The Pauli equation (seen here) contains its spin dependence in the term which reads(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\frac{1}{2m}\left[ \sigma\cdot\left(p-\frac{e}{c}A\right)\right]^2[/tex]

So let B be any vector. Then

[tex]\left( \sigma\cdot B\right)^2[/tex]

[tex]=\left(\sigma_1 B_1 +\sigma_2 B_2 + \sigma_3 B_3\right)\left(\sigma_1 B_1 +\sigma_2 B_2 + \sigma_3 B_3\right)[/tex]

[tex]=\sigma_1^2 B_1^2 +\sigma_2^2 B_2^2+\sigma_3^2 B_3^2 + (\sigma_1\sigma_2+\sigma_2\sigma_1)B_1 B_2 + (\sigma_1\sigma_3+\sigma_3\sigma_1)B_1 B_3 + (\sigma_3\sigma_2+\sigma_2\sigma_3)B_3 B_2[/tex]

[tex]=1\cdot B_1^2 + 1\cdot B_2^2 + 1\cdot B_3^2 + 0 + 0 +0[/tex]

[tex]=B^2[/tex]

So isn't the sigma-dependent term in the Pauli equation identically equal to

[tex]\frac{1}{2m}\left(p-\frac{e}{c}A\right)^2[/tex]

?

if yes, then in what sense is it spin-dependent? If no, then where did I go wrong?

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# Why isn't the Pauli equation equivalent to the Schrodinger equation?

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