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From Peskin-Schroeder, p.212:

The term $$ \langle \Omega | \phi(x) | \Omega \rangle \langle \Omega |\phi(y) | \Omega \rangle$$ is usually zero by symmetry; for higher-spin fields, it is zero by Lorentz invariance.

Where $|\Omega \rangle$ is the ground state and $\phi(x)$ a scalar field.

How can I prove this?

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  • $\begingroup$ all the annihilation operators will give you zero. The creation operators will create something that projects to 0 on the vacuum. $\endgroup$
    – Phoenix87
    Jan 10, 2015 at 23:35

2 Answers 2

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Here is another take making use of the symmetry, as hinted in the text.

The vacuum state is by definition symmetric with respect to any Lorentz transformation and translation. Let $U(\Lambda)$ be a unitary representation of the Lorentz transformation $\Lambda$ acting on the states, and let $$T(a) \equiv \exp ( -i Pa) \equiv \exp (- i P_\mu a^\mu )$$ be the unitary spacetime translation operator. Then by Lorentz invariance we have $$ \tag{A-1} |\Omega \rangle = U(\Lambda) |\Omega\rangle, $$ $$ \tag{A-2}\langle\Omega | = U(\Lambda)^{-1} \langle \Omega |,$$ and by translational invariance we have $$ \tag{B-1} | \Omega \rangle = T(a) | \Omega \rangle, $$ $$ \tag{B-2} \langle \Omega | = \langle \Omega | T(a)^{-1}. $$ Moreover, $U(\Lambda)$ and $T(a)$ act on a scalar field $\varphi(x)$ as $$ \tag{C-1} U(\Lambda)^{-1} \varphi(x) U(\Lambda) = \varphi(\Lambda^{-1}x), $$ $$ \tag{C-2} T(a)^{-1} \varphi(x) T(a) \equiv e^{iPa}\varphi(x) e^{-iPa} = \varphi(x-a). $$ Using these relations we have $$ \tag{D-1} \langle \Omega | \varphi(x) | \Omega \rangle = \langle \Omega | U(\Lambda)^{-1}\varphi(x)U(\Lambda) | \Omega \rangle = \langle \Omega | \varphi(\Lambda^{-1}x) \Omega \rangle, $$ $$ \tag{D-2} \langle \Omega | \varphi(x) | \Omega \rangle = \langle \Omega | \varphi(x+a) | \Omega \rangle, $$ for any Lorentz transformation (whose vector representation is) $\Lambda$ and for any 4-vector $a$. In particular from (D-2) you can see that $$ \tag{E} \langle \Omega | \varphi(x) | \Omega \rangle = c,$$ where $c$ does not depend on the space-time variable $x$. The answer of @lionelbrits explains why this constant can often be considered to be 0.

For a spinor field $\psi_\alpha(x)$ (or a vector field for that matter) you can use the same argument as above to obtain $$ \tag{F} \langle \Omega | \psi_\alpha(x) | \Omega \rangle = \langle \Omega | \psi_\alpha(0) | \Omega \rangle \equiv c_\alpha,$$ but now you have to remember the non trivial action of the Lorentz group on a spinor field: you have $$ \tag{G} U(\Lambda)^{-1} \psi_\alpha(x) U(\Lambda) = S(\Lambda)_\alpha^{\,\,\,\beta} \psi_\beta(\Lambda^{-1} x), $$ where $S(\Lambda)$ is the spinor representation of the Lorentz transformation. Using (G) in (F) you obtain: $$ \tag{H} c_\alpha \equiv \langle \Omega | \psi_\alpha(0) | \Omega \rangle = S(\Lambda)_\alpha^{\,\,\beta} \langle \Omega | \psi_\alpha(0) | \Omega \rangle \equiv S(\Lambda)_\alpha^{\,\,\beta} c_\beta, $$ and due to the arbitrariness in the choice of the Lorentz transformation this implies that $c_\alpha=0$, i.e. remembering (F), $$ \langle \Omega | \psi_\alpha(x) | \Omega \rangle = 0.$$

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It's a bit more involved than creation and annihilation operators. What is meant is that the field has no vacuum expectation value, and in a way this is by construction. You may think about why $\langle x \rangle = 0$ for a harmonic oscillator (or better yet, an an harmonic one). It's because $x\to -x$ is a symmetry of the Hamiltonian. It's not true for cubic potentials, though, but those theories can be sick.

In many situations, when we quantize a field, we perform perturbation theory around it's classical solution, and the dynamical field is the total field minus is vev, I.e. we write $\Phi = \langle \Phi \rangle + \phi$, and then work in terms of $\phi$. Thus the vev of the dynamical field vanishes by construction.

Edit:

To clarify, the reason you can't rely on creation and annihilation operators is because $\left\langle \Omega | a |\Omega\right\rangle$ is only zero if $\Phi$ has a zero vev, (circular argument). Imagine trying to naively quantize the Higgs field, which has a "Mexican hat" style potential where the minimum is located away from $\Phi = 0$. In that case, you do not have $\left\langle \Omega | \Phi |\Omega\right\rangle$ until you redefine the field to $\Phi = \langle \Phi \rangle + \phi$.

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    $\begingroup$ why doesn't the creation/annihilation operators argument work? If $\phi(x)$ connects states with $\pm 1$ particle how is the VEV not zero? $\endgroup$
    – glS
    Jan 11, 2015 at 0:03
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    $\begingroup$ Well, it works. It just obscures the physics. It also doesn't work well for interacting theories. $\endgroup$ Jan 11, 2015 at 11:22
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    $\begingroup$ Edited to clarify $\endgroup$ Jan 11, 2015 at 12:41

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