This relates to page 111 in Peskin and Schroeder.

We have the $\phi^4$ S-matrix for a 2-particle to 2-particle scattering reaction:

$$-i\frac{\lambda}{4!}\int d^4x \langle p_1p_2|\mathcal T\left(\phi(x)\phi(x)\phi(x)\phi(x)\right)|p_Ap_B\rangle \tag{4.90}$$

For which, using Wick's theorem we can rewrite the time-ordering as normal ordered contracted operators. This is all fine. To first order we obtain:

$$\langle p_1p_2|\mathcal N\left(-i\frac{\lambda}{4!}\int d^4x\text{ }\phi(x)\phi(x)\phi(x)\phi(x)+\text{contractions}\right)|p_Ap_B\rangle \tag{4.92}$$

Since we are evaluating S-matrix elements and not correlation functions we can no longer neglect terms which aren't fully contracted. We then note that if two fields are contracted with themselves and two fields remain, the normal-ordered product of the remaining two fields will be of the form $\mathcal N\left((a+a^\dagger)(a+a^\dagger)\right)=(a^\dagger a^\dagger+2a^\dagger a+aa)$, and as we commute these opeators past the $a$'s and $a^\dagger$'s of the initial and final states, we find that only a term with an equal number of $a$'s and $a^\dagger$'s can survive.

It is this last sentence that confuses me, I have written out in full what I expect this to look like, but I cannot see why an equal number of creation and annihilation operators is necessary for the term to not vanish.

  • $\begingroup$ For instance $\langle 0 | a_{p_1} a_{p_2} \, ( a^\dagger a^\dagger )\, a^\dagger_{P_A}a^\dagger_{P_B} | 0 \rangle =0$ $\endgroup$
    – nwolijin
    Dec 28, 2020 at 18:32
  • $\begingroup$ Ok I see that that is what the first term gives, but not why that has to evaluate to zero. $\endgroup$
    – Charlie
    Dec 28, 2020 at 18:34
  • $\begingroup$ Would you accept the corresponding statement for the regular quantum mechanical harmonic oscillator? $\endgroup$
    – kaylimekay
    Dec 28, 2020 at 18:47
  • $\begingroup$ Sure! Unless the answer depends on the QFT $a$/$a^\dagger$ commutation relations, but I'm not sure @kaylimekay $\endgroup$
    – Charlie
    Dec 28, 2020 at 18:49
  • $\begingroup$ @Charlie if you commute operators in the expression I wrote you'll see that it is indeed zero $\endgroup$
    – nwolijin
    Dec 29, 2020 at 17:55

1 Answer 1


Suppose we have a general term of the form $$\langle 0 | a_{p_1} a_{p_2} \cdots a_{p_{n-1}}a_{p_n} a_{p_1}^\dagger a_{p_2}^\dagger \cdots a^\dagger_{p_{m-1}}a_{p_m}^\dagger | 0 \rangle$$ We can eliminate $a_{p_1}^\dagger$ by commuting it past all the $a_{p_i}$ factors until it anhilitates the vacuum state. Along the way, we will end up with a bunch of terms of the form $$\langle 0 | a_{p_1} \cdots [a_{p_i},a_{p_1}^\dagger] \cdots a_{p_n} a_{p_2}^\dagger \cdots a_{p_m}^\dagger | 0 \rangle.$$ In each of these terms, we can use the commutation relation $$[a_p,a_{p'}^\dagger]=(2\pi)^3\delta^{(3)}(p-p')$$ to eliminate $a_{p_i}$ and $a_{p_1}^\dagger$ from the term. The net result is the sum of a bunch of terms, each with one less $a$ and one less $a^\dagger$. We can continue this process until we run out of either $a^\dagger$ factors or $a$ factors.

At this point, there are three possibilities. If there were more $a$ factors than $a^\dagger$ factors, we will have the sum of a bunch of terms, each the product of delta functions with a factor of the form $\langle 0|a_{p'_1}\cdots a_{p'_{n-m}}|0\rangle$. These terms are all zero because an annihilation operator appears next to $|0\rangle$. Similarly, if there were more factors of $a^\dagger$, we would end up with an $a^\dagger$ next to $\langle 0|$ in each term. The only way to avoid a trivial result is if there are an equal number of factors of $a$ and $a^\dagger$. Then we will end up with a sum of products of delta functions.

  • $\begingroup$ Got it, wrote out a few examples and I see what you mean, thanks! $\endgroup$
    – Charlie
    Dec 28, 2020 at 20:58

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