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. 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.

5 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 .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.

6 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. 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.

7 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. 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.

8 This can be cross-checked verifying the commutationrelationships [Jk,W] = 0. 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.

9 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).

10 In this formula we calledx=r/rBandrBis the Bohr radius. The integrationoverxcan be easily performed exploiting the formulas in Appendix. Theremaining expectation value can be calculated in a similar Exercise s consider a tridimensional isotropic oscillator. Determine the degeneracy of the first excited level. Assume that the particle is charged and placed into a uniform electricfield of intensityE0. Evaluate the first non vanishing perturbativecontribution to the energies of the first excited :Question AThe degeneracy of the levels can be studied in various ways (see, forexample, Konishi-Paffuti, ).The simplest one is to observe that, mathematically, the problem is equiv-alent to a system of three identical oscillators. The energy levels are thusgiven byEn1n2n3=h- (3/2 +n1+n2+n3) =h- (3/2 +N)and for a given energy level the degeneracy is provided by the set ofn1, n2,andn3such thatN=n1+n2+n3 ForN= 1 it is readily find2that the level is three-fold possibility is to point out that then-th energy eigenstates containthe angular momentum eigenstates up tol=nand furthermore the givenlevel has a definite parity.