su(2) group

Discussion in 'Physics & Math' started by oscmh3, Aug 26, 2015.

  1. brucep Valued Senior Member

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    4,098
    He has a problem with logic to start with. He doesn't have much respect for the mathematics to start with much less modern mathematical physics.
     
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  3. arfa brane call me arf Valued Senior Member

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    These days there are several inroads to some understanding of the Standard Model and its symmetries.

    The strictly mathematical approach from representations of Lie groups and Lie algebras happens to coincide with SM physics.
    But you can start with the informational approach too, see if you can find a satisfactory answer to the question: what is quantum information? That isn't a trivial question.
    In the fairly new discipline of Quantum Information Science, all particles with spin not zero are qubits and all qubits have spin symmetries; a spin-1/2 particle can be in a superposition of spin up/down, two spin-1/2 particles can be spin-entangled, but it isn't all about spin or conservation of angular momentum, it's about paths or "histories".

    Since a qubit is a Bloch sphere, mathematically. It might be interesting to construct a Bloch/Riemann sphere and show how the quaternions give the same construction.
     
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  5. QuarkHead Remedial Math Student Valued Senior Member

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    At least let me continue my ramble.

    Note first I typoed. My last basis for \(\mathfrak{su(2)}\) should be

    \(X_3 =\frac{1}{2}\begin{pmatrix}i&0\\0&-i \end {pmatrix}\).

    Anyway, the following is central to Lie theory:

    Given a Lie algebra \(\mathfrak{g}\) the associated Lie group \(G\) is found from the mapping

    \(\text{exp}:\mathfrak{g} \to G\)that is \(\text{exp}(\mathfrak{g})=G\). Explicitly this is

    \(I+X+\frac{X^2}{2!}+\frac{X^3}{3!}+\frac{X^4}{4!}+.......\)

    Applying this to my basis \(X_1=\frac{1}{2}\begin{pmatrix}0&i\\i&0 \end {pmatrix}\) and some Real number \(\alpha\) one gets the convergent series.

    \(\alpha X_1=\begin{pmatrix}1&0\\0&1\end{pmatrix}+(\frac{\alpha}{2})\begin{pmatrix}0&i\\i&0 \end {pmatrix}+\frac{1}{2}(\frac{\alpha}{2})^2\begin{pmatrix}-1&0\\0&-1 \end {pmatrix}+\frac{1}{6}(\frac{\alpha}{2})^3\begin{pmatrix}0&-i\\-i&0 \end {pmatrix}+......\)

    This converges to the matrix in \(SU(2)\)

    \(\begin{pmatrix}\cos \frac{\alpha}{2}&i\sin \frac{\alpha}{2}\\i\sin \frac{\alpha}{2}&\cos\frac{\alpha}{2}\end{pmatrix}\).

    So if we assume that the element \(\begin{pmatrix}1&0&0\\0&\cos \beta&-\sin\beta\\0&\sin \beta& \cos \beta\end{pmatrix}\) of \(SO(3)\)describes a symmetry around the Real \(p\)-axis of the form \(x=2 \pi p\), then we easily convince ourselves that the above element of \(SU(2)\) describes a symmetry of the form \( q=4 \pi q\)

    This is precisely the symmetry we expect for an object with spin \(\frac{1}{2}\). From here it is an easy matter to describe spin states. Later for that.......
     
    Last edited: Sep 4, 2015
    oscmh3 likes this.
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  7. QuarkHead Remedial Math Student Valued Senior Member

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    Another typo, I'm afraid. I said
    This should read (of course)

    "describes a symmetry around the Real \(x\)-axis of the form \(p=2 \pi p\)" Sorry
     

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