The relativistic scalar field operator is not a ladder operator. Its commutation relations are

$$\begin{align} \left[\hat{\phi}\left(\vec{x}\right), \hat{\phi}\left(\vec{y}\right)\right] = \left[\hat{\pi}\left(\vec{x}\right), \hat{\pi}\left(\vec{y}\right)\right] &= 0,\\ \left[\hat{\phi}\left(\vec{x}\right), \hat{\pi}\left(\vec{y}\right)\right] &= i \delta\left(\vec{x} - \vec{y}\right), \end{align}$$

and its ladder operators $$\begin{align} \hat{a}(\vec{k}) &\propto \left(\widetilde{\phi}(\vec{k}) + i\widetilde{\pi}(\vec{k})\right),\\ \hat{a}^\dagger(\vec{k}) &\propto \left(\widetilde{\phi}(\vec{k}) - i\widetilde{\pi}(\vec{k})\right), \end{align}$$


$$ [\hat{a}(\vec{k}_1), \hat{a}^\dagger(\vec{k}_2)] =\delta(\vec{k}_1-\vec{k}_2) $$

and commute elsewhere. The same holds for other normal field operators. The non-relativistic limit of any free field is supposedly two Schrödinger fields representing the particle and anti-particle (Padmanabhan 2018), but the Schrödinger field's commutation relations are

$$ \left[\hat{\psi}\left(\vec{x}\right), \hat{\psi}^\dagger\left(\vec{y}\right)\right] = \delta\left(\vec{x} - \vec{y}\right) $$

or equivalently

$$ \left[\tilde{\psi}\left(\vec{k}_1\right), \tilde{\psi}^\dagger\left(\vec{k}_2\right)\right] = \delta\left(\vec{k}_1 - \vec{k}_2\right), $$

and thus $\hat{\psi}$ is an annihilation operator used to define a non-relativistic Fock space.

If this is the case, how does the field operator become a ladder operator in the non-relativistic limit? (Edit 1: it isn't.)

Can one construct non-relativistic quantum field, that's the non-relativistic limit of a normal field, that commutes with its adjoint (like normal field operators), and that requires you to construct ladder operators separately?

Edit 1:

The Schrödinger field is the limit of the annihilation operator, not the field. I have some of it worked out, starting from the momentum-space annihilation operators $\hat{a}_\mathbf{p},\hat{b}_\mathbf{p}$,

$$\hat{a}(x)=\int d\Omega_\mathbf{p}\hat{a}_\mathbf{p}e^{-i p\cdot x},\quad \hat{b}(x)=\int d\Omega_\mathbf{p}\hat{b}_\mathbf{p}e^{-i p\cdot x}$$

$$\hat\phi(x)=\hat a(x)+\hat b^\dagger(x)$$

Defining $\hat{A}(x)$ and $\hat{B}(x)$ as

$$\hat{a}(x)=\frac{e^{-i m c^2 t/\hbar}}{\sqrt{2m}}\hat{A}(x),\quad \hat{b}(x)=\frac{e^{-i m c^2 t/\hbar}}{\sqrt{2m}}\hat{B}(x)$$

and substituting, the Lagrangian becomes

$$ \begin{align} L &= (\hbar c)^2\partial_a\phi\partial^a\phi^\dagger - (mc^2)^2\phi\phi^\dagger\\ &= (\hbar c)^2(\partial_a\hat{a}\partial^a\hat{a}^\dagger + \partial_a\hat{b}\partial^a\hat{b}^\dagger + \ldots) - (mc^2)^2(\hat{a}\hat{a}^\dagger + \hat{b}\hat{b}^\dagger + \ldots) \\ &= \frac{1}{2m}\left[(\hbar c)^2(\frac{-imc}{\hbar}\hat{A} + \partial_a\hat{A})(\frac{imc}{\hbar}\hat{A}^\dagger + \partial^a\hat{A}^\dagger) + \ldots - (mc^2)^2(\hat{A}\hat{A}^\dagger + \hat{B}\hat{B}^\dagger + \ldots)\right] \\ &= \frac{1}{2m}\left[ (\hbar c)^2\partial_a\hat{A}\partial^a\hat{A}^\dagger + (\hbar c)^2\partial_a\hat{B}\partial^a\hat{B}^\dagger + \ldots \right] \end{align} $$

where $\ldots$ are proportional to $e^{\pm 2 i m c^2 t/\hbar}$ and ignored. But where to go from here?

  • 1
    $\begingroup$ Why would you say that the field operator becomes a "ladder" operator? Just because the commutator is a delta, it doesn't make it a ladder operator. In order for an operator to be a ladder there must be a Fock representation upon which they act in a certain way onto the N-particle states, with $a^*a$ being diagonal onto those states. $\endgroup$ – gented Jun 19 '19 at 21:22
  • $\begingroup$ Yes, that is the case. It is a ladder operator in that sense. And it's taught that way in QFT courses. $\endgroup$ – alexchandel Jun 19 '19 at 21:24
  • $\begingroup$ I think I got it: it's the annihilation operator(s) of $\hat{\phi}$ that obey(s) the Schrödinger field equation as $c\to\infty$, not $\hat{\phi}$ itself. So $\hat{\psi}$ is necessarily a ladder operator. $\endgroup$ – alexchandel Jun 19 '19 at 22:29
  • $\begingroup$ The problem with that construction is that those fields that appear in the equations aren't operators in the strict sense, rather operator values distributions, therefore you must smear them with appropriate functions in order to get an operator in the standard sense. If you do so then you can precisely define the Fock space and the ladder operators, but there are a few steps that are "hidden under the carpet" both in that wiki page and in the notes. $\endgroup$ – gented Jun 20 '19 at 8:12
  • $\begingroup$ I know, I elided that detail (and the rest of the formalism) in order to get to the question without quoting all of Peskin & Schroeder. Even with appropriate treatment, the question still remained: why was the non-relativistic limit of $\hat{\phi}$ (suitably smeared) an annihilation operator, rather than a field operator from which one derived ladder operators? –– Like I said, I think my assumption was wrong. I'll leave this open for a bit if someone can/wants to answer it with a proof. $\endgroup$ – alexchandel Jun 28 '19 at 21:37

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