Why can we shift the field $\phi$, so that $\langle \Omega | \phi(x) | \Omega \rangle = 0$?

Question

Problem

Introduction

In different derivations of the LSZ reduction formula the author makes a shift of the field $\phi(x)$ $$ \phi'(x) = \phi(x) - \langle \Omega | \phi(x) | \Omega \rangle, $$ and it is also proven, that the second term is just a number (more precisely proportional to the identity), which doesn't depend on $x$,
so one can write $$ \phi'(x) = \phi(x) - c. $$ Then, the argument is the following.

"This is just a change of the name of the operator of interest, and does not affect the physics."
Mark Srednicki, http://web.physics.ucsb.edu/~mark/qft.html, p. 53.

Question

Why this does not affect the physics?

The $\phi^2$ and other, possibly higher order terms in the $H(\phi, \pi)$ hamilton operator, or in the $\mathcal{L}(\phi, \partial_\mu \phi)$ will look completely different. $$ \mathcal{L}(\phi, \partial_\mu \phi) =\ ...\ - \frac{1}{2} m^2 \phi^2,\\ \mathcal{L}(\phi', \partial_\mu \phi') =\ ...\ - \frac{1}{2} m^2 (\phi'^2 + 2 c \phi' + c^2). $$ This will correspond to a different theory, so we cannot use the usual results we had already for the fields $\phi$.
For example, we cannot really calculate $n$-point functions, Green functions, because the $H_{int}$ will be totally different this way.

Notes

I am not talking about spontaneous symmetry breaking. I don't think authors here talk about fields which are only the perturbations of some other field.
The operators and states are all in Heisenberg picture, $|\Omega\rangle$ is the interacting vacuum.

Other mentions

"Therefore, if for some reason $\langle \Omega | \phi(x) | \Omega \rangle$ is non-zero, we redefine the field $\phi(x)$ [...] which doesn't spoil any of the conditions"
AccidentalFourierTransform, https://physics.stackexchange.com/a/311876/254794

"We can now define a new field, $\phi$, which is normalized to have a [...], and a vanishing VEV (vacuum expectation value)"
Michael Luke, https://www.physics.utoronto.ca/~luke/PHY2403F/References_files/lecturenotes.pdf, p. 117

"Then, without loss of generality [...] we redefine $\phi \rightarrow \phi - c$"
Timo Weigand, https://www.thphys.uni-heidelberg.de/~weigand/QFT2-14/SkriptQFT2.pdf, p. 43

Related questions

How can I prove that $\langle\Omega\vert \phi(x) \vert\Omega\rangle \langle\Omega\vert\phi(y)\vert\Omega\rangle=0$ for a scalar field? (See Notes why I don't consider this an answer, or I don't understand it.)
Why does Srednicki insist on $\phi$ having zero VEV? (Does not have answer as far I saw.)
Assumptions in the LSZ reduction formula (Seems reasonable, but not sure, haven't seen anybody with the same derivation.)
The use of $a^\dagger(\mathbf{k}) = -i \int d^3x e^{ikx}\stackrel{\leftrightarrow}{\partial}_0 \phi(x)$ in the derivation of the LSZ-formula (Don't understand what it really means to "We should get rid of tadpoles".)

score 4 · Answer 1 · answered Apr 14 '20 at 16:03

I think you just misunderstood the claim of "this does not affect the physics". If shifting the field so that its VEV is zero causes linear (or other) terms in the Lagrangian, then we must deal with these additional terms - they are a feature of the theory, but in order to apply LSZ we still need to deal with the theory in terms of the field that has zero VEV.

No one claims that that the Lagrangian is invariant under the shift operation, just that the shift operation is allowed. To make an analogy, this is equivalent to the claim that e.g. a shift of the origin in classical point mechanics "does not affect the physics". Of course, unless the system is translation-invariant, this does change the form of the action. But it does not change the physics, the system is still the same, just expressed in different coordinates.

Your answer was very insightful, helped me to understand the problem. However I can give the "answer" to only one person, and gave it to JF132, who was more elaborate on some points. The "different coordinates" example is especially expressive. — Gabor, Apr 15 '20 at 18:11

Qmechanic · Answer 2 · 2024-04-01T03:37:34.503

First of all, kudos for a well-researched opening post.

Lorentz covariance implies that the 1-point function $$\langle \Omega | \phi^k | \Omega\rangle~=~\langle \phi^k \rangle_{J=0}$$ vanish for non-scalar fields. So let's assume that $\phi^k$ are scalar fields$^1$.

Similarly, a pertinent global symmetry implies that $\langle \phi^k \rangle_{J=0}$ vanishes, cf. e.g. my Phys.SE answer here. So let's assume there is no such global symmetry.
A field redefinition $$\phi^k~=~ \bar{\phi}^k + c^k$$ is just a change of coordinates, which is always possible and doesn't change the path integral, cf. the other answers.
The form of the Lagrangian density inside of the action $$\bar{S}[\bar{\phi}]~:=~ S[\phi]~=~S[\bar{\phi} + c]$$ would in general change accordingly. The shift would trickle down from higher-order terms to lower-order terms.

In particular, if before the field redefinition ${\cal L}_1= Y_k\phi^k$ denotes the Lagrangian density terms linear in the field, then after the field redefinition, the terms linear in the field would be $\bar{\cal L}_1= \bar{Y}_k\bar{\phi}^k$ for some (in general new) coefficient $\bar{Y}_k$.

(The $Y_k$ notation is inspired by Ref. 1. Note that the $Y_k\phi^k$ term looks similar to an external source term $J_k\phi^k$. The difference is that the $Y_k$ are assumed to be intrinsic to the model.)
A non-zero $\langle \Omega | \phi^k | \Omega\rangle$ reflects transitions between the vacuum and 1-particle states, cf. e.g. this Phys.SE post. (It should perhaps be stressed that the above field redefinition would change the definition of 1-particle states, but not the underlying physics, only the picture.)
One may show that the condition $\langle \phi^k \rangle_{J=0}=0$ simplifies the perturbation theory considerably, cf. a proposition in my Phys.SE answer here.
Alternatively, a shifting of the value of $\langle \phi^k \rangle_{J=0}$ can be related to a shifting of the $Y_k$ coefficients in the action: To the zeroth order ${\cal O}(g^0)$ in the coupling constants $g$ (or equivalently, if we turn off interactions $g=0$), then the condition $\langle \phi^k \rangle_{J=0}=0$ is equivalent to that $Y_k={\cal O}(g^1)$ vanish.

Conversely, in order to fulfill $\langle \phi^k \rangle_{J=0}=0$ to all orders in perturbation theory, there must be appropriate (possibly infinite) counterterms within the $Y_k$ coefficients.

See e.g. Ref. 1 for details.

References:

M. Srednicki, QFT, 2007; chapter 9. A prepublication draft PDF file is available here.

--

$^1$Here we use deWitt condensed notation

score 1 · Accepted Answer · answered Apr 14 '20 at 16:21

You are right that in term of the new field $\phi'(x)$, the Lagrangian (and Hamiltonian) has extra terms. But when talking about Green's functions, there is a distinction between two different types:

$G^{(n)}(x_1,...,x_n)$: The original Green's functions, computed for the canonical field $\phi(x)$.
$G'^{(n)}(x_1,...,x_n)$: Renormalized Green's functions, computed for the new (renormalized) field $\phi'(x)$.

(To be exact, the renormalized field $\phi'(x)$ here is the shifted field, as you described, scaled further to give the correct one-particle matrix element. This "scaling" process gives rise to even more extra terms in the Lagrangian).

For $G^{(n)}(x_1,...,x_n)$, we can use the original Lagrangian. But to compute $G'^{(n)}(x_1,...,x_n)$, it is easier to use the new Lagrangian, treating the extra terms as new interaction types.

Which of these two types of Green's functions are we more interested in? To apply the LSZ reduction formula, we need the renormalized Green's functions $G'^{(n)}(x_1,...,x_n)$, not the original Green's functions $G^{(n)}(x_1,...,x_n)$.

So that is the idea, we want to use the LSZ formula to compute scattering cross section. We can try to use the original Lagrangian to obtain $G^{(n)}(x_1,...,x_n)$, from which derive the required $G'^{(n)}(x_1,...,x_n)$. Or, we can do it the easier way, rewriting the Lagrangian in terms of the correctly renormalized field $\phi'(x)$, treating the extra terms as new kind of interactions, and compute the $G'^{(n)}(x_1,...,x_n)$ directly. Either way, the physics is the same.

Thanks. It's very interesting. I could only prove that we could do that (for example the Hamilton operator will do the same after the new interaction terms, with the usual process to construct H from L) on Lagrangians which have phi^n interaction terms, and the shifting is linear. This is enough for me now, I could maybe generalize for complicated interaction terms later. — Gabor, Apr 15 '20 at 18:08

score 1 · Answer 4 · answered Apr 16 '20 at 22:40

No, the different Lagrangian you have written does not correspond to a different physical theory. In fact, given a Lagrangian $\mathcal{L}$, you can redefine the field in any arbitrary way $\phi \to f(\phi)$ for any function $f$ such that $f'(0)=0$ where the derivative is with respect to $\phi$. Field redefinitions do not change physics - Lagrangians, actions and n-point Green's functions all change under field redefinitions, but none of these are physically relevant/observable quantities. In field theory, the physical observables are scattering amplitudes (or cross sections, really). Scattering amplitudes are invariant under field redefinitions. This makes sense, as mentioned in another answer by ACuriousMind, because it just amounts to shifting the origin of your field coordinates. Another way to see it is that scattering amplitudes are integrals over fields, and therefore, physics doesn't change when you use a different set of integration variables.

You can look at some other answers on the invariance of amplitudes under field redefinitions, like this one.