Quantum numbers and Angular Momentum Algebra

33 Pages·2017·0.43 MB·English

Quantum numbers and Angular
Momentum Algebra
Morten Hjorth-Jensen
National Superconducting Cyclotron Laboratory and Department of Physics and Astronomy, Michigan State University, East Lansing, MI 48824, USA
Jun 27, 2017
Quantum numbers
Outline.
• Discussionofsingle-particleandtwo-particlequantumnumbers, uncoupled
and coupled schemes
• Discussion of angular momentum recouplings and the Wigner-Eckart theo-
rem
• Applications to specific operators like the nuclear two-body tensor force
Forquantumnumbers,chapter1onangularmomentumandchapter5ofSuhonen
and chapters 5, 12 and 13 of Alex Brown. For a discussion of isospin, see for
example Alex Brown’s lecture notes chapter 12, 13 and 19.
Motivation
When solving the Hartree-Fock project using a nucleon-nucleon interaction in
an uncoupled basis (m-scheme), we found a high level of degeneracy. One sees
clear from the table here that we have a degeneracy in the angular momentum
j, resulting in 2j+1 states with the same energy. This reflects the rotational
symmetry and spin symmetry of the nuclear forces.
(cid:13)c 2013-2017,MortenHjorth-Jensen. ReleasedunderCCAttribution-NonCommercial4.0
license
Quantum numbers Energy [MeV]
0sπ -40.4602
1/2
0sπ -40.4602
1/2
0sν -40.6426
1/2
0sν -40.6426
1/2
0pπ -6.7133
1/2
0pπ -6.7133
1/2
0pν -6.8403
1/2
0pν -6.8403
1/2
0pπ -11.5886
3/2
0pπ -11.5886
3/2
0pπ -11.5886
3/2
0pπ -11.5886
3/2
0pν -11.7201
3/2
0pν -11.7201
3/2
0pν -11.7201
3/2
0pν -11.7201
3/2
We observe that with increasing value of j the degeneracy increases. For j =3/2
we end up diagonalizing the same matrix four times. With increasing value of j,
it is rather obvious that our insistence on using an uncoupled scheme (or just
m-scheme) will lead to unnecessary labor from our side (or more precisely, for
the computer). The obvious question we should pose ourselves then is whether
we can use the underlying symmetries of the nuclear forces in order to reduce
our efforts.
Single-particle and two-particle quantum numbers
In order to understand the basics of the nucleon-nucleon interaction and the
pertaining symmetries, we need to define the relevant quantum numbers and
how we build up a single-particle state and a two-body state, and obviously our
final holy grail, a many-boyd state.
• For the single-particle states, due to the fact that we have the spin-orbit
force,thequantumnumbersfortheprojectionoforbitalmomentuml, that
is m , and for spin s, that is m , are no longer so-called good quantum
l s
numbers. The total angular momentum j and its projection m are then
j
so-called good quantum numbers.
• This means that the operator Jˆ2 does not commute with Lˆ or Sˆ .
z z
• We also start normally with single-particle state functions defined using
say the harmonic oscillator. For these functions, we have no explicit
dependence on j. How can we introduce single-particle wave functions
which have j and its projection m as quantum numbers?
j
2
Single-particle and two-particle quantum numbers, brief re-
view on angular momenta etc
We have that the operators for the orbital momentum are given by
∂ ∂
L =−i(cid:126)(y −z )=yp −zp ,
x ∂z ∂y z y
∂ ∂
L =−i(cid:126)(z −x )=zp −xp ,
y ∂x ∂z x z
∂ ∂
L =−i(cid:126)(x −y )=xp −yp .
z ∂y ∂x y x
Single-particle and two-particle quantum numbers, brief re-
view on angular momenta etc
Since we have a spin orbit force which is strong, it is easy to show that the
total angular momentum operator
Jˆ=Lˆ+Sˆ
does not commute with Lˆ and Sˆ . To see this, we calculate for example
z z
[Lˆ ,Jˆ2] = [Lˆ ,(Lˆ+Sˆ)2] (1)
z z
= [Lˆ ,Lˆ2+Sˆ2+2LˆSˆ]
z
= [Lˆ ,LˆSˆ]=[Lˆ ,Lˆ Sˆ +Lˆ Sˆ +Lˆ Sˆ ]6=0,
z z x x y y z z
since we have that [Lˆ ,Lˆ ]=i(cid:126)Lˆ and [Lˆ ,Lˆ ]=i(cid:126)Lˆ .
z x y z y x
Single-particle and two-particle quantum numbers, brief re-
view on angular momenta etc
We have also
|Jˆ|=(cid:126)pJ(J +1),
with the the following degeneracy
M =−J,−J +1,...,J −1,J.
J
With a given value of L and S we can then determine the possible values of J
by studying the z component of Jˆ. It is given by
Jˆ =Lˆ +Sˆ .
z z z
The operators Lˆ and Sˆ have the quantum numbers L =M (cid:126) and S =M (cid:126),
z z z L z S
respectively, meaning that
M (cid:126)=M (cid:126)+M (cid:126),
J L S
or
M =M +M .
J L S
Since the max value of M is L and for M is S we obtain
L S
(M ) =L+S.
J maks
3
Single-particle and two-particle quantum numbers, brief re-
view on angular momenta etc
For nucleons we have that the maximum value of M =m =1/2, yielding
S s
1
(m ) =l+ .
j max 2
Using this and the fact that the maximum value of M =m is j we have
J j
1 1 3 5
j =l+ ,l− ,l− ,l− ,...
2 2 2 2
To decide where this series terminates, we use the vector inequality
(cid:12) (cid:12)
|Lˆ+Sˆ|≥(cid:12)|Lˆ|−|Sˆ|(cid:12).
(cid:12) (cid:12)
Single-particle and two-particle quantum numbers, brief re-
view on angular momenta etc
Using Jˆ=Lˆ+Sˆ we get
|Jˆ|≥|Lˆ|−|Sˆ|,
or
|Jˆ|=(cid:126)pJ(J +1)≥|(cid:126)pL(L+1)−(cid:126)pS(S+1)|.
Single-particle and two-particle quantum numbers, brief re-
view on angular momenta etc
If we limit ourselves to nucleons only with s=1/2 we find that
r
|Jˆ|=(cid:126)pj(j+1)≥|(cid:126)pl(l+1)−(cid:126) 1(1 +1)|.
2 2
It is then easy to show that for nucleons there are only two possible values of j
which satisfy the inequality, namely
1 1
j =l+ orj =l− ,
2 2
and with l=0 we get
1
j = .
2
Single-particle and two-particle quantum numbers, brief re-
view on angular momenta etc
Letusstudysomeselectedexamples. Weneedalsotokeepinmindthatparity
is conserved. The strong and electromagnetic Hamiltonians conserve parity.
Thus the eigenstates can be broken down into two classes of states labeled by
4
their parity π =+1 or π =−1. The nuclear interactions do not mix states with
different parity.
For nuclear structure the total parity originates from the intrinsic parity of
the nucleon which is π =+1 and the parities associated with the orbital
intrinsic
angular momenta π =(−1)l . The total parity is the product over all nucleons
l
π =Qiπintrinsic(i)πl(i)=Qi(−1)li
The basis states we deal with are constructed so that they conserve parity
and have thus a definite parity.
Note that we do have parity violating processes, more on this later although
our focus will be mainly on non-parity viloating processes
Single-particle and two-particle quantum numbers
Consider now the single-particle orbits of the 1s0d shell. For a 0d state we
have the quantum numbers l = 2, m = −2,−1,0,1,2, s+1/2, m = ±1/2,
l s
n = 0 (the number of nodes of the wave function). This means that we have
positive parity and
3 3 1 1 3
j = =l−s m =− ,− , , .
2 j 2 2 2 2
and
5 5 3 1 1 3 5
j = =l+s m =− ,− ,− , , , .
2 j 2 2 2 2 2 2
Single-particle and two-particle quantum numbers
Oursingle-particlewavefunctions,ifweusetheharmonicoscillator,dohowever
not contain the quantum numbers j and m . Normally what we have is an
j
eigenfunction for the one-body problem defined as
φ (r,θ,φ)=R (r)Y (θ,φ)ξ ,
nlmlsms nl lml sms
wherewehaveusedsphericalcoordinates(withasphericallysymmetricpotential)
and the spherical harmonics
s
(2l+1)(l−m )!
Y (θ,φ)=P(θ)F(φ)= l Pml(cos(θ))exp(im φ),
lml 4π(l+m )! l l
l
with Pml being the so-called associated Legendre polynomials.
l
Single-particle and two-particle quantum numbers
Examples are
r
1
Y = ,
00 4π
for l=m =0,
l
r
3
Y = cos(θ),
10 4π
5
for l=1 and m =0,
l
r
3
Y = sin(θ)exp(±iφ),
1±1 8π
for l=1 and m =±1,
l
r
5
Y = (3cos2(θ)−1)
20 16π
for l=2 and m =0 etc.
l
Single-particle and two-particle quantum numbers
How can we get a function in terms of j and m ? Define now
j
φ (r,θ,φ)=R (r)Y (θ,φ)ξ ,
nlmlsms nl lml sms
and
ψ (r,θ,φ),
njmj;lmlsms
as the state with quantum numbers jm . Operating with
j
ˆj2 =(ˆl+sˆ)2 =ˆl2+sˆ2+2ˆl sˆ +ˆl sˆ +ˆl sˆ ,
z z + − − +
on the latter state we will obtain admixtures from possible φ (r,θ,φ)
nlmlsms
states.
Single-particle and two-particle quantum numbers
To see this, we consider the following example and fix
3 3
j = =l−s m = .
2 j 2
and
5 3
j = =l+s m = .
2 j 2
It means we can have, with l = 2 and s = 1/2 being fixed, in order to have
m = 3/2 either m = 1 and m = 1/2 or m = 2 and m = −1/2. The two
j l s l s
states
ψ
n=0j=5/2mj=3/2;l=2s=1/2
and
ψ
n=0j=3/2mj=3/2;l=2s=1/2
willhaveadmixturesfromφ andφ .
n=0l=2ml=2s=1/2ms=−1/2 n=0l=2ml=1s=1/2ms=1/2
How do we find these admixtures? Note that we don’t specify the values of m
l
and m in the functions ψ since ˆj2 does not commute with Lˆ and Sˆ .
s z z
6
Single-particle and two-particle quantum numbers
We operate with
ˆj2 =(ˆl+sˆ)2 =ˆl2+sˆ2+2ˆl sˆ +ˆl sˆ +ˆl sˆ
z z + − − +
on the two jm states, that is
j
3
ˆj2ψ =α(cid:126)2[l(l+1)+ +2m m ]φ +
n=0j=5/2mj=3/2;l=2s=1/2 4 l s n=0l=2ml=2s=1/2ms=−1/2
p
β(cid:126)2 l(l+1)−m (m −1)φ ,
l l n=0l=2ml=1s=1/2ms=1/2
and
3
ˆj2ψ =α(cid:126)2[l(l+1)+ +2m m ]+φ +
n=0j=3/2mj=3/2;l=2s=1/2 4 l s n=0l=2ml=1s=1/2ms=1/2
p
β(cid:126)2 l(l+1)−m (m +1)φ .
l l n=0l=2ml=2s=1/2ms=−1/2
Single-particle and two-particle quantum numbers
This means that the eigenvectors φ etc are not eigen-
n=0l=2ml=2s=1/2ms=−1/2
vectors of ˆj2. The above problems gives a 2×2 matrix that mixes the vectors
ψ and ψ with the states
n=0j=5/2mj3/2;l=2mls=1/2ms n=0j=3/2mj3/2;l=2mls=1/2ms
φ and φ . The unknown coeffi-
n=0l=2ml=2s=1/2ms=−1/2 n=0l=2ml=1s=1/2ms=1/2
cients α and β are the eigenvectors of this matrix. That is, inserting all values
m ,l,m ,s we obtain the matrix
l s
(cid:20) (cid:21)
19/4 2
2 31/4
whose eigenvectors are the columns of
√ √
(cid:20) (cid:21)
2/ 5 1/ 5
√ √
1/ 5 −2/ 5
These numbers define the so-called Clebsch-Gordan coupling coefficients (the
overlaps between the two basis sets). We can thus write
X
ψ = hlm sm |jm iφ ,
njmj;ls l s j nlmlsms
mlms
where the coefficients hlm sm |jm i are the so-called Clebsch-Gordan coefffi-
l s j
cients.
7
Clebsch-Gordan coefficients
The Clebsch-Gordan coeffficients hlm sm |jm i have some interesting proper-
l s j
ties for us, like the following orthogonality relations
X
hj m j m |JMihj m j m |J0M0i=δ δ ,
1 1 2 2 1 1 2 2 J,J0 M,M0
m1m2
X
hj m j m |JMihj m0j m0|JMi=δ δ ,
1 1 2 2 1 1 2 2 m1,m01 m2,m02
JM
hj m j m |JMi=(−1)j1+j2−Jhj m j m |JMi,
1 1 2 2 2 2 1 1
and many others. The latter will turn extremely useful when we are going to
define two-body states and interactions in a coupled basis.
Clebsch-Gordan coefficients, testing the orthogonality rela-
tions
The orthogonality relation can be tested using the symbolic python package
wigner. Let us test
X
hj m j m |JMihj m j m |J0M0i=δ δ ,
1 1 2 2 1 1 2 2 J,J0 M,M0
m1m2
The following program tests this relation for the case of j =3/2 and j =3/2
1 2
meaning that m and m run from −3/2 to 3/2.
1 2
from sympy import S
from sympy.physics.wigner import clebsch_gordan
# Twice the values of j1 and j2
j1 = 3
j2 = 3
J = 2
Jp = 2
M = 2
Mp = 3
sum = 0.0
for m1 in range(-j1, j1+2, 2):
for m2 in range(-j2, j2+2, 2):
M = (m1+m2)/2.
""" Call j1, j2, J, m1, m2, m1+m2 """
sum += clebsch_gordan(S(j1)/2, S(j2)/2, J, S(m1)/2, S(m2)/2, M)*clebsch_gordan(S(j1)/2, S(j2)/2, Jp, S(m1)/2, S(m2)/2, Mp)
print sum
Quantum numbers and the Schroeodinger equation in rela-
tive and CM coordinates
Summing up, for for the single-particle case, we have the following eigenfunc-
tions
X
ψ = hlm sm |jm iφ ,
njmj;ls l s j nlmlsms
mlms
8
where the coefficients hlm sm |jm i are the so-called Clebsch-Gordan coefffi-
l s j
cients. The relevant quantum numbers are n (related to the principal quantum
number and the number of nodes of the wave function) and
ˆj2ψ =(cid:126)2j(j+1)ψ ,
njmj;ls njmj;ls
ˆj ψ =(cid:126)m ψ ,
z njmj;ls j njmj;ls
ˆl2ψ =(cid:126)2l(l+1)ψ ,
njmj;ls njmj;ls
sˆ2ψ =(cid:126)2s(s+1)ψ ,
njmj;ls njmj;ls
but s and l do not result in good quantum numbers in a basis where we use
z z
the angular momentum j.
Quantum numbers and the Schroedinger equation in rela-
tive and CM coordinates
Foratwo-bodystatewherewecoupletwoangularmomentaj andj toafinal
1 2
angularmomentumJ withprojectionM ,wecandefineasimilartransformation
J
in terms of the Clebsch-Gordan coeffficients
X
ψ = hj m j m |JM iψ ψ .
(j1j2)JMJ 1 j1 2 j2 J n1j1mj1;l1s1 n2j2mj2;l2s2
mj1mj2
We will write these functions in a more compact form hereafter, namely,
|(j j )JM i=ψ ,
1 2 J (j1j2)JMJ
and
|j m i=ψ ,
i ji nijimji;lisi
where we have skipped the explicit reference to l, s and n. The spin of a nucleon
is always 1/2 while the value of l can be deduced from the parity of the state.
It is thus normal to label a state with a given total angular momentum as jπ,
where π =±1.
Quantum numbers and the Schroedinger equation in rela-
tive and CM coordinates
Our two-body state can thus be written as
X
|(j j )JM i= hj m j m |JM i|j m i|j m i.
1 2 J 1 j1 2 j2 J 1 j1 2 j2
mj1mj2
DuetothecouplingorderoftheClebsch-Gordancoefficientitreadsasj coupled
1
to j to yield a final angular momentum J. If we invert the order of coupling we
2
would have
X
|(j j )JM i= hj m j m |JM i|j m i|j m i,
2 1 J 2 j2 1 j1 J 1 j1 2 j2
mj1mj2
9
and due to the symmetry properties of the Clebsch-Gordan coefficient we have
X
|(j j )JM i=(−1)j1+j2−J hj m j m |JM i|j m i|j m i=(−1)j1+j2−J|(j j )JM i.
2 1 J 1 j1 2 j2 J 1 j1 2 j2 1 2 J
mj1mj2
We call the basis |(j j )JM i for the coupled basis, or just j-coupled ba-
1 2 J
sis/scheme. Thebasisformedbythesimpleproductofsingle-particleeigenstates
|j m i|j m i is called the uncoupled-basis, or just the m-scheme basis.
1 j1 2 j2
Quantum numbers
We have thus the coupled basis
X
|(j j )JM i= hj m j m |JM i|j m i|j m i.
1 2 J 1 j1 2 j2 J 1 j1 2 j2
mj1mj2
and the uncoupled basis
|j m i|j m i.
1 j1 2 j2
The latter can easily be generalized to many single-particle states whereas the
first needs specific coupling coefficients and definitions of coupling orders. The
m-scheme basis is easy to implement numerically and is used in most standard
shell-model codes. Our coupled basis obeys also the following relations
Jˆ2|(j j )JM i=(cid:126)2J(J +1)|(j j )JM i
1 2 J 1 2 J
Jˆ|(j j )JM i=(cid:126)M |(j j )JM i,
z 1 2 J J 1 2 J
Components of the force and isospin
The nuclear forces are almost charge independent. If we assume they are, we
can introduce a new quantum number which is conserved. For nucleons only,
that is a proton and neutron, we can limit ourselves to two possible values which
allow us to distinguish between the two particles. If we assign an isospin value
of τ =1/2 for protons and neutrons (they belong to an isospin doublet, in the
same way as we discussed the spin 1/2 multiplet), we can define the neutron
to have isospin projection τ = +1/2 and a proton to have τ = −1/2. These
z z
assignements are the standard choices in low-energy nuclear physics.
Isospin
This leads to the introduction of an additional quantum number called isospin.
We can define a single-nucleon state function in terms of the quantum numbers
n, j, m , l, s, τ and τ . Using our definitions in terms of an uncoupled basis, we
j z
had
X
ψ = hlm sm |jm iφ ,
njmj;ls l s j nlmlsms
mlms
10