2
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/83260/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Dec 25, 2017 at 0:43
  • $\begingroup$ I looked at the links and don't see an answer to my question. $\endgroup$
    – Dwagg
    Dec 25, 2017 at 1:14
  • $\begingroup$ 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 $a^{\dagger}_p|0\rangle = \frac{1}{\sqrt{\omega_p}} |\vec{p}\rangle$. $\endgroup$
    – Dwagg
    Dec 25, 2017 at 1:15
  • $\begingroup$ 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? $\endgroup$
    – Dwagg
    Dec 25, 2017 at 1:18

2 Answers 2

Reset to default
4
$\begingroup$

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 $|\vec{p}\rangle$: $$ U(\Lambda)|\vec{p}\rangle = \sqrt\frac{(\Lambda p)^0}{p^0} |\vec{p}_{\Lambda}\rangle $$

Improve this answer
$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – Dwagg
    Dec 25, 2017 at 1:34
  • $\begingroup$ @Dwagg I edited, see if now this answers? $\endgroup$
    – Nogueira
    Dec 25, 2017 at 1:49
  • $\begingroup$ @Noguira In your last statement, are you talking about the first equation in my question? $\endgroup$
    – Dwagg
    Dec 25, 2017 at 1:55
  • 1
    $\begingroup$ 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? $\endgroup$
    – Dwagg
    Dec 25, 2017 at 2:16
  • 1
    $\begingroup$ @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. $\endgroup$
    – user178876
    Dec 25, 2017 at 5:02
3
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.