LOG#098. Group theory(XVIII).Posted: 2013/04/22
This and my next blog post are going to be the final posts in this group theory series. I will be covering some applications of group theory in Quantum Mechanics. More advanced applications of group theory, extra group theory stuff will be added in another series in the near future.
Angular momentum in Quantum Mechanics
Take a triplet of linear operators, . We say that these operators are angular momentum operators if they are “observable” or observable operators (i.e.,they are hermitian operators) and if they satisfy
The presence of an imaginary factor makes compatible hermiticity and commutators for angular momentum. Note that if we choose antihermitian generators, the imaginary unit is absorbed in the above commutators.
We can determine the full angular momentum spectrum and many useful relations with only the above commutators, and that is why those relations are very important. Some interesting properties of angular momentum can be listed here:
1) If are two angular momentum operators, and they sastisfy the above commutators, and if in addition to it, we also have that , then also satisfies the angular momentum commutators. That is, two independen angular momentum operators, if they commute to each other, imply that their sum also satisfy the angular momentum commutators.
2) In general, for any arbitrary and unitary vector , we define the angular momentum in the direction of such a vector as
and for any 3 unitary and arbitrary vectos such as , we have
3) To every two vectors we also have
4) We define the so-called “ladder operators” as follows. Take the angular momentum operator and write
These operators are NOT hermitian, i.e, ladder operators are non-hermitian operators and they satisfy
5) Ladder operators verify some interesting commutators:
6) Commutators for the square of the angular momentum operator
7) Additional useful relations are:
8) Positivity: the operators are indefinite positive operators. It means that all their respective eigenvalues are positive numbers or zero. The proof is very simple
In fact this also implies the positivity of . For the remaining operators, it is trivial to derive that
The general spectrum of the operators can be calculated in a completely general way. We have to search for general eigenvalues
The general procedure is carried out in several well-defined steps:
1st. Taking into account the positivity of the above operators , it means that there is some interesting options
A) is definite positive, i.e., . Then, we can write for all practical purposes
Specifically, we define how the operators and act onto the states, labeled by two parameters and in the following way
B) are positive, and we also have
That means that the following quantities are positive
Therefore, we have deduced that
2nd. We realize that
(1) is an eigenstate of and eigenvalue . Check:
(2) is an eigentstate of and eigenvalue . Check (using :
(3) is an eigenstate of with eigenvalue . Check:
(4) is an eigenvector of and is its eigenvalue. Check:
Therefore, we have deduced the following conditions:
1) if , equivalently if , then the eigenstates , equivalently , are the eigenstates of . The same situation happens if we have vectors or for any (positive integer numbers). Thus, the sucessive action of any of these two operators increases (decreases) the eigenvalue in one unit.
2) If or respectively if , the vectors , respectively are null vectors:
If we begin by certain number , we can build a series of eigenstates/eigenvectors and their respective eigenvalues
And thus, since , then . The number can be integer or half-integer. The eigenvalues have the same character but they can be only separated by one unit.
(1) The only possible eigenvalues for are with integer or half-integer.
(2) The only possible eigenvalues for are integer numbers or half-integer numbers, i.e.,
(3) If is an eigenvector for and , then
We have seen that, given an state , we can build a “complete set of eigenvectors” by sucessive application of ladder operators ! That is why ladder operators are so useful:
This list is a set of eigenvectors, all of them with the same quantum number and different . The relative phase of is not determined. Writing
from the previous calculations we easily get that
The modulus of is determined but its phase IS not. Remember that a complex phase is arbitrary and we can choose it arbitrarily. The usual convention is to define real and positive, so
Invariant subspaces of angular momentum
If we addopt a concrete convention, the complete set of proper states/eigentates is:
This set of eigenstates of angular momentum will be denoted by , the proper invariant subspace of angular momentum operators , with corresponding eigenvalues .
The previously studied (above) features tell us that this invariant subspace is:
a) Invariant with respect to the application of , the operators , and every function of them.
b) is an irreducible subspace in the sense we have studied in this thread: it has no invariant subspace itself!
The so-called matrix elements for angular momentum in these invariant subspaces can be obtained using the ladder opertors. We have
Example(I): Spin 0. (Scalar bosons)
This case is trivial. There are no matrices for angular momentum. Well, there are…But they are and they are all equal to cero. We have
Example(II): Spin 1/2. (Spinor fields)
Now, we have
The angular momentum operators are given by multiples of the so-called Pauli matrices. In fact,
and then and .
The Pauli matrices have some beautiful properties, like
i) The eigenvalues of these matrices are .
ii) , , . This property is related to the fact that the Pauli matrices anticommute.
iv) With the “unit” vector , we get
This matrix has only two eigenvalues for every value of the parameters . In fact the matrix has only an eigenvalue equal to zero, twice, and its eigenvector is:
And has only an eigenvalue equal to zero twice and eigenvector
Example(III): Spin 1. (Bosonic vector fields)
In this case, we get
The restriction to this subspace of the angular momentum operator gives us the following matrices:
A) are commutative matrices.
C) is a diagonarl matrix.
D) is a nilpotent matrix since with 3 equal null eigenvalues and one single eigenvector
Example(IV): Spin 3/2. (Vector spinor fields)
In this case, we have
The spin-3/2 matrices can be obtained easily too. They are
is nonnormal since and it is nilpotent in the sense that and its eigenvalues is zero four times. The only eigenvector is the vector
This vector is “interesting” in the sense that it is “entangled” and it can not be rewritten as a tensor product of two . There is nice measure of entanglement, called tangle, that it shows to be nonzero for this state.
Example(V): Spin 2. (Bosonic tensor field with two indices)
In this case, the invariant subspace is formed by the vectors
For the spin-2 particle, the spin matrices are given by the following matrices
Moreover, the following matrix
is nonnormal and nilpotent with . Moreover, it has 5 null eigenvalues and a single eigenvector
We see that the spin matrices in 3D satisfy for general s:
ii) The ladder operators for spin s have the following matrix representation:
Moreover, in the matrix sense and the above matrix could even be extended to the case of a non-bounded spin particle. In that case the above matrix would become an infinite matrix! In the same way, for spin s, we also get that would be (2s+1)-nilpotent and it would own only a single eigenvector with Z having null eigenvalues. The single eigenvector can be calculated quickly.
Example(VI): Rotations and spinors.
We are going to pay attention to the case of spin 1/2 and work out its relation with ordinary rotations and the concept of spinors.
Given the above rotation matrices for spin 1/2 in terms of Pauli matrices, we can use the following matrix property: if M is a matrix that satisfies , then we can write that
Then, we write
From these equations and definitions, we can get the rotations around the 3 coordinate planes (it corresponds to the so-called Cayley-Hamilton parametrization).
a) Around the plance (XY), with the Z axis as the rotatin axis, we have
b) Two sucessive rotations yield
This matrix has a strange periodicity! That is, rotations with angle don’t recover the identity but minus the identity matrix!
Imagine a system or particle with spin 1/2, such that the wavefunction is :
If we apply a rotation to this object, something that we call “spinor”, we naively would expect that the system would be invariant but instead of it, we have
The norm or length is conserved, though, since
These objects (spinors) own this feature as distinctive character. And it can be generalized to any value of j. In particular:
A) If is an integer number, then . This is the case of “bosons”/”force carriers”.
B) If is half-integer, then !!!!!!!. This is the case of “fermions”/”matter fields”.
Rotation matrices and the subspaces E(j).
We learned that angular momentum operators are the infinitesimal generators of “generalized” rotations (including those associated to the “internal spin variables”). A theorem, due to Euler, says that every rotation matrix can be written as a function of three angles. However, in Quantum Mechanics, we can choose an alternative representation given by:
Given a representation of J in the subspace , we obtain matrices as we have seen above, and these matrices have the same dimension that those of the irreducible representation in the subspace . There is a general procedure and parametrization of these rotation matrices for any value of . Using a basis of eigenvectors in :
and where we have defined the so-called Wigner coefficients
The reduced matrix only depends on one single angle (it was firstly calculated by Wigner in some specific cases):
Generally, we will find the rotation matrices when we “act” with some rotation operator onto the eigenstates of angular momentum, mathematically speaking:
See you in my final blog post about basic group theory!