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The scalar quantum field operator is defined

$$\phi_0(\vec x,t) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \left(a_p e^{-ipx}+a_p^{\dagger}e^{ipx}\right)$$

in Schwartz eq. 2.78. Here $\omega_p = \sqrt{\vec{p}^2+m^2}$. One can show that

$$\int \frac{d^3\vec{k}}{2\omega_k} = \int d^4k \delta(k^2-m^2)\theta(k^0)$$

is a Lorentz invariant measure. I don't see how the measure in the definition of the quantum field is Lorentz invariant because of that square root.

Qmechanic
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Dwagg
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  • Possible duplicates: https://physics.stackexchange.com/q/83260/2451 and links therein. – Qmechanic Dec 25 '17 at 00:43
  • I looked at the links and don't see an answer to my question. – Dwagg Dec 25 '17 at 01:14
  • Perhaps: The measure in the first integral above is is not necessarily Lorentz invariant, but the expression as a whole is because it is an operator that produces Lorentz invariant states when it acts upon kets. As in:

    $$\phi(\vec x) |0\rangle = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_p} e^{-ip\cdot x} |\vec{p}\rangle$$

    where we used $a^{\dagger}_p|0\rangle = \frac{1}{\sqrt{\omega_p}} |\vec{p}\rangle$.

     – Dwagg Dec 25 '17 at 01:15
  • Along with

    $$\langle 0 |\phi(\vec x)|\vec{p}\rangle = e^{ip\cdot x}.$$

    And since the $\phi$ operator produces Lorentz invariant things, it is Lorentz invariant.

    Is this true? Can this be made rigorous?

     – Dwagg Dec 25 '17 at 01:18

2 Answers2

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We want to construct the field such that $$ \langle\vec{p}|\phi(x)|0\rangle=e^{-i\vec{p}\cdot \vec{x}} $$ and the definition is $\langle \vec{p}|=\langle 0|\sqrt{2\omega_p} a_p$, i.e. the state $|\vec{p}\rangle$ have an invariant inner product: $$ \langle \vec{k}|\vec{p}\rangle=2\omega_p (2\pi)^3\delta^3(\vec{p}-\vec{k}) $$ such that the identity operator can be writen as: $$ \mathcal{1}=\int \frac{d^3 p}{(2\pi)^3}\frac{1}{2\omega_p}|p\rangle\langle p| $$ so the square root is just cancelling the square root of the $\langle \vec{p}|$ state.

All this is convention. The Weinberg texbook chooses a different one in witch $$ \langle \vec{k}|\vec{p}\rangle= (2\pi)^3\delta^3(\vec{p}-\vec{k}) $$ and the price is to introduce a new factor into the Lorentz transformation of the state $|\vec{p}\rangle$: $$ U(\Lambda)|\vec{p}\rangle = \sqrt\frac{(\Lambda p)^0}{p^0} |\vec{p}_{\Lambda}\rangle $$

Nogueira
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  • I think you may have misread my question. I am not asking about $\int \frac{d^3k}{2\omega_k}.$ Rather I am asking why $$\int \frac{d^3k}{\sqrt{2\omega_k}}$$ is Lorentz invariant. – Dwagg Dec 25 '17 at 01:34
  • @Dwagg I edited, see if now this answers? – Nogueira Dec 25 '17 at 01:49
  • @Noguira In your last statement, are you talking about the first equation in my question? – Dwagg Dec 25 '17 at 01:55
  • @Dwagg This $\sqrt{2\omega_p}$ that appears in the field expresion (your first formula), is just there to cancel the square root of the definition of the state $\langle \vec{p}|$ in order to give what you want, the $e^{ipx}$ – Nogueira Dec 25 '17 at 02:00
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    Okay. In particular, the first formula is not Lorentz invariant by itself? It is constructed to make $\langle 0 | \phi(x,t)|\vec{p}\rangle = e^{ipx}=$ Lorentz invariant? – Dwagg Dec 25 '17 at 02:16
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    @Dwagg The first line can be made Lorentz invariant by assigning the operators $a_{\vec p}$ and $a_{\vec p}^\dagger$ the corresponding transformation properties. Whether or not that's a beautiful convention is a different question. –  Dec 25 '17 at 05:02
  • @Dwagg, yes, after all, the field $\phi(x)$ is not Lorentz invariant, but transforms as $\phi(\Lambda x)$, it's a nontrivial representation (just the vacuum is a trivial representation). – Nogueira Dec 25 '17 at 05:24
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The measure you consider is not Lorentz invariant. The point is that, under Lorentz transformations, $a_p$ and $a_p^\dagger$ are not $p$-scalar fields but they take a factor which compensates the failure of the measure to be Lorentz invariant, and the integral defining the quantum field produces a scalar field. Nogueira's answer includes all information necessary to write down the transformation rule of $a_p$ and its Hermitian conjugate companion. However all that is matter of convention, since one could use the invariant measure from scratch.

Valter Moretti
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