Maybe I need this information.

Fine, I think if you calculate it for any wavelength, then it would be okay.

I will repeat the task, I will not solve it now because it's super simple.

The task was to calculate the mean number of photons per mode in sunlight.

Well, in my notes I wrote for the green part of the spectrum.

I didn't say this, I'm very sorry, but I think if you calculated it for any part of the spectrum,

so for any wavelength, that's okay.

And the idea was just to use the Planck's formula, which I will write because also we'll use it today.

So this N of omega is 1 e h bar omega over kT, where k or sometimes the right kappa is the Boltzmann's constant

and T is the temperature minus 1.

So this is the Planck's formula and the task was just to substitute omega corresponding to the visible,

to the green part of the spectrum and substitute the constants and then it was all.

And the result was 10 to the minus 2.

So the result for the wavelength about 500 nanometers, the N of omega was about 0.01.

That was the result.

Okay, that was very simple.

Today's lecture, I'm afraid, will still make our way through classical optics,

but today I'm going to show you that classical optics fails.

So I will finish the example that I started at the last lecture,

which is stimulated transitions in atoms and the paradox that appears after that as a result.

And then today we'll do the same for parametric downconversion.

So I will show you how if you try to describe parametric amplification in the classical optics,

in terms of classical optics, you fail to describe what we get in experiment.

And then there will be several new examples, several other examples of what is wrong with classical description of optics.

And finally, I will pass to quantization of the electromagnetic field,

and I will start with introducing the modes and describing every mode.

But we will not probably reach the quantum description today.

It will be the subject of the next lecture.

So I will remind you that if we had levels in atoms, level 1 and level 2 excited and ground states,

then the rate of transitions from level 1 to level 2 was proportional to the energy density of the field.

And the proportionality constant was b. Let's call it b.

So it's the Einstein coefficient, b times rho.

And the same for the transitions down.

So it was equal to the rate of transitions down.

By the way, I forgot to apologize at the last lecture, as one of you correctly pointed out.

While proving that W12 is equal to W21, I mentioned wrongly that the matrix element of the dipole moment D12 is equal to D12 complex conjugated.

And this is, of course, wrong.

So what is correct is D12 is D21 complex conjugated.

So this is wrong. This is correct.

And because the transition rate depended on its squared model, then W12 is equal to W21.

So the transition rate from up to down is equal to the transition rate from down to up.

And so we use this equation.

And then we know what rho is from the black body in the situation where you don't have any laser shining on this system.

You just have your equilibrium radiation, so thermal radiation.

If the temperature here is not zero, then there is always some radiation at any frequency.

And it's given, this rho given by the Planck formula.

I will write it explicitly.

So this rho is h bar k cubed and then over pi squared and this n of omega, which I wrote the Planck formula.

And then what happens in this equilibrium situation?

The number of photons on level two changes in time.