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 H_{0}.

(An example is a spin ½ particle in a magnetic field **B** ≈ B_{0}**k**. Here
H_{0} = ω_{0}S_{z}, ω_{0} = -γB_{0}.)

The eigenstate of H_{0} are |Φ_{1}>
and |Φ_{2}>, and the corresponding
eigenvalues are E_{1} and E_{2}.

(In our example |+> and |-> denote the
eigenstates and the eigenvalues are ±ħω_{0}/2.)

We have

H_{0}|Φ_{1}> = E_{1}|Φ_{1}>, H_{0}|Φ_{2}>
= E_{2}|Φ_{2}>, <Φ_{i}|Φ_{j}> = δ_{ij} .

We now introduce a time-independent external perturbation, which
changes H_{0} to H = H_{0} + W.

(In our example we introduce a magnetic field pointing in the x-direction,
**B** = B_{0}**k** + B_{1}**i**.

Here U = -**m**∙**B** = -γ**S**∙**B** _{0} = -γS_{z}B_{0}
- γS_{x}B_{1} = ω_{0}S_{z} + ω_{1}S_{x}
= H_{0} + W.)

We denote the eigenstates of H by |ψ_{±}>,
and the eigenvalues by E_{±}. In the {|Φ_{1}>,|Φ_{2}>} basis W is represented by
the Hermitian matrix

,

where W_{11}, W_{22} are real and W_{12} = W_{21}*.

We want to find the eigenvalues and eigenvector of H. In the {|Φ_{1}>,|Φ_{2}>}
basis the matrix of H is

.

The eigenvalues are

E_{±} = ½(E_{1} + W_{11} + E_{2} + W_{22}) ±
½((E_{1} + W_{11} - E_{2} - W_{22})^{2}
+ 4|W_{12}|^{2})^{½}.

The eigenvectors are

|ψ_{+}>_{ }= cos(θ/2)exp(-iφ/2)|Φ_{1}> + sin(θ/2)exp(iφ/2)|Φ_{2}>,

|ψ_{-}>_{ }= -sin(θ/2)exp(-iφ/2)|Φ_{1}> + cos(θ/2)exp(iφ/2)|Φ_{2}>,

where tanθ = 2|W_{12}|/(E_{1} + W_{11} - E_{2} -
W_{22}), W_{21} = |W_{12}|e^{iφ}.

In our example

E_{±} = ±[(ħω_{0})^{2} + (ħω_{1})^{2}]^{½},
tanθ = ħω_{1}/(ħω_{0}) = B_{1}/B_{0}, 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 E
_{1}and E_{2 }but E_{+}and E_{- }.If W

_{11}= W_{22}= 0, then E_{±}= ½(E_{1}+ E_{2}) ± ½((E_{1}- E_{2})^{2}+ 4|W_{12}|^{2})^{½}. (If W_{11}, W_{22}≠ 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(E_{1}+ E_{2}), ∆ = ½(E_{1}- E_{2}), then E_{±}= E_{m}± (∆^{2}+ |W_{12}|^{2})^{½}.Let us choose the origin of the total energy such that E

_{m}= 0. Then E_{1}= Δ, E_{2}= -Δ.

Let us plot E_{1}, E_{2}, E_{+}, and E_{-}vs. Δ, to see how these quantities depend on E_{1 }- E_{2}, i.e. on how far the energy level of H_{0}are separated from each other.2Δ represents the difference between the eigenvalues of E

_{1}and E_{2}. If Δ = 0, E_{1}= E_{2}, then the eigenstates |Φ_{1}> and |Φ_{2}> of H_{0}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 B

_{0}--> 0 in our example, ω_{0}--> 0, E_{1}= -E_{2}--> 0, the state becomes two-fold degenerate.

∆ = ½(E_{1}- E_{2}) = ħω_{0}/2 = |γ|ħB_{0}/2. E_{1}and E_{2}"cross" at Δ = B_{0}= 0.

E_{+}= ħω_{1}/2, E_{-}= -ħω_{1}/2 at B_{0}= 0.

E_{+}is always positive, E_{-}is always negative. E_{+}and E_{-}never "cross".) - (b) |Φ
_{1}> and |Φ_{2}> are no longer stationary states. The eigenstates of H are |ψ_{+}> and |ψ_{-}>.If |tanθ| = |W

_{12}|/|Δ| << 1, |Δ| >> |W_{12}|, then for Δ > 0 we have

θ ≈|W_{12}|/Δ, cos(θ/2) ≈ 1, sin(θ/2) ≈ |W_{12}|/(2Δ).

This is called**weak coupling**. Now we have

|ψ_{+}>_{ }= exp(-iφ/2)|Φ_{1}> + exp(iφ/2)(|W_{12}|/(2Δ))|Φ_{2}>

= exp(-iφ/2)[|Φ_{1}> + exp(iφ)(|W_{12}|/(2Δ))|Φ_{2}>],

|ψ_{-}>_{ }= -exp(-iφ/2)(|W_{12}|/(2Δ))|Φ_{1}> + exp(iφ/2)|Φ_{2}>

= exp(iφ/2)[|Φ_{2}> - exp(-iφ)(|W_{12}|/(2Δ))|Φ_{1}>].

Except for a global phase factor we have |ψ_{+}> ≈ |Φ_{1}>, |ψ_{-}> ≈ |Φ_{2}>.If |tanθ| = |W

_{12}|/|Δ| >> 1, |Δ| << |W_{12}|, 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)|Φ_{1}> + exp(iφ/2)|Φ_{2}>),

|ψ_{-}>_{ }= 2^{-½}(exp(-iφ/2)|Φ_{1}> - exp(iφ/2)|Φ_{2}>).

Now we have near complete mixing of |Φ_{1}> and |Φ_{2}>.

When |Φ_{1}> and |Φ_{2}> are degenerate, then Δ = 0 and any coupling is strong.

(In our example we have for B_{0}>> B_{1}

|ψ_{+}>_{ }= |+> + (B_{1}/(2B_{0}))|->, |ψ_{-}>_{ }= |-> - (B_{1}/(2B_{0}))|+>.

for B_{0}<< B_{1}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)> = |Φ_{1}> = 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

|Φ_{2}>
= exp(-iφ/2)[sin(θ/2)|ψ_{+}> + cos(θ/2)|ψ_{-}>]

is

P_{12}(t) = |<Φ_{2}|ψ(t)>|^{2} = sin^{2}θ sin^{2}((E_{+}
- E_{-})t/2ħ))

= [4|W_{12}|^{2}/(4|W_{12}|^{2} + (E_{1}
- E_{2})^{2})] sin^{2}((4|W_{12}|^{2} + (E_{1}
- E_{2})^{2})^{½}t/(2ħ)).

This is the **Rabi’s formula**.

P_{12}(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 |W_{12}| and of (E_{2} - E_{1})^{2}. If E_{1} = E_{2}, then a system which starts in |Φ_{1}> will at some later time be found in state
|Φ_{2}> with certainty.

Consider a system having only two states, i.e. H_{0}u_{n}
= E_{n}u_{n},
with n = 1,2. At t = 0 a perturbation U(t) = U_{0} , which has
no diagonal matrix elements, is turned on. Here U_{0} 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. E_{1} = E_{2}. For the initial conditions take P_{1}(0)
= 1,
P_{2}(0) = 0.

- Solution:

H_{0}|u_{1}> = E_{1}|u_{1}>, H_{0}|u_{2}> = E_{2}|u_{2}>, E_{1}= E_{2}= E.

H = H_{0}+ U_{0}, H|ψ_{+}>_{ }= E_{+}|ψ_{+}>_{, }H|ψ_{-}>_{ }= E_{-}|ψ_{-}>_{. }|ψ_{+}>_{ }= 2^{-½}(exp(-iφ/2)|u_{1}> + exp(iφ/2)|u_{2}>),

|ψ_{-}>_{ }= 2^{-½}(-exp(-iφ/2)|u_{1}> + exp(iφ/2)|u_{2}>).

where W_{21}= |W_{12}| exp(iφ). Here W_{21}= U_{0}and φ = 0.|u

_{1}> = 2^{-½}[ψ_{+}>_{ }- ψ_{-}>], |u_{2}> = 2^{-½}[ψ_{+}>_{ }+ ψ_{-}>].

|ψ(t)> = U(t,0)|ψ(0)> = exp(-iHt/ħ)|u_{1}>.

|ψ(t)> = 2^{-½}[exp(-iE_{+}t/ħ)|ψ_{+}>_{ }- exp(-iE_{-}t/ħ)|ψ_{-}>].

P_{12}(t) = <u_{2}|ψ(t)>|^{2}= sin^{2}θsin^{2}((E_{+}- E_{-})t/2ħ)).

P_{12}(t) = sin^{2}(U_{0}t/ħ),

P_{21}(t) = 1 - P_{12}(t).

**Problem:**

- Solution:
The matrix of H’ is

,

with H'

_{12}= H'_{21}*.

The matrix of H isThese are the exact eigenvalues of H. In the case of a two level system with a degenerate Hamiltonian H

_{0}first order perturbation theory requires finding the eigenvalues of H’ and adding them to E_{0}, which also gives the exact result.