M1L4c.txt

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I want to now use the density matrix formalism for arbitrary
two-level systems.

So, what is the most general Hamiltonian for the most
general two-level system?
Well, the most general Hamiltonian
is the most general Hamiltonian we can
construct with 2 by 2 matrices.
And the basis set to expand the 2 by 2 matrices
are the Pauli matrices.
So, if we expand the Hamiltonian into the unity matrix-- sigma
x, sigma y, and sigma z, we have four coefficients,
four amplitudes, which are complex in general-- omega
1, omega 2, omega 3.
And here is something which I've called omega bar.

By appropriately shifting what is the 0 point of energy.
We can always get rid of this.
So, this is just a definitional character.
So, therefore, the most general Hamiltonian
for any two-level system can be written
in this very compact way.
That it is the scalar product of the vector-- omega,
omega 1, omega 2, omega 3-- with the vector
sigma of the three Pauli matrices--
sigma x, sigma y, sigma z.

OK.
So, this is a way to write down the most general Hamiltonian
for a two-level system.
Now, we describe two-level systems by a density matrix
by statistical operator, which is also a 2 by 2 matrix.
And the most general density matrix
can also be expanded into its four components.
Sort of the basis set of matrices
is the unity matrix and the three Pauli matrices.
So r1, r2, r3.
Of course, this time we cannot throw away the unity matrix,
because otherwise the density matrix would have no trace,
and there would be no probability to find
the particle.
But we can, again, write it in a compact form
that it is one half-- yes, I'm using the fact now
that the trace of rho is r0.
And this is by definition, or by conservation of probability,
is 1.
So, therefore, r0 is not a free parameter.
It's just the sum of all the probabilities
to find the system in any state.
And the non-trivial part is then scalar
product of this vector r-- rx, ry,
rz-- with the vector of the three Pauli matrices.
So, we have our most general Hamiltonian,
we have our most general density matrix,
and now we can insert these into the equation of motion
for the density matrix.
Which, as I said before, is just a reformulation
of Schrodinger's equation.
And if we insert the Hamiltonian and the density matrix
into this equation, we find actually
something which is very simple.
It says that this vector r, which
we call the Bloch vector-- the derivative of the Bloch vector
is given by the cross product of the vector omega, which
were the coefficients with which we parametrize the Hamiltonian
cross r.
The derivation is straightforward.
And you will be asked to do that on your homework, assignment
number one.
But, it has a very powerful meaning.
It tells us that an arbitrary two-level system,
with an arbitrary Hamiltonian, can
be regarded as a system where we have a vector
r which undergoes precession.
This is the time evolution of the system.
So, this is the powerful generalization
from the result we discussed previously.
Where we found that if you have an arbitrary
quantum-mechanical spin, the time derivative
can be written in that way.
So, previously we found for a pure state,
but now we find it that it's even valid for general density
matrix and its time evolution.
So, what I've derived for you is the famous theorem, which
is traced back to Feynman, Vernon, and Hellwarth--
it's sort of a famous paper.
So this famous theorem-- and I've
summarized it here for you-- says
that the time evolution of the density matrix for the most
general two-level system is isomorphic to pure precession.
And that means it's isomorphic to the behavior
of a classical moment-- classical magnetic moment--
in a suitable time-dependent magnetic field.
So, when you have an Hamiltonian, which
is characterized by-- the most general
Hamiltonian is characterized by the three coefficients w1, w2,
w3.
But.
If you would create a classic system where w1, w2, and w3 are
the time-dependent components, xyz-component
of a magnetic field, then the precession of a magnetic moment
would be exactly the same as the time
evolution of a quantum-mechanical density
matrix.

So, in other words, we've started out
with rotating frames and rotation.
And now, we've gone as far as I will go.
Namely, I've in a way told you that
an arbitrary quantum-mechanical two-level system, the time
evolution is just precession.
It's a rotation.
There is nothing more complicated possible.
Well, unless we talk about decoherence.
If we have such an Hamiltonian, we
know, of course, that a pure state will stay pure forever.
And you can immediately verify that if you look
at the trace of rho square.

If the trace of rho square is 1, we have a pure state.
And now we have parametrized the density matrix
with a Bloch vector component-- r1, r2, r3.
So, in those components, the trace of rho square
can be written in this way.
And of course, r0 square was constant-- this
was our normalization of 1.
So the question is now when we have an arbitrary time
evolution, which we know now, according to the Feynman,
Vernon, Hellwarth theorem.
The arbitrary time evolution of the Bloch vector
can be written as omega cross r.
So, this equation tells us immediately
that the length of the vector r is constant
because r dot is always orthogonal to r.
And therefore, the length of the vector r is not changing.
So, what we have derived says that with the most general
Hamiltonian, the length of the vector r will be constant,
and therefore, the trace of rho square will be constant.
This is constant because r dot is perpendicular to r.
So, this will tell us that a pure state will just
precess with the constant length of its Bloch vector forever.
However, we know that in real life, some coherences are lost,
and now we have to introduce something else.
So this does not describe loss of coherence.