Tensor Force Operator Between Nucleons: Spin & Position

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SUMMARY

The tensor force operator between two nucleons is defined as ##S_{12}=3\sigma_1\cdot r\sigma_2\cdot r - \sigma_1\cdot \sigma_2##, where ##\sigma_1## and ##\sigma_2## are the Pauli matrices and ##r## is the distance between nucleons. The operator can be rewritten using the total spin operator ##S = \sigma_1 + \sigma_2##, leading to the expression ##S_{12} = \frac{6}{r^{2}} (S \cdot r)^{2} - 2S^{2}##. The derivation involves substituting expressions for the spin operators and applying the identity ##(\sigma\cdot a)(\sigma \cdot b) = a\cdot b + i(a\times b)\sigma##, ensuring that the cross product vanishes when both vectors are equal.

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kelly0303
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Homework Statement


The tensor force operator between 2 nucleons is defined as ##S_{12}=3\sigma_1\cdot r\sigma_2\cdot r - \sigma_1\cdot \sigma_2##. Where r is the distance between the nucleons and ##\sigma_1##and ##\sigma_2## are the Pauli matrices acting on each of the 2 nucleons. Rewrite ##S_{12}## only in terms of the spin operator S and relative position r.

Homework Equations

The Attempt at a Solution


For the second part of the equation I tried this. Using the fact that ##S=\sigma_1+\sigma_2## we have ##\sigma_1\sigma_2=(S^2-\sigma_1-\sigma_2)/2=(S^2-6)/2##. For the first part I was thinking to use this expression ##(\sigma\cdot a)(\sigma \cdot b) = a\cdot b + i(a\times b)\sigma## As in my case a and b are both r, the cross product would vanish and the first term would be just ##r^2##. But I am not sure if I can do that, as my ##\sigma## is not the same in both cases. Can someone tell me if I can use that formula or give me some hint or how to approach the first part of the equation? Thank you!
 
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kelly0303 said:

Homework Statement


The tensor force operator between 2 nucleons is defined as ##S_{12}=3\sigma_1\cdot r\sigma_2\cdot r - \sigma_1\cdot \sigma_2##. Where r is the distance between the nucleons and ##\sigma_1##and ##\sigma_2## are the Pauli matrices acting on each of the 2 nucleons. Rewrite ##S_{12}## only in terms of the spin operator S and relative position r.

Homework Equations

3. The Attempt at a Solution [/B]
For the second part of the equation I tried this. Using the fact that ##S=\sigma_1+\sigma_2## we have ##\sigma_1\sigma_2=(S^2-\sigma_1-\sigma_2)/2=(S^2-6)/2##. For the first part I was thinking to use this expression ##(\sigma\cdot a)(\sigma \cdot b) = a\cdot b + i(a\times b)\sigma## As in my case a and b are both r, the cross product would vanish and the first term would be just ##r^2##. But I am not sure if I can do that, as my ##\sigma## is not the same in both cases. Can someone tell me if I can use that formula or give me some hint or how to approach the first part of the equation? Thank you!
S = \frac{1}{2} (\sigma_{1} + \sigma_{2}) \ \ \Rightarrow \ \ S^{2} = \frac{1}{2} ( 3 + \sigma_{1} \cdot \sigma_{2} ) . \ \ \ \ (1)S \cdot r = \frac{1}{2} \left( ( \sigma_{1} \cdot r ) + ( \sigma_{2} \cdot r ) \right) \ \ \Rightarrow \ \ (S \cdot r)^{2} = \frac{1}{2} \left( r^{2} + ( \sigma_{1} \cdot r )(\sigma_{2} \cdot r ) \right) . \ \ \ \ (2) Substitute (1), (2) in S_{12} = \frac{3}{r^{2}} ( \sigma_{1} \cdot r )( \sigma_{2} \cdot r ) - \sigma_{1} \cdot \sigma_{2} , to find S_{12} = \frac{6}{r^{2}} ( S \cdot r )^{2} - 2S^{2} .
 
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