Physics 443, Solutions to PS 3 1

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1 Physics 443 Solutions to PS 3. Griffiths 3.3. It is easiest to first write the hamiltonian matrix. By inspection ( ) H ɛ We find the eigenvalues λ by setting det(h λi) Then λ ± ±ɛ. Let ( ) a v b be an eigenvector. Then Or in the representation. Griffiths 3.4. H v ± λ ± v ± ( ) ( ) a ɛ b ( ) a λ ± b ( v ± ± ) v ± ( ± ) Since the set of orthonormal vectors e n is complete any state can be written as a linear combination of those vectors. In paricular the state α can be written as α m a m e m () Then e n α m a m e n e m m a m δ mn a n () where we have used the orhthonormality of the eigenvectors. Finally substitute e n α a n from Equation 3 into Equation. α n e n e n α (3) Courtesy Shaffique Adam

2 So ˆQ α n ˆQ e n e n α q n e n e n α n ( ) q n e n e n α n 3. Griffiths 3.3. We are given that h ˆQ h ˆQh h for all states h. If we define h f g then f ˆQ f f ˆQ g g ˆQ f g ˆQ g ˆQf f ˆQf g ˆQg f ˆQg g By hypothesis f ˆQ f ˆQf f and similarly for g so Equation 3 reduces to f ˆQ g g ˆQ f ˆQf g ˆQg f (5) (4) Alternatively if we let h f i g we find that i f ˆQ g i g ˆQ f i ˆQf g i ˆQg f or f ˆQ g g ˆQ f ˆQf g ˆQg f (6) (7) The sum of equations (4) and (6) gives f ˆQ g ˆQg f and the difference gives g ˆQ f ˆQg f 4. Griffiths 3.3. We have: d dt xp ī p [ V (x) xp] h m ī p [ xp] [V (x) xp] h m ī pp [ x]p x[v (x) p] h m ī p i h h m x( hi V x ) T x V. x (8)

3 For a stationary state we see that T x x V which is the Virial Theorem. For the Harmonic Oscillator V (x) mω x / using the Virial Theorem we see that T mω x / V (x). 5. Griffiths We begin by using that a ± m (P ± imωx) P x m (a a ) i mω (a a ) a n i n hω n a n i n hω n n x n i (n a a n ) mω n p n i mω (δ n ni n hω δ nn (i n hω)) h mω (δ n n n δnn n ) m (n a a n ) hmω i (δ n n n δnn n ). We can then write these out in matrix notation as... h 3... x mω hmω 3... p i (9) 3

4 And you can verify that p /m(mω /)x is diagonal with the matrix element given by hω(n /). 6. Griffiths (a) By inspection the eignenvalues of H are E hω E E 3 hω and the eigenvectors are e e e 3 To find the eigenvaluesγ we set γ λ det(a γi) λ γ λ γ γ (λ γ) λ (λ γ) (b) Then the normalized eigenvalues of A are A λ A λ A 3 λ and the eigenvectors are a a a 3 The same strategy gives the eigenvalues B are B µ B µ B 3 µ and the eigenvectors are b 3 b b 3 H S H S hω ( c c ) hω( c ( c )) hω( c ) 4 c c

5 A λ ( c c ) λ(c c c c ) B µ ( c c ) µ( c c c 3c ) c c c c (c) S(t) c e iωt c e iωt e iωt The probability of measuring energies E E and E 3 is c c and respectively independent of time. The probability of measuring A i is a i S(t) a S(t) c ( ) c e iωt c e iωt a S(t) ( ) a 3 S(t) ( ) c c c c c c e iωt c e iωt The probability of measuring B i is b i S(t) b S(t) ( ) b S(t) ( ) b 3 S(t) ( ) c c c c c c c c e iωt e iωt c c e iωt e iωt c 5

6 7. Charmonium. The Schrodinger equation for charmonium is h m ψ αrψ Eψ Define u(r) rψ and for spherically syymetric wave functions the Schrodinger equation reduces to h d u αru Eu m dr Let r l z and E ɛe where z and ɛ are dimensionless and the Schrodinger equation becomes or h d u ml dz αl zu ɛe u d u dz αml h l zu ɛ ml E h u Set l ( ) h /3 mα and E h ml equation looks like d u dz ( h α m zu ɛu Now let y z ɛ and we have Airy s equation d u dy yu ) /3 and our differential Since u(r) ψ(r)/rthen it must be that u() so that ψ() is finite. Therefore u(z ɛ) u(ɛ). The energy eigenvalues ɛ are the zeros of the Airy function a i. The first two zeros are.3 and 4. so E.3E and E 4.E. We have that m s c m c c E m s c m c c E () where m c is the charmed quark mass and E and E are the binding energies. The difference of the two equations yields E (m s 6

7 m s )c /.8.35GeV. We find m c c.76gev. Finally the reduced mass m m c /. Meanwhile ( h α ) /3 E m α me3 h m cc E 3 h c And ( h c ) /3 ( l m c c α (.76GeV )(.35GeV )3 (.97GeV fm) (.97GeV fm) (.76GeV )(.GeV/f m) 8. Rotations. We define x il y θ/ h. Notice that x θ ( ) x ( ) ( ) θ x 3 ( ) 3 ( ) θ x ( ) 4 ( ) θ In particular.gev/fm ) /3.38fm. R(θ) e x x x x3 3!... ( ) ( ) ( θ...! ( ) cos( θ ( ) ) sin( θ ) ( cos( θ ) sin( θ) ) sin( θ) cos( θ). ) ( ) θ 3! ( ) 3 θ... You can see that L y L y and R(θ) T R(θ) making L y Hermitian and R(θ) unitary. 7