Tentamen i Kvantfysik I

Transcription

1 Karlstads Universitet Fysik Lösningar Tentamen i Kvantfysik I [ VT 7, FYGB7] Datum: Tid: 4 9 Lärare: Jürgen Fuchs Tel: Total poäng: 5 Godkänd / 3: 5 Väl godkänd: 375 4: 335 5: 4 Tentan består av delar som inlämnas separat: Del : p Del : 4 p Hjälpmedel: Del & : Ordbok engelska svenska Del efter del har inlämnats dessutom: Ett hskrivet A4 ark med valfritt innehåll skrivet på ena sidan, ej maskinskriven eller maskinkopierad inlämnas tillsammans med tentan Physics Hbook Mathematics Hbook FYGB7 Tentamen

2 Problem Operators 7 p a One possibility: show it separately for each Cartesian component p For p x = i the property follows by noting that applying the derivative x to the left amounts to partial integration using that the wave functions vanish at ±, up to a minus sign A second minus sign comes from i = i b One possibility: First write p zp z = Ψ zp z Ψ = zp z Ψ Ψ = Ψ zp z Ψ = Ψ p z z Ψ = Ψ p z z Ψ, then use the commutator of p z z c 3 p The product of two linear operators is linear: ABa φ +b ψ = ABa φ +b ψ = AaB φ +bb ψ = aab φ +bab ψ = aab φ +bab ψ The same holds for BA, thus for [A,B]=AB BA An analogous calculation shows that any power A n is linear, hence any suitably well behaved function FA If A B are hermitian, then [A,B] = AB BA = B A A B = BA AB = [A,B] A n = A n, hence FA = FA In particular, for FA=e ia one gets e ia = e ia = e ia = e ia, so e ia is unitary Further, e ib is unitary as well, so that e ia e ib = e ib e ia = e ib e ia = e ia e ib FYGB7 Tentamen

3 Problem 3 One-dimensional problems 8 p a 4 p The general solution of the Schrödinger equation in the region x b is ψx = A sinkx+b coskx with constants A,B k= me/ Outside this region the wave function must be zero Continuity of ψ at x = thus implies B =, then continuity of ψ at x = b implies that k = nπ/b with n=,, In terms of the energy E this means that E = E n = π mb n with n=,, The normalized eigenfunction to the energy eigenvalue E n is nπ ψ n = A sin b x with A = b b The function ψ can be exped in terms of the energy eigenfunctions as p b ψx = c n ψ n x with c n = ψnxψxdx n= The coefficients c n can be computed easily via partial integration The probability that a measurement of the energy yields the ground state energy E is P = N c = N cb 5/ 96 = π 3 π 6 Here the factor N, which is present because ψ is not normalized, is given by N = b ψx dx = c b 5 3 c The given range of energies only contains the eigenvalues E E, so that p the probability is P = N c + c But explicit calculation gives c =, hence P coincides with P FYGB7 Tentamen

4 Problem 4 One-dimensional problems 8 p a p Scattering states for E>, bound states for E< b For now we leave the value of E> arbitrary With the notations of of the book, the solution of the time-independent Schrödinger equation is Ae ikx +Be ikx for x< a, ψx = F e iαx +Ge iαx for x <a, Ce ikx for x>a 4 p with k= me/ α= mv +E/ To get the physical solution we must impose continuity conditions: ψ continuous at x= a = Ae ika +Be ika = F e iαa +Ge iαa, dψ dx continuous at x= a = kae ika Be ika = αf e iαa Ge iαa, ψ continuous at x=a = Ce ika = F e iαa +Ge iαa, dψ dx continuous at x=a = kceika = αf e iαa Ge iαa Adding subtracting the last two equalities yields F = e iαa+ika + k α C G = eiαa+ika k α Thus together with the first two equalities we get Ae ika +Be ika = e iαa+ika + k α ie C + eiαa+ika k α = e ika cosαa i k α sinαa C C C kae ika Be ika = [ α e iαa+ika + k α C eiαa+ika k ] α C = αe ika i sinαa+ k α C, cosαa Ae ika Be ika = Ce ika i α k sinαa+cosαa Adding subtracting the so obtained equalities yields [ Ae ika = Ce ika cosαa i k α + α ] k sinαa Be ika = i k Ceika α α k sinαa Inserting now finally E=3V, ie α=k, these formulas simplify to A = Ce ika [cos4ka 5i 4 sin4ka] B = 3i C sin4ka FYGB7 Tentamen

5 c 3 p For general values of E>, T = A C = cos αa+ k 4 α + α sin αa k = + k 4 α α sin αa = + k 4k α k α sin αa R = A k B = T α α sin αa = +4k α k α sin αa k For E=3V this reduces to T = sin 4ka R = sin 4ka FYGB7 Tentamen

6 Problem 5 Three-dimensional problems 8 p a One has 3 p ψ r d 3 r = 4π 6π A r 3 e r/aµ dr = 4π A 3!/a µ 4 R 3 = 9π a 4 µ A Thus in order that ψ is normalized to one needs A = 3πa µ b For all energy eigenstates of the hydrogen atom the radial part of the wave p function is a polynomial times an exponential in r Thus a wave function that involves a factor r cannot be an eigenstate c The normalized ground state wave function of the hydrogen atom is ψ,, r = π / aµ 3/ e r/aµ Thus the probability is taking A to be real positive, ie A=3πa µ P,, = ψ,, ψ = ψ,, rψ rd3 r R 3 3 p = 4π π / a 3/ µ /3πa µ r 5/ e r/aµ dr = 5 3π 6 FYGB7 Tentamen

7 Problem 6 Angular momentum spin 9 p a The operator A is realized as the -matrix 5 p A = σ x +σ y = i +i Thus the eigenvalue equation reads x i x iy a y = +i y = +ix The corresponding secular equation is a = i +i a = a is solved by a = a ± = ± The eigenvectors χ A ± are then determined by solving the eigenvalue equation for a=a ± Up to normalization chosen here such that the vectors are normalized to, with some arbitrary phase choice they are χ A ± = ±e iπ/4 b The probability is given by the square of the absolute value of the scalar product between χ A + the eigenvector to the eigenvalue of S z, which p is χ z + = Thus P = χ z + χ A + = e iπ/4 = Another way to obtain this result is to argue that as can eg be checked via using step operators in an eigenstate to A the expectation value of S z must be zero c The expectation value is p S x = χ A + S x χ A + = e iπ/4 = e iπ/4 FYGB7 Tentamen