# Tag Info

## Hot answers tagged notation

7

The wedge product has its roots in exterior algebra. Exterior algebra lets you talk about objects like planes or volumes as algebraic elements of their own, separate from ordinary vectors, but still obeying the same notions of being "vectors" in their own vector spaces. The wedge product of two vectors is a bivector, and many concepts you may have been ...

5

Canonically, the wedge product is distinct from the cross product and should not be confused, however in three dimensions they are inextricably linked. The wedge product (or outer product) comes from exterior algebra, first due to Grassmann who generalized vector products to arbitrary dimensions. This would later be extended by Clifford into the Clifford ...

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This is the same notation that you'll find in Weinberg's books. $$(\psi, \chi)$$ is the inner product of the two states $\psi$ and $\chi$, and corresponds to $$\langle \psi \mid \chi \rangle$$. So, the above corresponds literally to $$\frac{1}{\sqrt{\langle \psi_k \mid \psi_k \rangle}} \left| \psi_k \right>$$ This new object is just the normalized ...

3

What you really want to know are the definitions of the $\sigma_i$ --- these are the Pauli matrices: $$\sigma_x = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \qquad \sigma_y = \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} \qquad \sigma_z = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$ Hopefully you can see now how the equation ...

3

As stated, $\mathbf{n}$ is a unit vector and $n_x$, $n_y$ and $n_z$ are its cartesian components. $\mathbf{n}$ is just a vector pointing in an arbitrarily direction with magnitude 1. Taking $\mathbf{n} \cdot \mathbf{\sigma}$, we have \mathbf{n} \cdot \mathbf{\sigma} = n_x\sigma_x + n_y \sigma_y + n_z \sigma_z \\ = n_x \left(\begin{array}{cc} ...

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