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Quantum Mechanics: Basis

  1. Mar 3, 2015 #1
    1. The problem statement, all variables and given/known data

    How can I find the matrix representation of ##\mathbb{S}_+## and ##\mathbb{S}_-## in the ##|\pm y\rangle## or ##|\pm x\rangle## basis?


    2. Relevant equations

    ##
    \mathbb{\hat{S}}_+|s,m\rangle = \sqrt{s(s+1)-m(m+1)}\hbar|s,m+1\rangle
    ##



    3. The attempt at a solution

    The book almost always ask to find something in the ##z-##basis but rarely asks to find something in ##x## or ##y## basis. I know how this matrix representation will look with the ##z-##basis, so I am wondering how the matrix representation will look in the ##x## or ##y## basis instead of ##z##?
     
  2. jcsd
  3. Mar 3, 2015 #2

    Simon Bridge

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    You find the projections along the unit vectors like normal.
    If you already have the projection along the z axis, then what will the projections along the x and y axes look like?
     
  4. Mar 3, 2015 #3
    Can you elaborate please? Will it just then be a reflection about the axis?
     
  5. Mar 4, 2015 #4

    Simon Bridge

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    Well the way I'm reading it, he's asking for a component.
    "the z basis" does not really mean anything.

    You can find, for eg, the z component of angular momentum right?
    You know how the x and y components are given the z component.
    If you don't, look it up.
     
  6. Mar 4, 2015 #5
    So I will just use ##|+x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle +|-z\rangle)## as my basis and solve it the way I solved for the ##z## basis?
     
  7. Mar 4, 2015 #6

    Simon Bridge

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    Well, again, there is no such thing as a z basis that I know of.
    I think you need to revisit the definitions - what is it that S+ and S- do?
     
  8. Mar 4, 2015 #7

    DrClaude

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    It's a common expression when talking about spin. The z basis for a spin-1/2 particle is the set ##\left\{ |+z\rangle, |-z\rangle\right\}##, with "spin up" corresponding to a spin along +z, and "spin down" to a spin along -z.

    You can express an operator ##\hat{O}## in, e.g., the x basis, as a matrix using
    $$
    \mathbf{O}_x = \begin{pmatrix} \langle +x | \hat{O} | +x \rangle & \langle +x | \hat{O} | -x \rangle \\
    \langle -x | \hat{O} | +x \rangle & \langle -x | \hat{O} | -x \rangle \end{pmatrix}
    $$
    For that, you need of course to find how to apply the operator in the x basis, i.e., ##\hat{O} | +x \rangle##. This you can find by expressing ## | +x \rangle## in terms of ##| +z \rangle## and ##| -z \rangle##.

    Alternatively, you can calculate the rotation matrix ##\mathbf{R}## that brings you from the z basis to the x basis, and find the operator matrix using
    $$
    \mathbf{O}_x = \mathbf{R} \mathbf{O}_z \mathbf{R}^T
    $$
     
  9. Mar 4, 2015 #8
    Very helpful! Thank you very much!
     
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