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Expectation Value of Composite System

  1. Oct 21, 2014 #1

    Sci

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    1. The problem statement, all variables and given/known data
    System of 2 particles with spin 1/2. Let
    [tex]
    \vert + \rangle =
    \begin{pmatrix}
    0 \\
    1
    \end{pmatrix} \\
    \vert - \rangle =
    \begin{pmatrix}
    1 \\
    0
    \end{pmatrix}
    [/tex]

    singlet state [tex]
    \vert \Phi \rangle = \frac{1}{\sqrt{2}} \Big( \vert + \rangle \otimes \vert - \rangle - \vert - \rangle \otimes \vert + \rangle \Big)
    [/tex]
    observables:
    [tex]
    (2 \vec{a} \vec{S}^1) \otimes 1 \\
    (2 \vec{a} \vec{S}^1) \otimes (2 \vec{b} \vec{S}^2)
    [/tex]
    for arbitraty a,b
    2. Relevant equations
    [tex]
    S_x^i=
    \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix}
    [/tex]
    and similar for S_y and S_z
    3. The attempt at a solution
    I have to calculate
    [tex]
    \langle \Phi \vert(2 \vec{a} \vec{S}^1) \otimes 1 \vert \Phi \rangle
    [/tex]
    in the first task. Does the tensor product notation of Phi means that particle A is in state + and particle B is in state - or the other way round?
    Does the 1 in the observable means that the state of B is simply ignored in the meaurement?
    So does the first case simplify to
    [tex]
    \langle + \vert a_1 \hat S_x +a_2 \hat S_y +a_3 \hat S_z \vert + \rangle -
    \langle - \vert a_1 \hat S_x +a_2 \hat S_y +a_3 \hat S_z \vert - \rangle
    [/tex]
    and the expectation value becomes zero, as expected for the singlet state
     
  2. jcsd
  3. Oct 21, 2014 #2

    Orodruin

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    The state ##|+\rangle \otimes |-\rangle## is the state with the first particle in the + state and the second particle in the - state. The singlet state ##|\Phi\rangle## that you have is a linear combination of ##|+\rangle \otimes |-\rangle## and ##|-\rangle \otimes |+\rangle##.

    In general:
    $$
    (\hat A\otimes \hat B)(|a\rangle \otimes |b\rangle) = (\hat A |a\rangle) \otimes (\hat B |b\rangle).
    $$
    This should help you solve your problem.
     
  4. Oct 22, 2014 #3

    Sci

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    Thank you!
    I am still confused about the basic rules
    [tex]
    (\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert + \rangle \otimes \vert - \rangle ) \\
    -(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert - \rangle \otimes \vert + \rangle )\\
    -(\langle - \vert \otimes \langle + \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert + \rangle \otimes \vert - \rangle )\\
    +(\langle - \vert \otimes \langle + \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert - \rangle \otimes \vert + \rangle )
    [/tex]
    using your rule
    [tex]
    (\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a}\vert + \rangle \otimes 1\vert - \rangle ) + others
    [/tex]
    can I do the following step;
    [tex]
    (\langle + \vert \vec{S} \vec{a}\vert + \rangle \otimes \langle - \vert 1 \vert - \rangle) + others
    [/tex]

    the tensor product doesn't mke sense here...
     
    Last edited: Oct 22, 2014
  5. Oct 22, 2014 #4

    Orodruin

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    When taking the inner product between two states, you simply get the product of the inner products on the separate spaces:
    $$
    (\langle a'|\otimes\langle b'|)(|a\rangle\otimes |b\rangle)
    = \langle a'|a\rangle \langle b'|b\rangle
    $$
    There is no need to keep the tensor product and as you say it makes little sense to do so.

    Edit: that said, you are otherwise on the right track.
     
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