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A question for Sakurai's advanced qm

  1. Oct 30, 2011 #1
    On page 96, he defined [itex]\psi'(x')=S\psi(x)[/itex], but it seems what he later derived for S only transforms the spinor part not the space-time coordinate of the 4-component wave function, then how is the space-time coordinate primed after acted by S? I'm pretty sure it's not a typo according to what he did on page101, eqns (3.158)~(3.161).
     
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  3. Oct 30, 2011 #2
    I had Sakurai's advanced qm a few month ago.
    But now I don't. Sorry, I imagine from your sentence.

    First, Dirac's solution includes exponential function

    [tex]\psi = \int_{-\infty}^{\infty} \frac{d^3 p}{\sqrt{2E_p}} a_p u(p) e^{-ipx} \cdots[/tex]

    Here, both the p (momentum, energy) and x (time , spece) are 4 vectors.
    So "px" means scalar (= vector x vector). (px doesn't change under Lorentz transformation.)
    And the integration of p is from -infinity to +infinity, so the change from p to p' is meaningless.

    But of course, the differentiation by time and space coordinate in Dirac equation changes under Lorentz transformation ( x' = a x ).

    [tex]\frac{\partial }{\partial x'_{\mu}} = \sum_{\nu} \frac{\partial x_{\nu}}{\partial x'_{\mu}} \frac{\partial}{\partial x_{\nu}} = \sum_{\nu} a_{\mu \nu} \frac{\partial}{\partial x_{\nu}} [/tex]

    So under Lorentz transformation, Dirac equation becomes

    [tex](-i\hbar \gamma^{\mu} \partial_{\mu}' + mc) \psi' (x') = (-i\hbar \gamma^{\mu} a_{\mu\nu} \partial_{\nu} + mc) S \psi (x) = (-i\hbar \gamma^{\mu} \partial_{\mu} + mc) \psi (x) =0 [/tex]

    Is this what is discussed here ? (If not, I transfer your question to someone with advanced qm.)
     
  4. Oct 30, 2011 #3

    George Jones

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    I won't have access to my copy of Sakarai until tomorrow. Is [itex]S[/itex] the 4-component spinor versionof a Lorentz transformation [itex]L[/itex]? If so, then [itex]x' = Lx[/itex] and
    [tex]
    \begin{align}
    \psi \left( x' \right) &= S \psi \left( x \right) \\
    \psi \left( Lx \right) &= S \psi \left( x \right) \\
    \psi \left( x \right) &= S \psi \left( L^{-1} x \right)
    \end{align}
    [/tex]
     
  5. Oct 30, 2011 #4
    Yes, S is purely written in terms of gamma matrices and parameters like angle and rapidity, and the expression came into my mind first was [itex]\psi'(x')=S\psi({\Lambda}^{-1}x)[/itex]
     
  6. Oct 30, 2011 #5

    George Jones, I want to confirm this equation.
    According to p.60 of an introduction to quantum field theorey by Peskin, Dirac field change under Lorentz transformation,

    [tex] \psi (x) \quad \to \quad \ \psi' (x) = \Lambda_{1/2} \psi (\Lambda^{-1} x) \quad (S = \Lambda_{1/2}) \quad ( x' = \Lambda x ) [/tex]

    Changing x to x', this meaning is the same as

    [tex]\psi' (x') = \Lambda_{1/2} \psi (x) = S \psi (x) \qquad ( x = \Lambda^{-1} x') [/tex]

    S includes gamma matrices, so I think the form of psi changes under Lorentz trandformation.

    Sorry. the next equation is what you mean ?

    [tex] \psi' (x') = S \psi (\Lambda^{-1} x') = S \psi (x) \quad or \quad \psi' (x) = S \psi (\Lambda^{-1} x) \qquad x' = \Lambda x \quad ( x = \Lambda^{-1} x')[/tex]
     
  7. Oct 31, 2011 #6
    Ah..I see where my problem is, I was thinking [tex]\psi' (x) = S \psi (\Lambda^{-1} x)[/tex], which means an active transformation acting on the wavefunction not the reference frame, so [tex] \psi' (x') = S \psi (\Lambda^{-1} x') = S \psi (x)[/tex] means Sakurai transformed both the reference frame and the wavefunction? That's weird, why did he do this?
     
  8. Oct 31, 2011 #7
    I think it is easier to imagine "vector" ( ex, 4-vector potential A(x) ) instead of spinor, first.

    [tex]A^{\mu} (x) = ( A^0 (x), A^1 (x), A^2 (x), A^3 (x) ) \qquad x^{\mu} = (x^0, x^1, x^2, x^3)[/tex]
    This 4-vector A (x) means that there is a thing " vector A " at the position of x ( from the viewpoint of reference frame K ).
    Here we can consider 4-vector A as an thing such as "arrow".
    So there is one "arrow" ( which vector direction is (A0, A1, A2, A3) ) at the coordinate of ( x0, x1, x2, x3 ) in the reference frame of K.

    From the viewpoint of a different reference frame K', the position of the "arrow" changes (under Lorentz transformation),

    [tex]x'^{\mu} = \Lambda^{\mu}_{\nu} x^{\nu} \quad ( x' = \Lambda x ) \qquad x'^{\mu} = ( x'^0, x'^1, x'^2, x'^3 )[/tex]
    This means that from this reference frame K', the "arrow" exists at x'.
    ( x' from K' originally exists at x from K. )
    How do we see the direction of this arrow from this reference frame K' ?

    A is 4-vector, so this direction changes as 4-vector,

    [tex] A'^{\mu} = \Lambda^{\mu}_{\nu} A^{\nu} \quad ( A' = \Lambda A ) [/tex]
    As a result, from the reference frame K', there is an arrow which direction is A' at the coordinate of x'. So,

    [tex] A'^{\mu} (x') = \Lambda^{\mu}_{\nu} A^{\nu} ( \Lambda^{-1} x' ) \qquad x = \Lambda^{-1} x'[/tex]
    ( x' from K' originally exists at x from K. )
    In the case of spinor, the position (= x ) of spinor changes as 4-vector, which is the same as that of vector A above.
    But the change of "spinor direction" is different from the vector A.
    ( As you said, we use "S" instead of "Lambda" )

    [tex] \psi' (x') = S \psi (\Lambda^{-1} x') [/tex]
     
    Last edited: Oct 31, 2011
  9. Nov 1, 2011 #8
    Thanks, your explanation is very clear. I just get confused every now and then on this issue, now i get the moment of clarity, but probably someday I'll get confused again: (
     
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