Understanding Spherical Tensors & Their Applications

In summary, the conversation discusses the concept of spherical tensors and how they are applied, specifically in the case of combining a rank 1 tensor with a rank 1/2 tensor. The mathematical expressions for forming new tensors are provided, using Clebsch-Jordan symbols. The physical meaning of the expression in the first case is explained in terms of quantum mechanics, while the physical meaning in the second case is unclear as it appears to change the identity of the particle.
  • #1
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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  • #2
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.
 
  • #3
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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What are spherical tensors?

Spherical tensors are mathematical objects that describe the rotational properties of physical systems. They are used to represent the symmetry of a system with respect to rotations in three-dimensional space.

What are some common applications of spherical tensors?

Spherical tensors have a wide range of applications in physics, chemistry, and engineering. They are commonly used in quantum mechanics to describe the angular momentum of particles, in molecular spectroscopy to analyze the rotational properties of molecules, and in solid-state physics to study the symmetry of crystals.

How are spherical tensors represented mathematically?

Spherical tensors are typically represented as matrices or tensors with specific properties. In physics, they are often represented using the Wigner-Eckart theorem, which relates them to the Clebsch-Gordan coefficients of angular momentum.

What is the significance of the rank of a spherical tensor?

The rank of a spherical tensor determines the number of indices needed to represent it. It also determines the transformation properties of the tensor under rotations. Higher rank tensors have more complex transformation properties and are used to describe more complex systems.

How are spherical tensors related to spherical harmonics?

Spherical tensors and spherical harmonics are closely related. Spherical harmonics are a special type of spherical tensor with rank 0, which are used to describe the angular dependence of physical systems. They are often used in conjunction with spherical tensors to fully describe the symmetry of a system.

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