Deriving a vector identity using Pauli spin matrices

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SUMMARY

The discussion focuses on deriving the vector identity \(\left({\bf A} \cdot {\bf \sigma} \right) \left({\bf B }\cdot {\bf \sigma} \right) = {\bf A} \cdot {\bf B} I + i \left( {\bf A } \times {\bf B} \right) \cdot {\bf \sigma}\) using Pauli spin matrices. Participants emphasize the importance of expressing any 2x2 matrix in terms of spin matrices and utilizing the trace properties effectively. Key insights include the necessity of treating \({\bf A}\) and \({\bf B}\) as vectors of scalars and the application of commutation relations between Pauli matrices. The discussion concludes that constructing a 2-D matrix from the original equation is essential for progressing towards the right-hand side of the identity.

PREREQUISITES
  • Understanding of Pauli spin matrices and their properties
  • Familiarity with matrix representation of quantum states
  • Knowledge of trace operations in linear algebra
  • Ability to manipulate vector identities in quantum mechanics
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  • Study the properties of Pauli matrices, including their commutation relations
  • Learn about the trace operation and its applications in quantum mechanics
  • Explore the derivation of vector identities in quantum mechanics
  • Investigate the representation of 2x2 matrices in terms of spin matrices
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Students and researchers in quantum mechanics, particularly those focusing on spin systems, matrix representations, and vector identities involving Pauli matrices.

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Homework Statement



I'm supposed to derive the following:

\left({\bf A} \cdot {\bf \sigma} \right) \left({\bf B }\cdot {\bf \sigma} \right) = {\bf A} \cdot {\bf B} I + i \left( {\bf A } \times {\bf B} \right) \cdot {\bf \sigma}

using just the two following facts:

Any 2x2 matrix can be written in a basis of spin matrices:

M = \sum{m_\alpha \sigma_\alpha}

which means that the beta-th component is given by

m_\beta = \frac{1}{2}Tr(M \sigma_\beta)


Homework Equations



listed above...

The Attempt at a Solution



It should just be a left side= right side proof.

I started by saying \left({\bf A} \cdot {\bf \sigma} \right) \left({\bf B }\cdot {\bf \sigma} \right) = <br /> <br /> \left(\sum_\alpha a_\alpha \sigma_\alpha \cdot \sigma \right) \left(\sum_\alpha b_\alpha \sigma_\alpha \cdot \sigma \right) = <br /> <br /> \left( \sum_\beta a_\alpha \delta_{\alpha \beta} \right) \left( \sum_\gamma b_\alpha \delta_{\alpha \gamma} \right) = <br /> <br /> \sum_\beta a_\beta \sum_\gamma b_\gamma =<br /> <br /> \frac{1}{2} Tr(A \sigma_\beta) \frac{1}{2} Tr(B \sigma_\gamma)

Not sure if this is even the right way to start, and I can't see at all where I would go from here to get the appropriate RHS of the identity I'm proving. Any ideas? :confused:
 
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You can't expand A or B into Pauli matrices because neither of them are 2x2 matrices. They are both vectors of scalars.

Also, are you able to use any other info, such as the product of two Pauli matrices?
 
Thanks for the reply. I can use the usual commutation relations between the Pauli matrices. Also, I know the Pauli spin matrices are traceless... not sure if that's helpful or not.
 
Yes, that helps a lot. First you will need to make a 2-D matrix from the original equation. You had the right idea originally but expanded too much early on. You should do something like:

(A\cdot\sigma)(B\cdot\sigma) = \sum_{i} a_i \sigma_i \sum_{j} b_j \sigma_j<br /> = \sum_{ij}a_i b_j \sigma_i \sigma_j<br /> =M

That last term is now a 2d matrix. You will expand that in spin matrices including the identity. I suggest solving for the m_0 (identity piece) first, since that will give you your dot product. Be sure to use properties of the trace. For example, a_i and b_j are scalars. And you can break the sum of matrices in a trace up into a sum of traces of the individual matrices.
 
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