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:
- 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.
