4

I found that the taking the non-relativistic limit of the Lagrgangian for complex scalar fields gives

$$\mathcal{L} = i\dot{\psi}\psi^* -\frac{1}{2m}\nabla\psi \nabla\psi^*.\tag{1}$$

Now, when we quantise relativistic complex scalar fields, we must do so in terms of two types of creation/annihilation operators,

$$\psi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(b_pe^{-ipx}+c_P^\dagger e^{ipx}).\tag{2}$$

As per my understanding, the reason we need two such operators is that we are really quantising two independent fields $\phi_1$ and $\phi_2$ corresponding to $$\psi = \phi_1 + i\phi_2.\tag{3}$$

My question is why does this same argument not extend to the non-relativistic limit, where we quantize the complex field in terms of a single type of creation/annihilation operator:

$$\psi(x) = \int \frac{d^3 p}{(2\pi^3)}a_pe^{-ipx}.\tag{4}$$

Going back to the logic above, does this mean that for non-relativistic complex scalar fields, the components of the complex field are not independent, ie., $\phi_1$ and $\phi_2$ are somehow related?

Qmechanic
  • 201,751
Souroy
  • 153
  • 4

2 Answers2

3

The reason behind the two ''types'' of creation/annihilation operators comes from the non-Hermiticity of $\phi$ (cf. Zee's QFT in a Nutshell, Ch. I.8.). Indeed this can also be seen as the appearance of having 2 independent real scalar fields $\phi_1, \phi_2$ as you point out. The reason that we even have 2 terms for any momentum $p$ in the Fourier expansion is because we have ''positive'' and ''negative'' Fourier modes because of the Klein-Gordon equation being ($p^2 + m^2 = 0$), $p$ being the four-momentum. (cf. https://en.wikipedia.org/wiki/Klein–Gordon_equation#Solution_for_free_particle)

In the non-relativistic limit, the approximation $E \ll m$ breaks the symmetry of the K-G field and yields a field obeying Schroedinger's eq., which when expanded in modes has only ''positive'' Energy solutions ($E = \vec{p}^2/2m$) thus we have only ''positive'' Fourier modes. Which you point out as a single type of creation/annihilation operators.

Regarding quantizing the theory, canonically it corresponds to having $[\partial_0 \phi^\dagger(\vec{x}',t),\phi(\vec{x},t)] = -i\delta(x-x')$ regardless of which creation-annihilation operators we end up having in our Fourier expansion (again, cf. Zee's QFT in a Nutshell, Ch. I.8.).

Juan Fernandez
  • 81
2

That's a good question.

  1. It turns out that the corresponding Greens function for the Schrödinger action (1) is the retarded Greens function, which only has a pole for positive frequencies in the complex $k^0$-plane.

  2. This implies that the Fourier expansion (4) for $\psi$ can not contain an antiparticle creation operator mode with negative frequency, cf. e.g. my Phys.SE answer here.

  3. Moreover the real and imaginary parts of the complex field (3) become canonically conjugate pair (up to a $1/\sqrt{2}$ normalization), cf. e.g. this Phys.SE post.

  4. Interestingly, the above conclusions hold whether or not the Schrödinger field $\psi$ originates from a non-relativistic limit of a Klein-Gordon field as suggested in e.g. this Phys.SE post. (Although assuming such limit does indeed eliminate the negative frequency modes.)

Qmechanic
  • 201,751