As has been pointed out the energy comes in the form of visible light from the sun. The reaction written in summary or overview is
$$\displaystyle \ce{2H2O \overset{light}\to O2 + 4H^++ 4e^-}$$$$\displaystyle \ce{CO2 +4H^++ 4e^-\to (CH2O) + H2O}$$
where in the first step, the oxidation of water, the light does not react directly with water. In the second step, $\ce{CH2O}$ represents carbohydrate and the electrons are carried by proteins.
The overall reactions occur in proteins in the cell membrane of chloroplasts (in green plants) and the first step is absorption of a red photon typically of $700$ nm. This causes an electron to be transferred from a chlorophyll dimer, (the special pair) down an electron transfer chain involving cytochromes and other proteins that eventually reduces the carbon dioxide, the second reaction above.
As the special pair is now short of an electron it recovers this via a metalloprotein grabbing an electron from water. Using X-ray diffraction all these protein structure are now known and ultrafast spectroscopy has measured most if not all the rates of these primary processes.
In energy terms, the couple: CO2-glucose has a redox potential $\Delta V=-0.43$ eV and for $\ce{H2O - O2}$ is $+0.82$ eV so the total redox span is $1.25$ eV and as $4$ electrons are involved this is makes $5$ eV in total. A visible photon at $700$ nm has an energy of $1.77$ eV and as one electron is released /photon absorbed then the total energy supplied is $7.1$ eV, which is more than enough for this reaction.
However, there are losses in the process, and the efficiency rarely exceeds $0.36$ which makes the useable energy available $0.63$ eV/photon which is not enough thus we need $5/0.63\to 8$ photons in total. This suggest that two photosystems (see Z-scheme) are needed in green plants and this is indeed the case.