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In the book "Advanced Quantum Mechanics" by Franz Schwabl on page 280 you can find a formula (13.1.18c) which assumes linearity of time ordering operator $::$

$ :\phi\left(x\right)\phi\left(y\right):=:\left(\phi^{+}\left(x\right)+\phi^{-}\left(x\right)\right)\left(\phi^{+}\left(y\right)+\phi^{-}\left(y\right)\right):$

(here)

$=:\phi^{+}\left(x\right)\phi^{+}\left(y\right):+:\phi^{+}\left(x\right)\phi^{-}\left(y\right):+:\phi^{-}\left(x\right)\phi^{+}\left(y\right):+:\phi^{-}\left(x\right)\phi^{-}\left(y\right): $

$ = \phi^{+}\left(x\right)\phi^{+}\left(y\right)+\phi^{-}\left(y\right)\phi^{+}\left(x\right)+\phi^{-}\left(x\right)\phi^{+}\left(y\right)+\phi^{-}\left(x\right)\phi^{-}\left(y\right)$

You can find this formula in any advanced QFT textbook.

But we know $::$ is not linear because on the one hand we have

$ :a^{\dagger}aaa^{\dagger}:=a^{\dagger}a^{\dagger}aa $

But assuming linearity we get

$:a^{\dagger}aaa^{\dagger}: = :a^{\dagger}a\left(1+a^{\dagger}a\right):$

$= :a^{\dagger}a: + :a^{\dagger}aa^{\dagger}a:$

$= a^{\dagger}a + :a^{\dagger}\left(1+a^{\dagger}a\right)a: $

$=a^{\dagger}a + :a^{\dagger}a: + :a^{\dagger}a^{\dagger}aa: $

$=2a^{\dagger}a+a^{\dagger}a^{\dagger}aa $

Which implies 2=0.

Why do we assume linearity of $::$ if it's not consistent?

Alexandre Gurchumelia
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1 Answers1

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The problem is that "normal ordering" is not defined on operators but on "symbols". Within the normal ordering $::$ all symbols commute and hence the second part of your proof is not valid.

See the following Physics.SE question and its answers: How exactly is "normal-ordering an operator" defined?

NDewolf
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