Understanding Spherical Tensors & Their Applications

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Malamala
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Hello! I came across spherical tensors, and I am a bit confused about the way they are applied. For example, Pauli matrices, can be grouped together to form a rank 1 (vector) spherical tensor as ##(\sigma_-, \sigma_z, \sigma_+)##, which are the raising operator, the z projection operator and the lowering operator. When this acts on a spin state, say ##(1/2,1/2)##, we can think of it as normal angular momentum addition. For example, for ##\sigma_-##, which is, as a tensor, ##(1,-1)##, we would get overall ##(1/2,-1/2)## i.e. the spin down state, which is what you expect. The same applies for the other 2 operators. This makes sense. However, combining a rank 1 tensor with a rank 1/2 tensor would give both a rank 1/2 tensor, which is what I mentioned before i.e. the down state is still part of a rank 1/2, but it should also give a rank 3/2 tensor. What is the mathematical expression and physical meaning of this 3/2 rank tensor? Thank you!
 
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So, in your example you have one tensor ##S## of angular momentum (rank) ##j_1=1/2## and another one ##D## with ##j_2 = 1##. You can form the new tensors (using Clebsch-Jordan symbols)

$$T_{J=1/2, M} = \sum_{m_1 m_2} (1/2 m_1 1 m_2 | 1/2 M) S_{1/2, m_1}D_{1, m_2}$$,

and

$$T_{J=3/2, M} = \sum_{m_1 m_2} (1/2 m_1 1 m_2 | 3/2 M) S_{1/2, m_1}D_{1, m_2}$$,

where ##(1/2 m_1 1 m_2 | 3/2 M)## is a Clebsch-Jordan symbol.

The only example of spherical tensors of non-integer rank which I can think of are creation and annihilation operators.
 
eys_physics said:
So, in your example you have one tensor ##S## of angular momentum (rank) ##j_1=1/2## and another one ##D## with ##j_2 = 1##. You can form the new tensors (using Clebsch-Jordan symbols)

$$T_{J=1/2, M} = \sum_{m_1 m_2} (1/2 m_1 1 m_2 | 1/2 M) S_{1/2, m_1}D_{1, m_2}$$,

and

$$T_{J=3/2, M} = \sum_{m_1 m_2} (1/2 m_1 1 m_2 | 3/2 M) S_{1/2, m_1}D_{1, m_2}$$,

where ##(1/2 m_1 1 m_2 | 3/2 M)## is a Clebsch-Jordan symbol.

The only example of spherical tensors of non-integer rank which I can think of are creation and annihilation operators.
Thank you for your reply. So in the first case, I understand the meaning of the expression from a physical point of view, in terms of quantum mechanics. For example, for ##M=1/2##, it says that you can get that either by applying ##\sigma_z = \sigma_0## on the state ##(1/2,1/2)## or applying ##\sigma_+## on ##(1/2,-1/2)##. I.e. you have the same particle, of spin ##1/2## (say an electron), but you can change its spin orientation. I am not sure what is the physical meaning in the second case. It looks like you start with a particle of spin ##1/2## and end up with a particle of spin ##3/2## (and some direction of the spin on the z axis). So you basically change the identity of the particle. What is the physical meaning of this?
 
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