4.2 – REFLEXIÓN FINAL 4.2.1 - CONCLUSIÓN
4.2.2 - AUTOCRÍTICA
Multiple-Choice Problems
013 qmult 00100 1 1 4 easy memory: ang. mom. commutation relations
1. The fundamental angular momentum commutation relation and a key corollary are, respectively: a) [Ji, Jj] = 0 and [J2, Ji] = Ji. b) [Ji, Jj] = Jk and [J2, Ji] = 0.
c) [Ji, Jj] = 0 and [J2, Ji] = 0. d) [Ji, Jj] = i h−εijkJk and [J2, Ji] = 0.
e) [xi, pj] = i h−δij, [xi, xj] = 0, and [pi, pj] = 0.
013 qmult 00910 1 1 3 easy memory: vector model
2. In the vector model for angular momentum of a quantum system with the standard axis for the eigenstates being the z axis, the particles in the eigenstates are thought of as having definite z-components of angular momentum mj− and definite total angular momenta of magnitudeh
pj(j + 1)h−, where j can stand for orbital, spin, or total angular momentum quantum number and mj is the z-component quantum number. Recall j can be only be integer or half-integer
and there are 2j + 1 possible values of mj given by −j, −j + 1, . . . , j − 1, j. The x-y component
of the angular momementum has magnitudeqj(j + 1) − m2
j−, but it has no definite direction.h
Rather this component can be thought of as pointing all x-y directions in simultaneous: i.e., it is in a superposition state of all direction states. Diagramatically, the momentum vectors can be represented by
a) cones with axis aligned with the x-axis. b) cones with axis aligned with the y-axis. c) cones with axis aligned with the z-axis.
d) cones with axis aligned with the x-y-axis. e) the cones of silence. 013 qmult 01000 1 1 5 easy memory: rigid rotator eigen-energies
3. For a rigid rotator the rotational eigen-energies are proportional to:
a) ℓ h−. b) ℓ2−h2. c) h−2/[ℓ(ℓ + 1)]. d) h−2/ℓ2. e) ℓ(ℓ + 1) h−2.
013 qmult 02000 1 1 1 easy memory: added ang. mom. operators
4. Does the fundamental commutation relation for angular momentum operators (i.e., [Ji, Jj] =
i h−εijkJk) apply to angular momentum operators formed by summation from angular momentum
operators applying to individual particles or to spatial and spin degrees of freedom? The answer is:
a) Yes. b) No. c) Maybe. d) All of the above. e) None of the above. 013 qmult 02100 1 4 5 easy deducto-memory: Clebsch-Gordan coefficients
5. “Let’s play Jeopardy! For $100, the answer is: The name for the coefficients used in the expansion of a total angular momentum state for 2 angular momentum degrees of freedom in terms of products of individual angular momemtum states.”
What are the , Alex?
a) Racah W coefficients b) Wigner 6j symbols c) Buck-Rogers coefficients d) Flash-Gordon coefficients e) Clebsch-Gordan coefficients
82 Chapt. 13 General Theory of Angular Momentum
013 qmult 02200 1 4 5 easy deducto-memory: Clebsch-Gordan m rule
6. “Let’s play Jeopardy! For $100, the answer is: In constructing a set of |j1j2jmi states from
a set of |j1j2m1m2i states using Clebsch-Gordan coefficients, this is a strict constraint on the
non-zero coefficients.”
What is the rule , Alex?
a) of complete overtures b) of incomplete overtures c) m = m2 1+ m22
d) m = m1− m2 e) m = m1+ m2
Full-Answer Problems
013 qfull 00090 2 5 0 moderate math: kroneckar delta, Levi-Civita
1. There are two symbols that are very useful in dealing with quantum mechanical angular momentum and in many other contexts in physics: the Kronecker delta:
δij = 1 , i = j;
0 , i 6= j; and the Levi-Civita symbol
εijk=
1 , if ijk is a cyclic permutation of 123 (3 cases); −1 , if ijk is an anticyclic permutation of 123 (3 cases); 0 , if any two indices are the same.
NOTE: Leopold Kronecker (1823–1891) was a German mathematician although born in what is now Poland. Tullio Levi-Civita (1873–1941) was an Italian mathematician: the “C” in Civita is pronounced “ch”.
a) Prove δijδik= δjk, where we are using Einstein summation here and below, of course.
b) Now the toughie. Prove
εijkεiℓm= δjℓδkm− δjmδkℓ .
HINTS: I know of no simple one or two line proof. The best I’ve ever thought of was to consider cases where jkmℓ span 3, 1, and 2 distinct values and to show that the two expressions are equal in all cases.
c) Now the cinchy one. Prove
εijkεijm = 2δkm.
d) What does εijkεijk equal? Note there is Einstein summation on all indices now.
013 qfull 00100 2 5 0 moderate thinking: angular momentum operator identities
2. Prove the following angular momentum operator identities. HINT: Recall the fundamental angular momentum commutator identity,
[Ji, Jj] = i h−εijkJk , and the definition J± ≡ Jx± iJy .
a) [Ji, J2] = 0.
b) [J2, J ±] = 0.
c) [Jz, J±] = ±h−J±.
Chapt. 13 General Theory of Angular Momentum 83 e) Jx= 1 2(J++ J−) and Jy= 1 2i(J+− J−) . f) [J+, J−] = 2 h−Jz. g) J{2x y}= ± 1 4 J 2 ++ J−2 ± {J+, J−} ,
where the upper case is for J2
x and the lower case for Jy2 and where recall that {A, B} =
AB + BA is the anticommutator of A and B. h)
J2=1
2{J+, J−} + J
2 z .
013 qfull 00200 2 3 0 mod math: diagonalization of Jxfor 3-d
Extra keywords: diagonalization of the J-x angular momentum matrix for 3-d
3. The x-component angular momentum operator matrix in a three-dimensional angular momentum space expressed in terms of the z-component orthonormal basis (i.e., the standard basis with eigenvectors |1i, |0i, and | − 1i) is:
Jx= h − √ 2 0 1 0 1 0 1 0 1 0
(Co-659) and yes the 1/√2 factor is correct. Is this matrix Hermitian? Diagonalize this matrix: i.e., solve for its eigenvalues and normalized eigenvectors (written in terms of the standard basis ket eigenvectors) or, if you prefer in column vector form. Note the solution is somewhat simpler if you solve the reduced eigen problem. Just divide both sides of the eigen equation by h−/√2 and solve for the reduced eigenvalues. The physical eigenvalues are the reduced ones times h
−/√2. Verify that the eigenvectors are orthonormal.
NOTE: Albeit some consider it a sloppy notation since kets and bras are abstract vectors and columns vectors are from a concrete representation, its concretely useful to equate them at times. In the present case, the kets equate like so
|1i = 1 0 0 , |0i = 0 1 0 , | − 1i = 0 0 1 ,
and the bras, like so
h1| = (1, 0, 0)∗, h0| = (0, 1, 0)∗, h−1| = (0, 0, 1)∗ .
013 qfull 00300 2 3 0 mod math: diagonalization of J-y for 3-d
Extra keywords: diagonalization of the J-y angular momentum matrix for 3-d
4. The y-component angular momentum operator matrix in a three-dimensional angular momentum space expressed in terms of the z-component orthonormal basis (i.e., the standard basis with eigenvectors |1i, |0i, and | − 1i) is:
Jy = h − √ 2 0 −i 0 i 0 −i 0 i 0
84 Chapt. 13 General Theory of Angular Momentum
(Co-659) and yes the 1/√2 factor is correct. Is this matrix Hermitian? Diagonalize this matrix: i.e., solve for its eigenvalues and normalized eigenvectors (written in terms of the standard basis kets) or, if you prefer in column vector form. Verify that the eigenvectors are orthonormal.
Note the solution is somewhat simpler if you solve the reduced eigen problem. Just divide both sides of the eigen equation by h−/√2 and solve for the reduced eigenvalues. The physical eigenvalues are the reduced ones times h−/√2.
NOTE: Albeit some consider it a sloppy notation since kets and bras are abstract vectors and columns vectors are from a concrete representation, its concretely useful to equate them at times. In the present case, the kets equate like so
|1i = 1 0 0 , |0i = 0 1 0 , | − 1i = 0 0 1 ,
and the bras, like so
h1| = (1, 0, 0)∗, h0| = (0, 1, 0)∗, h−1| = (0, 0, 1)∗ .
013 qfull 00400 2 3 0 mod math: angular momemtum eqn. of motion Extra keywords: (Gr-150:4.21) torque
5. Let’s consider the angular momentum equation of motion in in the context of quantum mechanics.
a) Prove that
dh~L i
dt = h~τ i ,
where ~L = ~r × ~p is the angular momentum operator and ~τ = ~r × (−∇V ) is the torque operator.
b) Then prove that
dh~L i dt = 0
for any central potential system: i.e., a system where the potential depends on radius alone. HINTS: You’ll need to use the general time evolution equation—or equation of motion or derivative of expectation value: whatever one calls it—people do seem to avoid giving it a name. Then you will need to work out a commutation relation with a cross product operator. There are two approaches. First, show what the commutation relation is component by component. But that’s for pedestrians. The second way is to use the Levi-Civita symbol with the Einstein summation rule to prove the all commutation relations simultaneously. Part (a) is most easily done using Cartesian coordinates and part (b) using spherical polar coordinates.
013 qfull 00500 2 3 0 moderate thinking: orbital angular momentum
Extra keywords: expectation values, standard deviations, quantum and classical analogs 6. Consider a spinless particle in an eigenstate |ℓ, mi of the L2 and L
z operators: ℓ is the L2
quantum number and m the Lz quantum number. The set of |ℓ, mi states are a complete
orthonormal set for angular coordinates. Recall L2|ℓ, mi = ℓ(ℓ + 1)h−2|ℓ, mi , Lz|ℓ, mi = mh−|ℓ, mi ,
L±|ℓ, mi = h−pℓ(ℓ + 1) − m(m ± 1)|ℓ, m ± 1i ,
Chapt. 13 General Theory of Angular Momentum 85 L±= Lx± iLy .
a) Solve for expectation values hLxi, and hLyi, and standard deviations ∆Lx and ∆Ly.
HINTS: You will need expressions for Lx and Ly in terms of the given operators. Also
the everything can be done by operator algebra: there is no need to bring in the spherical harmonics or particular representations of the operators.
b) Let us now see if there are classical analogs to the results in part (a). Let classical Lz= m h− ,
Lx= h−pℓ(ℓ + 1) − m2cos(φ)
and
Ly= h−pℓ(ℓ + 1) − m2sin(φ) ,
where φ is the azimuthal angle of the angular momentum vector. Note L2
x+ L2y+ L2z =
ℓ(ℓ+1). Now solve for the classical hLxi and hLyi, and the classical ∆Lxand ∆Lyassuming
(i) that φ has a random uniform distribution the range [0, 2π] and (ii) that φ = wt where ω is a constant angular frequency.
013 qfull 00600 2 5 0 moderate thinking: orb. ang. mon. commutator proofs
7. You are given the basic commutator identity [ri, pj] = i h−δij and the correspondence principle
result ~L = ~r × ~p.
a) Prove [Li, rj] = i h−εijkrk.
b) Prove [Li, pj] = i h−εijkpk.
c) Prove [Li, Lj] = i h−εijkLk. HINT: Remember the old subtract and add the same thing
trick.
d) Prove [Li, qjqj] = 0, where ~q is any of ~r, ~p, and ~L.
e) Prove [Li, Qj] = i h−εijkQk with ~Q = A~qB, where A and B are any scalar combination of
~r, ~p, and ~L: e.g., A = r2kp4mL2n. . ., where k, m, and n are integers.
f) Prove [ˆα · ~L, ~Q] = i h− ~Q × ˆα, where ˆα is a constant c-number unit vector. g) Show that d ~Q dα = i h −[ˆα · ~L, ~Q]
is the differential equation for a right-hand-rule rotation by α of operator ~Q about the axis in the direction ˆα. HINT: I’m not looking for mathematical rigor—but if you can do that it’s OK.
h) Show that the solution of
d ~Q dα = i h −[ˆα · ~L, ~Q] is ~ Q = ei~L·~α/− ~Qh 0e−i~L·~α/−h ,
where ~α = α ˆα is a general angle in vector form and ~Q0 is the inital operator ~Q.
013 qfull 02000 1 5 0 easy thinking: ang. mom. fundamental commutation Extra keywords: for addition. Reference Ba-332.
86 Chapt. 13 General Theory of Angular Momentum
8. The fundamental commutation relation of angular momentum can be generalized for multiple degrees of freedom. The degrees of freedom could be the angular momenta of multiple particles or the spatial and spin angular momenta of a particle or combinations thereof. Say we have degrees of freedom f and g, then the relation is
[Jf i, Jgj] = δf gi h−εijkJf k .
We see that component operators refering to different degrees of freedom commute: this is true even in the case of indistinguishable particles. The total angular momentum operator ~J for a set of degrees of freedom with individual angular momentum operators ~Jf is, by the correspondence
principle, ~ J =X f ~ Jf .
a) Prove that the fundamental commutation relation holds for the components of ~J: i.e., prove [Ji, Jj] = i h−εijkJk .
What does this result imply for summed angular momenta? b) Now let ~J = ~J1+ ~J2. Prove
J±= J1±+ J2± .
c) Prove
J2= J12+ J22+ J1+J2−+ J1−J2++ 2J1zJ2z .
013 qfull 02100 3 5 0 hard thinking: Clebsch-Gordan ell plus 1/2 Extra keywords: See Ba-341 and CDL-1020
9. One special case of great interest for which general formulae can be found for all Clebsch-Gordan coefficients is that of a general angular momentum added to a spin 1/2 angular momentum. Let the original angular momentum operators be labeled J2
1, J1z, J22, and J2z: the corresponding
eigenvalues are j1(j1+ 1), m1, (1/2)(1/2 + 1), and ±1/2. The set of product states of the
original operators is {|j1, 1/2, m1, m2 = ±1/2i}. The summed operators are J2 and Jz:
the corresponding eigenvalues are j and m. The set of eigenstates of ~J2
1, J22, J2, and Jz is
{|j1, 1/2, j = j1± 1/2, mi}. The expression for the Clebsch-Gordan coefficients is
hj1, 1/2, m1, m2= ±1/2|j1, 1/2, j = j1± 1/2, mi .
a) For a given j1 what are the possible j values?
b) Consider the trivial subspace for j1= 0. What are all the Clebsch-Gordan coefficients for
this subspace.
c) Now consider the general subspace for j1≥ 1/2. First find the expression for the summed
state with the largest m value. HINT: Recall
hj1, j2, m1, m2|Jz|j1, j2, j, mi = hj1, j2, m1, m2|(J1z+ J2z)|j1, j2, j, mi ,
and so
mhj1, j2, m1, m2|j1, j2, j, mi = (m1+ m2)hj1, j2, m1, m2|j1, j2, j, mi .
Thus, the Clebsch-Gordan coefficient is zero unless m = m1+ m2.
d) Determine the expression for
Chapt. 13 General Theory of Angular Momentum 87 e) Now show that general expression for Clebsch-Gordan coefficient
hj1, 1/2, m1= m − 1/2, m2= 1/2|j1, 1/2, j = j1+ 1/2, mi is given by hj1, 1/2, m1= m − 1/2, m2= 1/2|j1, 1/2, j = j1+ 1/2, mi = s j1(j1+ 1) − (m + 1/2)(m − 1/2) (j1+ 1/2)(j1+ 3/2) − (m + 1)m s j1(j1+ 1) − (m + 3/2)(m + 1/2) (j1+ 1/2)(j1+ 3/2) − (m + 2)(m + 1) . . . s j1(j1+ 1) − j1(j1− 1) (j1+ 1/2)(j1+ 3/2) − (j1+ 1/2)(j1− 1/2) .
HINT: What is mostly needed is a word argument. f) Now show that
s j1(j1+ 1) − (m + 1/2)(m − 1/2) (j1+ 1/2)(j1+ 3/2) − (m + 1)m s j1(j1+ 1) − (m + 3/2)(m + 1/2) (j1+ 1/2)(j1+ 3/2) − (m + 2)(m + 1) . . . s j1(j1+ 1) − j1(j1− 1) (j1+ 1/2)(j1+ 3/2) − (j1+ 1/2)(j1− 1/2) = s j1+ m + 1/2 2j1+ 1 . HINTS: Simplify s j1(j1+ 1) − (m + 1/2)(m − 1/2) (j1+ 1/2)(j1+ 3/2) − (m + 1)m
by dividing top and bottom by a common factor. You might try (Aj1+ Bm + . . .)(Cj1+
Dm + . . .) factorizations of the top and bottom.
g) Now show that the general expressions for the Clebsch-Gordan coefficients are
hj1, 1/2, m1= m ∓ 1/2, m2= ±1/2|j1, 1/2, j = j1+ 1/2, mi = s j1± m + 1/2 2j1+ 1 hj1, 1/2, m1= m ∓ 1/2, m2= ±1/2|j1, 1/2, j = j1− 1/2, mi = ∓ s j1∓ m + 1/2 2j1+ 1 .
HINTS: Use the parts (e) and (f) answers and the normalization and orthogonality conditions.
h) The operator ~J1· ~S turns up in the spin-orbit interaction where ~J1 = ~L. Show that the