Download Jaynes-Cummings model

Jaynes-Cummings model
AQM Lecture 4
(Dated: 15 March 2005)
I.
INTERACTION OF A 2-LEVEL ATOM WITH A RESONANT OPTICAL MODE
A.
Cavity Mode as a Quantum Harmonic Oscillator
Consider an optical (or microwave) cavity and a single frequency ω that is resonant with the cavity. The optical
field at this frequency (mode) can be described by a quantum harmonic oscillator.
Recall that a photon at frequency ω has energy ~ω, which is the energy spacing of the oscillator levels. Then we
can interpret the energy eigenstate |ni as the state of the cavity mode with precisely n photons.
↠: create a photon
â : annihilate a photon
(1)
(2)
The ground state |0i corresponds to the vacuum (no photons).
B.
Refresher: 2-level atom
A 2-level atom has a Hilbert space spanned by two energy eigenstates: a lower-energy “ground” state |gi and
an excited state |ei. Because it consists of only a 2-dimensional Hilbert space, it is mathematically equivalent to a
spin-1/2 particle. Just as for the spin-1/2 particle, we can think of a state of a 2-level atom as a 2-vector:
µ ¶
α
α|ei + β|gi ≡
.
(3)
β
Some useful operators on this Hilbert space were the Pauli spin matrices:
µ
¶
µ
¶
µ
¶
0 1
0 −i
1 0
σx =
, σy =
, σz =
,
1 0
i 0
0 −1
µ
¶
µ
¶
0 1
0 0
σ+ =
, σ− =
.
0 0
1 0
(4)
(5)
We will prefer to use “abstract” representations of these operators acting on states in the Dirac bra-ket notation.
The operators σ̂z , σ̂+ and σ̂− we will use act on the energy eigenstates as follows:
σ̂z |ei = |ei ,
σ̂+ |ei = 0 ,
σ̂− |ei = |gi ,
σ̂z |gi = −|gi ,
σ̂+ |gi = |ei ,
σ̂− |gi = 0 .
(6)
(7)
(8)
Let ∆ be the energy difference between the ground and excited state. The Hamiltonian for the two-level atom,
then, is
Ĥ =
up to an arbitrary constant.
∆
σ̂z .
2
(9)
2
C.
Interaction of a Two-Level Atom with a Single Optical Mode on Resonance
We place the atom into the optical cavity and position it such that it can interact with our optical mode. We say
that the 2-level atom (with energy difference ∆) is in resonance with the field mode at frequency ω if ∆ = ~ω. That
is, if the energy of a photon is equal to the difference in energy between the ground and excited states of the atom.
First, we describe the Hilbert space for the composite system (atom + field), and find an orthonormal basis. The
Hilbert space is given by the tensor product of the Hilbert spaces of the atom and the oscillator:
Htotal = Hatom ⊗ Hfield .
(10)
With a basis {|ei, |gi} for Hatom and a basis {|ni , n = 0, 1, 2, . . .} for Hfield , a basis for the total Hilbert space is given
by
{|ei|ni, |gi|ni , n = 0, 1, 2, . . .} .
(11)
The state |ei|ni can be interpreted as “the atom is in the state |ei (i.e., it is excited) and the field is in the state |ni
(i.e., it has n photons)”, and similarly for those with the ground state. Such states are called product states. Not all
states in the total Hilbert space Htotal are product states, though, such as
´
1 ³
|Ψi = √ |ei|ni + |gi|n + 1i .
(12)
2
States that cannot be written as a product state, but only as superpositions of product states, are entangled.
We can express the Hamiltonian for the atom-field system as:
Ĥtotal = Ĥfield + Ĥatom + Ĥint ,
(13)
where Ĥint is the interaction Hamiltonian. We can create a simple model for this interaction by considering what
physical process(es) can happen.
Clearly, one interaction process is for the atom to absorb a photon and thus excite from its ground state. (Because
we are in resonance, this is an energy-conserving process). The operator that describes this process is σ̂+ â, where
the σ̂+ part takes the atom from the ground to excited state, and the â part annihilates a photon. There is a reverse
process as well, where the atom emits a photon and goes to its ground state, described by the operator σ̂− ↠. Note
that each process operator is the Hermitean conjugate of the other. They can be put together into an interaction
Hamiltonian of the form:
Ĥint = ~Ω(σ̂+ â + σ̂− ↠) ,
(14)
where Ω is a coupling constant that determines the strength of the interaction. (In practice, this coupling constant is
determined by several factors, such as the location of the atom in the optical mode.)
The total Hamiltonian is then
Ĥtotal = ~ω(↠â + 1/2) + 12 ~ωσ̂z + ~Ω(σ̂+ â + σ̂− ↠) .
(15)
This model is called the Jaynes-Cummings model.
Determine the energy eigenstates of this Hamiltonian.
D.
Eigenstates of the Interaction
In this section, we will identify the eigenvalues and eigenstates of the total Hamiltonian of the Jaynes-Cummings
model. These eigenstates will allow us to easily calculate the evolution of any state. (See Question 3 on Problem Set
1.)
First, we note that we already have a basis of energy eigenstates for the harmonic oscillator Hamiltonian Ĥfield ,
being the number states {|ni, n = 0, 1, 2, . . .} with eigenvalues ~ω(n + 1/2). We also have a basis of energy eigenstates
for the 2-level atom Hamiltonian Ĥatom , being the states |gi, |ei with eigenvalues ± 12 ∆. Both of these bases are
non-degenerate; i.e., every state in the basis has a distinct eigenvalue.
Consider the combined Hamiltonian for the field + atom, where we “turn off” the interaction:
Ĥfree = Ĥfield + Ĥatom ,
(16)
3
which we denote the free Hamiltonian. The states |ei|ni and |gi|ni on Htotal = Hatom ⊗ Hfield are eigenstates of Ĥfree .
Prove this. However, they may be degenerate.
Let the atom be in resonance with the field, ∆ = ~ω. It is easy to see that the two states
|ei|ni ,
|gi|n + 1i ,
(17)
have the same energy under Ĥfree , i.e., they are both eigenstates with the same eigenvalue, ~ω((n + 1) + 1/2). That’s
because they both have the same number of energy quanta, either n in the field and one in the atom, or n + 1 in the
field and none in the atom. Thus, any superposition of these states,
α|ei|ni + β|gi|n + 1i ,
(18)
with |α|2 + |β|2 = 1 is also an energy eigenstate with eigenvalue ~ω((n + 1/2) + 1/2). Thus, all the eigenstates of Ĥfree
are 2-fold degenerate.
We now “turn on” the interaction, and find that it lifts the degeneracy. Note that
[Ĥfree , Ĥint ] = 0 .
(19)
Prove this. As a result, the eigenstates of Ĥtotal = Ĥfree + Ĥint as linear combinations of the degenerate eigenstates
of Ĥfree . It is straightforward to show that the states |ei|ni and |gi|ni are not eigenstates of Ĥint , though. The
eigenstates of Ĥint must have the form of Equation (18) for some values of α, β.
Show that the eigenstates of Ĥint in the 2-dimensional degenerate subspaces given by Equation (18)
are:
1
|Ψ±
(20)
n i = √ (|ei|ni ± |gi|n + 1i) .
2
These states are called dressed states. They are energy eigenstates satisfying
√
±
Ĥint |Ψ±
n i = ±~Ω n + 1|Ψn i .
(21)
Thus, they are also energy eigenstates of the total Hamiltonian satisfying
³
´
√
±
Ĥtotal |Ψ±
n i = ~ω((n + 1/2) + 1/2) ± ~Ω n + 1 |Ψn i .
(22)
The set of states {|Ψ±
n i} form a nondegenerate energy eigenstate basis for the Hilbert space. Thus, we can use these
states to determine the evolution of any state.
E.
The Interaction Picture
Because the free and interaction Hamiltonians commute (see Equation (19)), we can perform a “trick” that allows
us to effectively ignore the free Hamiltonian evolution, which is not of interest to us. The Schrödinger equation for
the total Hamiltonian is
i~
d
|Ψ(t)i = (Ĥfree + Ĥint )|Ψ(t)i .
dt
(23)
Define the free evolution operator to be
i
Ufree (t) = exp(− Ĥfree t) ,
~
(24)
|Ψ(t)iI = Ufree (t)−1 |Ψ(t)i .
(25)
and the interation picture state to be
Then it is straightforward to show that the interaction picture state satisfies the equation
i~
d
|Ψ(t)iI = Ĥint |Ψ(t)iI .
dt
(26)
Thus, states in the interaction picture evolve only under the interaction Hamiltonian.
The intuitive way to view the interaction picture is to think of our basis as “rotating with the free evolution” of
the oscillator. Relative to this rotating frame, we only see changes due to the interaction.
In the following, we will work exclusively in the interaction picture, and will drop the subscript I.
4
F.
Rabi oscillations
We now consider evolution (in the interaction picture), which is simple now that we have a basis of energy eigenstates
for the problem. We know that this basis evolves as:
±
i
|Ψ±
n (t)i = exp(− ~ Ĥint t)|Ψn i
√
= exp(− ~i (±~Ω n + 1)t)|Ψ±
ni
√
±
= exp(∓iΩ n + 1t)|Ψn i .
(27)
(As they are energy eigenstates, they do not change with time but simply pick up a phase.)
Consider the following initial condition: the atom is prepared in the excited state |ei, and the cavity field in the
state with n photons, |ni. Thus, the combined system is in the initial state |ψ(t = 0)i = |ei|ni. To determine the
time evolution of this state, we expand |ei|ni in terms of our basis of energy eigenstates:
1
−
|ψ(t = 0)i = √ (|Ψ+
n i + |Ψn i) .
2
(28)
Thus, the state after time t is
1
−
|ψ(t)i = √ (|Ψ+
n (t)i + |Ψn (t)i)
2
√
√
1
iΩ n+1t
= √ (e−iΩ n+1t |Ψ+
|Ψ−
ni + e
n i)
2
√
√
1
= √ (e−iΩ n+1t √12 (|ei|ni + |gi|n + 1i) + eiΩ n+1t √12 (|ei|ni − |gi|n + 1i)
2
√
√
√
√
n+1t
= 21 (e−iΩ n+1t + eiΩ n+1t )|ei|ni + 12 (e−iΩ n+1t − eiΩ
√
√
= cos(Ω n + 1t)|ei|ni − i sin(Ω n + 1t)|gi|n + 1i .
)|gi|n + 1i
(29)
Imagine we measured the state of atom at time t. Precisely, we mean that we measure the quantum observable σ̂z ,
with eigenstates |ei and |gi. Importantly, we are not measuring the state of the cavity mode. The probabilities of
getting either of these outcomes are:
√
Pe (t) = cos2 (Ω n + 1t)
√
Pg (t) = sin2 (Ω n + 1t) .
(30)
Thus, if the cavity mode √
is initially in a number eigenstate (a state of precise photon number) then there are times
(corresponding to cos2 (Ω n + 1t) = 1 when the atom will be found in the excited state with certainty. Similarly for
the ground state.