Time-Dependent Perturbation Theory - UMass Amherst

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Time-Dependent Perturbation Theory
Time-evolution operator as a product of elementary operators
Let U(t ,t ) be the time-evolution operator evolving the density matrix
1 0
ρˆ(t ) into ρˆ(t ) [see Eq. (22) in the Density Matrix chapter of Quantum
0 1
Mechanics I]:
ρˆ(t ) = U(t ,t )ρˆ(t )[U(t ,t )]† (t > t ). (1)
1 1 0 0 1 0 1 0
By its very definition, the evolution operator satisfies the identity
U(t ,t )U(t ,t ) ··· U(t ,t )U(t ,t ) ≡ U(t ,t ), (2)
n n−1 n−1 n−2 2 1 1 0 n 0
where we assume
t > t > ... > t > t . (3)
n n−1 1 0
On the other hand, if the time step
t −t = (cid:15) (j = 1,2,3,...,n) (4)
j j−1
is small enough, then, up to small corrections in (cid:15)2, the evolution operator
U(t ,t ) can be related to the Hamiltonian as (we set (cid:126) = 1)
j j−1
U(t ,t ) = 1−i(cid:15)H +O((cid:15)2), (5)
j j−1 tj
where H stands for the Hamiltonian at the time moment t. To check the
t
validity of (5), substitute it into (1) and make sure that the result satisfies
the equation of motion for ρˆ [equation (5) in the Density Matrix chapter of
Quantum Mechanics I]. We thus arrive at the representation
U(t ,t ) = (1−i(cid:15)H )···(1−i(cid:15)H )(1−i(cid:15)H )(1−i(cid:15)H ) + O((cid:15)). (6)
n 0 tn t3 t2 t1
Perturbative expansion for time-evolution and statistical operators
1
Let the Hamiltonian H consist of two parts:
t
H = H +V , (7)
t 0 t
where H is time-independent and V is a certain perturbation. Our goal is
0 t
to expand U(t,t ) in powers of V . Speaking practically, such an expansion
0 t
becomes useful when V is appropriately small allowing one to truncate the
t
perturbative series to one or two first terms.
Observing that
1−i(cid:15)H = (1−i(cid:15)H )(1−i(cid:15)V )+O((cid:15)2), (8)
tj 0 tj
wesubstitutether.h.s. of(8)for(1−i(cid:15)H )in(6)andthenopenthebrackets
tj
with (1−i(cid:15)V ). This naturally introduces different powers of V along with
tj tj
the necessity to sum over the time moments t . Taking the limit (cid:15) → 0, we
j
replace summation over t with integration. Simultaneously, we replace the
j
products of all the terms (1−i(cid:15)H ) with exponentials:
0
··· V (1−i(cid:15)H )···(1−i(cid:15)H )V ··· → ··· V e−i(tb−ta)H0V ··· .
tb 0 0 ta tb ta
This allows us to take the limit of (cid:15) → 0 rendering our expansion exact.
Finally, we decompose the exponentials to associate them with V ’s:
t
···e−i(tc−tb)H0V e−i(tb−ta)H0V e−i(ta−td)H0··· =
tb ta
= ··· e−itcH0V(t )V(t )eitdH0 ··· ,
b a
where we define
V(t) = ei(t−t0)H0V e−i(t−t0)H0. (9)
t
This brings us to the result
(cid:20) (cid:90) t (cid:90) t (cid:90) t1
U(t,t ) = e−i(t−t0)H0 1−i dt V(t )− dt dt V(t )V(t )+...
0 1 1 1 2 1 2
t0 t0 t0
(cid:90) t (cid:90) t1 (cid:90) tn−1 (cid:21)
+(−i)n dt dt ··· dt V(t )V(t )···V(t )+ ... , (10)
1 2 n 1 2 n
t0 t0 t0
Pay attention to the limits of integrating over t ’s. Those limits reflect the
j
fact that V ’s emerge from the expression (6) in the strict chronologic order.
tj
We can readily establish similar expression for the statistical operator
e−βH(cid:48) by treating β as an (imaginary) time interval.
e−βH(cid:48) = (1−(cid:15)H(cid:48))n + O((cid:15)), (cid:15) = β/n. (11)
2
H(cid:48) = H +V. (12)
0
(1−(cid:15)H(cid:48)) = (1−(cid:15)H )(1−(cid:15)V) + O((cid:15)2). (13)
0
(cid:20) (cid:90) β (cid:90) β (cid:90) τ1
e−βH(cid:48) = e−βH0 1− dτ V(τ )+ dτ dτ V(τ )V(τ )+...
1 1 1 2 1 2
0 0 0
(cid:90) β (cid:90) τ1 (cid:90) τn−1 (cid:21)
+(−1)n dτ dτ ··· dτ V(τ )V(τ )···V(τ )+ ... , (14)
1 2 n 1 2 n
0 0 0
V(τ) = eτH0V e−τH0. (15)
The chronologic order in (14) has the same origin as the chronologic order in
(10).
Linear Response: Kubo formula
By linear response of an observable A to the perturbation V one means the
t
leading (in powers of V ) effect of the perturbation on expectation value of A
t
at a given moment of evolution.1 The linear response is given by the Kubo
formula:
(cid:90) t
(cid:104)A(cid:105) = (cid:104)A(t)(cid:105) − i dt (cid:104)[A(t),V(t )](cid:105) , (16)
0 1 1 0
t0
where
A(t) = ei(t−t0)H0Ae−i(t−t0)H0, (17)
and
(cid:104)(...)(cid:105) = Tr(...)ρˆ(t ). (18)
0 0
Problem 11. Derive Kubo formula from (10).
The similarity of the forms of Eqs. (9) and (17), as well as the deep mean-
ing of this form will become clear in the next section, when we introduce the
interaction picture.
1In condensed matter physics, typical external perturbations—used to probe the prop-
erties of various systems—are periodic electric and magnetic fields, electromagnetic radi-
ation, beams of particles (electrons, neutrons, etc.).
3
Time-dependent unitary transformations: Heisenberg and interac-
tion pictures, “rotating frames”
Let S be some time-dependent unitary operator:
t
S†S = S S† = 1.
t t t t
If for each observable A, we perform the unitary transformation
A˜(t) = S†AS ⇔ A = S A˜(t)S†,
t t t t
and do the same for the density matrix:
ρ˜(t) = S†ρ(t)S ⇔ ρ(t) = S ρ˜(t)S†,
t t t t
we will only change the form of the density matrix and the operators of all
the observables, but not the physical properties of the system. This is be-
cause the measurement postulates of quantum mechanics are invariant with
respect to any unitary transformation.
Problem 12. Verify the above statement for both parts of the measuring
postulate: (i) for the probability part and (ii) for the projection part.
By performing this or that unitary transformation, we simply change repre-
sentation of the very same physics.
The change in the form of the density matrix and observables naturally
implies a change in the form of the evolution operator. Let us see how the
new evolution operator will look like. We have
U˜(t,t )ρ˜(t )U˜†(t,t ) = ρ˜(t) = S†ρ(t)S = S†U(t,t )ρ(t )U†(t,t )S
0 0 0 t t t 0 0 0 t
= S†U(t,t )S ρ˜(t )S† U†(t,t )S .
t 0 t0 0 t0 0 t
Hence
U˜(t,t ) = S†U(t,t )S . (19)
0 t 0 t0
We also want to relate the new and the old Hamiltonians. To this end we use
thegenericrelationbetweentheHamiltonianandevolutionoperator(implied
by the evolution equation for the density matrix):
∂ ∂
˜ ˜ ˜
i U(t,t ) = H U(t,t ), i U(t,t ) = H U(t,t ).
0 t 0 0 t 0
∂t ∂t
4
In view of the unitarity of evolution operators, these relation imply
(cid:20) (cid:21) (cid:20) (cid:21)
∂ ∂
H = i U(t,t ) U†(t,t ), H˜ = i U˜(t,t ) U˜†(t,t ). (20)
t 0 0 t 0 0
∂t ∂t
Applying (20) to (19) brings us to the result
∂S†
H˜ = S†H S + i t S . (21)
t t t t ∂t t
We are especially interested in the cases when
S = e−iH∗(t−t0), (22)
t
where H is a certain time-independent Hamiltonian. Here we have
∗
H˜ = eiH∗(t−t0)H e−iH∗(t−t0) − H . (23)
t t ∗
If H = H (since H does not depend on time, this implies that H is also
∗ ∗
˜
time-independent), we are dealing with the Heisenberg picture, where H = 0
t
and the density matrix does not evolve.
Problem 13. Show that for the second-quantized harmonic Hamiltonian
(either bosonic or fermionic)
(cid:88)
H = (cid:15) aˆ†aˆ ,
s s s
s
the creation and annihilation operators in the Heisenberg picture have the
form (here we use t = 0):
0
aˆ (t) = aˆ e−i(cid:15)st aˆ†(t) = aˆ†ei(cid:15)st. (24)
s s s s
Hint. The easiest way to proceed is to derive a certain differential equation
for aˆ (t) and aˆ†(t) by differentiating
s s
aˆ (t) = eiHtaˆ e−iHt, aˆ†(t) = eiHtaˆ†e−iHt
s s s s
with respect to time and then using the particular form of the Hamiltonian
and the properties of creation and annihilation operators.2 After that, one
2Different for bosons and fermions!
5
simply verifies that (24) satisfy the obtained equations, with appropriate ini-
tial conditions.
Yet another important class of pictures takes place when
H = H +V , [H ,H ] = 0.
t 0 t 0 ∗
Here we have
H˜ = H −H + eiH∗(t−t0)V e−iH∗(t−t0).
t 0 ∗ t
In particular, if H is time-independent, we can chose H = H to get the
0 ∗ 0
so-called interaction picture, in which only the perturbation term survives in
the Hamiltonian
H˜ ≡ V(t) = eiH0(t−t0)V e−iH0(t−t0) (interaction picture). (25)
t t
Comparing (25) to (9) and (17), one can guess that the interaction picture is
especially convenient for considering perturbative effects. This is indeed the
case as illustrated by the following problem.
Problem 14. Derive Eq. (10) employing the interaction picture. In the in-
teraction picture, find the evolution operator as a series of integrals by using
expansion (6). Then return back to the original picture. In particular, make
sure that Kubo formula can be derived in the interaction picture as well.
Rotating frame. Suppose our Hilbert space consists of two subspaces, I and
II, and the perturbation V has non-zero matrix elements V and V only
ab ba
if the state |a(cid:105) belongs to the subspace I, while the state |b(cid:105) belongs to the
subspace II. In this case, the following transformation—the so-called rotating
frame—might prove very useful.
ˆ
H = ωP (rotating frame), (26)
∗ I
ˆ
whereP istheprojectoronthesubspaceIandω isacertainfrequency/energy.
I
In this case (below we set t = 0),
0
e−iH∗t = Pˆ +e−iωtPˆ = 1+(cid:0)e−iωt −1(cid:1)Pˆ , (27)
II I I
6
ˆ ˆ
where P = 1 − P is the projector on the subspace II. In the “rotating
II I
frame,” the matrix elements become:
V˜ = eiωtV , V˜ = e−iωtV . (28)
ab ab ba ba
Problem 15. Derive Eqs. (27) and (28).
We conclude that in the rotating frame, all the energies in the subspace I
get shifted by ω, while all the matrix elements from subspace I to subspace
II acquire periodic phase factor eiωt, and all the matrix elements from sub-
space II to subspace I acquire periodic phase factor e−iωt. This is extremely
convenient for treating periodic perturbations. Periodic time dependence of
matrix elements can be readily eliminated at the expense of simply shifting
the energies of one of the two subspaces.
The Golden Rule
Suppose that at the initial time moment t = 0, the density matrix ρˆ(0)
0
corresponds to either a pure eigenstate of the Hamiltonian H , or a statisti-
0
cal mixture of some—but not all—eigenstates of the Hamiltonian H . The
0
actual Hamiltonian of the system is (7), so that ρˆ evolves in time.3 Assum-
ing that the perturbation V is small, we want to answer the following very
important question.
At the time moment t > 0, what is the probability, p (t), to find the system
Ω
in a certain span Ω of eigenstates of H orthogonal to the subspace of the
0
initial state?
As a typical example, think of and excited state of an atom or nucleus as
the initial state ρˆ(0). Then p (t) is the probability to establish the decay of
Ω
the excited atom/nucleus into a certain span Ω of the final states, provided
corresponding measurement is performed at time t.4 Correspondingly, Ω
3Excluding the trivial case of [V ,ρˆ](cid:54)=0.
t
4Specifying the time of measurement matters, because, speaking generally, we cannot
continuously observe quantum system without changing its properties (recall quantum
Zeno effect).
7
is associated with certain areas in the momentum space of the products of
decay. If Ω includes all possible momenta of the products of decay, then p (t)
Ω
is the probability for the excited state to decay by the time t.
In accordance with the measurement axiom,
(cid:88)
ˆ ˆ
p (t) = TrP ρˆ(t), P = |f(cid:105)(cid:104)f|, (29)
Ω Ω Ω
|f(cid:105)∈Ω
where |f(cid:105) is an eigenstate of H . The notation f comes from “final” to
0
distinguish those states from the “initial” eigenstates of H contributing to
0
ρˆ(0):
(cid:88)
ρˆ(0) = w |i(cid:105)(cid:104)i|. (30)
i
i
All the final states are orthogonal to all the initial states, and thus
ˆ ˆ
P ρˆ(0) = ρˆ(0)P = 0. (31)
Ω Ω
In view of Eq. (31), the leading term of the perturbative expansion of ρˆ(t)
that yields non-vanishing contribution to p (t) is the second-order term
Ω
(cid:90) t (cid:90) t
ρˆ(2)(t) = e−itH0 dt(cid:48)V(t(cid:48))ρˆ(0) dt(cid:48)(cid:48)V(t(cid:48)(cid:48))eitH0,
0 0
V(t) = eitH0V e−itH0.
t
Incidentally, note the convenience of the interaction picture where
(cid:90) t (cid:90) t
ρˆ(2)(t) = dt(cid:48)V(t(cid:48))ρˆ(0) dt(cid:48)(cid:48)V(t(cid:48)(cid:48)) (interaction picture).
0 0
The term ρˆ(2) is the minimal term in which ρˆ(0) is sandwiched between V’s
ˆ
and the expression is thus protected from vanishing upon tracing with P .
Ω
ˆ
Note that P commutes with H , thus commuting with the exponentials
Ω 0
e±itH0. This, in particular, means that it stays invariant when going to the
interaction picture. We thus have
(cid:90) t (cid:90) t
p (t) ≈ TrPˆ ρˆ(2)(t) = dt(cid:48) dt(cid:48)(cid:48)(cid:104)V(t(cid:48)(cid:48))Pˆ V(t(cid:48))(cid:105) , (32)
Ω Ω Ω 0
0 0
where
(cid:104)(...)(cid:105) = Tr(...)ρˆ(0). (33)
0
8
General expression (32) simplifies significantly when V is either time-
t
independent, or a harmonic function of time (normally referred to as peri-
odic perturbation). And the latter case actually reduces to the former one
in the appropriate rotating-frame picture, where the frequency of the frame
matches the frequency of the perturbation.
Time-independent perturbation. Consider the case of time-independent per-
turbation. The first observation here is that
(cid:104)V(t(cid:48)(cid:48))Pˆ V(t(cid:48))(cid:105) = (cid:104)V(0)Pˆ V(t(cid:48) −t(cid:48)(cid:48))(cid:105) ,
Ω 0 Ω 0
resulting in (we shift the integration variable t(cid:48) → t(cid:48) −t(cid:48)(cid:48))
(cid:90) t (cid:90) t−t(cid:48)(cid:48)
p (t) = dt(cid:48)(cid:48) dt(cid:48)(cid:104)V(0)Pˆ V(t(cid:48))(cid:105) .
Ω Ω 0
0 −t(cid:48)(cid:48)
The next observation is that for p (t) to be appreciably different from zero,
Ω
the time t has to be much larger than the typical time t of variation (decay)
∗
of the function (cid:104)V(0)Pˆ V(t(cid:48))(cid:105) . In essence, this is the criterion of applica-
Ω 0
bility of the perturbative treatment for p (t). Vanishing (cid:104)V(0)Pˆ V(t(cid:48))(cid:105) at
Ω Ω 0
certain appropriately large t(cid:48) allows one—at t (cid:29) t(cid:48)—to extend the limits of
integration over t(cid:48) to plus/minus infinity. This brings us to the result
(cid:90) ∞
p (t) = tW , W = dt(cid:48)(cid:104)V(0)Pˆ V(t(cid:48))(cid:105) (t (cid:29) t ), (34)
Ω Ω Ω Ω 0 ∗
−∞
known as Fermi’s Golden Rule. According to the Golden Rule, the probabil-
ity to find the system in one of the“final” states grows linearly with time.5
The linearity of increasing p with time is a crucial result on its own, allow-
Ω
ing one to speak of the decay of the initial state and transitions to the final
states, thus interpreting W as the transition rate. In particular, extending
Ω
Ω to all possible finite states, one gets the total transition/decay rate. With
a natural requirement that V does not have non-zero matrix elements be-
tween any two initial states,6 the total transition/decay rate acquires a very
compact form
(cid:90) ∞
W = dt(cid:48)(cid:104)V(0)V(t(cid:48))(cid:105) . (35)
tot 0
−∞
5Within a wide range of times where t(cid:29)t while p (t)(cid:28)1.
∗ Ω
6Otherwise, those elements can be simply added to H .
0
9
Analternative—andmostfrequentlyused—formoftheGoldenRuledeals
with the situation when ρˆ(0) is a pure state (called the initial state):
ρˆ(0) = |i(cid:105)(cid:104)i|.
In this case we have
(cid:88)
(cid:104)V(0)Pˆ V(t(cid:48))(cid:105) = (cid:104)i|V(0)Pˆ V(t(cid:48))|i(cid:105) = (cid:104)i|V(0)|f(cid:105)(cid:104)f|V(t(cid:48))|i(cid:105).
Ω 0 Ω
|f(cid:105)∈Ω
Taking into account
H |i(cid:105) = E |i(cid:105), H |f(cid:105) = E |f(cid:105),
0 i 0 f
the time-dependence can be explicitly factored:
(cid:104)f|V(t)|i(cid:105) = ei(Ef−Ei)tV , V ≡ (cid:104)f|V|i(cid:105), (36)
fi fi
and the integration over time performed:
(cid:90) ∞
dtei(Ef−Ei)t = 2πδ(E −E ).
f i
−∞
This yields the following form of the Golden Rule (here we restore (cid:126))
2π (cid:88)
W = |V |2δ(E −E ). (37)
Ω (cid:126) fi f i
|f(cid:105)∈Ω
In particular, for the total rate we have
2π (cid:88)
W = |V |2δ(E −E ). (38)
tot (cid:126) fi f i
f
The presence of the delta-function implies that the summation over the final
states is understood as an integral over the continuous quantum numbers—
usually, momenta—specifying the final states.
It is absolutely straightforward to generalize (37) to the case when ρˆ(0) is
not a pure state. Using (30), we readily reduce the problem to the previous
one, with the final answer
2π (cid:88) (cid:88)
W = w |V |2δ(E −E ). (39)
Ω (cid:126) i fi f i
i |f(cid:105)∈Ω
10