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How $$[\phi_1^+,:\phi_2\phi_3:]=:[\phi_1^+,\phi_2^-]\phi_3:+:\phi_2[\phi_1^+,\phi_3^-]:$$ with $\phi_i=\phi(x_i)$ field operators ($\phi_i^+$ is the annihilation part while $\phi_i^-$ is the creation part), can be demonstrated?

This formula seems to be used in the demonstration of the Wick theorem in Peskin&Schroeder's book (in the second line of the middle page equation (page 90)).

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    $\begingroup$ The equation that you wrote is not correct and is not the same equation that is in Peskin&Scroeder (if you are talking about the set of equations that comes after (4.41). Notice that you are missing some $\pm$ on top of $\phi_1,\phi_2$. $\endgroup$
    – Gonenc
    Commented Feb 26, 2018 at 23:34
  • $\begingroup$ Of course, in P&S it is $ϕ^+_1$ but I still have a problem to demonstrate that. $\endgroup$
    – ketherok
    Commented Feb 28, 2018 at 8:41
  • $\begingroup$ At a first look, even with the corrections, I’m not sure either if the equation can be correct. On the rhs, all creation ops are to the left via the normal ordering. But on the lhs this need not be true. $\endgroup$
    – CAF
    Commented Mar 1, 2018 at 19:11
  • $\begingroup$ Therefore, what formula is used in P&S? $\endgroup$
    – ketherok
    Commented Mar 2, 2018 at 15:05
  • $\begingroup$ @ketherok I don't have access to P&S right now to check $\endgroup$
    – CAF
    Commented Mar 4, 2018 at 15:10

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If you haven't done such a computation before, try splitting the fields $$\phi_i = \phi_i^+ + \phi_i^-,$$ a decomposition into its positive and frequency parts containing, respectively, the annihilation and creation operators. Using the definition of normal ordering should give you your result after a little plug and chug.

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  • $\begingroup$ Are you sure that the formula in the question is correct? $\endgroup$
    – ketherok
    Commented Feb 23, 2018 at 14:45
  • $\begingroup$ The equation is equivalent to $[\phi_1,:\phi_2\phi_3:]=:[\phi_1,\phi_2\phi_3]:$ This is not because there is not much freedom for placing the normal order symbol, that this equation should be right. $\endgroup$
    – ketherok
    Commented Feb 24, 2018 at 8:22
  • $\begingroup$ @ketherok Yes the equivalence is there, I just meant terms like $:[A,B]:$ on the right hand side can’t be there because they vanish. Actually from what you wrote you can simply derive a formula for $[A,BC]$ and insert between the normal ordered signs. $\endgroup$
    – CAF
    Commented Feb 24, 2018 at 8:38
  • $\begingroup$ Actually, I did not demonstrate the last equation I wrote. I have just written the right-hand side of the original equation in an other way. Why the right-hand side $:[\phi_1,\phi_2\phi_3]:$ is non zero ? $\endgroup$
    – ketherok
    Commented Feb 24, 2018 at 10:18
  • $\begingroup$ @ketherok Ah it seems I misunderstood your last comment. I thought you were considering another case. Have you tried what I suggested in my answer? $\endgroup$
    – CAF
    Commented Feb 26, 2018 at 16:50

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