Universidad Veracruzana
M AESTR ÍA EN F ÍSICA
M ATHEMATICAL M ETHODS
G ROUP THEORY
Homework 6
Autor:
Jorge Alberto Orduña Mendoza
December 06, 2021
Group theory
Universidad Veracruzana
Problem 1
Suppose that f, g ∈ Γ (the phase space). Consider de bracket
{{f, g}} :=
n
X
Ω
ij
i,j=1
∂g ∂f
∂f ∂g
−
∂qi ∂pj
∂q1 ∂pj
,
(1)
in place of the standard Poisson bracket.
• Which property must the matrix Ω possess in order that this bracket satisfies the Jacobi
identiy?
• Can it depend on p and q?
Solution:
From the definition the original Poisson Bracket, we know that in order of (1) to satisfy the Jacobi
identity it must be antisymmetric. This condition implies that Ωij = −Ωji , that is, Ω is also antisymmetric.
For direct substitution of eq. (1) into Jacobi identity yields very long expressions from which it
is difficult to get useful information about the explicit form of the matrix Ω. This is known as the
sympletic matrix. It is introduced in the Hamiltonian mechanics as follows.
First, define a 2n-state vector ⃗z, whose first n entries are the qi coordinates and the rest, the other
n momentum coordinates pi . According to the equations of Hamilton,
q̇i =
∂H
,
∂pi
ṗi = −
∂H
∂qi
⇒
żi =
2n
X
j=1
Ωij
∂H
,
∂zj
(2)
where the coefficients Ωij are


 1 if j = i + n
Ωij = −1 if i = j + n


0 in otherwise.
(3)
Or, in matrix form
Ω=
0n In
,
−In 0n
(4)
where, of course, In and 0n denote the n×n identity and zero matrix. This matrix is antysimmetric
and satisfies Ω2 = −I2n and it is independent of the q and p coordinates.
Página 1
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This notation can be written in matrix form and then rewritten the expression as follows,
 ∂g 
2n
X
∂f
∂f
∂g
Ωij
=
∂qi
∂zi
∂zj
i,j=1
=
∂f
∂qi
∂q 
0n In 
 j
−In 0n  ∂g 
∂pj
 ∂g 
−
∂pj 
∂f 



∂g 
∂pi
∂qj
∂f
∂pi
2n X
∂f ∂g
∂g ∂f
=
−
.
∂qi ∂pj
∂q1 ∂pj
i
(5)
2n
X
∂f
∂g
Ωij
∂zi
∂zj
i,j=1
(6)
This means that
{f, g} =
is a valid definition of the Poisson Bracket. We would like to prove that this notation also satisfies
the Jacobi identity. That is, that for any f, g, h functions on the phase space, it is true that
{f, {g, h}} + {g, {h, f }} + {h, {f, g}} = 0.
(7)
We can check it,
∂{g, h}
∂f
Ωij
∂zi
∂zj
∂f
∂h
∂g
∂
=
Ωkl
Ωij
∂zi
∂zj ∂zk
∂zl
∂f ∂ 2 h ∂g
∂f ∂ 2 g ∂h
,
= Ωij Ωkl
+
∂zi ∂zj zk ∂zl ∂zi ∂zj zl ∂zk
{f, {g, h}} =
∂g
∂{h, f }
Ωij
∂zi
∂zj
∂g
∂
∂h
∂f
=
Ωij
Ωkl
∂zi
∂zj ∂zk
∂zl
∂g ∂ 2 h ∂f
∂g ∂ 2 f ∂h
= Ωij Ωkl
+
∂zi ∂zj zk ∂zl ∂zi ∂zj zl ∂zk
(8)
{g, {h, f }} =
(9)
Página 2
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Universidad Veracruzana
∂h
∂{f, g}
Ωij
∂zi
∂zj
∂h
∂
∂f
∂g
=
Ωij
Ωkl
∂zi
∂zj ∂zk
∂zl
∂h ∂ 2 f ∂g
∂h ∂ 2 g ∂f
= Ωij Ωkl
+
.
∂zi ∂zj zk ∂zl ∂zi ∂zj zl ∂zk
{h, {f, g}} =
(10)
Now we will show that by adding all the terms up, they will vanish. In order to do that, we group
in pairs with the same second derivatives. For example, termis with ∂ 2 g are
∂f ∂ 2 g ∂h
∂h ∂ 2 g ∂f
.
(11)
Ωij Ωkl
+
∂zi ∂zj zk ∂zl ∂zi ∂zj zl ∂zk
There is antisymmetry under the change of indices i ↔ j and k ↔ l. So, we interchange k ↔ l in
the second term to get
∂h ∂ 2 g ∂f
∂h ∂f
∂ 2g
∂f ∂ 2 g ∂h
∂f ∂h
Ωij Ωkl
−
= Ωij Ωkl
−
.
(12)
∂zi ∂zj zk ∂zl ∂zi ∂zj zk ∂zl
∂zi ∂zl ∂zi ∂zl ∂zj ∂zk
The last expression is a funcion in the four indices that can be expressed as Ωij Ωkl Aijkl . This
function is antisymmetric under the change on the indices i ↔ l. Furtherthemore we know that
Ωij Ωkl Aijkl = Ωlk Ωji Alkji by renaming dummy indices
= Ωkl Ωij Alkji making i ↔ j and k ↔ l in Ω
= −Ωkl Ωij Aijkl making l ↔ i and j ↔ k in the funcion A.
(13)
(14)
(15)
Therefore the quantity Ωij Ωkl Aijkl equals itw own negative so it must be zero as we wanted to
prove. The same happends to the other two pairs of derivatives.
Problem 2
Consider again the group G = SO(4).The infinitesimal operators of this group may be written in
terms of the variables (x, y, z, t), as
∂
∂
∂
∂
∂
∂
−y
M2 = x − z
M3 = y
−x
M1 = z
∂y
∂z
∂z
∂x
∂x
∂y
∂
∂
∂
∂
∂
∂
N1 = x − t
N2 = y − t
N3 = z − t
∂t
∂x
∂t
∂y
∂t
∂z
• Show that these infinitesimal operators satisfy the commutation relations
[Mi , Mj ] = εijk Mk ,
[Mi , Nj ] = εijk Nk ,
[Ni , Nj ] = εijk Mk
(i, j, k = 1, 2, 3)
• Consider now the linear transformation to a new basis given by
Mi + Ni
Mi − Ni
Ji :=
,
Ki :=
2
2
Compute the commutation relations among these new infinitesimal operators.
• What is the geometrical interpretation arising from the resulting commutation relations?
Página 3
Group theory
Universidad Veracruzana
Solution:
We start by computing the commutation relations
∂
∂
∂
∂
∂
∂
∂
∂
[M1 , M2 ] = z
−y
x −z
− x −z
z
−y
∂y
∂z
∂z
∂x
∂z
∂x
∂y
∂z
∂
∂
∂
∂
∂
∂
x −z
−x
z
−y
=−y
∂z
∂z
∂x
∂z ∂y
∂z
2
2
∂
∂
∂
∂
= − xy 2 + y
−x
+ xy 2
∂z
∂x
∂y
∂z
∂
∂
−x
= M3
=y
∂x
∂y
∂
∂
∂
∂
∂
∂
∂
∂
[M1 , M3 ] = z
−y
y
−x
− y
−x
z
−y
∂y
∂z
∂x
∂y
∂x
∂y
∂y
∂z
∂
∂
∂
∂
∂
∂
=z
y
−x
+x
z
−y
∂y ∂x
∂y
∂y ∂y
∂z
2
2
∂
∂
∂
∂
=z
− xz 2 + xz 2 − x
∂x
∂y
∂y
∂z
∂
∂
=− x −z
= −M2
∂z
∂x
∂
∂
∂
∂
∂
∂
∂
∂
−x
−x
[M2 , M3 ] = x − z
y
− y
x −z
∂z
∂x
∂x
∂y
∂x
∂y
∂z
∂x
∂
∂
∂
∂
∂
∂
=−z
−x
y
−y
x −z
∂x ∂x
∂y
∂x ∂z
∂x
2
2
∂
∂
∂
∂
= − yz 2 + z
−y
+ yz 2
∂x
∂y
∂z
∂x
∂
∂
=z
−y
= M1
∂y
∂z
∂
∂
∂
∂
∂
∂
∂
∂
[M1 , N2 ] = z
−y
y −t
− y −t
z
−y
∂y
∂z
∂t
∂y
∂t
∂y
∂y
∂z
∂
∂
∂
∂
∂
∂
=z
y −t
+t
z
−y
∂y ∂t
∂y
∂y ∂y
∂z
2
2
∂
∂
∂
∂
=z − zt 2 + zt 2 − t
∂t
∂y
∂y
∂z
∂
∂
=z − t
= N3
∂t
∂z
(16)
(17)
(18)
(19)
Página 4
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Universidad Veracruzana
∂
∂
∂
∂
∂
∂
∂
∂
[M1 , N3 ] = z
−y
z −t
− z −t
z
−y
∂y
∂z
∂t
∂z
∂t
∂z
∂y
∂z
∂
∂
∂
∂
∂
∂
z −t
+t
z
−y
=−y
∂z ∂t
∂z
∂z ∂y
∂z
2
2
∂
∂
∂
∂
= − y + yt 2 + t
− yt 2
∂t
∂z
∂y
∂z
∂
∂
= −N2
=− y −t
∂t
∂y
∂
∂
∂
∂
∂
∂
∂
∂
[M2 , N3 ] = x − z
z −t
− z −t
x −z
∂z
∂x
∂t
∂z
∂t
∂z
∂z
∂x
∂
∂
∂
∂
∂
∂
=x
z −t
+t
x −z
∂z ∂t
∂z
∂z
∂z
∂x
2
2
∂
∂
∂
∂
=x − xt 2 + xt 2 − t
∂t
∂z
∂z
∂x
∂
∂
=x − t
= N1
∂t
∂x
∂
∂
∂
∂
∂
∂
∂
∂
[N1 , N2 ] = x − t
y −t
− y −t
x −t
∂t
∂x
∂t
∂y
∂t
∂y
∂t
∂x
∂
∂
∂
∂
∂
∂
y −t
−y
x −t
=x
∂t ∂t
∂y
∂t ∂t
∂x
2
2
∂
∂
∂
∂
=xy 2 − x
− xy 2 + y
∂t
∂y
∂t
∂x
∂
∂
=y
−x
= M3
∂x
∂y
∂
∂
∂
∂
∂
∂
∂
∂
[N1 , N3 ] = x − t
z −t
− z −t
x −t
∂t
∂x
∂t
∂z
∂t
∂z
∂t
∂x
∂
∂
∂
∂
∂
∂
=x
z −t
−z
x −t
∂t ∂t
∂z
∂t ∂t
∂x
2
2
∂
∂
∂
∂
=xz 2 − x − xz 2 + z
∂t
∂z ∂t
∂x
∂
∂
=− x −
= −M2
∂z ∂x
(20)
(21)
(22)
(23)
Página 5
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Universidad Veracruzana
∂
∂
∂
∂
∂
∂
∂
∂
[N2 , N3 ] = y − t
z −t
− z −t
y −t
∂t
∂y
∂t
∂z
∂t
∂z
∂t
∂y
∂
∂
∂
∂
∂
∂
z −t
−z
y −t
=y
∂t ∂t
∂z
∂t ∂t
∂y
2
2
∂
∂
∂
∂
=yz 2 − y
− yz 2 + z
∂t
∂z
∂t
∂y
∂
∂
=z
−y
= M1
∂y
∂z
(24)
In this way,
[M1 , M2 ] = M3 ,
[M3 , M1 ] = M2 ,
[M2 , M3 ] = M1 ,
[Mi , Mj ] = εijk Mk
(25)
[M1 , N2 ] = N3 ,
[M1 , N3 ] = −N2 ,
[M2 , N3 ] = N1 ,
[Mi , Nj ] = εijk Nk
(26)
[N1 , N2 ] = M3 ,
[N3 , N1 ] = M2 ,
[N2 , N3 ] = M1 ,
[Ni , Nj ] = εijk Mk
(27)
Now we consider the proposed transformation
Ji =
Mi + Ni
;
2
Ki =
Mi − Ni
,
2
(28)
so
1
1
[Ji , Jj ] = [Mi + Ni , Mj + Nj ] = [(Mi + Ni )(Mj + Nj ) − (Mj + Nj )(Mi + Ni )]
4
4
1
= [Mi Mj + Mi Nj + Ni Mj + Ni Nj − Mj Mi − Mj Ni − Nj Mi − Nj Ni ]
4
1
= [Mi Mj − Mj Mi + Ni Nj − Nj Ni + Mi Nj − Nj Mi + Ni Mj − Mj Ni ]
4
1
= {[Mi , Mj ] + [Ni , Nj ] + [Mi , Nj ] − [Ni , Mj ]}
4
1
1
= {[Mi , Mj ] + [Ni , Nj ] + 2[Mi , Nj ]} = {2εijk Mk + 2εijk Nk }
4 4
Mk + Nk
=εijk
= εijk Jk
2
(29)
1
1
[Ki , Kj ] = [Mi − Ni , Mj − Nj ] = [(Mi − Ni )(Mj − Nj ) − (Mj − Nj )(Mi − Ni )]
4
4
1
= [Mi Mj − Mi Nj − Ni Mj + Ni Nj − Mj Mi + Mj Ni + Nj Mi − Nj Ni ]
4
1
= [Mi Mj − Mj Mi + Ni Nj − Nj Ni − Mi Nj + Nj Mi − Ni Mj + Mj Ni ]
4
1
= {[Mi , Mj ] + [Ni , Nj ] − [Mi , Nj ] + [Ni , Mj ]}
4
1
1
= {[Mi , Mj ] + [Ni , Nj ] − 2[Mi , Nj ]} = {2εijk Mk − 2εijk Nk }
4 4
Mk − Nk
=εijk
= εijk Kk
(30)
2
Página 6
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Universidad Veracruzana
1
1
[Ji , Kj ] = [Mi + Ni , Mj − Nj ] = [(Mi + Ni )(Mj − Nj ) − (Mj − Nj )(Mi + Ni )]
4
4
1
= [Mi Mj − Mi Nj + Ni Mj − Ni Nj − Mj Mi − Mj Ni + Nj Mi + Nj Ni ]
4
1
= [Mi Mj − Mj Mi − Ni Ni + Nj Ni − Ni Mj + Nj Ni + Ni Mj − Mj Ni ]
4
1
1
= {[Mi , Mj ] − [Ni , Nj ]} = εijk (Mk − Mk ) = 0
(31)
4
4
Finally, we have
[Ji , Jj ] = εijk Jk ,
[Ki , kj ] = εijk Kk ,
[Ji , Kj ] = 0.
(32)
Geometrically, the vector product of Ji and Jj produce an operator Jk orthogonal to them. The
same happens with the Ni operators that are parallel to Jj .
Problem 3
Compute the casimir operators for SO(4)
• By using the basis {Mi , Ni }.
• By using the basis {Ji , Ki }.
Solution:
• From the previous problem we know the commutation relations between the infinitesimal
operators Mi and Nj , which are
[Mi , Mj ] = εijk Mk ,
[Mi , Nj ] = εijk Nk ,
[Ni , Nj ] = εijk Nk
(33)
and form the lie algebra for SO(4). Now, to obtain the casimir operator we first need to
compute the killing form or metric tensor that is given by
N
X
gij =
s
,
Cisr Cjr
(34)
r,s=1
where N is the number of elements of the basis and Cijk are the structure constants. In this
case the structure constants are in terms of the Levi-Civita tensor that appears in each of the
commutation relations. In order to calculate the metric tensor gij we are going to use the
next convention:
M1 = X1 ,
M2 = X2 ,
M3 = X3 ,
N1 = X4 ,
N2 = X5 ,
N3 = X6 .
(35)
So the metric tensor is
gij =
6
X
s
Cisr Cjr
(36)
r,s=1
Página 7
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To simplify the expansion we need to recall that the metric tensor is antisymmetric and in
the case of having i = j, j = k or i = k the Levi-Civita tensor is equal to 0. To save space
and time we are just going to write the terms that are not equal to zero:
g11
g22
g33
g44
g55
g66
g14
g25
g36
3
2
6
2
2
3
5
3
3
5
6
5
2
6
5
6
= C12
C13
+ C12
C16
+ C13
C12
+ C13
C15
+ C15
C13
+ C15
C16
+ C16
C12
+ C16
C15
4
1
6
4
6
6
4
3
3
4
3
1
1
6
1
3
C26
+ C26
C21
+ C26
C24
+ C24
C23
+ C24
C24
+ C23
C21
+ C23
C26
+ C21
C23
= C21
2
1
5
1
1
2
4
2
2
4
5
4
1
5
4
5
= C31
C32
+ C31
C35
+ C32
C31
+ C32
C34
+ C34
C32
+ C34
C35
+ C35
C31
+ C35
C34
3
2
6
2
2
3
5
3
3
5
6
5
2
6
5
6
= C42
C43
+ C42
C46
+ C43
C42
+ C43
C45
+ C45
C43
+ C45
C46
+ C46
C42
+ C46
C45
4
1
6
4
6
6
4
3
3
4
3
1
1
6
1
3
C56
+ C56
C51
+ C56
C54
+ C54
C53
+ C54
C54
+ C53
C51
+ C53
C56
+ C51
C53
= C51
2
1
5
1
1
2
4
2
2
4
5
4
1
5
4
4
= C61
C62
+ C61
C65
+ C62
C61
+ C62
C64
+ C64
C62
+ C64
C65
+ C65
C61
+ C65
C65
3
2
6
2
2
3
5
3
3
5
6
5
2
6
5
6
= C12
C43
+ C12
C46
+ C13
C42
+ C13
C45
+ C15
C43
+ C15
C46
+ C16
C42
+ C16
C45
3
1
6
1
1
3
4
3
3
4
6
4
1
6
4
6
= C21
C53
+ C21
C56
+ C23
C51
+ C23
C54
+ C24
C53
+ C24
C56
+ C26
C51
+ C26
C54
2
1
5
1
1
2
4
2
5
4
2
4
1
5
4
5
= C31
C62
+ C31
C65
+ C32
C61
+ C32
C64
+ C34
C65
+ C34
C62
+ C35
C61
+ C35
C64
Using the antisymmetric propertie we can deduce that g41
−g36 . The metric tensor in this case has the form:

1
0
0 1 0
0
1
0 0 1

0
0
1 0 0
gij = −8 
−1 0
0 1 0

 0 −1 0 0 1
0
0 −1 0 0
= −8
= −8
= −8
= −8
= −8
= −8
= −8
= −8
= −8.
= −g14 , g52 = −g25 and g63 =

0
0

1
.
0

0
1
(37)
Since the expression for the casimir operator needs the inverse of gij , i.e. g ij , we use the
software Mathematica to compute the inverse of this matrix and the result is


1 0 0 −1 0
0
0 1 0 0 −1 0 


1 
0 0 1 0
0 −1
ij

.
g =− 
(38)
0
0
16 1 0 0 1

0 1 0 0
1
0
0 0 1 0
0
1
The Casimir operator is defined as,
C=
6 X
6
X
g ij Xi Xj .
(39)
i=1 j=1
Using (38) in the last expression,
C = g 11 X1 X1 + g 22 X2 X2 + g 33 X3 X3 + g 44 X4 X4 + g 55 X5 X5 + g 66 X6 X6
+ g 14 X1 X4 + g 25 X2 X5 + g 36 X3 X6 + g 41 X4 X1 + g 52 X5 X2 + g 63 X6 X3 .
Página 8
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Finally, the casimir operator is,
C=−
1
M12 + M22 + M32 + N12 + N22 + N32 + N1 M1 + N2 M2 + N3 M3
16
−M1 N1 − M2 N2 − M3 N3 ) .
(40)
(41)
• As before, from the previous problem we know the commutation relations between the infinitesimal operators Ji and Kj , which are
[Ji , Jj ] = εijk Jk ,
[Ki , Kj ] = εijk Kk .
(42)
Again we are going to use a convention to refer to the operators
J1 = X1 ,
J2 = X2 ,
J3 = X3 ,
K1 = X4 ,
K2 = X5 ,
K3 = X6 .
(43)
Employing (42) and (43) we can calculate the metric tensor, which non zero elements are
g11
g22
g33
g44
g55
g66
3
2
2
3
= C12
C13
+ C13
C12
3
1
1
3
= C21
C23
+ C23
C21
2
1
1
2
= C31
C32
+ C32
C31
6
5
5
6
= C45
C46
+ C46
C45
6
4
4
6
= C54
C56
+ C56
C54
5
4
4
5
= C64
C65
+ C65
C64
= −2
= −2
= −2
= −2
= −2
= −2.
Then the metric tensor is just

1
0

0
gij = −2 
0

0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0

0
0

0

0

0
1
(44)
0
1
0
0
0
0
0
0
1
0
0
8
0
0
0
1
0
0
0
0
0
0
1
0

0
0

0
.
0

0
1
(45)
and its inverse is

1
0

1
0
gij = − 
2
0
0
0
Using the last result into the formula for the casimir operator we obtain
C=−
1 2
J1 + J22 + J32 + K12 + K22 + K32 .
2
(46)
Página 9
Group theory
Universidad Veracruzana
Problem 4
Compute the Casimir operator for SO(2, 1).
Solution:
In the previous homework, we obtained three infinitesimal operators X1 ,X2 and X3 , given as
∂
∂
∂
∂
∂
∂
−z
, X2 = − z
+x
, X3 = − y
+x
(47)
X1 = y
∂z
∂y
∂x
∂z
∂x
∂y
which satisfy the commutation relations,
[X1 , X2 ] = X3 ,
[X2 , X3 ] = −X1 ,
[X3 , X1 ] = X2
(48)
That is, there is a commutation relation of the form,
[Xi , Xj ] = Cijk Xk , with i, j, k = 1, 2, 3.
(49)
where Cijk are the structure constants.
Now, we want to find the Casimir operator from the operators Xi with i = 1, 2, 3, and for this we
must first obtain the metric tensor for SO(2,1), which is defined as,
gij =
3
X
s
Cisr Cjr
(50)
r,s=1
So, for this case the non-zero structure constants are,
3
3
C12
= −C21
= 1,
1
1
C23
= −C32
= −1,
2
2
C31
= −C13
= 1.
In the metric tensor gij its components must fulfill the conditions g im gil = δlm , due to the structure
constants for i ̸= j, the components of the metric tensor gij = 0 and the non-zero components are,

P3
r
s
3
2
2
3
g
=

11
r,s=1 C1s C1r = C12 C13 + C13 C12 = (1)(−1) + (−1)(1) = −2



P
r
s
3
1
1
3
gii = g22 = 3r,s=1 C2s
C2r
= C21
C23
+ C23
C21
= (−1)(−1) + (−1)(−1) = 2



P

r
s
2
1
1
2
g33 = 3r,s=1 C3s
C3r
= C31
C32
+ C32
C31
= (1)(1) + (1)(1) = 2
So, the metric tensor can be written as a matrix given as


−2 0 0
gij =  0 2 0
0 0 2
(51)
(52)
Página 10
Group theory
Universidad Veracruzana
Using g im gil = δlm to calculate the inverse matrix,
 
 11 12 13  
 

−2 0 0
1 0 0
g
g
g
−2g 11 2g 12 2g 13
g im gil = g 21 g 22 g 23   0 2 0 = −2g 21 2g 22 2g 23  = 0 1 0
g 31 g 32 g 33
0 0 2
−2g 31 2g 32 2g 33
0 0 1
1
⇒ g 11 = − ,
2
1
g 22 = g 33 = ,
2
and g ij = 0 for i ̸= j.
So that, the inverse matrix g ij is


−1 0 0
1
g ij =  0 1 0
2
0 0 1
(53)
Finally, the Casimir operator is calculated as,
C=
3 X
3
X
i=1
1
1
1
g ij Xi Xj = g 11 X1 X1 + g 22 X2 X2 + g 33 X3 X3 = − X12 + X22 + X32
2
2
2
j=1
C=
1
− X12 + X22 + X32
2
(54)
Or explicitly in terms of partial derivatives, substituting the operators (47) in eq. (54)
2 2 2 i
1h
∂
∂
∂
∂
∂
∂
C=
− y
−z
+x
+x
+ z
+ y
2
∂z
∂y
∂x
∂z
∂x
∂y
1h
∂2
∂
∂2
∂2
∂
=
− y 2 2 + y + 2yz
+ z − z2 2
2
∂z
∂y∂z ∂z
∂y
∂y
C=
+ z2
2
∂
∂2
∂
∂2
2 ∂
− 2xz
−
−
x
+
x
z
∂x2 ∂z
∂x∂z
∂x
∂z 2
+ y2
2 i
∂2
∂
∂2
∂
2 ∂
−
y
−
2xy
−
x
+
x
∂x2 ∂y
∂x∂y
∂x
∂y 2
1h 2
∂2
∂2
∂2 i
(y + z 2 ) 2 + (x2 − z 2 ) 2 + (x2 − y 2 ) 2
2
∂x
∂y
∂z
∂2
∂2
∂2
∂ − xy
+ xz
− yz
− 2x
(55)
∂x∂y
∂x∂z
∂y∂z
∂x
Página 11
Group theory
Universidad Veracruzana
References
[1] Sakurai, J. J. and Commins, E. D. (1995). Modern quantum mechanics, revised edition.
[2] Zettili, N. (2003). Quantum mechanics: concepts and applications, Pag. 7.
[3] Jürgen Fuchs and Christoph Schweigert (2003), Symmetries, Lie algebras and Representations, Cambridge University Press.
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