Solved problems in quantum mechanics - Unife

1 Solved problems in quantummechanicsMauro Moretti and Andrea Zanzi AbstractThis is a collection of Solved problems in quantum exercises have been given to the students during the past Email: E-mail: are kindly requested to report typos and mistakes to the authors1 Contents1 Recommended books and resources32 February 1, Exercise .. Exercise .. Exercise .. Exercise .. 103 Exercise , February 22, Exercise .. 154 Exercise , June 26, Exercise .. Exercise .. Exercise .. Exercise 2 .. Exercise 3 .. Exercise 3 .. Exercise 4 .. Exercise 4 .. 3621 Recommended books and resourcesLectures closely follow: Cohen-Tannoudji, Diu, Laloe; quantum mechanics Sakurai; Modern quantum mechanics Schiff; quantum MechanicsOther useful references: quantum mechanics - a new introduction , K.

2 Konishi and G. Paf-futi. Oxford Univ. Press (2009).An extremely useful textbook. Verystrongly recommended. It contains detailed explanations and also somechapters that are not easy to find in other books. Remarkably, a veryinteresting collection of problems is included. The solutions to all theexercises are given in a CD but they can be found also in the websiteof Kenichi Konishi: Lectures on quantum mechanics , 2nd edition , S. Weinberg. Cam-bridge Univ. excellent book written by the famous Nobellaureate. This book can be considered the first of a set of books. In-deed, S. Weinberg wrote excellent books about quantum field theory,gravitation, cosmology and these lectures on quantum mechanics arebasically the first step into the particle-sector of his books. I fondamenti concettuali e le implicazioni epistemologiche della mec-canica quantistica , Ghirardi in Filosofia della fisica.

3 EdizioneBruno Mondadori, Milano, very interesting paper about theconceptual foundations of quantum mechanics written by one of themasters of the subject. Very strongly recommended. Esercizi di meccanica quantistica elementare , C. Rossetti. e Bella - Torino. All the exercises are Solved step-by-step. Itis a very useful collection of problems . Istituzioni di fisica teorica - 2nd edition , C. A classic textbook on the February 1, Exercise shall label as|n,l,m the eigenstates of the hydrogen atom. Let be = n= 3,l= 2,m= 2|xy|n= 3,l= 0,m= 0 .(1)Compute, as a function of , n= 3,l= 2,m|Oj|n = 3,l = 0,m = 0 whereOj=xy, xz, yz, xx, yy, :In this problem we must evaluate matrix elements for various will not follow the path of the direct evaluation. Indeed, we will exploitthe Wigner-Eckart theorem over and over of all, let us analyze the transformation properties of the operatorsunder the operators are components of a rank two cartesian tensor.

4 Sincethe Wigner Eckart theorem applies to a spherical tensor, let s first recall howa rank two cartesian tensor is mapped into a spherical tensor. The cartesiantensor is symmetric therefore it decompose into a rank 0 spherical tensorT0= (x2+y2+z2)(2)and a rank 2 spherical tensorT2= x2+ 2ixy y2 2z(x+iy)1 6( 2x2 2y2+ 4z2)+2z(x iy)x2 2ixy y2 (3)Notice that the above definitions are unique up to two overall irrelevant con-stants:Tr krTr, whereris the rank. The Wigner-Eckart theorem providesinformation about the ratio of matrix elements which is insensitive to to the Wigner-Eckart theorem, we shall have (the Clebsch-Gordan is 1 because the composition of angular momenta is trivial) n= 3,l= 2,m|T2q|n = 3,l = 0,m = 0 = mq(4)4 n= 3,l= 2,m|T0|n = 3,l = 0,m = 0 = 0(5)From eqn. 3xy=i4(T2 2 T22)(6)and from eqns.

5 1, 4 and 6 we have = 4i .Indeed we can write =<322|xy|300>=<322|[i4(T2 2 T22)]|300>=<322|[i4( T22)]|300>= i4 .where the compact notation|n0,l0,m0 |n=n0,l=l0,m=m0 has been eqns. , 5 and 6 we have n= 3,l= 2,m= 2|xy|n = 3,l = 0,m = 0 = and all remaining matrix elements ofxyvanish. Indeed, we can write<32 2|xy|300>=i4<32 2|T2 2|300>=i4 = .Analogouslyxz=14(T2 1 T21)which implies<321|xz|300>=<321|14(T2 1 T21)|300>=<321|14( T21)|300>= 4= i and<32 1|xz|300>=<32 1|14(T2 1 T21)|300>=<32 1|14(T2 1)|300>= 4=i with all other matrix elements (T2 1+T21)5implies n= 3,l= 2,m= 1|yz|n = 3,l = 0,m = 0 = 2 with all other matrix elements ofyzvanishing. Indeed we have 32 1|yz|300 =<32 1|i2(T2 1+T21)|300>=<32 1|i2(T2 1)|300>=i2 = 2 and 321|yz|300 =<321|i2(T2 1+T21)|300>=<321|i2(T21)|300>=i2 = 2.

6 Z2=1 6T20+13T0which implies n= 3,l= 2,m= 0|z2|n = 3,l = 0,m = 0 =<320|1 6T20|300>=4i 6 with all other matrix elements ofz2vanishing;x2= 12 6T20+14(T22+T2 2) +13T00which implies n= 3,l= 2,m= 2|x2|n = 3,l = 0,m = 0 =i n= 3,l= 2,m= 0|x2|n = 3,l = 0,m = 0 = i 23 with all other matrix elements ofx2vanishing;y2= 12 6T20 14(T22+T2 2) 12T00which implies n= 3,l= 2,m= 2|y2|n = 3,l = 0,m = 0 = i n= 3,l= 2,m= 0|y2|n = 3,l = 0,m = 0 = i 23 6with all other matrix elements notice that the operators are the sum of two spherical tensors,namelyT0andT2. Consequently, for generic matrix elements, we would needthe evaluation of two distinct non zero matrix elements to use the Wigner-Eckart theorem to evaluate the remaining ones. In the present case, however,one of the two tensors (namelyT0) always vanishes and one matrix elementis Exercise shall label as|n,l,m the eigenstates of the hydrogen atom.

7 Compute thematrix elementsajk= n= 2,l= 1|rjpk|n = 3,l = 0,m = 0 ,whererjandpkare thej-th component of the position operator and thek-thcomponent of the momentum operator, :rjandpkare both parity-odd. Consequently, the operatorrjpkis parity-even and, therefore, it must connect states with the same parity. Since theparity of the wave functions is ( 1)l, all the matrix elements are Exercise hydrogen atom is subjected to a perturbation WW= S rEvaluate if and how the degeneracy of then= 2level is :We will neglect the fine-structure splitting. The degeneracy is 8: we havea degeneracyn2= 4 without spin and then we take into account the twopossible spin states (up and down) in the basis|L2,S2,Lz,Sz .Our intention is to use time-independent perturbation theory for the de-generate case.

8 We must diagonalize the perturbation matrix (it is an 8 8matrix).The first step is to evaluate the matrix elements and, as usual, we im-pose the selection rules coming from parity and Wigner-Eckart theorem. Letus start with parity: the perturbation is parity-odd (it is a pseudo-scalar7operator). Hence the perturbation must connect states with opposite par-ity. In particular the matrix elements of the form<2s|W|2s >and<2p|W|2p >are all vanishing. We are left with<2s|W|2p >expectation diagonalize the matrix, it is better to change the basis. Let us furtherelaborate this point. TheWoperator is ascalarproduct of twovectoropera-tors, hence is ascalarunder rotations and it commutes with thetotalangularmomentum operators. This can be cross-checked verifying the commutationrelationships [Jk,W] = 0.

9 It is therefore convenient to move to theJ,l,s,Jzbasis. From the above consideration only matrix elements with J= 0, Jz= 0 and l= 1 (parity) are non the above constraints the only non vanishing matrix elements are j= 1/2,l= 0,s= 1/2,jz= 1/2|W|j= 1/2,l= 1,s= 1/2,jz= 1/2 = and the hermitian conjugates. Taking into account the Wigner-Eckart theo-rem we have += = since the operator is a scalar ( the Clebsch-Gordan coefficient is trivial).We infer that the fourJ= 3/2 states don t receive any contribution. TheJ= 1/2,jz= 1/2 subspaces are invariant subspaces for the matrix element restricted to these subspaces are (as discussed we havetwo identical such matrices, one forjz= 1/2 and one forjz= 1/2)WJ=1/2,jz= 1/2=(0 0)with eigenvalues (energy shift) E= | |and (four) eigenvectors given by1 2(|j= 1/2,l= 1,jz= 1/2 |j= 1/2,l= 0,jz= 1/2 )and1 2(|j= 1/2,l= 1,jz= 1/2 |j= 1/2,l= 0,jz= 1/2 )8 Let us evaluate.

10 To perform the calculation we move back to thel,s,lz,szbasis exploiting the Clebsch-Gordan coefficients. Indeed, we canwrite|l= 0,s= 1/2,J= 1/2,Jz= 1/2>=|l= 0,s= 1/2,lz= 0,sz= 1/2>and|l= 1,s= 1/2,J= 1/2,Jz= 1/2>= 2/3|lz= 1,sz= 1/2> 1/3|lz= 0,sz= 1/2> .Consequently, to evaluate =< l= 0,s= 1/2,J= 1/2,Jz= 1/2|W|l= 1,s= 1/2,J= 1/2,Jz= 1/2>we have to calculate two expectation values. Let us consider, for example,the direct evaluation of<2s+|W|2p1 >. We have<2s+|W|2p1 >=<2s+| S r|2p1 >=<2s+| [Sxx+Syy+Szz]|2p1 >==<2s+| [S++S 2x+S+ S 2iy+Szz]|2p1 >==<2s+| [S+2x+S+2iy+Szz]|2p1 >==<2s+| [x2+y2i]|2p1+>where in the last step we exploited the factorization of the wave function andthe orthogonality of up and down states. Now only the orbital part remainsand we write<2s| [x2+y2i]|2p1>= R d (Y00) [cos sin 2+sin sin 2i]Y11== R d d sin3 1 4 38 12= 2 R3 3/8,whereRis the radial part of the integral and it is given byR= 0drr2R 2srR2p=rB 0dxx4e x14 12(2 x).