In many treatments in current journals you find authors using the **interaction
picture**. Assume the Hamiltonian of an arbitrary system is H_{0}(t),
and the corresponding evolution operator is U_{0}(t,t_{0}).
We have

(iħ∂/∂t)U_{0}(t,t_{0}) = H_{0}(t)U_{0}(t,t_{0}),
U_{0}(t_{0},t_{0}) = I.

Now assume that the system is perturbed in such a way that the Hamiltonian becomes

H(t) = H_{0}(t) + W(t).
For such a system the state vector in the interaction picture, |Ψ_{I}(t)>
is defined from the state vector in the Schroedinger picture through

|Ψ_{I}(t)>
= U_{0}^{}(t,t_{0})|Ψ_{S}(t)>.

**How does |Ψ _{I}(t)>
evolve?
**iħ∂|Ψ

= iħ[∂/∂t(U

= -U

Here we used

iħ∂/∂t U

We now can write

iħ∂|Ψ_{I}(t)>/∂t = -U_{0}^{}H_{0S}(t)U_{0}U_{0}^{}|Ψ_{S}(t)>
+ U_{0}^{}H_{s}(t)U_{0}U_{0}^{}|Ψ_{S}(t)>

= -H_{0I}(t)Ψ_{I}(t)> + (H_{0I}(t) + W_{I}(t))|Ψ_{I}(t)>
= W_{I}(t))|Ψ_{I}(t)>.

Therefore

iħ∂|Ψ_{I}(t)>/∂t = W_{I}(t))|Ψ_{I}(t)>.

We can rewrite this differential equation in the form of an integral equation.

d|Ψ_{I}(t)> = (iħ)^{-1}W_{I}(t))|Ψ_{I}(t)>dt.

∫_{t0}^{t }d|Ψ_{I}(t)> = (iħ)^{-1}∫_{t0}^{t
}W_{I}(t')|Ψ_{I}(t')>dt'.

|Ψ_{I}(t)> = |Ψ_{I}(t_{0})> + (iħ)^{-1}∫_{t0}^{t
}W_{I}(t')|Ψ_{I}(t')>dt'

ŻŻ_______________________________ŻŻ

This integral equation can be solved by iteration.

The ket |Ψ_{I}(t)> can be
expanded in a power series of the form

|Ψ_{I}(t)> = {I + (iħ)^{-1}∫_{t0}^{t
}dt'W_{I}(t') + (iħ)^{-2}∫_{t0}^{t
}dt'W_{I}(t') ∫_{t0}^{t' }dt''W_{I}(t'')
+ ... }|Ψ_{0}(t)>.

The interaction picture assigns part of the time dependence to the state vectors, and
part to the operators. Following the method used to derive the equation for the evolution
of the operators in the Heisenberg picture, we can derive the equation for the evolution
of operators in the interaction picture.

dA_{I}/dt = (∂A_{S}/∂t)_{I}+ (iħ)^{-1}[A_{I},H_{0I}].

**When do we use the interaction picture?
**The interaction picture is often used when H

Assume that the problem governed by H

Assume W

|Ψ

with W

Let {|n>} be an orthonormal eigenbasis of H_{0} and let the
system be in the eigenstate |m> at t = 0, i.e. |Ψ_{I}(0)> = |m>.
We have H_{0}|m> = E_{m}|m>,

The probability P(E_{k},t) of finding the system in the eigenstate |k>
of H_{0} at time t, i.e. the probability of measuring the
eigenvalue E_{k}, is |<k|Ψ_{I}(t)>|^{2.}.
(The predictions of quantum mechanics are independent of the representation.)

<k|Ψ_{I}(t)> = <k|m> + (iħ)^{-1}∫_{0}^{t
}<k|W_{I}(t')|m>dt'

= <k|m> + (iħ)^{-1}∫_{0}^{t
}exp(i(E_{k} - E_{m})t/ħ) <k|W_{S}(t')|m>dt'.

With <k|m> = 0 we have

P(E_{k},t) = (1/ħ^{2})|∫_{0}^{t }exp(i(E_{k}
- E_{m})t/ħ) <k|W_{S}(t')|m>dt'|^{2}.

This is the result **of first order time dependent perturbation
theory**. P(E_{k},t) is the transition probability.

If W_{S}(t) is independent of time, i.e. if a constant
small term is added to the Hamiltonian at t = 0 then

P(E_{k},t) = (1/ħ^{2})|<k|W_{S}|m>|^{2}|^{
}(exp(i(E_{k} - E_{m})t/ħ - 1)/(i(E_{k} - E_{m})/ħ)|^{2}

= (1/ħ^{2})|<k|W_{S}|m>|^{2}|^{ }(exp(-iE_{m}t/ħ
- exp(-iE_{k}t/ħ)/(i(E_{k} - E_{m})/ħ)|^{2
}= (|<k|W_{S}|m>|^{2}/(E_{k} - E_{m})^{2})
(2 sin(ω_{km}t/2))^{2
}= (|<k|W_{S}|m>|^{2}/(E_{k} - E_{m})^{2})^{ }4
sin^{2}(ω_{km}t/2).

Here ω_{km} = (E_{k} - E_{m})/ħ and we have used |e^{iθ} - e^{iφ}| = 2sin((θ - φ)/2).

**Problem:**

Consider a one-dimensional, infinitely deep well. Let U(x) = 0, for
0 < x < a, and U(x) = ∞ everywhere else. The eigenstates of H_{0} = p^{2}/(2m) +
U(x) are Φ_{n}(x) = (2/a)^{1/2}sin(nπx/a)
with eigenvalues E_{n} = n^{2}π^{2}ħ^{2}/(2ma^{2}).

Assume that at t = 0 you put a coin on the bottom of the well.

W(x) = W_{0}, a/4 < x < 3a/4 for t > 0, W(x) = 0 everywhere
else, H = H_{0 }+ W.

If at t = 0 the system is in the state Φ_{3}(x),
what is the probability of finding it in Φ_{1}(x)
at time t?

Solution:

- Concepts:

The eigenfunctions of the infinite square well, time-dependent perturbation theory - Reasoning:

W is a small correction to H_{0}. The problem governed by H_{0}is already solved.

W = 0 for t < 0. The probability P(E_{k},t) of finding the system in the eigenstate |k> of H_{0}at time t, i.e . the probability of measuring the eigenvalue E_{k}, is

P(E_{k},t) = (1/ħ^{2})|∫_{0}^{t}exp((i/ħ)(E_{k}-E_{3})t')<Φ_{k}|W|Φ_{3}>dt'|^{2}.

P(E_{k},t) is the transition probability.

[Note: We are calculating the probability of finding the system in the ground state of the unperturbed Hamiltonian H_{0}, not of the perturbed Hamiltonian H. We are calculating the probability that we find the system in the ground state after we take the coin out at time t.] - Details of the calculation:

P(E_{1},t) = (1/ħ^{2})|∫_{0}^{t}exp((i/ħ)(E_{1}-E_{3})t')<Φ_{1}|W|Φ_{3}>dt'|^{2}

= |(<Φ_{1}|W|Φ_{3}>/ħ)∫_{0}^{t}exp(iω_{13}t')dt'|^{2}

[∫_{0}^{t}exp(iω_{13}t')dt' = (-i/ω_{13})(exp(iω_{13}t) - 1)

= (-i/ω_{13})exp(iω_{13}t/2)(exp(iω_{13}t/2) - exp(-iω_{13}t/2))

= (2/ω_{13})exp(iω_{13}t/2)sin(ω_{13}t/2)]

P(E_{1},t) = (|<Φ_{1}|W|Φ_{3}>|^{2}/(ħω_{13})^{2}) 4 sin^{2}(ω_{13}t/2).

ω_{13}= 8π^{2}ħ/(2ma^{2}).

<Φ_{1}|W|Φ_{3}> = (2W_{0}/a)∫_{a/4}^{3a/4}dx sin(πx/a)sin(3πx/a)

= (2W_{0}/π)∫_{π/4}^{3π/4 }dx' sin(x')sin(3x')

= (2W_{0}/π)[(sin(2x')/4 - sin(4x')/6))|_{π/4}^{3π/4}= -W_{0}/π.

P(E_{1},t) = (2W_{0}/(ħπω_{13}))^{2}sin^{2}(ω_{13}t/2) = (m^{2}a^{4}W_{0}^{2}/(4ħ^{4}π^{6}))sin^{2}(2π^{2}ħt/(ma^{2})).

P(E_{1},t) oscillates with time t.

For t > 0 the Hamiltonian is H. At t = 0 the system is not in an eigenstate of H but in a eigenstate of H_{0}, an operator that does not commute with H. The probability of finding the system in an eigenstate of an operator that does not commute with the Hamiltonian changes with time.