In my opinion, one of the largest standing problems in physics is called the Vacuum Catastrophe. The Quantum Field theory leads us believe that a perfect vacuum innately has a constant energy. The catastrophe arises due to a discrepancy between the theorized value and the measured zero-point energy, energy contained in the vacuum.


In Quantum Field theory, every elementary particle has an associated field and, when taking time into account, they are quantum harmonic oscillators. We can then use the Time-independent Schrödinger equation to find the theoretical zero-point energy. $$\hat{H} \ket{\Psi}=E \ket{\Psi}$$ The hamiltonian, basically, is the sum of the kinetic energy and the potential energy. The kinetic energy of our quantum osillator is $\frac{1}{2}m \omega^2 \hat{x}^2$. where $\hat{x}$ is the position operator and $\omega$ is the angular frequency. The potential energy would be $\frac{\hat{p}^2}{2m}$, where $\hat{p}=-i\hbar\nabla$ is the momentum operator. So, $$\hat{H}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m \omega^2 \hat{x}^2$$ For simplicity, let's restrict our problem to a 1D problem allowing us to easily use the hamiltonian operator on the wave equation, $\Psi$. $$\frac{-\hbar^2}{2m}\frac{d^2 \Psi}{dx^2} +\frac{1}{2}\omega^2 x^2\Psi=E\Psi$$ The solution to the differential equation is explained here. Anyway, the solution for the ground state ends up being $$\Psi_0(x)=\Big(\frac{m\omega}{\pi\hbar}\Big)^{\frac{1}{4}}e^{-\frac{m\omega x^2}{2\hbar}}$$ And, its corresponding energy of $$E_0=\frac{1}{2}\hbar\omega=\pi\hbar f=\frac{hf}{2}=\frac{hc}{2\lambda}$$ So, the lowest possible energy for a specific frequency in a vacuum for the photon field is $\frac{hf}{2}$, where $f$ is the frequency and $\lambda$ is the wavelength. To find the total energy we must integrate the function with respect to $f$ for all frequencies. That integral would diverge into infinity. The task now is to find an upper limit to the frequency of a photon field.
There is however a limit to which our knowledge is absent which is Plank's Energy. At any energy above plank's energy, quantum gravity is dominate. Plank's energy is $$E_p=\sqrt{\frac{h c^5}{2\pi G}}$$ And the corresponding photon frequency would be $$\frac{E}{h}=\frac{\sqrt{\frac{h c^5}{2\pi G}}}{h}=f$$ Our Integral now is $$\int_{0}^{\frac{\sqrt{\frac{h c^5}{2\pi G}}}{h}} \frac{hf}{2} df=\frac{c^5}{8\pi G}=1.44\times 10^{51} \frac{J}{m^3}$$


There is no direct measurement of the zero-point energy, but the zero-point energy would, according to General Relativity, have a gravitational influence permeating through out the universe, like dark energy. This energy, like dark energy, would accelerate the expansion rate of the universe, and this can be measured from stellar observation. Anyway, the energy density ends up being $$9.47\times 10^{-27}\frac{kg}{m^3}$$ and converted to energy units via mass-energy equivalence $$8.51\times 10^{-10}\frac{J}{m^3}$$


The discrepancy between the measured and the predicted is about 60 orders of magnitude large or, exactly, $1.69\times 10^{60}$ times, which as "General Relativity: An introduction for physicists" by MP Hobson, et al. puts it, "the worst theoretical prediction in the history of physics."

In my opinion, we are missing something, maybe relating to quantum gravity and interoperability between General Relativity and Quantum Physics.

For more information, check out "Vacuum catastrophe: An elementary exposition of the cosmological constant problem" by Adler, RonaldJ, et al. (link) and "Dark Energy and the Accelerating Universe" by Joshua Frieman, et al. (link)