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 = ω₀S_z, 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
H₀|Φ₁> = E₁|Φ₁>, H₀|Φ₂> = E₂|Φ₂>, <Φ_i|Φ_j> = δ_ij .
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 = -m∙B = -γS∙B ₀ = -γS_zB₀ - γS_xB₁ =  ω₀S_z + ω₁S_x = 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
|ψ₊> = cos(θ/2)exp(-iφ/2)|Φ₁> + sin(θ/2)exp(iφ/2)|Φ₂>,
|ψ_-> = -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^iφ.

In our example

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

• (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 E_i' = E_i + W_ii, and write the above formula in terms of E_i'.) 
  Let us define E_m = ½(E₁ + E₂),  ∆ = ½(E₁ - E₂),  then  E_± = E_m ± (∆² + |W₁₂|²)^½.

  Let us choose the origin of the total energy such that E_m  =  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
  |ψ₊> = exp(-iφ/2)|Φ₁> + exp(iφ/2)(|W₁₂|/(2Δ))|Φ₂>
  = exp(-iφ/2)[|Φ₁> + exp(iφ)(|W₁₂|/(2Δ))|Φ₂>],
  |ψ_-> = -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
  |ψ₊> = 2^-½(exp(-iφ/2)|Φ₁> + exp(iφ/2)|Φ₂>),
  |ψ_-> = 2^-½(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
  |ψ₊> = |+> + (B₁/(2B₀))|->,  |ψ_-> = |-> - (B₁/(2B₀))|+>.
  for B₀ << B₁ we have
  |ψ₊> = 2^-½(|+> + |->),  |ψ_-> = 2^-½(|+> - |0>).

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
|Φ₂> = 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.

Consider a system having only two states, i.e. H₀u_n = E_nu_n, 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 P_n(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₊|ψ₊>_,  H|ψ_-> = E_-|ψ_->_.
  |ψ₊> = 2^-½(exp(-iφ/2)|u₁> + exp(iφ/2)|u₂>),
  |ψ_-> = 2^-½(-exp(-iφ/2)|u₁> + exp(iφ/2)|u₂>).
  where W₂₁ = |W₁₂| exp(iφ).  Here W₂₁ = U₀ and φ = 0.

  |u₁> = 2^-½[ψ₊> - ψ_->],  |u₂> = 2^-½[ψ₊> + ψ_->]. 
  |ψ(t)> = U(t,0)|ψ(0)> = exp(-iHt/ħ)|u₁>.
  |ψ(t)> =  2^-½[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:

• 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.