Equal-time Canonical Commutation Relation for a scalar field

In chapter 2 of Quantum Field Theory and the Standard Model, Schwartz derives the equal-time commutation relations of the second-quantised field. Using $$\phi(\vec{x}) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \left(a_p e^{i\vec{p}\cdot\vec{x}} + a_p^{\dagger} e^{-i\vec{p}\cdot\vec{x}}\right),\tag{2.75}$$ he arrives at $$[\phi(\vec{x}), \phi(\vec{y})] = \int\frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_p} \left(e^{i\vec{p}\cdot(\vec{x}-\vec{y})} - e^{-i\vec{p}\cdot(\vec{x}-\vec{y})} \right).\tag{2.89}$$ Now he argues that the integral measure and $$\omega_p$$ are symmetric under $$\vec{p} \rightarrow -\vec{p}$$ and hence the commutator vanishes. I cannot see why $$d^3 p$$ is symmetric. From my understanding, it is just the 3-momentum and should be $$-d^3p$$ under the substitution. Am I missing something here?

Also, in an earlier section he defines, $$\phi_0(\vec{x},t) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \left(a_p e^{-ipx} + a_p^{\dagger} e^{ipx}\right).\tag{2.78}$$ The $$p$$'s and $$x$$'s here are 4-vectors. Since the integral is over only the 3-momentum, is this Lorentz invariant? (I can see that the integrand is Lorentz invariant).

References:
Section 2.3.2 & 2.3.3, Quantum Field Theory and the Standard Model, M.D. Schwartz

• The flip of the integration limits matches that of the differentials. Surely your text has proven the Lorentz invariance of the 3-p measure involving the energy downstairs... Jul 28, 2022 at 18:27

You can write the integrals in question as $$\int_{\mathbb{R}^3} f(\vec{p}) d^3 p = \int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\! f(p_x,p_y,p_z) dp_x dp_y,dp_z$$ where the function $$f(\vec{p})$$ can be defined to include the $$1/\omega_p = \sqrt{m^2 + \vec{p}^2}$$ prefactor.
Now start with just one variable switch $$p'_z = -p_z$$. This changes the differential to $$dp'_z = - dp_z$$ and the bound $$p_z = \infty$$ to $$p'_z = -\infty$$, and the same with a flipped sign for the second bound. So the integral is now $$\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\! f(p_x,p_y,p_z) dp_x dp_y,dp_z = -\int_{\infty}^{-\infty}\!\!\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\! f(p_x,p_y,-p'_z) dp_x dp_y,dp'_z$$ However, you can flip the integration bounds on the $$p'_z$$ integration at the cost of changing the sign of the integral, so you have $$\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\! f(p_x,p_y,p_z) dp_x dp_y,dp_z = \int_{-\infty}^{\infty}\!\!\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\! f(p_x,p_y,-p'_z) dp_x dp_y,dp'_z$$
You can do this with every variable $$p_x,p_y,p_z$$, drop the primes, and finally you will obtain $$\int_{\mathbb{R}^3} f(\vec{p}) d^3 p = \int_{\mathbb{R}^3} f(-\vec{p}) d^3 p$$