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A Angular momentum of Dirac particle

  1. Apr 6, 2016 #1

    ShayanJ

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    I'm reading Sakurai's "Advanced Quantum Mechanics" (which is different from his "Modern Quantum Mechanics"). In chapter 3, which is about the Relativistic Quantum Mechanics of spin 1/2 particles, after discussing the covariance of the Dirac equation, he goes on to give some examples to clarify the significance of the operator S (as in ## \psi'(x')=S \psi(x) ##). In the first example, he considers an infinitesimal rotation around the z-axis and so he considers the transformation:

    ## \psi'(x')= \left[ 1+i\Sigma_3 \frac{\delta \omega}{2} \right] \psi(x) \ \ \ \ (1)##

    where ## \vec{x}'=\vec x+\delta \vec{x} ## and ## \delta \vec x=(x_2 \delta \omega,-x_1 \delta \omega,0)##.

    He then notes that:
    ## \psi'(x')=\psi'(x)+\delta x_1 \frac{\partial \psi'}{\partial x_1}+\delta x_2 \frac{\partial \psi'}{\partial x_2} \ \ \ \ (2)##

    Then from all the above, he concludes that:

    ## \psi'(x)=\left[ 1+ i \Sigma_3 \frac{\delta \omega}{2}-\left(x_2 \delta \omega \frac{\partial}{\partial x_1}-x_1 \delta \omega \frac{\partial}{\partial x_2} \right) \right] \psi(x) \ \ \ \ (3)##

    It seems he rearranged equation 2 and then used equation 1 to substitute for ## \psi'(x') ##. But there still remain the derivative terms which are derivatives of ## \psi'(x) ## which should somehow be related to the derivatives of ## \psi(x) ##. But he seems to just assumes that ## \frac{\partial \psi'(x)}{\partial x_i}=\frac{\partial \psi(x)}{\partial x_i }## which doesn't seem justified at all! In fact using equation 1, you can easily show how far this is from truth! Am I missing something here?

    Thanks
     
  2. jcsd
  3. Apr 6, 2016 #2

    vanhees71

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    You only take contributions up to order ##\mathcal{O}(\delta \omega)##. So there's no mixing of derivatives with the spin matrix ##\Sigma_3##.
     
  4. Apr 6, 2016 #3

    dextercioby

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    Sakurai uses the unfortunate x_4 = ict, the only flaw of this book. If you are into well-written but old books, you've got from the 60s either the monograph by S. Schweber, or the two books by Bjorken and Drell.
     
  5. Apr 6, 2016 #4

    ShayanJ

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    Yeah...I like Sakurai's too, but x=ict really hurts sometimes!
    I just wish my first QFT course were with someone else so I didn't have to meet this book in this situation! I'm starting to feel I have a very bad luck regarding the professors I have to take my courses with!
    And I appreciate any well-written book because I like to search for insights that are present only in a few books.
     
  6. Apr 6, 2016 #5

    dextercioby

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    Yes, unfortunately Japanese books are marred by this annoying issue, but most of them are still good. I particularly like Umezawa's QFT book, North Holland, 1956. It has nice sections on general relativistic equations.
     
  7. Apr 6, 2016 #6

    samalkhaiat

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    It is extremely important to know about the calculus of infinitesimals. When you do infinitesimal coordinate transformation [tex]\bar{x} = x + \epsilon \delta x ,[/tex]
    where [itex]\epsilon \ll 1[/itex] is the infinitesimal parameter, and [itex]\delta x[/itex] is some arbitrary function of x, the word infinitesimal means you can always put [itex]\epsilon^{2} = 0[/itex]. This leads to many useful relations. The first and most important one is the chain rule operator:
    [tex]\epsilon \frac{\partial}{\partial x} = \epsilon \frac{\partial \bar{x}}{\partial x} \frac{\partial}{\partial \bar{x}} = \epsilon \frac{\partial}{\partial x}(x + \epsilon \delta x) \frac{\partial}{\partial \bar{x}}.[/tex]
    Since [itex]\epsilon^{2} = 0[/itex], we get
    [tex]\epsilon \frac{\partial}{\partial x} = \epsilon \frac{\partial}{\partial \bar{x}} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/tex]
    Now, consider the following infinitesimal transformation of some object [itex]\varphi (x)[/itex]
    [tex]\bar{\varphi}(\bar{x}) = \varphi (x) + \epsilon \Sigma \varphi (x) ,[/tex] which is Eq(1) in your post. Using (1), and [itex]\epsilon^{2} = 0[/itex], you find the following important relation (no coordinate expansion yet)
    [tex]\epsilon \frac{\partial}{\partial \bar{x}} \bar{\varphi}(\bar{x}) = \epsilon \frac{\partial}{\partial x} \varphi(x) . \ \ \ \ \ \ \ \ \ \ (2)[/tex]
    Now, if we do the coordinate expansion
    [tex]\bar{\varphi}(\bar{x}) = \bar{\varphi}( x + \epsilon \delta x ) = \bar{\varphi}(x) + \epsilon \delta x \ \frac{\partial}{\partial x} \bar{\varphi}(x) ,[/tex] substitute in (2) and use [itex]\epsilon^{2} = 0[/itex] again, we get
    [tex]\epsilon \frac{\partial}{\partial \bar{x}} \bar{\varphi}(x) = \epsilon \frac{\partial}{\partial x} \varphi(x) . \ \ \ \ \ \ \ \ \ \ (3)[/tex]
    On the left hand side, if we use (1) again, we obtain the following very important relation (which justify Sakurai’s action in replacing [itex]\bar{\varphi}(x)[/itex] with [itex]\varphi (x)[/itex])
    [tex]\epsilon \frac{\partial}{\partial x} \bar{\varphi}(x) = \epsilon \frac{\partial}{\partial x} \varphi(x) . \ \ \ \ \ \ \ \ \ \ (4)[/tex]
    So, always remember to use the equations (1)-(4), when you deal infinitesimal transformations.
     
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