# Srednicki LSZ reduction formula motivation once more

I have seen this question asked before, but have not really found a satisfying answer, or maybe more exactly, I have not seen an answer that I can really understand.

I refer to section 5 of his book, p 39. He imposes the condition $\langle0| \varphi(0)|0 \rangle =0$ as one of the conditions needed for the LSZ formula to be valid. His motivation is that

"This is because we would like $a^\dagger_1(\pm\infty)$, when acting on $|0 \rangle$, to create a single-particle state. We do not want $a^\dagger_1(\pm\infty)$ to create a linear combination of a single particle state and the ground state. But this is precisely what will happen if $\langle 0| \varphi(0)|0 \rangle$ is not zero."

I understand everything, expect the last sentence. Why would $a^\dagger_1(\pm\infty)|0 \rangle$ be such a linear combination if $\langle 0| \varphi(0)|0 \rangle$ is not zero. I just can't work this out.

• Seen where before? – Qmechanic Nov 7 '17 at 21:41
• In this forum. If you search for “Srednicki LSZ” you will find several related threads. – Oбжорoв Nov 7 '17 at 21:43
• – AccidentalFourierTransform Nov 7 '17 at 21:44
• Since the ground state is orthgonal to all other states, getting a nonzero expectation means $\phi$ must include some grounds state contribution. This is basic quantum mechanics, not field theory. – KF Gauss Nov 7 '17 at 22:14

First, it's important to know what kind of states there are. I'll call them ground state $|0\rangle$, one-particle state $|k\rangle$ and multi-particle state $|n,k\rangle$. Here, $k$ specifies the momentum of a particle and $n$ is a collection of other quantum numbers that may arise (e.g. relative momentum between some other momenta, ...).

We want $a^\dagger|0\rangle$ to create a one-particle state. In Eq. (3.19), Srednicki defines $$\varphi(x) = \int \widetilde{\text{d}k} \big[ a(k)\, \text{e}^{\text{i} kx} + a^\dagger(k)\, \text{e}^{-\text{i} kx} \big].\tag{1}\label{a}$$

Assume, that $a^\dagger$ creates both a one-particle state, as well as the ground state, like this $$a^\dagger |0\rangle = c_1 |k\rangle + c_2 |0\rangle,$$ with some complex numbers $c_{1,2}$. Now we calculate the vacuum expectation value of $\varphi$ (and only consider the $a^\dagger$-part for notational simplicity): \begin{align*} \lim\limits_{x\to 0}\langle 0|\varphi(x)|0\rangle &= \lim\limits_{x\to 0}\int \widetilde{\text{d}k} \langle 0| a^\dagger(k)\, \text{e}^{-\text{i} kx} |0\rangle \\ &= \int \widetilde{\text{d}k} \langle 0| a^\dagger(k) |0\rangle \,\text{e}^{0} \\ &= \int \widetilde{\text{d}k} \langle 0| \big[ c_1 |k\rangle + c_2 |0\rangle \big]\\ &\propto c_1 \underbrace{\langle 0|k\rangle}_{0,\text{ orthogonal!}} + c_2 \underbrace{\langle 0|0\rangle}_{1}\\ &= c_2 \neq 0 \end{align*} This means, if we let $a^\dagger$ create both a one-particle state and the ground state, then $\langle 0|\varphi(0)|0\rangle$ does not vanish. But if we define $\langle 0|\varphi(0)|0\rangle\equiv 0$, we must set $c_2=0$, which means that $a^\dagger$ truly only creates a one-particle state in Eq. \eqref{a}.

Ad comment: (1) If we want to be precise and include the $a$ term, we should also not perform the limit so quickly. Inside the integration, there's a $\text{e}^{\text ikx}$ and a $\text{e}^{-\text ikx}$. They are linearly independent, so each should vanish. (Also note that while $x\to 0$, $k\to \infty$, so that's ill-defined within the exponential)

(2) $a$ being evaluated at $t=\pm\infty$ helps us make it non-interacting. If this doesn't answer your question, I am sorry I did not understand it. If so, please restate it.

• Thanks. This is probably it indeed, but I still have a few questions. (1) if you add the "a" part then acting on <0| one will get a c_2* <0|0> contribution. So the requirement is that c_2* + c_2 =0, i.e. c_2 is a purely imaginary number? (2) The "a" are at infinite past or future, but phi is not, or is that why we use the energy momentum operator? – Oбжорoв Nov 8 '17 at 8:52
• I've edited my answer, does this help? – Stephan Nov 8 '17 at 9:17
• (1) understood - agreed. (2) Not sure that is my question. You expand phi(0) in terms of a and a^dagger, which are time-dependent and also taken at (t=0) then? So how do you relate this expansion to the "a"s at infinity? – Oбжорoв Nov 8 '17 at 9:25
• My first guess is that since $\langle 0|\phi(x)|0\rangle$ is invariant under Poincaré transformations, it does not matter if we set $t=0$, $t=t_0$ or $t=\pm\infty$ for the VEV. But I'm not 100% certain on that one.. – Stephan Nov 9 '17 at 1:46