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  B ≈ B0k. Here H0 = ω0Sz, ω0  = -γB0.)

The eigenstate of H0 are |Φ1> and |Φ2>, and the corresponding eigenvalues are E1 and E2.
(In our example |+> and |-> denote the eigenstates and the eigenvalues are ±ħω0/2.)
We have
H01> = E11>, H02> = E22>, <Φij> = δij .
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, B = B0k + B1i.
Here U = -mB = -γSB 0 = -γSzB0 - γSxB1 =  ω0Sz + ω1Sx = H0 + W.)
We denote the eigenstates of H by |ψ±>, and the eigenvalues by E±.  In the {|Φ1>,|Φ2>} basis W is represented by the Hermitian matrix

,

where W11, W22 are real and W12 = W21*.

We want to find the eigenvalues and eigenvector of H.  In the {|Φ1>,|Φ2>} basis the matrix of H is

.

The eigenvalues are
E± = ½(E1 + W11 + E2 + W22) ± ½((E1 + W11 - E2 - W22)2 + 4|W12|2)½.

The eigenvectors are
+> = cos(θ/2)exp(-iφ/2)|Φ1> + sin(θ/2)exp(iφ/2)|Φ2>,
-> = -sin(θ/2)exp(-iφ/2)|Φ1> + cos(θ/2)exp(iφ/2)|Φ2>,
where tanθ = 2|W12|/(E1 + W11 - E2 - W22),  W21 = |W12|e.

In our example

E± = ±[(ħω0)2 + (ħω1)2]½,  tanθ = ħω1/(ħω0) = B1/B0,  φ = 0.
+> = cos(θ/2)|+> + sin(θ/2)|->,   |ψ-> = -sin(θ/2)|+> + cos(θ/2)|->.

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 E± = ½(E1 + E2) ± ½((E1 - E2)2 + 4|W12|2)½.  (If W11, W22 ≠ 0, then we can define Ei' = Ei + Wii, and write the above formula in terms of Ei'.)
Let us define Em = ½(E1 + E2),  ∆ = ½(E1 - E2),  then  E± = Em ± (∆2 + |W12|2)½.

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

2Δ represents the difference between the eigenvalues of E1 and E2.  If Δ = 0, E1 = E2, then the eigenstates |Φ1> and |Φ2> 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 Δ = 0. In the presence of a perturbation the energy levels "repel" each other. The diagram above is often called an "anti-crossing" diagram.

(As B0 --> 0 in our example, ω0 --> 0, E1 = -E2  --> 0, the state becomes two-fold degenerate.
∆ = ½(E1 - E2) =  ħω0/2 = |γ|ħB0/2.  E1 and E2 "cross" at Δ = B0 = 0.
E+ = ħω1/2,  E- = -ħω1/2  at B0 = 0.
E+ is always positive, E- is always negative. E+ and E- never "cross".)

• (b) |Φ1> and |Φ2> are no longer stationary states. The eigenstates of H are |ψ+> and |ψ->.

If  |tanθ| = |W12|/|Δ| << 1,  |Δ| >> |W12|, then for Δ > 0 we have
θ ≈|W12|/Δ,  cos(θ/2) ≈ 1,  sin(θ/2) ≈ |W12|/(2Δ).
This is called weak coupling.  Now we have
+> = exp(-iφ/2)|Φ1> + exp(iφ/2)(|W12|/(2Δ))|Φ2>
= exp(-iφ/2)[|Φ1> + exp(iφ)(|W12|/(2Δ))|Φ2>],
-> = -exp(-iφ/2)(|W12|/(2Δ))|Φ1> + exp(iφ/2)|Φ2>
= exp(iφ/2)[|Φ2> - exp(-iφ)(|W12|/(2Δ))|Φ1>].
Except for a global phase factor we have  |ψ+> ≈ |Φ1>,  |ψ-> ≈ |Φ2>.

If  |tanθ| = |W12|/|Δ| >> 1,  |Δ| << |W12|, then for Δ > 0 we have
θ ≈ π/2,  cos(θ/2) ≈ sin(θ/2) ≈ 1/√2.
This is called strong coupling.  Now for Δ > 0 we have
+> = 2(exp(-iφ/2)|Φ1> + exp(iφ/2)|Φ2>),
-> = 2(exp(-iφ/2)|Φ1> - exp(iφ/2)|Φ2>).
Now we have near complete mixing of |Φ1> and |Φ2>.
When |Φ1> and |Φ2> are degenerate, then Δ = 0 and any coupling is strong.
(In our example we have for B0 >> B1
+> = |+> + (B1/(2B0))|->,  |ψ-> = |-> - (B1/(2B0))|+>.
for B0 << B1 we have
+> = 2(|+> + |->),  |ψ-> = 2(|+> - |0>).

The evolution of the state vector

Let |ψ(0)> = λ|ψ+> + μ|ψ->. If H does not explicitly depend on time then
|ψ(t)> = U(t,0)|ψ(0)> = exp(-iHt/ħ)|ψ(0)> = λ exp(-iE+t/ħ)|ψ+> + μ exp(-iE-t/ħ)|ψ->.
Let |ψ(0)> = |Φ1> = exp(iφ/2)[cos(θ/2)|ψ+> - sin(θ/2)|ψ->].
Then |ψ(t)> = exp(iφ/2)[cos(θ/2)exp(-iE+t/ħ)|ψ+> - sin(θ/2)exp(-iE-t/ħ)|ψ->].

The probability of finding the system at time t in state
2> = exp(-iφ/2)[sin(θ/2)|ψ+> + cos(θ/2)|ψ->]
is
P12(t) = |<Φ2|ψ(t)>|2 = sin2θ sin2((E+ - E-)t/2ħ))
= [4|W12|2/(4|W12|2 + (E1 - E2)2)] sin2((4|W12|2 + (E1 - E2)2)½t/(2ħ)).
This is the Rabi’s formula.

P12(t) oscillates with frequency f+- = (E+ - E-)t/h, 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 |Φ1> will at some later time be found in state |Φ2> with certainty.

Problem:

Consider a system having only two states, i.e. H0un = Enun, with n = 1,2.  At t = 0 a perturbation U(t) = U0 , which has no diagonal matrix elements, is turned on.  Here U0 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:
H0|u1> = E1|u1>, H0|u2> = E2|u2>,  E1 = E2 = E.
H = H0 + U0,  H|ψ+> = E++>H|ψ-> = E-->.
+> = 2(exp(-iφ/2)|u1> + exp(iφ/2)|u2>),
-> = 2(-exp(-iφ/2)|u1> + exp(iφ/2)|u2>).
where W21 = |W12| exp(iφ).  Here W21 = U0 and φ = 0.

|u1> = 2+> - ψ->],  |u2> = 2+> + ψ->].
|ψ(t)> = U(t,0)|ψ(0)> = exp(-iHt/ħ)|u1>.
|ψ(t)> =  2[exp(-iE+t/ħ)|ψ+> - exp(-iE-t/ħ)|ψ->].
P12(t) = <u2|ψ(t)>|2 = sin2θsin2((E+ - E-)t/2ħ)).
P12(t) = sin2(U0t/ħ),
P21(t) = 1 - P12(t).

Problem:

Consider the Hermitian Hamiltonian H = H0 + H’, where H’ is a small perturbation. Assume that the exact solutions H0|ψ> = E0|ψ> 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

,

with H'12 = H'21*.
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.