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Peskin page 46 to 47

  1. Sep 12, 2014 #1
    Hi, all

    I'm reading peskin by myself.
    I can't understand from eq(3.50) to eq(3.53).

    i) What should I interpret [itex]\sqrt{p\cdot\sigma}[/itex]?
    I guess below, but I can't understand [itex]\sqrt{\;\;}[/itex] of matrices.

    p\cdot\sigma=E \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) - p^3\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) \\
    = \left(\begin{array}{cc} E-p^3 & 0 \\ 0 & E+p^3 \end{array}\right)

    And why is it the same as (3.49)?

    ii)How can I confirm (3.50) is a solution of the Dirac equation?

    iii)What's meaning of "large boost" in (3.52) and (3.53)?
    If I understand the Dirac spinor more, is it easy transform?
    When so, where can I study Dirac spinor easily?

    Thanks in advance!
    Last edited: Sep 12, 2014
  2. jcsd
  3. Sep 12, 2014 #2


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    i. In Peskin they immediately explain what they mean by sqrt of a matrix right after its use...
    then it's just algebra...

    ii. you just put it in dirac's equation...

    iii. Large boost means that you are doing a large boost... a boost is described by the parameter [itex]\eta[/itex] (in the same way rotations are described by [itex]\theta[/itex]). At the limit [itex]\eta \rightarrow \infinity[/itex] you get that result...
  4. Sep 12, 2014 #3
    Thanks ChrisVer.
    I'm not native, so a little difficult to understand that sentense. Sorry,but I ask in another words.

    If [itex]A=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)[/itex],
    [itex]\sqrt{A}=\left(\begin{array}{cc} \sqrt{eigenvalue1} & 0 \\ 0 & \sqrt{eigenvalue2} \end{array}\right)[/itex]?
    I want to know the component expression.
  5. Sep 12, 2014 #4


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    then you have to find the eigenvalues of your matrix A.... :) except for the case the A is in diagonal form...
    for the eigenvalues of a general matrix, you find them by solving the characteristic polynomial of [itex]\alpha[/itex]:
    [itex] det(A-I \alpha)=0[/itex]
    [itex]\alpha[/itex] are the eigenvalues. [itex]I[/itex] the unitary matrix... det=determinant...
  6. Sep 12, 2014 #5
    I understand.
    Thank you!
  7. Sep 12, 2014 #6


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    sorry I just saw Peskin did a boost only along the 3 direction ok...
    Last edited: Sep 12, 2014
  8. Sep 20, 2014 #7


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    Hello, I am sorry to come up here again, but I just read these parts in Peskin.
    I don't understand your question now about the "Easy to transform spinor", But I can answer better the question about the large boosts.
    You have the quantities [itex]\sqrt{E \pm p_3} [/itex] multiplying the 2-component spinors.
    Now a large boost means that you are letting [itex]p_3[/itex] become large... In this case [itex]E = \sqrt{m^2 + p_3 ^2} \approx p_3[/itex]
    So the [itex]\sqrt{E - p_3}= 0[/itex] and [itex]\sqrt{E + p_3}= \sqrt{2E} [/itex]
  9. Sep 20, 2014 #8


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    Just let me make some remarks about the square root of a matrix. It's not so simple! It's not even necessary for the issue it's applied in (3.50). Everything can derived with the Dirac-[itex]\gamma^{\mu}[/itex] matrices without taking roots.

    To define the square root of matrices, let's discuss only hermitean positive semidefinite matrices. As you know from linear algebra, a hermitean matrix can also be diagonalized by a unitary transformation, and all eigenvalues are real. The eigenvectors are orthogonal to each other and can be normalized, so that you have a unitary transformation from the original basis to the so defined eigenbasis. The matrix is called positive semidefinite, if all eigenvalues are [itex]\geq 0[/itex].

    Now to define [itex]\sqrt{\hat{A}}[/itex] for such a matrix, of course you like to have [itex](\sqrt{\hat{A}})^2=\hat{A}[/itex]. Now you can diagonalize the original matrix with a unitary transformation,
    [tex]\hat{A}'=\hat{U} \hat{A} \hat{U}^{\dagger}=\mathrm{diag}(\lambda_1,\ldots,\lambda_n),[/tex]
    where [itex]n[/itex] is the dimension of our unitary vector space (Hilbert space of finite dimension).

    Now for this diagonal matrix, it's easy to find [itex]\sqrt{\hat{A}'}[/itex], but it's not unique. One solution possibility is the one Peskin and Schroeder choose: Just take the positive roots of all the eigenvalues:
    Of course you can also choose the negative roots or the positive and negative roots for the different eigenvalues. All together you have [itex]2^n[/itex] square roots of such a postive semidefinite diagonal matrix.

    Each of these square roots is uniquely mapped back to the original basis by
    [tex]\sqrt{\hat{A}}=\hat{U}^{\dagger} \sqrt{\hat{A}'} \hat{U}.[/tex]
    Indeed you directly verify
    [tex](\hat{U}^{\dagger} \sqrt{\hat{A}'} \hat{U})^2 = \hat{U}^{\dagger} \sqrt{\hat{A}'} \hat{U}\hat{U}^{\dagger} \sqrt{\hat{A}'} \hat{U} = \hat{U}^{\dagger} (\sqrt{\hat{A}'})^2 \hat{U} = \hat{U}^{\dagger} \hat{A} \hat{U} = \hat{U}^{\dagger} \hat{U} \hat{A} \hat{U}^{\dagger} \hat{U} = \hat{A}.[/tex]
    In this way you define arbitrary functions of hermitean matrices (or even operators in infininte-dimensional Hilbert space).
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