I have some trouble understanding what is the physical meaning of the vacuum expectation value (VEV) when applied in a QFT context.

Question: What is the physical meaning of a VEV? I understand that it is the expectation value of some operator on a specific state, so the question can also be casted in the form:

What does the expectation value for QFT operators tell me about the theory?

Is it the average energy of the field in that state or something like that? For example, if you find it useful, consider the scalar field $\langle \hat{\phi}\rangle$ or for quark fields $\langle \hat{\bar{q}}\hat{q}\rangle$, what does the expectation value tell me about the physics of the theory?

How, then, is the VEV related to the vacuum/0-particle state of my field? What I mean is, do the VEV tell me the average energy of the 0-particle state, or something like the average field content in the 0-particle state? Like, if the VEV is 5GeV, this means that my 0-particle state has an average energy of 5GeV?

This is driving me crazy, I can't find direct answers to these questions, only some workarounds. Also sorry for my English, not a native speaker.

  • $\begingroup$ Does looking at the Wiki page for propagators help you? Basically, VEVs of certain operators give the probability amplitude of a particle going from point x to another point x'. The intuitive sense of a field operator $\phi (x)$ is that it "creates a particle at point x". $\endgroup$ Apr 7 at 13:07
  • 1
    $\begingroup$ I know how propagators work and what field operators do, the fact is that the 1 point correlation function, $\langle 0 | \hat{\phi} | 0\rangle$ isn't the probability amplitude of going x->x', you need a 2 point correlator for evaluating the probability of the x->x' transition. $\endgroup$
    – LolloBoldo
    Apr 7 at 13:15
  • $\begingroup$ Also, VEV are constant everywhere due to Lorentz invariance, so they cannot depend on x $\endgroup$
    – LolloBoldo
    Apr 7 at 13:16
  • $\begingroup$ A 2-point function (propagator) is also Lorentz invariant, but it isn't constant. Being conserved is different from being constant. $\endgroup$
    – Avantgarde
    Apr 7 at 13:32
  • $\begingroup$ Yes right, i meant that the VEV with a single operator is constant everywhere due to Lorentz invariance: $\langle \hat{\phi}(x)\rangle = \langle e^{-ipx}\hat{\phi}(0)e^{ipx}\rangle = \langle \hat{\phi}(0)\rangle$ since the phases on the vacuum states give the product $e^{-itE_0}e^{+itE_0}$ which is 1 $\endgroup$
    – LolloBoldo
    Apr 7 at 13:38

3 Answers 3


Suppose the VEV is a constant for simplicity. Then a field taking a certain value as a VEV tells you that this field value minimizes the quantum effective potential \begin{equation} \frac{\partial V_{\rm eff}}{\partial \phi}\Big|_{\phi=\langle \phi \rangle} = 0 \end{equation} The quantum effective potential is the term in the quantum effective action with no derivatives acting on the fields.

In other words, knowing the VEV tells you one solution of the equations of motion of the theory.

Now, often in particle physics we use perturbation theory to understand what's going on. In the context of a field with a VEV, we can also perform perturbation theory about the solution $\phi=\langle \phi \rangle$. This means you take the full action, set $\phi = \langle \phi \rangle + \delta \phi$, and expand to some order in the fluctuation $\delta \phi$ about the VEV.

All of the usual story of perturbation theory you learned carries through. You can approximately treat the fields (which includes the fluctuation field $\delta \phi$) as harmonic oscillators. The ground state of the whole system corresponds to the situation when all the harmonic oscillators are in their ground state, so for example the fluctuations about the VEV have only zero point energy. We often call this ground state the vacuum state. There are also particle states that correspond to excitations in the fluctuation fields.

If the VEV were to take a different value, then the perturbation expansion would of course be different. We say that this different VEV corresponds to a different vacuum state, since it corresponds to a different starting point for perturbation theory, which in general will lead to different parameters for the perturbative theory of the fluctuations about the ground state (eg, masses, couplings).

Conceptually, it is important to separate out which of these statements are exact, and which are a way of talking about perturbation theory. In particle physics, perturbation theory is so prevalent that it is sometimes difficult to remember that a lot of the language is fundamentally derived from an approximation.

  • $\begingroup$ Yes i know that stuff, it's what we usually do in QFT. My question is different: what kind of solution is $\phi = \langle \hat{\phi}\rangle$ ? Being $\phi$ the "quanta creation/annihilation operator" what does it mean that this operator is a constant like 9GeV? That is the part which i don't understand. What is the meaning of the VEV as an operator in the particle/QFT context. I know we later use it as the point to expand about in perturbation theory, but what does it mean to have a constant field as a quantum operator? $\endgroup$
    – LolloBoldo
    Apr 7 at 14:16
  • $\begingroup$ @LolloBoldo $\phi$ being the "quanta creation/annhilation operator" is a perturbative way of looking at things. $\phi$ is an operator. An exact statement is that there is a state such that $\langle \hat{\phi} \rangle = VEV$ which minimizes the quantum effective potential. In other words this state is an exact solution of the theory. $\endgroup$
    – Andrew
    Apr 7 at 14:42
  • $\begingroup$ You can also view the new vacuum state as a coherent state of $\phi$ particles. $\endgroup$
    – Andrew
    Apr 7 at 14:43
  • $\begingroup$ I don't think is a perturbative way of describing $\phi$, it's just it's fourier decomposition. You are choosing a basis in the hilbert space to express the operator, nothing to do with the perturbative theory. The perturbative theory enters the game via the choice of operator, namely if i choose $\phi$ rather then a different $\chi$ or $\psi$ $\endgroup$
    – LolloBoldo
    Apr 7 at 16:05
  • 1
    $\begingroup$ @LolloBoldo The interpretation of the Fourier coefficients in terms of a creation and annihilation operator comes from perturbation theory since it is based on the creation and annihilation operators in the harmonic oscillator problem. $\endgroup$
    – Andrew
    Apr 7 at 16:46

Part of the confusion may be a result of the terminology. Perhaps it would help if we distinguish between a vacuum state and the ground state. The vacuum state is a formal concept which is associated with an eigenstate of the number operator with a zero eigenvalue. The ground state is the lowest energy state of a theory. In free field theories, the ground state is a vacuum state. When there are interactions in a theory its ground state may not be a vacuum state.

A classical example is the Higgs mechanism in the electroweak theory. The potential for the scalar Higgs field has the famous sombrero hat shape, which shows that the lowest energy of the theory does not appear at $\phi=0$, but at some nonzero value. Therefore, we end up with a ground state with a nonzero vacuum expectation value for the scalar.

What does it mean? Are there particles floating around in the "vacuum"? That would be problematic because it would break Lorenz invariance. Remember, the best we have is the formal description provided by the theories that we have to study these scenarios. We cannot perform an experiment to see what is sitting inside that VEV. So, in that sense, perhaps the best way to view it is simply to say that the field of the ground state is nonzero.


After doing some research i might have found an answer, i will post it hoping to help others and to see if someone disagree. I will start stating that we have 2 different states regarding a field $\hat{\phi}$ which are important there:

  1. The zero $\phi-$particles state, with associated energy $E_0$. I'll call it $|0\rangle_{\phi}$

  2. The vacuum state for $\phi$, namely the one which minimizes the energy of the field, $E_v$. I'll call it $|V\rangle_{\phi}$

Usually we have that $|0\rangle_{\phi} = |V\rangle_{\phi}$, that means that the zero $\phi-$particles state is the vacuum one, ie the one with the minimun energy.

Sometimes that's not true, see Higgs field or Cooper Pairs, and we have that the zero $\phi-$particles state actually has more energy than the real vacuum state, so that the state of minimum energy $|V\rangle_{\phi}$ differs from the zero $\phi-$particles state by an amount v. This difference is the so called vacuum expectation value. Since we lose energy in the transition $|0\rangle_{\phi} \rightarrow |V\rangle_{\phi}$ this extra energy is used to excite uniformly (in space) the field $\hat{\phi}$ to create a "soup" of coherent states with zero momentum (for translational invariance). This is what is called a condensation. Regard to this, the VEV v is the energy difference between the two states which is used to condensate the field and generate that soup of coherent states. As an operator, it acts on the Fock space as $v\mathcal{I}$. We can then take a new field variable, for which the vacuum and the zero particles state are the same, considering the "condensed field" operator $\hat{\Phi} = \hat{\phi} - v\mathcal{I}$.

In this sense the zero $\Phi-$particles state is full of coherent $\phi-$particles states with vanishing momentum.

As a pure mnemonic analogy, the $\hat{\phi}$ is like a "water vapor" operator which at 300K describes a gas state condensed to a liquid giving extra energy, while the $\hat{\Phi}$ is the liquid droplets operator which at 300K is a pure liquid, with no "free" energy to release (in the lagrangian the free energy generates masses in fields interacting with the field condensed)

  • $\begingroup$ What is a "soup of coherent state"? $\endgroup$ Jun 9 at 3:13
  • $\begingroup$ Is was a simple jargon to say that in each point of spacetime in my Hilbert space the state $|V\rangle_\phi$ can be expandend into a superposition of coherent states with zero momentum. $\endgroup$
    – LolloBoldo
    Jun 9 at 8:08
  • $\begingroup$ By soup i meant that superpositioning of those vanishing momentum coherent states $\endgroup$
    – LolloBoldo
    Jun 9 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.