Let $\phi(\textbf{x}) \neq \phi^\dagger(\textbf{x})$ be a complex scalar field, and let $\varphi(\textbf{x}) = \varphi^\dagger(\textbf{x})$ be a real scalar field.

$\phi(\textbf{x})$ destroys a particle and/or creates an antiparticle at $\textbf{x}$. Equivalently, $\phi(\textbf{x})$ destroys a single charge quantum at $\textbf{x}$, while $\phi^\dagger(\textbf{x})$ creates it. Reinterpreting this in terms of "particles" and "antiparticles" gives the same results as above because to create a charge I could equally well create a particle with that charge or destroy the corresponding antiparticle.

This is perfectly compatible with the fact that complex/non-hermitian fields represent charged particles, and from this reasoning we can expect that a chargeless particle must be represented by a real/hermitian field $\varphi$, as in this case there is no charge to create or destroy hence no way to distinguish $\varphi$ from $\varphi^\dagger$, which are thus the same thing (which is related to the lack of $U(1)$ gauge simmetry in the real case, and that of the related conserved charge). These reasonings raise a few questions:

  1. The fact that $\phi$ represents creation and destruction of a particle can be naturally interpreted saying that $\phi$ destroys a charge quantum. What does the operator $\varphi$ represents then? What quantity does $\varphi$ destroy that is equivalently achieved by destruction and creation of a chargeless "particle"?
  2. Consider the Fourier expansion of $\varphi$ (neglecting constants ecc.): $$ \varphi(\textbf{x}) = \sum_{\textbf{k}} \left( e^{i\textbf{k} \cdot \textbf{x}} a_\textbf{k} + e^{-i\textbf{k} \cdot \textbf{x}} a_\textbf{k}^\dagger \right)$$ For this to be hermitian we must have $a_{-\textbf{k}} = a_\textbf{k}^\dagger$, which I can interpret saying that "adding a momentum quantum $\textbf{k}$" is equivalent to "removing a momentum quantum $-\textbf{k}$". This makes me thing that the $a_\textbf{k}$ operators do not represent the "creationg of a particle with momentum $\textbf{k}$" but rather the addition of a momentum quantum $\textbf{k}$ to the field. However, I can think of a situation in which I have 2 chargeless particles each one with momentum $\textbf{k}$ (we are talking of bosons so this should be acceptable), and this reasoning leads me to the conclusion that this situation would be equivalent to a single particle with momentum $2\textbf{k}$, which is not what I would expect. I suspect that this is due to the fact that the particle interpretation here is misleading, and we should just talk of "field excitations" throught creation/destruction of quanta of physically measurable quantities. I cannot measure "a particle" but rather its momentum (or whatever), so I should just talk of momentum quanta and forget the "particles". Are this reasonings correct? (SPOILERS: NO)
  • $\begingroup$ Gotcha! Well then, I should delete my comment. $\endgroup$
    – BMS
    Sep 17 '14 at 13:27
  1. The field theory is fully analogous for Hermitian and non-Hermitian fields

The Hermitian operator $\varphi$ still creates and/or annihilates particles and the number of these particles $N$ is still well-defined (at least if we ignore interactions and problems with loops and divergences).

The only difference from the non-Hermitian field is that the particles and the antiparticles are the same things, so instead of two different "counts" of a particle species – one for an electron and one for a positron, for example – we only have one species (e.g. the Higgs boson) and one operator $N$.

If the operator $\varphi$ acts on an $N$ eigenstate with the eigenvalue $n$, it produces a superposition of two states, one of which is an $N$ eigenstate with the eigenvalue $n+1$ and the other one has the eigenvalue $n-1$.

In the charged, non-Hermitian case, there is usually a simple quantity that is conserved, the "charge" – it's the difference of the number of positrons minus the number of electrons, for example. In the case of the Hermitian field, there is no analogous operator (except for the discrete $(-1)^N$) that is conserved. But otherwise there is no difference between the two cases.

  1. It is not true that the modes with opposite momenta are linked

Your conclusion that $a_k=a^\dagger_{-k}$ is simply incorrect. If you rewrite the sum you wrote as the sum of two sums, then the second sum is self-evidently the Hermitian conjugate of the first sum (I didn't want to talk about the terms for a fixed value of $\vec k$, to avoid discussions about the difference between $\vec k$ and $-\vec k$), so the total expression for $\varphi(x)$ is Hermitian without any additional conditions just like $z+\bar z$ is real without any additional conditions on a complex $z$.

Obviously, it can't be true that a creation operator is equal to an annihilation operator. They are qualitatively different (each annihilation operator annihilates the vacuum, while no creation operator does so) so they cannot be equal.

  • $\begingroup$ Fantastic, thanks! Could you also elaborate just a little on the discrete $(-1)^N$ conserved quantity you mentioned? Is it the parity or what? $\endgroup$
    – glS
    Sep 19 '14 at 12:30
  • 2
    $\begingroup$ It is a quantum number with eigenvalues $+1$ and $-1$, so it's a form of parity, but an unrelated parity to the usual reflection in space. It reflects the sign of $\varphi$. Equivalently, it's $+1$ if the number of the quanta is even, and $-1$ if it is odd. The operator $\varphi$ may create or destroy but whatever it is, it changes an even number to an odd one or vice versa. So the parity of the number of particles always changes by $\varphi$, and if the Hamiltonian is even in $\varphi$ or its derivative, this parity is conserved. $\endgroup$ Sep 19 '14 at 19:34
  • $\begingroup$ @glS This "conservation of boson number parity" leads immediately to the consequence that any correlation function for an odd number of free real scalar fields is automatically zero, e.g. $\langle \varphi(x) \rangle = \langle \varphi(x) \varphi(y) \varphi(z) \rangle = 0$. $\endgroup$
    – tparker
    May 25 '16 at 1:18

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