Derivation of equations using tensor

In summary: This happens because the two states ##|3/2,3/2\rangle## and ##|1/2,-1/2\rangle##, which are different in the tensor product basis, are actually the same state in the ##|J,m\rangle## basis. This is due to the orthonormality of the basis states.In summary, the conversation discusses the derivation of equations (41), (42), and (43) from the tensor product basis to the ##|J,m\rangle## basis. Equation (41) represents the states ##|1/2,1/2\rangle## and ##|1/2,-1/2\rangle## in the tensor product basis, while (42) is
  • #1
TheCanadian
367
13
http://hitoshi.berkeley.edu/221a/tensorproduct.pdf

I was following the above pdf and got through most of it but am not quite understanding how (41), (42), and (43) are derived.

It appears that (31) and (41) are representing the same states and are still orthogonal, but how exactly is (41) derived? Aside from a scalar multiplication, the equation in (40) works perfectly well when applying (31) instead of (41).

For (42), I understand this is a unitary matrix, but what's the motivation and how exactly is it derived from the basis states? To clarify, how many basis states are there?

For (43), I tried doing the matrix multiplication, and I'm not sure if it's a typo, but how are there two possible solutions for the m = 3/2 and j = 3/2 states when multiplied by U?

These final equations have given me a bit of trouble in understanding the material, so any help with understanding their derivations would be greatly appreciated!
 
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  • #2
TheCanadian said:
how (41), (42), and (43) are derived.
As for (41), I must admit that the author didn't give a clearer explanation about the transition he was making from equation (40) to (41).
Equation (41) are just the states ##|1/2,1/2\rangle ## and ##|1/2,-1/2\rangle## written in the tensor product basis ##|m_1\rangle \otimes |m_2\rangle##. To obtain those column matrices, you can employ the rule ##m = m_1+m_2##. For example take the ##|1/2,1/2\rangle ## state (the left in (41)), here ##m=1/2##, therefore to write it as
$$
|1/2,1/2\rangle = \sum\sum_{1/2=m_1+m_2} C(m_1,m_2) |m_1\rangle \otimes |m_2\rangle
$$
you have to find ##|m_1\rangle \otimes |m_2\rangle## which satisfies ##1/2 = m_1+m_2##. The required ##|m_1\rangle \otimes |m_2\rangle## should clearly be the same as that required in expanding ##|3/2,1/2\rangle## (eq. (38)) but the Clebsch-Gordan coefficient ##C(m_1,m_2)## cannot be the same between them, in particular they should be orthogonal, i.e. ##\langle1/2,1/2|3/2,1/2\rangle = 0##. From this it's easy to prove (41).

TheCanadian said:
For (42), I understand this is a unitary matrix, but what's the motivation and how exactly is it derived from the basis states? To clarify, how many basis states are there?
As the sentence following (41) says, this matrix was composed by sequentially stacking the (conjugate) transpose of the column matrices (37), (38), and (41). For instance, the first four rows of matrix (41) are (conjugate) transpose of ##|3/2,3/2\rangle##, ##|3/2,1/2\rangle##, ##|3/2,-1/2\rangle##, and ##|3/2,-3/2\rangle##, in their column matrix forms respectively. The motivation of deriving this matrix is to determine the change of coordinate matrix from the tensor product basis ##|m_1\rangle \otimes |m_2\rangle## to ##|J,m\rangle## basis.

TheCanadian said:
For (43), I tried doing the matrix multiplication, and I'm not sure if it's a typo, but how are there two possible solutions for the m = 3/2 and j = 3/2 states when multiplied by U?
Yes that should be a typo, the last one should be ##|1/2,-1/2\rangle##.
 
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FAQ: Derivation of equations using tensor

What is a tensor?

A tensor is a mathematical object that can be used to represent and manipulate multivariate data. It is a generalization of vectors and matrices and has components that represent the relationships between different sets of variables.

How are tensors used in the derivation of equations?

Tensors are used in the derivation of equations because they allow for a more concise and efficient representation of the mathematical relationships between variables. They can also help to simplify complex calculations and make the derivation process more intuitive.

What is the importance of tensors in physics and engineering?

Tensors are crucial in physics and engineering as they provide a powerful tool for representing and understanding physical phenomena. They are used extensively in fields such as mechanics, electromagnetism, and general relativity to describe the relationships between physical quantities.

How do tensors help to describe the properties of materials?

Tensors are used in material science to describe the anisotropy of materials, which refers to their directional dependence. By using tensors, we can represent the different mechanical, electrical, and thermal properties of materials in different directions, allowing for a more accurate understanding of their behavior.

What are some applications of tensor calculus?

Tensor calculus has numerous applications in various fields, including physics, engineering, computer science, and economics. It is used to derive equations and models for physical systems, design optimal control systems, and develop machine learning algorithms. It also plays a crucial role in understanding and analyzing complex networks and systems.

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