 3 +,, ;, d /, (θ) d /, / (θ), ;, 3, 3 +,, ;, d /, (θ) d /,/ (θ), ;, 3, 3 d, / (θ) d /, / (θ)+ ( ) d / 3, (θ) d /,/ (θ)+d /, (θ) d /, / (θ) / + 3 d, (θ) d /,/ (θ) 3 cos3 θ 3 sin θ sin θ + 3 cos θ cos θ θ 3 cos3 + 3θ cos 3 The other coponents of d 3/ (θ) can be worked out in a siilar fashion. Proble 3 We can represent V q as a colun vector: V q V V V Vx+iVy V z V x iv y Thus, Ṽ q q d qq (β)v q is given by the atrix equation: +cos β cos β sin β Ṽ q sin β cos β sin β sin β cos β ivy Vx cos β Vz sin β V x sin β + V z cos β ivy + Vx cos β Vz sin β + cos β Vx+iVy V z V x iv y V z cos β V x sin β (Vx cos β + Vz sin β) + ivy (V x cos β + V z sin β) iv y Therefore, Ṽ x V x cos β + V z sin β Ṽ z V z cos β V x sin <nonesep>β Ṽ x Ṽ y Ṽ z cos β sin β V x V y sin β cos β V z 3

4 Physics 57 This is clearly a rotation about the y axis by an angle β; one can even check that the signs are correct using the right-h rule. Proble 4 This proble is a straightforward application of (3..) in Sakurai. a) Given vectors U (U x, U y, U z ) V (V x, V y, V z ), we construct the usual cross product: W U V (U y V z U z V y, U z V x U x V z, U x V y U y V x ) We construct a spherical tensor out of W by the stard procedure, as in the previous proble: W ± W x ± iw y, W W z We then have: T q () (U V ) q i W q i [V z (U x + iu y ) U z (V x + iv y )] (U x V y U y V x ) i [V z (U x iu y ) U z (V x iv y )] Uy Vz Uz Vy + i (Uz Vx Ux Vz) i U x V y U y V x i U y V z U z V y i (U z V x U x V z) i where the overall noralization is erely conventional. c) Fro Saurai (3..), we find: () T ± U ± V ± (U x ± iu y ) (V x ± iv y ) (U x V x U y V y ) ± i (U x V y + U y V x ) () T ± U ± V +U V ± [V z (U x ±iu y )+U z (V x ± iv y )] [V z U x + U z V x ] i [V zu y +U z V y ] T () U + V + U V + U V + U zv z (U x + iu y )(V x iv y )/ (U x iu y )(V x + iv y )/ U zv z U x V x U y V y We recognize various coponents of the syetric/traceless piece of T i U i V i. 4

5 Physics 57 Proble 5 a) We define the spherical tensor: x ± iy X ±, X z Thus, the atrix eleents we are asked to related are those of X ±,. We apply the Wigner-Eckart theore, (3..3), to obtain: n, l, X q n, l, l, ;, q l, M X (n, n, l, l ) where M X (n, n, l, l ) is the rotationally invariant piece of the atrix eleent, independent of, q. The above expression relates the atrix eleents of X q for different q. We see that the atrix eleent vanishes unless: + q, l l l + since the Clebsch-Gordon coefficient vanishes unless these conditions are satisfied by conservation of angular oentu. In certain other special cases the Clebsch-Gordan coefficient will vanish even when these conditions are obeyed, e.g. for l l q. b) Instead of applying the Wigner-Eckart theore, we rewrite the atrix eleent in the position basis by stard ethods: n, l, X q n, l, d 3 x n, l, X q x x n, l, d 3 x n, l, x x n, l, X q (x) d 3 x Ψ n,l, (x) Ψ n,l, (x)x q (x) The wavefunction Ψ n,l, (x) should take the for Ψ n,l, (x) Y l (θ, φ)r nl (r) for soe unknown radial wavefunction R nl (r). Moreover, X q (x) x + iy z x iy r sin θ e iφ r cos θ r sin θ e iφ 3 ry q (θ, φ) Thus, since d 3 xr dr dω, we find n, l, X q n, l, [ 3 ] R n l (r) R nl (r) r 3 dr dω Y l (θ, φ) Y l (θ, φ) Y q (θ, φ) 5

6 Physics 57 We now apply (3.8.73) fro Sakurai, naely dω Y l (θ, φ) Y l (θ, φ) Y l (θ, φ) N l; l,l l l ; l l ; l where N l,l,l is a rotationally invariant nuerical factor which we do not write. Thus, n, l, X q n, l, [ 3 N l ; l, R n l (r) R nl (r) r 3 dr ] l, ;, q l we recover our result fro part (a), where M X (n, n, l, l ) 3 N l ; l, R n l (r) R nl (r) r 3 dr the exact for of N l ;l, is given by Sakurai (3.8.73). Proble a) This is a siple application of proble 4, with U V x. We find: () Q ± () Q ± X ± (x ± iy) (x y ) ± ix y X± X z (x ± iy) xz i yz Q () X X +X + X z (x+iy)(x iy) ( z x y ) Therefore, x y Q + + Q z x y xy i [Q + Q ] xz [Q + Q ] yz i [Q + + Q ] Q b) We have: Q e α,, (3z r ) α,, e α,, Q α,, e, ;,, MQ (α,, α, )

7 Physics 57 e α,, x y α,, e α,, Q + + Q α,, e α,, Q α,, e, ;,, M Q (α,, α, ), ;,, Q, ;,, where the Q + ter vanishes, since + is ipossible. We readily see that the atrix eleent is only nonvanishing for, in which case: e α,, x y α,,, ;,,, ;,, Q Proble 7 a) We find d (β), e i Jy β,, e i Jy β,, e i Jy β,, e i Jy β,, e i Jy β J z e i Jy β, where d (β) is defined as d (β), e i Jy β, we use the fact that, e i Jy β, vanishes for, together with the spectral decoposition of J z : J z,,,,, To interpret the resulting expression, recall that J is the generator of rotations, in that U(nˆ, θ) e i θ (J nˆ) represents a rotation by an angle θ about the axis nˆ, with direction of rotation deterined by the righth rule, e.g. U(ŷ, β) x x, x (x cos β + z sin β) xˆ + y ŷ + (z cos β x sin β)ẑ 7

8 Physics 57 Applying the operator x to both sides, we see that U can also be interpreted as rotating the operator x: x U(ŷ, β) x U(ŷ, β), x (x cos β + z sin β)xˆ+ y ŷ + (z cos β x sin β) ẑ (These two viewpoints are siilar to the active passive interpretations of rotations, respectively, are analogous to the Schrödinger Heisenberg pictures for tie translation.) This result generalizes to other vector operators. Thus, U(ŷ, β) J z U(ŷ, β) e i Jy β J z e i βjy J z cos β J x sin β Alternately, we could have established this identity by brute force, e.g. by applying the Cabell-Baker- Hausdorff equation. Thus, d since, J x, is only nonvanishing for ±. (β), (J z cos β J x sin β), cos β For the special case /, we have / d cos β sin β sin β cos β Thus,,, / d,/ / d, / (β) β sin + β cos cos β (β) β cos + β sin cos β as expected. b) We have d (β), e i Jy β,, e i Jy β,, e i Jy β,, e i Jy β,, e i Jy β J z e i Jy β, siilar to before. Note that e i Jy β J z e i Jy β ( e i Jy β J z e i Jy β) 8

9 Physics 57 siilar to what we found in part (a). Thus, e i Jy β J z e i Jy β J z cos β + J x sin β We then find e i Jy β J z e i Jy β J z cos β + (J x J z +J z J x ) cos β sin β + J x sin β d (β) cos β + sin β, J x, since, J x, as before. To evaluate the last ter, we write J x in ters of ladder operators: Thus, J x (J + + J ), J x, 4, J + J + J J +, since, J +,, J,. However, J + J + J J + (J x + ij y )(J x ij y )+(J x ij y )(J x + ij y ) ( J x + J y ) ( J J z ) Thus,, J x, [ ] ( +) so d which is the desired result. Proble 8 We have where (β) cos β + sin β [ ( + ) ] ( +) sin β + ( 3 cos β ) D (α, β, γ), e i Jz α e i Jy β e i Jz γ, e i ( α+γ)/ d (β) d (β), e i Jy β, Thus, since dα dγ e i ( α+γ)/ δ, δ, 9

10 Physics 57 we obtain dα dγ π sin β dβ D (α, β, γ) δ, δ, d (cos β) d (β) However, d (β) P (cos β) where P (x) is the th Legendre polynoial ( cannot be half-integral, since.) Thus, dα dγ π sin β dβ D (α, β, γ) δ, δ, dx P (x)p (x) where we insert P (x). However, the Legendre polynoials obey the orthogonality condition: dx P (x)p (x) + δ Thus, dα dγ π sin β dβ D (α, β, γ) δ, δ, δ, as desired.