Where Does the Problem Stem from in Decomposing Spinor Products?

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Homework Help Overview

The discussion revolves around the decomposition of spinor products in the context of quantum mechanics, specifically focusing on the tensor product of two spin-1/2 particles. Participants are exploring the mathematical representation of these spinor products and their relation to vector quantities.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive the decomposition of spinor products using tensor notation and are discussing the symmetric and antisymmetric components of the spinors. Questions arise regarding the comparison of these components to vector quantities and the potential for sign errors in the calculations.

Discussion Status

Some participants have offered guidance on extracting vector components from the spinor products and have noted discrepancies in signs and conventions used. There is an ongoing exploration of how to reconcile different forms of the expressions and the implications of these differences.

Contextual Notes

There are mentions of potential sign errors and differing conventions regarding the use of the spinor metric and Levi-Civita tensor, which may affect the interpretation of the results. Participants are also considering the implications of index raising and lowering in their calculations.

Slereah
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I am currently trying to find how to derive the decomposition for two particles via the tensor notation, for instance for the product of two particles of spin 1/2 :

\frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0

Giving the components of spin 1 and 0. So to do it, I write down the product of two spinors and write it as its symmetric and antisymmetric part :

\psi_a \chi_b = \frac{1}{2} (\psi_{[a} {\chi_{b}}_] + \psi_{\{a} {\chi_{b}}_\})

The antisymmetric part is the scalar part, being simply

\psi_{[a} {\chi_{b}}_] = \left( \!\!\begin{array}{cc}<br /> 0&amp;-1\\<br /> 1&amp;0<br /> \end{array}\! \right) (\psi^+ \chi^- - \psi^- \chi^+)

And the symmetric part is something of the form

\psi_{\{a} {\chi_{b}}_\} = \left( \!\!\begin{array}{cc}<br /> 2\psi_+ \chi_+&amp;\psi_+ \chi_- + \psi_- \chi_+\\<br /> \psi_+ \chi_- + \psi_- \chi_+&amp;2\psi_- \chi_-<br /> \end{array}\! \right)

Which looks like it contains all the right components, but then I try to compare it to the actual vector quantity :

V^{ab} = \varepsilon^{ca} V_c^b = \varepsilon^{ca} (\sigma^\mu)_c^b V_\mu = \left( \!\!\begin{array}{cc}<br /> - V_x - i V_y &amp;V_z\\<br /> V_z&amp;V_x - i V_y<br /> \end{array}\! \right)

There may be a sign wrong in here somewhere because of the spinor metric (\varepsilon[\tex], the levi-civita tensor), but even up to a sign, the symmetric part does not really lend itself to being put in vector form, and it would seem to mix the states (+,+) with (-,-) as well. I tried using directly the product of a contravariant and covariant spinor, but it did not help. So where does the problem stem from?
 
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you can extract the vector by using

V^{ \mu } = (\varepsilon \sigma^{ \mu })^{ab} V_{ab}

and shouldn't one of your i's have a positive coefficient
 
Extracting the vector isn't the hard part, but merely comparing the two gives me something like

\psi_+ \chi_+ \propto - V_x - i V_y

and

\psi_- \chi_- \propto V_x - i V_y

But the vector should be something like

(\psi_+ \chi_+, \psi_- \chi_-,\frac{1}{\sqrt{2}}(\psi_+ \chi_- + \psi_- \chi_+)

Plus maybe some factors.

One of the i's does have a + when using the transformation

\sigma^\mu V_\mu

But this disappears after index raising with the spinor metric epsilon (I also now realize that this should be index lowering and not raising, but the difference between the two is just a sign, which does not solve the problem)
 
I calculated there to be a +i but i guess you are using different conventions . Anyway if you did have the +i you can reduce to the form you want

using 2 \psi_{+} \chi_{+} = -V_x - i V_y and 2 \psi_{-} \chi_{-} = V_x + i V_y you can get \psi_{+} \chi_{+} = - \psi_{-} \chi_{-}

from which you can write 2 \psi_{+} \chi_{+} = \psi_{+} \chi_{+} - \psi_{-} \chi_{-} which is in the form you want up to some minus signs
 
Last edited:

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