Second quantisation of the scalar field leads to an algebra of quantum field operators $$ [\phi(x),\phi(y)] = 0, \ \ [\pi(x), \pi(y)] = 0, \ \ [\phi(x),\pi(y)] = i\hbar \delta(x-y). $$ Where the field operators are given by $$ \phi(x) = \int \frac{\text{d}^3 k}{(2\pi)^3 2 \omega_{\mathbf{k}}}\left( a_k e^{i k x} +a^{\dagger}_k e^{-ikx}\right) , \qquad\pi(x)=\partial_0 \phi(x) $$ These are made of two counter rotating terms. However in many body theory I have also seen field operators defined as $$ \phi(\mathbf{x}) = \frac1{\sqrt{V}} \sum_\mathbf{k} a_\mathbf{k} e^{i \mathbf{k}. \mathbf{x}} $$ I can understand why outside of the context of relativistic physics we drop the requirement for covariance, but the counter rotating term has also been dropped and so this definition has a different algebra altogether. The former definition seems like a generalisation of the $\hat{x}$-operator of a 1D harmonic oscillator, whereas the latter seems like a generalisation of the creation operator $\hat{a}$.

This would be fine, and I could accept it as a choice of convention were it not for the fact that I have seen Greens functions defined using both. The greens functions as I understand them allow for the calculation of time ordered operator products by giving a general form for the amplitude of measuring different field values at different times. However this does not seem to enlighten me to why this understanding of the Greens function is conflated with the latter definition.

I have clearly missed something important, but going through the relevant textbooks has failed to uncover what that is. If anyone can give an explanation of what a Green's function is and also explain how these two definitions can be reconciled I would be greatly appreciative.

  • $\begingroup$ See Feynman and Hibbs (1965), Quantum Mechanics and Path Integrals, Chapters 4 and 5. $\endgroup$
    – user28355
    Commented Feb 2, 2014 at 16:27
  • $\begingroup$ I fail to understand the point you're trying to make about the Green's function. It is just a matter of convention. Would it help if in the field theory context I wrote $\phi = \phi_+ + \phi_-$ (see P&S)? Obviously $\phi, \pi$ will obey different commutation relations from $\phi_+, \phi_-$, but they are related to one another. Then $\phi_+, \phi_-$ in the field theory context will be equivalent to $\phi, \phi^\dagger$ in the many-body context. $\endgroup$
    – nervxxx
    Commented Feb 2, 2014 at 20:19
  • $\begingroup$ Agreed, they are related, even trivially related, but the values $\langle T \phi(t) \phi(t')^\dagger \rangle$ are different quantities if you take the different definitions, and you will be evaluating different quantities. $\endgroup$ Commented Feb 2, 2014 at 21:03

2 Answers 2


I don't think the existing answer is getting at the right point. Instead, you've stumbled on the key difference between relativistic and nonrelativistic quantum field theory: you always need the counterrotating term in relativity.

For example, a nonrelativistic particle satisfies $E = p^2/2m$, which upon quantization gives the dispersion relation $\omega = k^2/2m$. When one describes multiple particles of this type using a field, the Heisenberg equation of motion of the field formally obeys the single-particle Schrodinger equation, which means that its frequency components satisfy $\omega = k^2/2m$. The sum over these $(\omega, k)$ pairs is what is performed in your many-body theory equation, with the time dependence implicit in $a_{\mathbf{k}}$, assuming we're in Heisenberg picture.

In relativity, a relation like $E = p^2/2m$ is forbidden because it doesn't put time and space on the same footing. The simplest relation that does work is $E^2 = p^2 + m^2$ which becomes the dispersion relation $\omega^2 = k^2 + m^2$. For each value of $k$, there are two valid values of $\omega$, which is why the field operator has two terms. This doubling of values for $\omega$ occurs for any relativistic dispersion relation.

(Of course, it should be noted that the existence of positive and negative frequency solutions don't mean that relativistic QFT has positive and negative energy particles. Due to nice tricks performed during quantization, which differ for bosonic and fermionic fields, you can make the negative frequency solutions correspond to positive energy particles, at the cost of flipping all their charges. This is why relativistic QFT predicts antimatter.)

  • $\begingroup$ I think the last paragraph is somewhat naive. Restricting energy to be positive is perfectly Lorentz-invariant -- and indeed otherwise we wouldn't have energy spectra bounded from below. The "negative energy" term is really required for microcausality (commutation at spacelike separation). Justifying microcausality is actually tricky; Weinberg, e.g., argues that it is required for Lorentz-invariance of the $S$-matrix. $\endgroup$ Commented Aug 29, 2019 at 3:40
  • $\begingroup$ In non-relativistic theory there is no reason to require microcausality from fields that one uses to construct the interaction density, and so one can get away with a single sign (i.e. containing only creation or annihilation operators). $\endgroup$ Commented Aug 29, 2019 at 3:42

It is a matter of 4D versus 3D notation. The point is that $kx=-\omega_\mathbf{k} t +\mathbf{k}\cdot\mathbf{x}$ and $\omega_\mathbf{k}>0$, so that two terms $\pm kx$ are needed to cover both signs of the prefactor of $t$ in a valid solution of the free relativistic field equations.

In the relativistic case, the negative energy terms must have creation operators as coefficients to ensure causality.

In the nonrelativisitc case, the negative energy terms must be dropped in order to have equivalence between the first- and second-quantized description.


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