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$\lambda$ stands for the eigenvalue. Eigenvalue equation is: $S_xX=\lambda X$ $S_xX-\lambda X=0$ $(S_x-\lambda I)X=0$ Since X is eigenfunction, we seek solutions for $det(S_x-\lambda I)=0$ \begin{align} (S_x-\lambda I)= \begin{bmatrix} 0 & \frac{\hbar}{2} \\ \frac{\hbar}{2} &0 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & ...

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Take case for an $n\times n$ matrix $A$. To find its eigenvalues, first you write the eigenvalue equation for it. $$Au=\lambda u$$ where $u$ are its eigenvectors. This can be rewritten in the following way $$Au-\lambda u=(A-\lambda I)u=0$$ with $I$ the identity matrix. Let $A-\lambda I=B$, and we know that the equation $Bu=0$ has a non zero solution $u$ ...

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The entirety of the modern quantum mechanics literature uses inner products that are linear in the second argument, and antilinear in the first one. Mathematicians often use the other convention, but I've never seen it used in physics. This is of course pure convention, but you will find grief, at least when you try to publish, if you go against the flock ...

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How do we prove that any directions are orthogonal? [...] we can use the pythagorean theorem. This involves of course a definition of (how to measure or compare) "angle(s)" in the first place; such that one may comprehend statements about (distinct) angles being "equal" (or else: "not equal") for instance in Euclid's 4th axiom (on "right angles") or in ...

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It depends on how you define orthogonality, or, as OSE puts it in his comment, "Orthogonality is usually tested using some defined inner product." I'll expand on this a bit. In order to mathematically answer the question Is direction A orthogonal to direction B? we need a definition of the terms "direction" and "orthogonal." The standard ...

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Mark Mitchison is right. The concept of entanglement in systems of indistinguishable particles is more controversial than it is in the case of systems composed of distinguishable subsystems. You need to define first what do you mean by it when it comes, for example, to fermions. Do you mean entanglement between particles (connected with single Slater ...

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A vector in a (polynimial) $gl(3)$ representation is highest weight if it is annihilated by the raising root operators $a_jk = b_j^{\dagger}b_k$, $k>j$. In our case, the relevant operators are $a_{12}$, $a_{23}$, and $a_{13}$. We do not need to check the third case, because $a_{13} = [ a_{12}, a_{23}]$ is given by the commutator of the two other ...

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