Let $\phi$ be a scalar field in an interacting theory ($\phi^3$ or $\phi^4$, for example). If $|0\rangle$ is the vacuum of the interacting theory and $P^\mu$ is the four-momentum operator, we have that

$$\langle 0 | \phi(x) | 0 \rangle = \langle 0 | e^{-iPx} \phi(0) e^{iPx} | 0 \rangle = \langle 0 | \phi(0) | 0 \rangle $$

In chapter 5 of his QFT book, Srednicki says that

We would like $ \langle 0 | \phi(0) | 0 \rangle$ to be zero. This is because we would like $a_1^\dagger (\pm \infty)$, when acting on $|0\rangle$, to create a single particle state. We do not want $a_1^\dagger (\pm \infty)$ to create a linear combination of a single particle state and the ground state.

Here $a_1^\dagger (\pm \infty)$ is the creation operator $a^\dagger$ for a momentum $\mathbf{k}_1$ taken at time $\pm \infty$, which (according to the book) guarantees that the particle is located away from the origin. In other words, we define $|k\rangle = \lim\limits_{t \to -\infty} a^\dagger(\mathbf{k}, t) |0\rangle$.

It seems to me that Srednicki wants $\langle 0 | k \rangle = 0$, which sounds reasonable. But applying the LSZ formula for the special case of one initial particle and zero final particles, I get $\langle 0 | k \rangle = i (2\pi)^4 m^2 \langle 0 | \phi(0) | 0 \rangle \delta^4(k)$ (the $2\pi$'s might be off). This is nonzero only when $k^\mu=0$, a.k.a. never, since $k$ must be on shell. So why must we ask that the fields's VEV be zero, when it seems that $\langle 0 | k \rangle = 0$ anyway?

  • $\begingroup$ I think the problem might lie in the proper definition of $\vert 0 \rangle$. Properly defining the vacuum can be a tricky thing to do and we can find things like the Unruh effect that comes from properly defining a vacuum. $\endgroup$
    – Neuneck
    Commented May 8, 2015 at 12:21
  • $\begingroup$ See also physics.stackexchange.com/q/43714 $\endgroup$
    – innisfree
    Commented Dec 10, 2015 at 13:06
  • 1
    $\begingroup$ @Neuneck - The Unruh effect is only important to understand when considering accelerated observers. For inertial observers in Minkowski spacetime there is a unique definition of the vacuum. Of course defining the vacuum in interacting theory is often tricky for other reasons, but not for the one you mentioned. $\endgroup$
    – Prahar
    Commented Mar 18, 2016 at 10:08

2 Answers 2


I think I have hit upon the main reason, which in hindsight was really obvious. The assumption that the creation operators work as in the free theory (that is, that $a^\dagger(\pm\infty)$ won't create something that overlaps with the vacuum) is built into the LSZ formula, so it doesn't make sense to use the LSZ formula in this case; we have to go farther back. I'm still not clear on the relation between $\phi(0)$ and $a^\dagger(\pm\infty)$, since I would expect $\phi(0)$ to be related to $a^\dagger(0)$. Hopefully later I'll edit in something more complete.


I can provide you with two reasons that I have learned in my studies. One $k^{\mu} $ need not be zero. The delta function implies $\omega_{k} = \tilde {k} $. This forces the single particle states to lie on the mass shell. Thus the only term that can be zero now is the vacuum expectation value. The reason this is zero is due to the assumption that the vacuum itself is Lorentz invariant. If this term was not zero then the vacuum expectation value would not be Lorentz invariant, which can be seen by allowing it to be nonzero and applying a Lorentz transformation to the field $\phi (0) $. I hope this helps, and good luck with your studies!

  • $\begingroup$ A scalar field does not transform under Lorentz transformations and can therefore have a nonzero expectation value - as experienced in the Higgs mechanism. $\endgroup$
    – Neuneck
    Commented May 8, 2015 at 12:18
  • $\begingroup$ But that's not the on-shell-forcing delta function, which would be $\delta(k^2+m^2)$ (in $-+++$ signature). This is just a four dimensional delta function, which forces $k$ to be the zero four-vector. $\endgroup$
    – Javier
    Commented May 8, 2015 at 13:01

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