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States, operators and matrices Starting with the most basic form of the Schrödinger equation, and the wave function (): jm jm c j ,m jm Aei kx t Be i kx t j ,m The state of a quantum mechanical system is . This state can be expanded to a vector and the system can be more complicated (e.g. with spin, etc.): jm Mˆ jm jm jm Note: to extract the expectation value of a property, one needs to sum over the expanded state: 2 2 i t 2m x 2 Dirac notation Mˆ c j ,m 2 j ,m * jm M jm c 2j ,m jm j ,m jm Mˆ jm c 2j ,m jm j ,m j ,m Combining operators Operator T, transforms the original state. M jm jm jm Here, a hermitian matrix M was introduced, to extract properties from the wave function (hermitian matrices have real eigenvalues). c 2 j ,m jm transformed Tˆ jm original Hence, to get the expectation value: trf jm Mˆ jm trf org jm Tˆ 1Mˆ Tˆ jm Mˆ TˆMˆ Tˆ 1 org Adding spin (1) • Projection is known (m quantum number) • Length of the two spins is known (j1 and j2) • Several possibilities to construct projection by adding the two spins Example: m = jmax - 2 j1 - 2 j2 < j1, m1 | j2, m2 > = < j1, j1 - 2 | j2, j2 > jmax - 2 m Adding spin (1) • Projection is known (m quantum number) • Length of the two spins is known (j1 and j2) • Several possibilities to construct projection by adding the two spins Example: m = jmax - 2 j1 - 1 j2-1 < j1, m1 | j2, m2 > = < j1, j1 - 1 | j2, j2 - 1 > jmax - 2 m Adding spin (1) • Projection is known (m quantum number) • Length of the two spins is known (j1 and j2) • Several possibilities to construct projection by adding the two spins Example: m = jmax - 2 j1 j2- 2 < j1, m1 | j2, m2 > = <j1, j1 | j2, j2 - 2 > jmax - 2 m Adding spin (2) • Projection is known (m quantum number) • Length of the maximum total spin known jmax = j1 + j2 • Several possibilities to construct projection from different sizes of total spin j j1 j2 | j1 –j2 | ≤ j ≤ j1 + j2 j Note: j1, j2 and j can be interchanged. However, changing the composite state with one of the constituent states is not trivial and requires re-weighting of the constituent states j = j1 + j2 -2 jmax - 2 m < j, m > = < j1 + j2, j1 + j2 - 2 > < j, m > = < j1 + j2 - 1, j1 + j2 - 2 > < j, m > = < j1 + j2 - 2, j1 + j2 - 2 > } m≥j Adding spin (3) • m=jmax m1=j1 & m2=j2 j=j1+j2 & m=jmax m=jmax-1 m1=j1 & m2=j2-1 m1=j1-1 & m2=j2 j=j1+j2 & m=jmax-1 j=j1+j2-1 & m=jmax-1 m=jmax-2 m1=j1 & m2=j2-2 m1=j1-1 & m2=j2-1 m1=j1-2 & m2=j2 j=j1+j2 & m=jmax-2 j=j1+j2-1 & m=jmax-2 j=j1+j2-2 & m=jmax-2 Note: - j1 m1 j1 , thus m1 has (2 . j1+1) possible values and m2 (2 . j2 +1). Each combination shows up exactly ones in the second column of the table so the total number of states is (2 . j1+1) (2 . j2 +1). • The third column has the same amount of states as the second column. The quantum number j is a vector addition, thus it will never be lower than |j1 – j2|, which is called jmin. • For m < jmin the amount of states is jmax – jmin + 1 = (j1+j2) – |j1 – j2| + 1 for each m. This situation occurs for - |j1 – j2| m |j1 – j2| , thus (2 |j1 – j2| + 1) times. Number of states: • For m -jmin and m jmin the amount of states is jmax m +1. This results in the following sum: Number of states: • (2 |j1 – j2| + 1) . [(j1+j2)-|j1 – j2| + 1] 2 m j1 j2 j 1 m j1 j2 1 j2 m 1 Together these contributions add up to (2 . j1 + 1) (2 . j2 + 1) Adding spin – Clebsch Gordan Example: m = jmax - 2 j = j1 + j2 -2 j1 jmax - 2 j2- 2 jmax - 2 m JM C J , M , m1 , m2 m1 , m2 m with : J M and : M m1 m2 m1 , m2 or : JM C J , M , m1 m1 , M m1 m1 , m2 e.g. J 1, J 2 C1 j1 , j2 2 C2 j1 1, j2 1 C3 j1 2, j2 From symmetry relations and ortho-normality, the C coefficients can be calculated. The first few: m1,m2 JM ½x½ 1, 1 1, 0 0, 0 1, -1 +½, +½ 1 0 0 0 +½, -½ 0 ½ ½ 0 -½, +½ 0 ½ ½ 0 -½, -½ 0 0 0 1 Combining spin and boost Lorentz transformations: cosh( ) 0 1 0 Lz p 0 0 sinh( ) 0 p tanh( ) sinh( ) E 0 sinh( ) 0 0 1 0 0 cosh( ) p E cosh( ) w w w E 2 p2 p E, p p p R , ,0 zˆ For: Jackson (section 11.7) calculates the corresponding operator: ip ˆ ˆ Extracting rotations: L p e L p Rˆ , ,0Lˆ z p Rˆ 1 , ,0 ˆ Allowing definition of the canonical state: L p jm p, jm Intermezzo: rotation properties – The total spin commutes with rotation 2 i nJ i n J 2 i n J J e jm e J jm j j 1 e jm – However, the projection is affected with a phase. Consider the rotation around the quantization axis R jm jm' e m' e.g. jm' e i n J i nz J jm jm' jm jm' e iJ z jm m'm e im – Euler rotations, convention: z y’ z’’ Advantage: quantization axis used twice for rotation iJ z '' iJ y ' iJ z ˆ Re e e – Rotations are unitary operators. The rotation around y’ includes a transformation of the previous rotation. UU 1 e i J y e iJ z e i J y – This results in: R ˆ eiJ z eiJ y eiJ z eiJ z e iJ y ' i.e. Rotations can all be carried out in same coordinate system when order is inverted. Intermezzo: the rotation matrix Summary of the result from the previous page R jm jm' e iJ z e iJ y e iJ z jm jm' m' Dmj 'm jm' Dmj 'm jm' m' m' – Euler z y z convention makes left and right term easy jm' e iJ z e e im ' jm' e i J y i J y e iJ z jm jm e im e im ' d mj 'm e im – The expression for djm´m is complicated, but is used mainly for the deduction of symmetry relations: d mj 'm m' m j d mm ' Note 1: Inverse rotation is accomplished by performing the rotations through negative angles in opposite order: ( Djm’m(, , ) )-1 = Djm’m(-,-,-) d mj 'm d jm, m' Note 2: Since rotations are hermitic, the conjugate matrix is: Djm’m(-,-,-)=Dj*mm’() Note 3: Combination of all this gives: Dj*m’m()=(-)m’-mDj-m’,-m() Rotations, boosts and spin Describing the rotation of a canonical state: ˆ ˆ ˆ R1 p, jm R1L p jm Rˆ1Rˆ0 Lˆ z p Rˆ01 jm Rˆ1 Rˆ 0 Lˆ z p Rˆ 01 Rˆ11 Rˆ1 jm Rˆ1,0 Lˆ z p Rˆ1,01 Rˆ1 jm j j ˆ ˆ LR1 p Dm 'm,1 jm' Dm 'm,1 LR1 p jm' D m' j m 'm ,1 m' j R1 p, jm' Dm 'm,1 R1 p, jm' m' Remember: R1 jm Rˆ1 jm Dmj 'm,1 jm' m' i.e. rotation of a canonical state rotates the boost and affects the spin state in the same way as it would in the `particle at rest’ state. Helicity ˆ ˆ The helicity state is defined with: p, j L p R0 j p, j ˆ p Rˆ j Rˆ Lˆ Rˆ 1Rˆ j Rˆ Lˆ j Note: L 0 0 z 0 0 0 z Compared to the same operations on the spin state: ˆ j ˆ L p R jm D p, jm' 0 m 'm , 0 m 1 ˆ j ˆ ˆ ˆ ˆ ˆ R0 Lz jm R0 Lz R0 R0 jm Dm 'm, 0 p, jm' m j Hence: p, j Dm ,0 p, jm m In other words: helicity () is the spin component (m) along the direction of the momentum. Note that the helicity state does not change with rotations: Rˆ1 p, j Rˆ1Lˆ p Rˆ 0 j Rˆ1 Rˆ 0 Lˆ z j Rˆ10 Lˆ z Rˆ101 Rˆ10 j Lˆ R1 p Rˆ10 j Rp, j Note that the helicity state does ˆ ˆ Lˆ p Rˆ 1 Rˆ Lˆ p Rˆ 1 Rˆ j L p p , j R 1 0 0 z 1 0 0 z 0 0 0 not change with boosts (as long ˆ as the direction is not reversed): Rˆ Lˆ p Rˆ 1 Rˆ j Lˆ p R j R p 0 z 10 0 0 10 0 10 , j Discrete symmetries Parity x x ˆ dx p p dt J x p J Commutation relations: ˆRˆ Rˆ ˆ ˆ ˆ ˆL p L p ˆ Charge conjugation Cˆ particles antiparticles i Dirac: Conjugating+transposing gives: i T C T C 1 Resulting in: Left-handed M T 0 C is the matrix doing the transformation: Mirror analogy Left-handed particle M 0 C C i 1 T M C 1C T 0 C 1 i CMC 1 C T 0 i M C T 0 e.g. C b b c antiparticle Time reversal e.g. b Vcb* W+ Vcs s c ˆ c c c b Vcb* Vcs W+ s Note that time reversal changes t in –t and input states in output states (in other words: < bra | to | ket > ). Another way to show this: i.e. the transformed state does not obey the description of motion of the Hamiltonian, it needs an extra ‘–’ sign. ˆ H i t ˆ ˆH ˆi i ˆ t t The solution is to make time reversal anti-unitary: Note: this can also be shown with the commutation relation: ˆ p, x i p, x i ˆHˆ ˆi i * i t t ' t ' ˆ t * Time reversal continued Next, the the time reversal operator is split in a unitary part and a complex conjugation. The m states consist of real numbers, i.e. projections. Hence a time reversed expectation ~ value can be described with: ~ Calculating the expectation value of operator Â: A ˆ And combining with the time reversal operator: K K m m m m K m m m * m m m m' m' 1 m m,m ' A i.e. Aˆ * 1 ~ ~ ˆ ˆ ˆ A Aˆ A Since a second time reversal should restore the original equation: Aˆ ~ ~ ~ ˆAˆ * ~ ˆAˆ *ˆ 1ˆ ~ ~ ˆAˆ ˆ 1 Aˆ Aˆ * ~ ˆ Hermitic ~ ~ 2 ~ ˆ ˆ Aˆ ~ Aˆ A ~ ~ ~ ~ 2 ˆ2 Aˆ ˆ2 Aˆ ˆAˆ ˆ1 Aˆ Time reversal and spin From the commutation relation: J , J i i j ˆ ijk x Consistent with: J k J i , J j i ijk J k x ˆ dx p p dt J x p J Rˆ ˆ jm Rˆ jm ˆRˆ j ˆDmm ' jm The equation: j mm' T j j Tmm D ' m 'm" j * j mm' m 'm" D T t t ˆ jm' D j m ' m" j * mm ' jm' D j * j mm' m 'm" D T Djm, m' j m' jm holds if: m', m Dmj 'm" j m Djm,m" mm' ˆ, Rˆ 0 j j ˆ j j Rˆ Tmm jm ' T R jm ' T D ' mm ' mm ' m 'm" jm" j * mm ' D j m * And: * t t j mm' T j * j mm ' m 'm" D T jm" j m m', m d mmj ' m", m' j m2 m' Djm,m" j m Djm,m" Time reversal and spin continued Time reversal of a canonical state: ˆ p , jm j m * j m j p, jm t t Tmm p , jm ' p , jm' ' m ', m j m p, j,m or : ˆ p , , jm p , , j,m Time reversal of a helicity state: j j * p, j Dm p, jm p, jm Dm p, j m j * ˆ p, jm ˆDm p, j Dmj ˆ p , j and * j m j m j p , , j,m Dm, , p , , j, * j m, with : D Hence: , m Dmj , m ei j Dmj , j m m i j Dmj ˆ p, j e Dmj , p , , j, Should give: ˆ p, j j m D j i m D e p , , j Parity and spin The parity operation on a canonical state: ˆ p, jm p, jm p , , jm The parity operation on a helicity state: j j * p, j Dm p, jm p, jm Dm p, j m j * j * ˆ p, jm ˆDm p, j Dm ˆ p , j m Djm,ˆ p , j j * p , , jm Dm , p , , j and , m Djm, , m ei j Dmj , m Djm,ˆ p , j m ei j Dmj , p , , j j * m with : D Hence: Should give: Note: ˆ i j p, j e p , , j, p, jm p , , jm Since helicity states include rotational properties: but: p, j p , , j ˆ ˆ p, j L p R0 j Rˆ0 Lˆ z j Composite states SM mm 1 2 SM m1m2 m1 , m2 0,0 s-quark 1 This ground state can decay to two vector mesons: Bs-meson b-quark c-quark c-quark s-quark Isospin = 0 Spin = 0 Parity = - m1m2 Note: M = m1 + m2 1 , 1 1 1 , 1 2 2 2 2 2 2 Bs-meson W+ S , m1 , m2 m1 , m2 Bs ground state: b-quark C s-quark s-quark J/-meson (easy to detect) Isospin = 0 Spin = 1 Parity = (C-parity = -) -meson Isospin = 0 Spin = 1 Parity = (C-parity = -) Two body decay properties SM mm 1 2 SM m1m2 m1 , m2 Spin states of Bs, J/ and : C S , m1 , m2 m1m2 m1 , m2 0,0 1 1,1 1 0,0 1 1,1 3 3 3 Not the complete story… Consider momentum in two particle decay (Bs rest frame): m1 , m2 a p | m1 p | m2 Normalization = Hence: , m1 , m2 , m m C D , SM ' , SM Remember: Rˆ jm D j m 'm jm' Rˆ , SM m' Still no complete story… Consider angular momentum: lm, SM dY , SM l m With: S , m1 , m2 1 2 m1 , m2 j m 'm m' 2l 1 l * Y Dm0 ,0 4 l m Angular Total spin momentum Missing: Formalism that describes angular momentum and Yml states. Rotation with angular momentum Rˆ lm, SM dYml Rˆ , SM dYml DMS 'M ' , SM ' dDMS 'M Yml ' , SM ' 2l 1 S l * d D D M 'M m 0 ,0 ' , SM ' 4 Split up the Yml state, to match with new rotation: l * m0 D D ,0 D l mm' l * mm' ? D ' ,0 D l * m '0 l * mm' R D ' ,0 1 l * m '0 4 l Ym ' 2l 1 Rˆ lm, SM dDMS 'M Dml 'm Yml ' ' , SM ' DMS 'M Dml 'm lm ' , SM ' i.e. the transformation is a product of the rotation of two rest states: Rˆ lm, SM Rˆ lm , SM DMS 'M Dml 'm lm' , SM ' C m1 , m2 S , m1 , m2 , m1m2 Angular momentum and spin We can also express them as states with a sum of angular momentum and spin: lm, SM JM J , lS S JM J lm, SM S m,M S C m,M S J ,m, M S lm SM S C J ,m,M S lm, SM S Note: MJ = m + MS m,M S C J ,m, M S lm m,M S m1m2 M S m1 , m2 C J ,m , M S CS ,m1 ,m2 dYml , m1m2 m,M S m1 , m2 Since: Rˆ lm, SM S DMS S 'M S Dml 'm lm' , SM S ' And: DMS S 'M S Dml 'm , m1m2 Note: MS = m1 + m2 (see 1 page back) M J ' M S ' m' M J M S m DMJ J ',M J M J ', M J The state can be rotated with: Rˆ JM J , lS DMS J 'M J JM J ' , lS Note that (as expected) the sum of angular momentum and spin (J) is not affected by the rotation, neither are the angular momentum (l) and the total spin (S). This result is the equivalent of the non-relativistic L-S coupling. Two body decay and helicity (1) As with spin, we need to consider momentum in the helicity state: ˆ ˆ ˆ ˆ 12 a p | 1 p | 2 L p R 1 L p R 1 Rˆ Lˆ z p 1 Rˆ Lˆ z p 1 Rˆ Lˆ z p 1 Lˆ z p 1 Rˆ 00 | 12 Angles are zero. Particles are boosted back to back along the positive and negative Z-axis The link with previous page is provided via the relation between canonical and helicity states: , m1m2 a p | m1 p | m2 s1 * s2 * a Dm11 Dm22 p , 1 p , 2 a SM S ,M S , C s S ,M S , m1m2 S 12 D S * MS S * MS Cm12 , S , D Rˆ 00 | 12 = 1 - 2 Why? MS = m1 + m2 12 , S , M S , 12 w a 4 p Single state normalization. Rest mass: w Momentum: p 1 , 12 a Two body decay and helicity (2) Expressing states with the sum of angular momentum and spin in helicity states: JM J , lS C J ,m, M S m,M S C J ,m,M S m,M S m1 , m2 C J ,m,M S m,M S l C d Y S ,m1 ,m2 m , m1m2 m1 , m2 CS ,m1 ,m2 dYml CS ,m1 ,m2 m1 , m2 C12 ,S ,M S Cm12 ,S , d S , M S , * lm C J ,m,M S m,M S m,M S C S , m1 , m2 m1 , m2 C J ,m,M S m1 , m2 * S , M S , DMJ J DMS S C S C C D 12 ,S ,M S m12 ,S , M S , 12 * * 2l 1 l * 2l 1 l * Dm 0 C Dm 0 DMS S DMJ J 2J 1 2J 1 lm Intermezzo, check… C 12 , S , M S Cm12 ,S , S ,M S , CS ,m1 ,m2 * 2l 1 l * Dm 0 DMS S , 12 4 C12 ,S ,M S Cm12 ,S , S , M S , 2l 1 J * d f (l , m) DM J , 12 4 2l 1 2J 1 J * f ' (l , m) d DM J , 12 2J 1 4 Normalization (later, easier the other way around) NJ JM J , 12 Two body decay and helicity (3) Check the transformation properties for rotations: S * ˆ R JM J , 12 N J d DM J Rˆ Rˆ 00 | 12 N J d D S * MJ N J d D ,M J ' D S MJMJ ' Rˆ 00 | 12 * S MJMJ ' D * S M J ' N J d D * S M J ' 'Rˆ ' 00 | 12 'Rˆ ' 00 | 12 DMS J M J ' JM J ' , 12 Transforms as it should… Remember: ˆ R1 p, j Rp, j i.e. Mj transforms, but 1 and 2 do not. Canonical versus helicity states JM J , 12 d N J DMJ J , 12 d * C , m1m2 Cm12 ,S , DMS S , 12 * 12 , S , M S 2J 1 J * DM J , 12 (2 pages back) 4 (3 pages back) S ,M S , s1 * m11 D D s2 * m2 2 , 12 , 12 Dms111 Dms222 , m1m2 JM J , 12 d N J DMJ J , 12 d * Dms111 Dms222 M S Combine 2J 1 J * DM J Dms111 Dms222 , m1m2 4 m1m2 12 DMS S S ,M S , DMS S DMJ J * lm * 2l 1 M S lmM J l 0 Dml 0 2J 1 4 l D Ym 2l 1 l * m0 JM J , 12 S , M S lm JM J , 12 Sl JM J , lS (4 pages back) Normalization 2l 1 2 J 1 4 M S lmM J l 0 M S m1m2 12 dYml , m1m2 2 J 1 4 2l 1 2l 1 l 0 12 2J 1 M S ,m M S lmM J M S m1m2 dYml , m1m2 Decay amplitudes Remember: The transition amplitude to a helicity state is calculated with the matrix element: A p , 1 | p , 2 a 12 m JM J m JM , 12 a p, 1 | p, 2 p, 1 | p, 2 a , 12 J w 4 12 p m JM Momentum of decay products: p Resonance rest mass: w J And: Complete set 4 w 12 JM J 12 JM J 12 p m JM J , 12 JM J , 12 , 12 d N J D With: d N J D J * MJ Hence: J * MJ A NJ D , 12 J * MJ , 12 N J D 4 w JM J 12 p w a 4 p , 12 J * MJ m JM J Check… J * MJ NJ D FJ 1 2 Helicity amplitude Helicity amplitude Remember: JM J , 12 Sl 2l 1 l 0 12 JM J , lS 2J 1 2J 1 And: N J 4 Switch to canonical states: J F12 w 4 JM J , 12 p m JM J Complete set 4 w JM J , 12 JM J , lS JM J , lS p m JM w 2l 1 4 l 0 12 JM J , lS p Sl 2 J 1 N J F12 J m JM J w 2J 1 2l 1 4 l 0 12 JM J , lS p 4 Sl 2 J 1 2l 1 l 0 12 Sl J w 4 JM J , lS p m JM J m JM Partial wave amplitude: als Canonical states J Bs0 → J/ , tree level Vcb* b Vcs Bs0 W+ s b 0 Bs s W W s Bs0 b Oscillation W- Vcs Vcb* Decay c c J/ s s s s J/ c c The basics a X0 b X0 d a i a i M dt b 2 b Ma 0 a 0 0 Mb a b 0 b No mixing, just 2 states 1 0 0 1 Particle and anti-particle: Ma Mb Schrödinger Eigen states of the Hamiltonian: iM t 2 a a0 e iM t b 2 Time evolution of eigen states b0 e 0 M en 0 0 0 M The basics a Bs0 b Bs0 d a i a i M dt b 2 b Mixing: Hermitic: 12 21 M M12 M M 21 * M 12 M 21 12 *21 M * M12 M12 M * 12 12 Note: to obtain properties with real values, the matrix needs to be Hermitic. The basics a Bs0 b Bs0 M * M 12 M 12 i * M 2 12 12 a b i i * i * M M 12 12 M 12 12 2 2 2 0 0 a b i i i * M M 12 12 M 12* 12 2 2 2 The matrix equation is 0 if: a p b q i * M 12 q 2 i p M 12 12 2 * 12 Eigen-states: p p q q The basics Eigen-states: p p q q p Bs0 q Bs0 Note: The Hamiltonian describes the quantum mechanical system. Only for the eigen-states of the Hamiltonian Mass and Decay time have meaning B1 p Bs0 q Bs0 B2 p Bs0 q Bs0 B1,2 (t ) B1,2 e p q 1 2 i i M1,2 1,2 t 2 2 Note 1: No particle and antiparticle. Two different masses and decay times Note 2: What is the meaning of time (t) at this point. This is a composite system, one particle contains two states. The time is calculated in the rest frame of the particle… On the next slide it becomes obvious. The basics Bs0 t 0 s t B p B q Bs0 e p B q B 0 s B1 B2 2p 0 s B1 B2 2q PB B 0 s 0 s t P B B 0 s 0 s f t e 2 e 0 s i i M1 1 t 2 e e i i M 2 2 t 2 p Bs0 q Bs0 e p B q B 0 s 0 s 2q B B t t i i M 2 2 t 2 2p 0 s 1 t i i M1 1 t 2 0 s B B t 0 s 2 t e 0 s 1 2 t 2 2 2 f t f t Bs0 f t f t 2 q f t p f e q 0 Bs p p 0 Bs f t Bs0 q i i M1 1 t 2 2 cos M 2 M1 t Note: So time (t) is just the decay time of the measured particle. e 2 i i M 2 2 t 2