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Griffiths Quantum 4.58

  1. Sep 11, 2007 #1
    [SOLVED] Griffiths Quantum 4.58

    From Griffiths, Intoduction to Quantum Mechanics, 2nd Ed, p198

    1. The problem statement, all variables and given/known data
    Deduce the condition for minimum uncertainty in [itex]S_x[/itex] and [itex]S_y[/itex] (that is, equality in the expresion [itex]\sigma_{S_x} \sigma_{S_y} \geq (\hbar/2)|\langle S_z \rangle | [/itex]), for a particle of spin 1/2 in the generic state.

    Answer: With no loss of generality we can pick [itex]a[/itex] to be real; then the condition for minimum uncertainty is that [itex]b[/itex] is either pure real or else pure imaginary.

    2. Relevant equations

    [tex]\sigma_{S_x} = \sqrt{\langle S_x^2 \rangle - \langle S_x \rangle^2}[/tex]

    [tex]\sigma_{S_y} = \sqrt{\langle S_y^2 \rangle - \langle S_y \rangle^2}[/tex]

    3. The attempt at a solution

    Without loss of generality, let [itex]a[/itex] be real.

    We get that:

    [tex]\langle S_x^2 \rangle = \frac{\hbar^2}{4}(a^2 + |b|^2)[/tex]

    [tex]\langle S_x \rangle = \frac{\hbar}{2}a(b + b^*) = \hbar a Re(b)[/tex]

    [tex]\langle S_x \rangle^2 = \hbar^2 a^2 Re(b)^2[/tex]

    [tex]\langle S_y^2 \rangle = \frac{\hbar^2}{4}(a^2 + |b|^2)[/tex]

    [tex]\langle S_y \rangle = \frac{\hbar}{2}a(b^* - b) = \hbar a Im(b)[/tex]

    [tex]\langle S_y \rangle^2 = \hbar^2 a^2 Im(b)^2[/tex]


    [tex]\sigma_{S_x} = \hbar \sqrt{\frac{1}{4}(a^2 + |b|^2) - a^2 Re(b)^2}[/tex]

    [tex]\sigma_{S_y} = \hbar \sqrt{\frac{1}{4}(a^2 + |b|^2) - a^2 Im(b)^2}[/tex]


    [tex]\sigma_{S_x}\sigma_{S_y} = \hbar^2 \sqrt{\frac{1}{4}(a^2 + |b|^2)(\frac{1}{4}(a^2 + |b|^2) - a^2 Re(b)^2 - a^2 Im(b)^2) + a^4 Im(b)^2 Re(b)^2 }[/tex]

    For the other side of the equation we have:

    [tex]\frac{\hbar}{2}|\langle S_z \rangle | = \frac{\hbar^2}{4}|a^2 - |b|^2 |[/tex]

    I'm not sure how to show that you need either Re(b) or Im(b) to be zero in order for these two things to be equal.
  2. jcsd
  3. Sep 12, 2007 #2
  4. Sep 13, 2007 #3

    George Jones

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    Hint: these expressions for the expectation values are true only if

    [tex]1 = |a|^2 + |b|^2 .[/tex]
    Last edited: Sep 13, 2007
  5. Sep 13, 2007 #4
    Thanks, I knew I must have overlooked something simple.
  6. Sep 13, 2007 #5
    With that substitution we get:

    [tex] \sigma_{S_x}\sigma_{S_y} = \hbar^2 \sqrt{\frac{1}{4}(\frac{1}{4} - a^2 Re(b)^2 - a^2 Im(b)^2) + a^4 Im(b)^2 Re(b)^2 }[/tex]


    [tex] \frac{\hbar}{2}|\langle S_z \rangle | = \frac{\hbar^2}{4}\left| 1-2|b|^2\right|[/tex]

    Substituting in that either [itex]Re(b)[/itex] or [itex]Im(b) = 0[/itex] and setting the two equal gives

    [tex]\left| 1-2|b|^2\right| = \sqrt{1 - 4a^2 |b|^2}[/tex]

    which doesn't look equal to me...
  7. Sep 13, 2007 #6
    nevermind, got it
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