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The Pauli Equation is given by

$$\left[\frac{1}{2m}\left[({\bf\hat{p}}-q {\bf A})^2-q\hbar{\bf\sigma}\cdot {\bf B}]+q\phi\right]\right]|\psi\rangle=i\hbar\frac{\partial}{\partial t}|\psi\rangle.$$

This contains a component ${\bf\hat{p}}^2 |\psi\rangle$. However, according to Wikipedia:

The state of the system, |ψ⟩ (written in Dirac notation), can be considered as a two-component spinor wavefunction

So $|\psi\rangle$ is a vector with two elements, as a function of time. Therefore, I think that $\bf{\hat{p}}$ can be viewed as a $2\times2$ matrix. What are the matrix elements?

If it is just $-i\hbar$ times the identity matrix, why is the equation not simplified to

$$\left[\frac{1}{2m}\left[(-i\hbar\bf{I}-q\bf{A})^2-q\bar{h}\bf{\sigma}\cdot \bf{B}]+q\phi\right]\right]|\psi\rangle=i\hbar\frac{\delta}{\delta t}|\psi\rangle,$$ with $\bf{I}$ the identity?

If it is $-i\hbar\nabla$ times the identity matrix, then how can we take the laplacian of an element of the two-spinor? After all, each element is just a complex number.

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    $\begingroup$ As usual, if $\psi\in L^2\otimes \mathbb C^2$, for example, then if something like $p^2$ is written, people usually mean $p^2\otimes \mathbb I$, where $\mathbb I$ is the identity matrix on $\mathbb C^2$ $\endgroup$ Jan 14, 2023 at 19:02
  • $\begingroup$ So we can replace the momentum operator by $-i\bar{h}$? $\endgroup$
    – Riemann
    Jan 14, 2023 at 19:04
  • $\begingroup$ Because Wikipedia says: The state of the system, $|\psi\rangle$ (written in Dirac notation), can be considered as a two-component spinor wavefunction $\endgroup$
    – Riemann
    Jan 14, 2023 at 19:06
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    $\begingroup$ Indeed, the momentum may be represented by $-i\hbar \nabla$ when acting on any component of the 2-spinor: there is an implicit 2-d identity matrix in it, when acting on spinor space. Is this your question? $\endgroup$ Jan 14, 2023 at 19:11
  • $\begingroup$ Yes that is my question $\endgroup$
    – Riemann
    Jan 14, 2023 at 19:16

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As mentioned in the comments, it is understood that $p^2$ stands for $p^2 \otimes \mathbb{I}$. Since $p^2 = - \hbar^2 \nabla^2$, this means you just apply $p^2$ to each component of the spinor separately. In other words, the Laplacian is just applied to each entry separately.

Furthermore, don't forget that each component of the spinor is a function. The state of the system is a two-component spinor wavefunction. This means it is composed of two wavefunctions stacked on a spinor.

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