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I Derivation of equations using tensor

  1. May 11, 2016 #1
    http://hitoshi.berkeley.edu/221a/tensorproduct.pdf

    I was following the above pdf and got through most of it but am not quite understanding how (41), (42), and (43) are derived.

    It appears that (31) and (41) are representing the same states and are still orthogonal, but how exactly is (41) derived? Aside from a scalar multiplication, the equation in (40) works perfectly well when applying (31) instead of (41).

    For (42), I understand this is a unitary matrix, but what's the motivation and how exactly is it derived from the basis states? To clarify, how many basis states are there?

    For (43), I tried doing the matrix multiplication, and I'm not sure if it's a typo, but how are there two possible solutions for the m = 3/2 and j = 3/2 states when multiplied by U?

    These final equations have given me a bit of trouble in understanding the material, so any help with understanding their derivations would be greatly appreciated!
     
  2. jcsd
  3. May 11, 2016 #2

    blue_leaf77

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    As for (41), I must admit that the author didn't give a clearer explanation about the transition he was making from equation (40) to (41).
    Equation (41) are just the states ##|1/2,1/2\rangle ## and ##|1/2,-1/2\rangle## written in the tensor product basis ##|m_1\rangle \otimes |m_2\rangle##. To obtain those column matrices, you can employ the rule ##m = m_1+m_2##. For example take the ##|1/2,1/2\rangle ## state (the left in (41)), here ##m=1/2##, therefore to write it as
    $$
    |1/2,1/2\rangle = \sum\sum_{1/2=m_1+m_2} C(m_1,m_2) |m_1\rangle \otimes |m_2\rangle
    $$
    you have to find ##|m_1\rangle \otimes |m_2\rangle## which satisfies ##1/2 = m_1+m_2##. The required ##|m_1\rangle \otimes |m_2\rangle## should clearly be the same as that required in expanding ##|3/2,1/2\rangle## (eq. (38)) but the Clebsch-Gordan coefficient ##C(m_1,m_2)## cannot be the same between them, in particular they should be orthogonal, i.e. ##\langle1/2,1/2|3/2,1/2\rangle = 0##. From this it's easy to prove (41).

    As the sentence following (41) says, this matrix was composed by sequentially stacking the (conjugate) transpose of the column matrices (37), (38), and (41). For instance, the first four rows of matrix (41) are (conjugate) transpose of ##|3/2,3/2\rangle##, ##|3/2,1/2\rangle##, ##|3/2,-1/2\rangle##, and ##|3/2,-3/2\rangle##, in their column matrix forms respectively. The motivation of deriving this matrix is to determine the change of coordinate matrix from the tensor product basis ##|m_1\rangle \otimes |m_2\rangle## to ##|J,m\rangle## basis.

    Yes that should be a typo, the last one should be ##|1/2,-1/2\rangle##.
     
    Last edited: May 11, 2016
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