Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normalization of free scalar field states

  1. Aug 26, 2015 #1

    if we adopt the convention, [itex] a^{\dagger}_\textbf{p} |0\rangle = |\textbf{p}\rangle [/itex]

    then we get a normalization that is not Lorentz invariant, i.e. [itex] \langle \textbf{p} | \textbf{q} \rangle = (2\pi)^3 \delta^{(3)}(\textbf{p} - \textbf{q}) [/itex].

    How do I explicitly show that this delta function (of 3-vectors) is not lorentz invariant, say for a boost in the p^3 direction ? What is [itex] \langle \textbf{p'} | \textbf{q'} \rangle [/itex] where the primed frame is the boosted frame?

    I thought since [itex] \int d^{3}\textbf{p} \hspace{2mm} \delta^{(3)}(\textbf{p} - \textbf{q}) = 1 [/itex] is lorentz invariant (integral over all space, so includes q)

    and as [itex] \hspace{3mm} d^3 \textbf{p} = \gamma^{-1} d^3 \textbf{p'} [/itex], (primed frame is boosted frame)

    so [itex] \hspace{5mm} \delta^{(3)}(\textbf{p} - \textbf{q}) = \gamma\,\delta^{(3)}(\textbf{p'} - \textbf{q'}) [/itex].

    so that [itex] \int \gamma\,\delta^{(3)}(\textbf{p'} - \textbf{q'}) \gamma^{-1} d^3 \textbf{p'} = 1 [/itex] stays Lorentz invariant.

    However, Peskin gets for the same calculation (p.22-23, eqn.2.34) the result

    [itex] \delta^{(3)}(\textbf{p'} - \textbf{q'}) = \delta^{(3)}(\textbf{p} - \textbf{q}) (\frac{E}{E'}) [/itex] using a method that I don't really understand well.

    Have I made a mistake somewhere?
  2. jcsd
  3. Aug 26, 2015 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    You can't just impose this because you want it to be true. The mathematical relationship between ##\delta^{(3)}(\textbf{p} - \textbf{q})## and ##\delta^{(3)}(\textbf{p'} - \textbf{q'})## has to be investigated mathematically. Peskin and Schroeder does this at the top of page 23.

    What don't you understand about the math at the top of page 23.
  4. Aug 26, 2015 #3
    Sorry, I don't know what I was thinking. Lack of sleep. Of course, [itex] \int d^{3}\textbf{p} \hspace{2mm} \delta^{(3)}(\textbf{p} - \textbf{q}) [/itex] does not have to be Lorentz invariant. I deserve a slap for that, please disregard that trail of thought.

    Here's what P&S do: They consider a boost in the 3-direction, with the unprimed frame as the boosted frame as:

    [itex] p'_3 = \gamma\,(p_3 + \beta E), \hspace{4mm} E' = \gamma\,(E + \beta\,p_3 ) [/itex]

    and then they use the identity [itex] \delta(f(x) - f(x_0)) = \frac{1}{|f'(x_0)|}\,\delta(x - x_0) \hspace{4mm} [/itex] to get:

    [itex] \delta^{(3)}(\textbf{p} - \textbf{q}) = \delta^{(3)}(\textbf{p'} - \textbf{q'}) \frac{dp'_3}{dp_3} [/itex]

    The subsequent steps, I follow. It's just the one above that I don't get. Here is my attempt:

    First, isn't the delta function identity [itex] \delta[g(x)] = \sum_i \frac{\delta(x-x_i)}{|f'(x_i)|} [/itex], where the xi are roots of the function g(x)?

    To get a relation between the boosted [itex] \delta^{(3)}(\textbf{p} - \textbf{q}) [/itex] and the unboosted [itex] \delta^{(3)}(\textbf{p'} - \textbf{q'}) [/itex], I suspect P&S have used the identity on the former to bring out p'3.

    I take [itex] g(\textbf{p'}) = \textbf{p} - \textbf{q}, \hspace{4mm} \textbf{p} = (p_1, p_2, p_3) = (p'_1, p'_2, \gamma(p'_3 - \beta E_p')), \hspace{4mm} \textbf{q} = (q_1, q_2, q_3) = (q'_1, q'_2, \gamma(q'_3 - \beta E_q')) [/itex]. We want the values of p' that will give a zero for g(p') as defined above.

    Now, [itex] \textbf{p} - \textbf{q} = (p'_1 - q'_1, p'_2 - q'_2, \gamma[(p'_3 - q'_3) - \beta (E'_p - E'_q)]) [/itex]

    So [itex] \textbf{p'} = \textbf{q'} [/itex] isn't a sufficient condition for a zero of g(p'), is it? One must also have E'p = E'q, isn't it so?

    Are there any flaws in my working? I suspect there obviously are, for I don't know how to get the factor [itex] \frac{dp'_3}{dp_3} [/itex] out.
  5. Aug 26, 2015 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes. (Typo in the denominator.) What happens if you set ##g\left(x\right) = f\left(x\right) - f\left(x_0 \right)##?

    But the fact that ##\left( \Delta E \right)^2 - \left( \Delta p \right)^2## is frame-invariant gives you .... ?
  6. Aug 26, 2015 #5
    Sorry about the typo. [itex] \hspace{2mm} \delta[g(x)] = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|} [/itex]. I get it now:

    Setting [itex] g(x) = f(x) - f(x_i) [/itex] gives the zeros of g(x) as xi and [itex] g'(x) = f'(x) [/itex] so one obtains the form of the identity used by P&S:

    [itex] \delta[f(x) - f(x_i)] = \sum_i \frac{\delta(x-x_i)}{|f'(x_i)|} [/itex].

    Ah, of course. [itex] \Delta E^2_p - \Delta |\textbf{p}|^2 = \Delta E^2_q - \Delta |\textbf{q}|^2 [/itex], and [itex] \textbf{p'} = \textbf{q'} [/itex] implies [itex] E'_p = E'_q [/itex]. So, p' = q' is indeed a sufficient condition for a zero of g(x).

    As for the obtaining the factor [itex] \frac{dp'_3}{dp_3} [/itex], I've managed to come up with it using intuition and some mathematics that I'm unfamiliar with. Perhaps you would care to comment if it is incorrect, or if there is a better way?

    Computing the factor [itex] |g'(\textbf{p'}=\textbf{q'})| [/itex] :

    Now, [itex] \textbf{g}(\textbf{p'}) = \textbf{p} - \textbf{q} = (p'_1 - q'_1, p'_2 - q'_2, p_3) [/itex].

    According to the definition in the 'addendum' of the first answer (by user 'joshphysics') on this page (linked), one has:

    [itex] \left| \frac{d\textbf{g}}{d\textbf{p'}} \right| = \left| \begin{array}{ccc} \frac{dg_1}{dp'_1} & \frac{dg_1}{dp'_2} & \frac{dg_1}{dp'_3} \\ \frac{dg_2}{dp'_1} & \frac{dg_2}{dp'_2} & \frac{dg_2}{dp'_3} \\ \frac{dg_3}{dp'_1} & \frac{dg_3}{dp'_2} & \frac{dg_3}{dp'_3} \end{array} \right| = \left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{dp_3}{dp'_3} \end{array} \right| = \frac{dp_3}{dp'_3} [/itex].

    (Actually, the link above defines the total derivative of a vector valued function w.r.t a vector as a matrix of various partial derivatives. I just assumed that the bars in the denominator when applied to a square matrix meant the determinant.)

    Did I get everything right?

    Even if I did, I confess that there are two pieces of math here that I haven't studied before (I'm a 2nd year undergrad). One is the identity involving the delta function (of a function), and the other is the definition of the derivative of a vector valued function w.r.t to another vector. Could you perhaps point me to resources (preferably a book for the latter) that discuss these concepts?

    Thanks a ton for the guidance!
  7. Aug 28, 2015 #6
    Sorry, disregard my questions in the last post. I proved the identity myself and found a good book on the latter subject - Hubbard's Vector Calculus, Linear Algebra, & Differential Forms (A Unified Approach).

    Thanks again.
  8. Aug 28, 2015 #7
    Corrected: [itex] \textbf{g}(\textbf{p'}) = \textbf{p} - \textbf{q} = (p'_1 - q'_1, p'_2 - q'_2, p_3 - q_3) [/itex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook