Two-Level Systems

A general study of a two-level system

Consider a physical system whose state space is two-dimensional.
(Usually this is an approximation).
Assume that if the system is not externally perturbed, its Hamiltonian is H0.
(An example is a spin ½ particle in a magnetic field  . Here .)

The eigenstate of H0 are |f1> and |f2>, and the corresponding eigenvalues are E1 and E2.
(In our example |+> and |-> denote the eigenstates and the eigenvalues are .)
We have

H0|f1>=E1|f1>, H0|f2>=E2|f2>, <fi|fj>=dij .

We now introduce a time-independent external perturbation, which changes H0 to H=H0+W.

(In our example we introduce a magnetic field pointing in the x-direction, .
Here .)

We denote the eigenstates of H by |y±>, and the eigenvalues by E±. In the {|f1> , |f2>} basis W is represented by the Hermitian matrix .

We want to find the eigenvalues and eigenvector of H. In the {|f1> , |f2>} basis the matrix of H is

.

The eigenvalues are

.

The eigenvectors are

,

where

.

(In our example

.

.

.)

Consequences of the perturbation ("the coupling")
(a) The possible results of a measurement are no longer E1 and E2 but E+ and E- .

If W11=W22=0, then . (If , then we can define and write the above formula in terms of .) Let us define

.

Let us choose the origin of the total energy such that Em = 0. Then E1=D, E2=-D, . Let us plot E1, E2, E+, and E- vs. D, to see how these quantities depend on E1-E2, i.e. on how far the energy level of H0 are separated from each other.

Image2981.gif (2890 bytes)

2D represents the difference between the eigenvalues of E1 and E2 . If D=0, E1=E2, then the eigenstates |f1> and |f2> of H0 are degenerate. The perturbation removes that degeneracy. If the ground state of a two-level system is twofold degenerate, than any perturbation will remove that degeneracy and produce a state with a lower energy, (i.e. a more tightly bound state). In the absence of a perturbation the energy levels "cross" at D=0. In the presence of a perturbation the energy levels "repel" each other. The diagram above is often called an "anti-crossing" diagram.

(As the state becomes twofold degenerate.

. E1 and E2 "cross" at D=B0=0.

 at B0=0.

E+ is always positive, E- is always negative. E+ and E- never "cross".)

(b) |f1> and |f2> are no longer stationary states. The eigenstates of H are |y+> and |y->.
If , .

This is called weak coupling. Now we have

,

.

Except for a global phase factor we have .

If  , .

This is called strong coupling. Now for D>0 we have

,

.

Now we have near complete mixing of |f1> and |f2>.

When |f1> and |f2> are degenerate, then D=0 and any coupling is strong.

(In our example we have for

, .

For we have , .)

The evolution of the state vector

Let |y(0)>=l|y+>+m|y->. If H does not explicitly depend on time then

.

Let .

Then .

The probability of finding the system at time t in state |f2>

(with ,) is

.

This is the Rabi’s formula.

P12(t) oscillates with frequency , the Bohr frequency of the system. It varies between 0 and some maximum value, which is a function of |W12| and of (E2-E1)2. If E1=E2, then a system which starts in |f1> will at some later time be found in state |f2> with certainty.

Problems:

Consider a system having only two states, i.e. H0un=Enun, with n=1,2.  At t=0 a perturbation V(t)=V0 , which has no diagonal matrix elements, is turned on. Here V0 is independent of time and real. Derive equations for the occupation populations Pn(t) of the two states as functions of time for the special case that the states are degenerate, i.e. E1=E2. For the initial conditions take P1(0)=1, P2(0)=0.
Solution:

.

.

,

where . Here .

.

.

, P21(t)=1-P12(t)

 

Consider the Hermitian Hamiltonian H=H0+H’, where H’ is a small perturbation. Assume that the exact solutions H0|y>=E0|y> are known, that there are two of them, and that they are orthogonal and degenerate in energy. Derive from first principles an expression to first order in H’ for the energies of the perturbed levels in terms of the matrix elements of H’.
Solution:

The matrix of H’ is .

The matrix of H is .

,

.

These are the exact eigenvalues of H. In the case of a two level system with a degenerate Hamiltonian H0 first order perturbation theory requires finding the eigenvalues of H’ and adding them to E0, which also gives the exact result.

Unstable States

Consider an isolated hydrogen atom. Solving the eigenvalue equation H0|y>=E|y> we find the eigenstate |ynlm> and the associated eigenvalues En. These energies En are in very good agreement with experimentally measured energies. However, quantum mechanics predicts, that if a system is at t=0 in an eigenstate |yn> of the Hamiltonian H0, then it will remain in that eigenstate. In experiments we observe that the excited states of the hydrogen atom decay to the ground state by emission of a photon. They are therefore not really stationary states.

The hydrogen atom is constantly interacting with the electromagnetic field. The Hamiltonian H0 of the isolated hydrogen atom is therefore not the Hamiltonian of the system. The actual Hamiltonian must describe the energy of the system, i.e. the energy of the atom and the field. While the total energy is a constant of motion, it can be lost by the hydrogen atom and be taken away by a photon.

How can we, in non relativistic quantum mechanics, deal with the instability of states?

Experimentally one observes that for a system which at t=0 is in an unstable state |yn>, the probability of finding it in this state at time t is .

Let N0 be the number of systems in the state |yn> at t=0. Then is the number of systems still in the state |yn> at time t.

.

 is the probability of leaving the state |yn> during the time interval dt for each of the N(t) systems still in the state |yn>. is the probability per unit time of leaving the unstable state |yn>. The mean time the system remains in the unstable state is , where is the probability that the system remains in the unstable state until time t and is the probability that it decays in the next time interval dt.

If |yn> is an eigenstate of a time-independent Hamiltonian H0 and |y(0)>=|yn> then and the probability of finding the system at time t in the state |yn> is .

What happens if we replace En by , a complex quantity. We then have . The probability of finding the system in the unstable state |yn> decreases exponentially with time, as is observed experimentally.

We can therefore take into account phenomenologically the instability of a state |yn> whose lifetime is t by adding an imaginary part to its energy and by setting .

Note: H0 is then no longer Hermitian, and the norm of the state vector varies with time as . This is due to the fact that the system under study is part of a larger system which is not completely described by H0.