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3

$\vert+\rangle$ and $\vert-\rangle$ are really just shorthand notations for the two eigenvectors of the diagonal spin operator $\sigma_z$. This means concretely: $$\vert+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$ $$\vert-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$ Therefore the action of the sigma operator gives you simply the corresponding ...


3

Before going further, I would suggest you to read Chapter 13 ("Spinors") of R.Wald's book "General Relativity". In that chapter, you will see that 2-spinors are simply vectors living in a two-dimensional complex vector space. The capital letters in the indices are simply the abstract index notation for these vectors (see Section 2.4 in Chapter 2 of the same ...


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$\newcommand{\ket}[1]{| #1 \rangle}$ An arbitrary spin state $\ket{s}$ can be broken down as a sum of the $\ket{+}$ and $\ket{-}$ eigenstates: $$ \ket{s} = \alpha \ket{+} + \beta \ket{-} $$ Where $\alpha$ and $\beta$ are complex numbers. We'll write the overall vector as: $$ \ket{s} = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} $$ Where we remember that ...


2

Application 1: It's the $z$-component of the vector valued angular momentum observable for a spin $\frac{1}{2}$ particle, when the basis states are the $z$-component angular momentum eigenstates. If this sounds a bit circular and tautological, it is the reason why $\sigma_z$ is diagonal. So the $n^{th}$ moment of the probability distribution of an angular ...


2

I haven't thought about this one before, so here is an approach that will work if you work hard enough at it. Before I begin banging on, point number 1: Should I assume a spin 3/2 system (4x4 Matrix) or an entangled Hilbert space with spin 1/2 and spin 1 (6x6 Matrix)? Unquestionably the latter. It is a bipartite system and its state space is the ...


1

Four component formalism is the "right" formalism, but it has negative energy eigenstates corresponding to the antiparticles. Most chemists and solid state physics are not interested in the antiparticles, and such negative energy solution causes trouble for conventional variational methods, where you might end up falling to negative infinity energy. It is ...


0

Its the first one. This is exactly what the "dagger" does. It transposes the spinor, converting it from a column spinor to a row spinor, and takes every entry to its complex conjugate, i.e: $$ \psi=\begin{pmatrix}\psi_L\\\psi_R\end{pmatrix} \xrightarrow{\dagger} \begin{pmatrix}(\psi^T_L)^* (\psi^T_R)^*\end{pmatrix} = \begin{pmatrix}\psi_L^\dagger ...


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If we choose the signature of the metric $\eta$ to be $(1,-1,-1,-1)$ and choose the gamma matrices $\gamma^{\mu}$ to be unitary (as they can be because they form a representation of a finite group), then it would follow from the commutation relations $\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu}$ that $\gamma^{0}$ would be Hermitian and ...


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The matrix exponential for any Pauli matrix using the general formula [see general formula][1]. This procedure gives, $$\exp[-iHt] = I_{2\times2}\cos \nu_{F}t - i {\bf\sigma}\cdot\left(q-By\hat x\right) \sin \nu_{F}t$$ I tryed to solve using this method but seems not convincing. Any comments would be appreciated.


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EDIT I can now include the other cases, not just when the roots are distinct. Under the transformation $(o,i) \mapsto (o + b i , i)$, we have $\hat{\Psi}_0 (b) = \Psi_0 + 4 b \Psi_1 + 6 b^2 \Psi_2 + 4 b^3 \psi_3 + b^4 \Psi_4 \\ \hat{\Psi}_1 (b) = \Psi_1 + 3 b \Psi_2 + 3 b^2 \Psi_3 + b^3 \psi_4 \\ \hat{\Psi}_2 (b) = \Psi_2 + 2 b \Psi_3 + \Psi_4 \\ ...



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