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In calculating the total-momentum operator of the real Klein-Gordon field, I end up with an equation like $$ \vec P = \frac{1}{(2\pi)^3}\int d^3p \mspace{9mu} \vec p\Big(a_pa_{-p} + a_pa^\dagger_{p}-a_{-p}^\dagger a_{-p} - a_{-p}^\dagger a_p^\dagger\Big) $$

and it seems the first and the last integrals vanish because they are antisymmetric w.r.t $ \vec p \rightarrow -\vec p $. I am unable to see this anti-symmetry. Also I am not able to see how to relate $ a_{-p} $ to $ a_p $ or $ a_p^\dagger $ other than some implicit integral relations like(which also am not very sure how to prove),

$$ \int d^3p a_p^\dagger e^{-i\vec p.\vec x} = \int d^3p a_p e^{i\vec p.\vec x}$$

Thanks for any inputs.

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I am not a 100% sure, but I think they mean the following: \begin{equation} \begin{aligned} \mathbf{P} & \sim \int \mathrm{d}^3 p \; \mathbf{p}a_p a_{-p} \\& = \int \mathrm{d}^3 p \; \left(\frac{1}{2} \mathbf{p}a_p a_{-p} + \frac{1}{2} \mathbf{p}a_p a_{-p} \right) \\& = \int \mathrm{d}^3 p \; \left(\frac{1}{2} \mathbf{p}a_p a_{-p} - \frac{1}{2} \mathbf{p}a_{-p} a_{p} \right) \\& = \int \mathrm{d}^3 p \; \frac{1}{2} [a_p, a_{-p}] \\& = 0 \end{aligned} \end{equation} where we performed a change of integration variable $\mathbf{p} \rightarrow - \mathbf{p}$ to get to the third line. In order to understand this, note that for $\mathbf{p} \rightarrow - \mathbf{p}$ we get: \begin{equation} \int\limits_{-\infty}^{\infty} \mathrm{d}p \rightarrow \int\limits_{\infty}^{-\infty} \left(-\mathrm{d}p\right)= - \int\limits_{-\infty}^{\infty} \left(-\mathrm{d}p\right)=\int\limits_{-\infty}^{\infty} \mathrm{d}p \end{equation} I think this may also explain your second question: $a_p$ is related to $a_{-p}$ through this change of integration variable.

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  • $\begingroup$ Thanks !! I made a terrible mistake of taking Jacobian to be -1 when $ \vec p \rightarrow -\vec p $. Now I see why the integral vanishes. Regarding the second question, your answer seems reasonable, but what is unclear is, what will I do with terms like $ a_pa_pa_{-p} $ (am not sure from where I will get them though). $\endgroup$
    – user35952
    Jan 19, 2014 at 3:27
  • $\begingroup$ @user35952 I am not sure. Do you have a particular example? $\endgroup$
    – Hunter
    Jan 19, 2014 at 3:29
  • $\begingroup$ I don't have an example of that kind of a term, but say for instance something like the integral $$ \int d^3p a_pa_pa_{−p} $$. Which we can't reduce to anything of the form that has only $ a_p $ or $a_p^\dagger$ alone. But that apart, I think your explanation sounds perfectly alright to me, thanks !! $\endgroup$
    – user35952
    Jan 19, 2014 at 3:33
  • $\begingroup$ @user35952 I don't think so. At least for the bosonic case; once you will work with the Dirac equation, then you will work with anti-commutation rule and so you will be to manipulate the integral. $\endgroup$
    – Hunter
    Jan 19, 2014 at 3:40

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