# Vacuum expectation value in presence of a source

If a vacuum is translationally invariant i.e., $P^\mu|0\rangle=0$ or $e^{(\pm ip\cdot x)}|0\rangle=0$, we can express the the vacuum expectation value of a field as $\langle 0|\phi(x)|0\rangle$ as $$\langle 0|e^{(-ip\cdot x)}\phi(0)e^{(+ip\cdot x)}|0\rangle=\langle 0|\phi(0)|0\rangle=\text{a constant}$$

In presence of a source $J(x)\neq 0$, it is asserted (as in Quantum Field theory by Lewis Ryder) that $\langle 0|\phi(x)|0\rangle^J$ is in general a function of spacetime, and reduces to a constant value only when $J\rightarrow 0$. I tried to understand this spacetime dependence starting from the solution of $(\Box+m^2)\phi(x)=j(x)$ given by (Peskin and Schroeder page 32, eqn. 2.64)$$\phi(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\{(a_p+\frac{i}{\sqrt{2E_p}}\tilde{j}(p))e^{-ip\cdot x}+(a^\dagger_p-\frac{i}{\sqrt{2E_p}}\tilde{j}^*(p))e^{_ip\cdot x}\}\tag{2.64}$$ where $$\tilde j(p)=\int d^4y e^{ip.y}j(y).$$ However, even with this expression $\langle 0|\phi(x)|0\rangle^J$ turns out to be independent of $x$.

My question is how does the presence of nonzero $J(x)$ results in a non-trivial spacetime dependent value of $\langle 0|\phi(x)|0\rangle^J$?

• hint: what is $|0\rangle_J$? (BTW, this is probably easier to show with path integrals...) – AccidentalFourierTransform May 23 '16 at 15:58

My question is how does the presence of nonzero $J(x)$ results in a non-trivial spacetime dependent value of $\langle 0|\phi(x)|0\rangle$?

The equation $\phi(x)=\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}$ works both for $J=0$ and $J\neq 0$. Therefore, $$\langle \phi(x)\rangle_J= {}_J\langle 0|\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}|0\rangle_J$$

What it is not true when $J\neq 0$ is that $P|0\rangle_J\stackrel{\text{no}}{=}0$ (because the source breaks the invariance), and therefore we cannot conclude that $$\langle \phi(x)\rangle_J\stackrel{\text{no}}{=} {}_J\langle 0|\phi(0)|0\rangle_J$$

Therefore, if $J\neq 0$ the vev depends on position $x$.

To find the explicit dependence of $\langle \phi(x)\rangle_J$ with $x$, instead of using operators, it is easier to work with path integrals: $$\langle \phi(x)\rangle_J=\frac{\delta}{\delta J(x)}\exp\left[-i\int \mathrm dy\,\mathrm dz\ J^*(y)\Delta(y-z)J(z)\right]$$ which I believe you can calculate yourself (note that the result is proportional to $J(x)$ and so the vev goes to zero as $J\to 0$, as expected).

## The (somewhat) bigger picture

The first thing we have to do is to differentiate from internal sources and external ones:

• An internal source is a term in the lagrangian that only includes dynamical fields, that is, fields that are part of the equations of motion. For example, you can have a KG theory, $$\mathcal L\sim (\partial\phi)^2-m^2\phi^2+g\phi^3$$ where the last term can be said to be an internal source (though the usual terminology is just interaction). This term is internal because it only depends on $\phi$, which is itself a dynamical field (determined from the EoM's). Another (more illustrative) example is the lagrangian for QED, $$\mathcal L\sim \bar\psi(i\not\partial-m)\psi-F^2+eA_\mu \bar\psi\gamma^\mu\psi$$ Again, the last term is an internal source, because it only depends on dynamical fields, $\psi$ and $A$, which are determined from the EoM. I would like to stress that in general people don't say "internal source" but "interaction" instead.

• An exernal source is a function in the lagrangian that is externally determined (fixed), that is, a function that is not dynamical (there is not an equation of motion for that function). Typical examples are the $J$'s that are used in path integrals, $$\mathcal L\sim (\partial\phi)^2-m^2\phi^2+g\phi^3+\phi(x)J(x)$$ and fixed (background) functions in effective theories, such as, for example, the electromagnetic field in a low energy treatment of the Hydrogen atom: $$\mathcal L\sim \bar\psi(i\not\partial-m)\psi+eA_\mu \bar\psi\gamma^\mu\psi$$ (here, $A_\mu$ is an external source, because there is not a kinetic term $F^2$ for it, and so the value of $A$ has to be written by hand, say, a Coulomb potential $A_0\sim e/r$.

Note that external sources break the translational invariance of the theory (because of the obvious reason: an external source has a fixed dependence on position, and so the "physics don't look the same everywhere"). Therefore, if there are external sources, $P_\mu|0\rangle\neq 0$ and vev's depend on position, as discussed in the first part of this answer.

On the other hand, internal sources don't break the translational invariance of the theory, because the sources themselves transform together with the fields. This might be easier to understand with an example. Consider first a theory with only internal sources: $$S=\int\mathrm dx\ (\partial\phi(x))^2-m^2\phi(x)^2-g\phi(x)^3$$ which, upon a translation $x\to x-a$ transforms into $$S_a=\int\mathrm dx\ (\partial\phi(x-a))^2-m^2\phi(x-a)^2-g\phi(x-a)^3$$ which is the same as before, $S_a=S$, because we integrate over all space and $\mathrm d(x-a)=\mathrm dx$.

On the other hand, consider a theory with an external source: $$S=\int\mathrm dx\ (\partial\phi(x))^2-m^2\phi(x)^2-\phi(x)J(x)$$ which, upon a translation $x\to x-a$ transforms into $$S_a=\int\mathrm dx\ (\partial\phi(x-a))^2-m^2\phi(x-a)^2-\phi(x-a)J(x)$$ which is not the same as before, because of the $J(x)$ term. The action is not the same as before, and so the translation changed the theory. At this point, you might want to read this answer of mine. In the notation of that post, the $(2)$ derivative of a lagrangian with external sources is non-zero.

To recapitulate,

• If there are only internal sources, then the theory is translationally invariant, and so all the vev's are position independent (as can be easily shown using $P_\mu|0\rangle=0$ and $Q_\alpha(x)=\mathrm e^{-iPx}Q_\alpha(x)\mathrm e^{iPx}$, where $Q_\alpha(x)$ is any field). Most of the times we redefine every field $Q_\alpha(x)\to Q_\alpha(x)-\langle Q\rangle$ so that all the vev's are zero (this is relevant for renormalisation). In some cases (e.g., in the case of the Higgs field) a non-zero vev is physically relevant (but only makes sense because of the form of the lagrangian for the Higgs field, and wouldn't make sense for, say, a standard KG field). In any case, if the sources are internal then vev's are constant.

• If there are only external sources, then the theory is free. Therefore, the vev's depend on position, but in the limit $J\to 0$ we must have $\langle\phi\rangle\to 0$, as it must be for a free theory.

• If there are internal and external sources, the vev's are position-dependent and don't go to zero as the external sources go to zero (and therefore we must renormalise the fields).

• @AccidentalFourierTransform- Is it always true that presence of a source breaks the translational invariance of the vacuum? In the $\phi^4-$theory, the $\phi^3$ term acts as the source of for the free Klein-Gordon equation. Right? But we still assume $P^\mu|0\rangle=0$. – SRS May 23 '16 at 17:54
• @SRS it depends. If the source is part of the EoM then it does not break the invariance (because the source itself transforms together with all the other fields, so "the physics look the same everywhere"). If the source is not part of the EoM (that is, if it is external) then it does break the invariance, because if you transform/translate the other fields, the source remains fixed. This is the reason $J$ breaks the invariance while $\phi^3$ does not. (If you need me to add more details or something is not clear enough, say so and I'll try to improve my answer) – AccidentalFourierTransform May 23 '16 at 17:58
• @AccidentalFourierTransform-Ryder says that as $J\rightarrow 0$, $\langle\phi(x)\rangle_J$ approach a constant value. It is not apriori obvious to me why should the VEV necessarily go to zero as $J\rightarrow 0$ and not to any nonzero constant value. We have spontaneously broken theories in which the VEV is indeed nonzero. – SRS May 23 '16 at 18:03
• @SRS my bad: it is true that unrenormalised fields may have non-zero (but constant) vev. It is just that most of the times it is zero, and when it is not, we usually take $\phi\to\phi-\langle\phi\rangle$ so that the vev is zero (this is kind of irrelevant for your question so you can forget about it if you haven't studied renormalisation yet). Ryder's statement is the general result: the vev is constant; but in the specific case of your question, that is, a free KG field+an external source the vev is zero, as can be checked from the last expression from my answer. – AccidentalFourierTransform May 23 '16 at 18:11
• @ AccidentalFourierTransform- I understand. Can you kindly elaborate your answer to distinguish between the $\phi^3$-source term and an external source term, and explain why an external source breaks the invariance of the vacuum but not the $\phi^3-$term? – SRS May 23 '16 at 18:15