Okay, so today we are going to run in on quantization of the transmission line, and that should be

relatively straightforward after all our preparation.

So the last time we wrote down the Lagrangian of the transmission line, let's take this

discrete representation where we know it's simply something like electric minus magnetic

energy.

So that would be a sum over all these cells, the electric energy which is q squared divided

by two times the capacitance.

And then taking the magnetic energy that is stored inside the magnetic field generated

by these currents.

So that would be m half i n squared.

And now we found that in order to have a good variable for describing this transmission line,

it's not so convenient to choose the charges for example because then the currents are

a non-local expression in terms of the charges.

It's much better to introduce another quantity which is something like the local magnetic

flux associated with each node.

And we defined this simply by saying the time derivative of this local magnetic flux is

given by the voltage at this node and that would be q n over c.

So this then has the advantage that we can easily write down the Lagrangian in terms

of this variable.

Namely, you already see that we can express the charge in terms of the time derivative,

so that would be the kinetic energy, so to speak, c half phi n dot square.

And then there's the magnetic energy and we can express that in terms of finite differences

of the phi's on the j-cant nodes.

So that would be phi n minus one minus phi n squared over two times n.

Okay, so that's the Lagrangian written with respect to a convenient variable, which is

these local fluxes.

And then in order to quantize things, of course, we want to switch to a Hamiltonian.

And we do that in the usual way.

So first we get ourselves the momentum, and that is easy enough.

If you know the Lagrangian, you know that the momentum should be obtained by taking

the partial derivative with respect to the time derivative of the variable, which means

taking the partial derivative with respect to the velocity.

So in this case that would be phi n dot.

And so you immediately read off that as c times phi n dot.

And that is obviously equal to the charge.

So yes, the momentum associated with each node, with each kelp is the charge.

And then you know that the Hamiltonian is given by taking something like momentum times velocity minus the Lagrangian,

and then re-expressing everything in terms of momentum and coordinate.

So in our case we have many variables, so we sum over m pn phi n dot minus the Lagrangian,

and if you work that out, it's basically the expression up here with q expressed as the momentum, that's the same,

and the only thing that changes as usual is that now you get a plus sign in the total energy.

So the Hamiltonian for us is sum over all cells pn squared over 2c, which is the same as q squared over 2c,

plus phi n minus 1 minus phi n squared over 2n.

So that's the classical Hamiltonian, the Hamiltonian function, and now it's a simple matter of quantizing.

You just take momenta positions and demand that they fulfill the usual commutation relation.

So now we take phi n becoming an operator, and the same goes for pn becoming an operator,

and what we know is that the commutator of position with momentum is always equal to i times h bar.

So that is what Heisenberg told us.

And more generally you could say what's the commutator between phi and p evaluated at different points,

for example if I plug in n prime here, and the rule is simply that if they belong to different points,