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I have some questions about the first part of the answer by Arnold Neumaier dealing with the interpretation of VEV (vacuum expectation value). Let me quote the parts I would like to understand better:

As is easily checked, fields linear in creation and annihilation operators (and hence amenable to a particle interpretation) have zero vacuum expectation value. Thus the $\phi$ field with its nonvanishing vacuum expectation value cannot be given a particle interpretation. But the field $\psi=\phi−v $ has such an interpretation as its vacuum expectation value is zero. This works only if $v$ is the vacuum expectation of $\phi$.

Questions:

  1. Why are only fields linear in creation and annihilation operators, i.e. fields which can be expanded as

$$ \hat{\phi}(\vec{x},t) = \int c \cdot d^3p\left[\hat{a}(\vec{p}) \mathrm{e}^{-i(\vec{p}\cdot\vec{x}-E_pt)} + \hat{b}(\vec{p}) \mathrm{e}^{+i(\vec{p}\cdot\vec{x}-E_pt)}\right],$$

'amenable to a particle interpretation'? In other words, why does a field which doesn't have such a linear expansion possibly not allow this 'particle interpretation'?

As far as I understand the issue, the 'particle interpretation' of a field means simply that the excited states become identified with 'particles' in excited states (classically analogous to an electron with energy level above ground state).

Why is it not possible to declare, in the same manner, that the 'excited states' are exactly the 'excited particles' for such a field, like in case of fields linear in creation/annihilation operators mentioned above? What are the obstructions?

This leads to question 2:

Seemingly, according to the quoted answer, the problem is that such a field may have a non-vanishing vacuum expectation value. But I don't understand why it is necessary in order to give a field a 'particle interpretation' with excited states that the field's vev vanishes.

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By definition, creation operators applied to the vacuum state define the 1-particle states of the field theory, and by iteration multiparticle states. All particle terminology in QFT is based on this definition.

In order that the particle interpretation works and produces a multiparticle Fock space, the creation and annihilation operators must satisfy specific (anti)commutation rules in momentum space. A nonzero VEV spoils these rules when one defines these operators by the standard rules, whereas they are valid when one first subtracts the VEV from the field before applying these rules.

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  • $\begingroup$ Yes, I understand that if we start with a field linear in creation and annihilation operators (ie it has the shape as in question 1. above), then the application of creation operator to vacuum state $| 0 \rangle $ $1 \times $ gives by definition exactly the 'thing' we can call a 'particle'. But I still not see why the concept cannot be generalized to the case where the field isnn't linear in creation and annihilation operators and possibly the VEV not vanishes. $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 23:30
  • $\begingroup$ In case of a field $\phi$ linear in creation and annihilation operator we can express the annihilation and creation operators all integrals over $dx$ in $\phi(x)$ and $\Pi(x) := \dot{\phi(x)}$; see e.g. here users.physik.fu-berlin.de/~kleinert/b6/psfiles/… Eq. (7.14) or here pas.rochester.edu/assets/pdf/undergraduate/… Eq (6) & (7) in case of Klein-Gordon field. $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 23:31
  • $\begingroup$ Now the question is what if we start with arbitrary field $\psi$, and try to mimic the case above. That is we define pure formally the operators $$ a^{\pm}_p := c \cdot \int d^3x e^{\mp px} [\pm \Pi(x) + \psi(x)] $$, take an arbitrary state $| s \rangle $ of the (possibly degenerated) vacuum and define $a_p | s \rangle $ as a 'particle'. Seemingly this is independent of the assumption if VEV of $\phi$ isn't zero or if vacuum is degenerated. Why this approach do generalize the concept of 'particles' for fields with nonvanishing vev and/or degeneraed vacua fails? $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 23:33
  • $\begingroup$ @katalaveino: see the new paragraph in my answer $\endgroup$
    – Arnold Neumaier
    Commented Jan 21, 2021 at 8:59
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    $\begingroup$ @CBBAM: Because one get infinities when carrying out the calculations. $\endgroup$
    – Arnold Neumaier
    Commented Apr 26 at 17:24

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