Well and etwas ab 측 Nagział

All right. So, today I will continue the story about a squeezed vacuum. At the last lecture, I

consider the case where we have one mode, and there was single mode Hamiltonian. Today,

has the form of ih bar, this I drag out for the convenience, then gamma is the notation

for the rest of the Hamiltonian, and then most importantly it contains the photon creation

operators. So as I explained at the last lecture, it depends. First we derive this Hamiltonian

in the general form where there were two creation operators. We denoted them a1 dagger, a2 dagger.

For simplicity, for convenience, I will denote them a and b. a dagger and b dagger.

So two photons are generated in two modes. So again, we pump the crystal. We have two

different now modes. I denote them as different indirection modes. They can differ in frequency

as well or in polarization. So these modes are a and b. And of course, by the way, there

is the Hermitian conjugated part. So this Hamiltonian obviously generates photon pairs

now in two different modes. So how to deal with such a problem and what are the observable

features of non-classicality of this radiation? Obviously, it will also be non-classical,

but what to measure? In the previous lecture, I considered a Hamiltonian containing a dagger

squared. And then we were supposed to measure quadrature squeezing because this Hamiltonian

led to the production of squeezed states. But it will be different in this case. So

let's describe the evolution. And again, we'll use the Heisenberg approach. So I will use

the equation i h bar a dot. So I use dot for time derivative. Or I hope you understand

that dot is a time derivative. i h bar a dot is the commutator of a with the Hamiltonian.

And we can easily calculate this commutator because the Hamiltonian contains a dagger

once, and the rest commutes with a. So it will be i h bar gamma. Then there will be

a commutator a a dagger and b dagger. And the Hermitian conjugated part, of course,

commutes. And this commutator is 1. So I get a very simple equation. a dot is gamma b dagger.

And because the Hamiltonian is symmetric with respect to a and b, obviously, I obtain the

same equation for the b dot. So this will be gamma a dagger. And so how to solve these

equations? The most convenient case, and it's actually very instructive, the most convenient

way to solve these equations is to introduce operators that are superpositions of these

two operators. So a plus minus b over square root of 2. So I consider a superposition of

a and b operators. And I denote it c or d operator. So c and d are photon annihilation

operators in molds formed like this. This is not so trivial. What are these molds for?

But in the case of polarization, probably you will guess that if a is horizontally polarized

mold, b is vertically polarized mold, then a plus minus b over square root of 2 are diagonal,

diagonally polarized mold. So just keep this analogy in mind. And then so by summing and

subtracting these two equations, I obtain equations for c and d. If I sum these two

equations, I get c dot is gamma. So I sum left hand parts and right hand parts, and

I get c dagger here. Square root of 2 disappears. Yeah, so for c, I get this. And for subtracting,

for the minus sign, I get a minus b over square root of 2. So it will be d dot. And on the

right, I will obtain minus d dagger, minus gamma d dagger. But this equation we already

know. We already know. This equation describes single mold Hamiltonian. So this equation

would be for a Hamiltonian of like this. We considered this equation at the last lecture.

And we know what's the result, what's the solution. We know that the solution is the

Bogoliubov transformation, which is c is cosh gamma t c0 plus sinh gamma t c0 dagger. Right?

We derived this equation at the last lecture. Now if we look at this equation, we see that

it differs only by the sine of gamma. It will be the same as gamma change the sine in the

first equation. So we write the solution changing the sine of gamma. And it gives us d is cosh

gamma t because hyperbolic cosine doesn't change the sine when the argument changes

the sine d0 plus minus. Because if we change the sine of gamma, then the sine in front

of hyperbolic sine changes minus sinh gamma t d0 dagger. d0 and c0 are the initial operators.

And then again, we do the summation. So to pass to a and b, we have to sum and to subtract