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Expectation value of composite observable in singlet state

  1. Jul 13, 2014 #1
    1. The problem statement, all variables and given/known data

    I've been reading Leonard Susskind's Theoretical Minimum volume on QM, and enjoying it quite a bit - the book doesn't include exercise solutions at the end though, and if they exist online for this volume I haven't been able to find them. (Perhaps if such solutions were accumulated somewhere they'd be of interest to readers of the book).

    One exercise asks you, given a system of two spins in the singlet state ([tex] |sing \rangle = \frac{1}{\sqrt{2}} (|ud \rangle - |du \rangle) [/tex] to calculate the expectation value for the observable [tex]\sigma_x\tau_y[/tex] (with one observer measuring their spin along the x axis and the other observing along the y).

    I've been getting some anomalous results trying to do this calculation, I'm wondering if someone might be able to spot where things are going wrong.

    2. Relevant equations

    [tex]
    |sing \rangle = \frac{1}{\sqrt{2}} (|ud \rangle - |du \rangle) \\
    \langle sing| = \frac{1}{\sqrt{2}} (\langle ud| - \langle du|) \\

    \tau_y|ud \rangle = -i|uu \rangle \\
    \tau_y|du\rangle = i|dd\rangle \\

    \sigma_x|dd\rangle = |ud\rangle \\
    \sigma_x|uu\rangle = |du\rangle

    [/tex]

    3. The attempt at a solution

    First,
    [tex]
    \langle\sigma_x\tau_y\rangle = \langle sing|\sigma_x\tau_y|sing\rangle \\
    [/tex]

    [itex]
    \begin{align}

    \sigma_x\tau_y|sing\rangle &= \sigma_x[\frac{1}{\sqrt{2}}(-i |uu\rangle + i |dd\rangle)]\\
    &=\frac{i}{\sqrt{2}}\sigma_x(|dd\rangle - |uu\rangle)\\
    &=\frac{i}{\sqrt{2}}(|ud\rangle - |du\rangle)\\
    \end{align}
    [/itex]

    Then completing the inner product and taking the product of othogonal vectors as 0,

    [tex]
    \begin{align}
    \langle\sigma_x\tau_y\rangle &= \frac{i}{\sqrt{2}}\frac{1}{\sqrt{2}}(\langle ud| - \langle du|)(|ud\rangle - |du\rangle)\\
    &=\frac{i}{2}(1-0-0+1)\\
    &=i
    \end{align}
    [/tex]

    That can't be right since the expectation value of an observable surely has to be real. So I decided to try an alternate method, using the trace of the projection operator of sing> and the observable, i.e.

    [tex]
    \begin{align}
    \langle sing|\sigma_x\tau_y|sing\rangle &= Tr~ |sing\rangle~ \langle sing|~\sigma_x\tau_y\\
    &=\sum_{i}\langle i|sing\rangle\langle sing|\sigma_x\tau_y|i\rangle\\
    \end{align} \\
    \text{i ranging over the basis vectors uu,ud,du,uu}
    [/tex]

    Since <uu|sing> and <dd|sing> are both 0, this has two terms:
    [tex]
    \langle sing|\sigma_x\tau_y|sing\rangle = \langle ud|sing\rangle\langle sing|\sigma_x\tau_y|ud\rangle + \langle du|sing\rangle\langle sing|\sigma_x\tau_y | du\rangle\\
    [/tex]

    [tex]
    \langle ud|sing\rangle = \frac{1}{\sqrt{2}} \\
    \langle du|sing\rangle = \frac{-1}{\sqrt{2}} \\

    \begin{align}
    \langle sing|\sigma_x\tau_y &= \frac{1}{\sqrt{2}}(\langle dd| - \langle uu|) \tau_y \\
    &= \frac{1}{\sqrt{2}}(-i \langle du| - i\langle ud|)\\
    &= \frac{-i}{\sqrt{2}}(\langle ud| + \langle du|)\\
    \end{align}
    [/tex]

    [tex]
    \begin{align}
    \langle sing|\sigma_x\tau_y|sing\rangle &= \frac{1}{\sqrt{2}}\frac{-i}{\sqrt{2}} \langle(\langle ud| + \langle du|) | ud\rangle\\
    &+ \frac{1}{\sqrt{2}}\frac{-i}{\sqrt{2}}\langle(\langle ud| + \langle du|) | du\rangle\\
    &= \frac{-i}{2}(1+1)\\
    &= -i
    \end{align}
    [/tex]

    That still doesn't look right. Finally, I tried working out the matrix representation of [tex]|sing\rangle \langle sing | \sigma_x\tau_y[/tex] directly:

    By Kronecker product,

    [tex]
    \sigma_x\tau_y =
    \begin{pmatrix}
    0 & 0 & 0 & i\\
    0 & 0 & i & 0\\
    0 & -i & 0 & 0\\
    i & 0 & 0 & 0\\
    \end{pmatrix}
    [/tex]

    For the matrix representation of [tex]|sing\rangle \langle sing|[/tex], the i,jth entry in the matrix is
    [tex]\langle sing|j\rangle \langle i|sing\rangle [/tex]

    Since |sing> is a superposition of |ud> and |du>, the inner products of |sing> with |uu> and |dd> are all 0,
    [tex]
    \langle sing|ud\rangle \langle ud|sing\rangle = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{1}{2} \\
    \langle sing|ud\rangle \langle du|sing\rangle = \frac{1}{\sqrt{2}} \frac{-1}{\sqrt{2}} = \frac{-1}{2} \\
    \langle sing|du\rangle \langle ud|sing\rangle = \frac{-1}{\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{-1}{2} \\
    \langle sing|du\rangle \langle du|sing\rangle = \frac{-1}{\sqrt{2}} \frac{-1}{\sqrt{2}} = \frac{1}{2} \\
    [/tex]

    So [tex]|sing\rangle \langle sing| =
    \begin{pmatrix}
    0 & 0 & 0 & 0\\
    0 & \frac{1}{2} & \frac{-1}{2} & 0\\
    0 & \frac{-1}{2} & \frac{1}{2} & 0\\
    0 & 0 & 0 & 0\\
    \end{pmatrix}
    [/tex]

    [tex]

    |sing\rangle \langle sing|\sigma_x\tau_y =
    \begin{pmatrix}
    0 & 0 & 0 & i\\
    0 & 0 & i & 0\\
    0 & -i & 0 & 0\\
    i & 0 & 0 & 0\\
    \end{pmatrix}

    \begin{pmatrix}
    0 & 0 & 0 & 0\\
    0 & \frac{1}{2} & \frac{-1}{2} & 0\\
    0 & \frac{-1}{2} & \frac{1}{2} & 0\\
    0 & 0 & 0 & 0\\
    \end{pmatrix} =
    \begin{pmatrix}
    0 & 0 & 0 & 0\\
    0 & \frac{i}{2} & \frac{i}{2} & 0\\
    0 & \frac{-i}{2} & \frac{-i}{2} & 0\\
    0 & 0 & 0 & 0\\
    \end{pmatrix}

    [/tex]

    So the trace consisting of the diagonal elements, Tr = [tex]\frac{i}{2} - \frac{i}{2}[/tex] = 0

    So 0 at least makes sense as an expectation value, although I'm not too sure now if it's the correct one.
     
  2. jcsd
  3. Jul 13, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, ostrich2. Welcome to PF!

    Check your signs on the right hand side of this equation.
     
  4. Jul 15, 2014 #3
    Aha - thanks, just looking at [tex]\tau_y|sing\rangle[/tex], I think the correct result is:

    [tex]\tau_y|sing\rangle = \frac{-i}{\sqrt{2}}(|uu\rangle + |dd\rangle)[/tex]

    So, the eventual result is

    [tex]\frac{1}{\sqrt{2}}\frac{-i}{\sqrt{2}}(\langle ud| - \langle du|)(|du\rangle + |ud\rangle)[/tex]

    which gives 0. Let me go back to the other calculation - there must be some similar sign problem there as well.

    Now I'm scratching my head a bit about the result - this seems to imply that even though the system is in an entangled state, and the individual expectation values [tex]\langle\sigma_x\rangle[/tex] and [tex]\langle\tau_y\rangle[/tex] are still 0, the correlation is also 0, almost as if it was a product state?
     
  5. Jul 30, 2014 #4
    Hi ostrich2, sorry for this rather late reply :frown:

    I'd certainly expect a correlation between [itex]\sigma_z[/itex] and [itex]\tau_z[/itex] for the singlet state, but I don't see why you think there would be a correlation between [itex]\sigma_x[/itex] and [itex]\tau_y[/itex].

    Roughly speaking: knowing what [itex]\sigma_x[/itex] is gives you no information whatsoever about what [itex]\tau_y[/itex] is.

    [disclaimer - I didn't check your calculations, but I think your answer is correct]
     
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