Chapter 3. Quantum Dynamics
3.1 Time-Evolution and Time-dependent Schrödinger Equation
In last chapter we have learned solving the time-independent Schrödinger Equation to obtain eigenstates (eigenfunctions) and eigenvalues (eigenenergies), which are required to study quantum
dynamics, i.e., solving the time-dependent Schrödinger Equation:
i
∂
ψ(t) = H(t) ψ(t)
∂t
(3.1)
Unfortunately, this equation is notoriously difficult to solve analytically, except for the simplest case of time-independent Hamiltonian. For a time-dependent H(t) , time-dependent perturbation theory is used to solve Eq. (3.1) approximately. Here we discuss the general theoretical framework of quantum dynamics using the time-evolution operator U(t,t0 ) :
ψ(t) =U(t,t0 ) ψ(t0 ) .
(3.2)
Since the initial state ψ(t0 ) is independent of time t, Eq. 3.1 of state ψ(t) is changed to an
equation for the time-evolution operator:
i
∂
U(t,t0 ) = H(t)U(t,t0 ) .
∂t
(3.3)
Case 1: Hamiltonian operator is time-independent. Then it is straightforward to show that
⎡ i
⎤
U(t,t0 ) = exp ⎢− (t −t0 )H ⎥ .
⎢ 
⎥
⎣
⎦
(3.4)
Case 2: Hamiltonian operator is time-dependent, but H(t) at different times always commute, then the time-evolution operator becomes
⎡ i t
⎤
U(t,t0 ) = exp ⎢− ∫ H(t ′)dt ′⎥ .
⎢  t0
⎥
⎣
⎦
(3.5)
Case 3: Hamiltonian operator at different times don’t commute, then
∞
n
U(t,t0 ) = 1 + ∑ (−i /  )
n=1
∫
t
t0
t1
dt1 ∫ dt2  ∫
t0
tn−1
t0
dtnH(t1 )H(t2 )H(tn ).
(3.6)
This is the Dyson series (by Freeman Dyson) in quantum field theory. We’ll derive this when we
discuss the time-dependent perturbation theory.
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Let’s consider case 1, where H is independent of time. The eigenkets of H are denoted by
n : H n = En n . For simplicity, set t0 = 0 , then the time-evolution operator is
⎛
⎞
⎛
⎞
−iE t/
U(t) = e −iHt/ = ⎜⎜∑ m m ⎟⎟e −iHt/ ⎜⎜∑ n n ⎟⎟ = ∑ n e n n .
⎟⎠
⎟⎠
⎜⎝ m
⎜⎝ n
n
(3.7)
At t = 0 , ψ(0) = ∑ n n ψ(0) = ∑ cn n ; then, at an arbitrary time t ,
n
n
ψ(t) =U(t) ψ(0) = ∑ n e
−iEnt/
n
n ψ(0) = ∑ e
−iEnt/
n
cn n .
(3.8)
One can also write Eq. (3.8) as
ψ(t) = ∑ cn (t) n , with cn (t) = e
n
−iEnt/
cn (0) .
(3.9)
Question: How can we distinguish ψ(t) and ψ(0) by measurements?
If n are eigenstates of operator A , A n = an n and we have [A,H ] = 0 , then
⎛
⎞
iE t/ ⎞ ⎛
−iE t/
A = ⎜⎜∑ m cm* e m ⎟⎟⎟ A ⎜⎜∑ e n cn n ⎟⎟ = ∑ an |cn |2 ,
⎟⎠
⎜⎝ m
⎠ ⎜⎝ n
n
(3.10)
Which is time-independent, so measurement A cannot distinguish ψ(t) from ψ(0) .
On the other hand, if B and H don’t commute, i.e., n are NOT eigenstates B , then
⎛
⎞
iE t/! ⎞ ⎛
−iE t/!
B = ⎜⎜∑ m cm* e m ⎟⎟⎟ B ⎜⎜∑ e n cn n ⎟⎟ = ∑ cm* cn exp(iωmnt) m B n ,
⎟⎠ m,n
⎜⎝ m
⎠ ⎜⎝ n
(3.11)
ωmn = (Em − En ) /  .
(3.12)
where
Now the expectation value B changes with time, therefore ψ(t) is a nonstationary state if it
is a superposition of energy eigenkets n . The cross terms cm* cn exp(iωmnt) lead to interference,
i.e., the oscillation of observables.
If ψ(t) is an energy eigenket, say ψ(0) = n and ψ(t) = e
B = ne
iEnt/
Be
−iEnt/
−iEnt/
n = n Bn ,
n , B is given by
(3.13)
which is also time-independent, and ψ(t) (or any energy eigenket n ) is a stationary state.
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3.2 Dynamics of Two-Level Systems
The simplest quantum systems have two levels, e.g., spin-half systems, and the quantum dynamics leads to interference, manifesting itself in spin procession and neutrino oscillations, etc.
Here the energy eigenkets are denoted by 1 and 2 : H 1, 2 = E1,2 1, 2 . Suppose at t = 0 ,
ψ(0) = c1 1 +c2 2 , the state ket at some time later is
ψ(t) = e
−iE1t/
c1 1 +e
−iE 2t/
c2 2 .
(3.14)
Measurements A ( A 1, 2 = a1,2 1, 2 ) lead to time-independent expectation value
A = a1 |c1 |2 +a 2 |c2 |2 .
(3.15)
Obviously, the probabilities of finding eigenkets of A are |c1 |2 and |c2 |2 , respectively, which
are also time-independent.
For a Hermitian operator B that doesn’t commute with H , the expectation value becomes
B =|c1 |2 1 B 1 + |c2 |2 2 B 2 + 2Re ⎡⎢c1*c2 exp(iω1,2t) 1 B 2 ⎤⎥ ,
⎣
⎦
(3.16)
where ω1,2 = (E1 − E 2 ) /  . Denoting eigenkets of B by b1 and b2 with eigenvalues of b1 and
b2 , the probabilities of finding the system in these two states at time t are p j (t) =| bj ψ(t) |2 ,
with j = 1, 2. Explicitly, state kets b1 and b2 can be expressed by linear combinations of 1
and 2 using the mixing angle θ :
b1 = cos θ 1 − sin θ 2
(3.17)
b2 = sin θ 1 + cos θ 2
(3.18)
The probabilities for the system to be found in b1 and b2 are
p1(t) =| b1 ψ(t) |2 =|c1 |2 cos2 θ+ |c2 |2 sin 2 θ − Re ⎡⎢c1*c2 sin(2θ)exp(−iω1,2t)⎤⎥ ,
⎣
⎦
2
2
2
2
2
*
p2(t) =| b2 ψ(t) | =|c1 | sin θ+ |c2 | cos θ + Re ⎡⎢c1c2 sin(2θ)exp(−iω1,2t)⎤⎥ ,
⎣
⎦
(3.19)
(3.20)
respectively, and the expectation value is
B = b1p1(t) +b2p2(t) = (b1 cos2 θ +b2 sin 2 θ) |c1 |2 +(b1 sin 2 θ +b2 cos2 θ) |c2 |2
+ (b1 −b2 )Re ⎡⎢c1*c2 sin(2θ)exp(−iω1,2t)⎤⎥ .
⎣
⎦
39
(3.21)
The two-level systems we often encounter are the spin-half particles. Consider a spin-half
particle, specifically here an electron, with magnetic moment

S
µ = gS µB ,

(3.22)
where the Landé g-factor for spin gS ≈ 2 , the Bohr magneton µB =
e
with e = −1.602×10−19
2mec
C, me the electron mass and c the speed of light. The Hamiltonian of an electron subject to an
external magnetic field B is
⎛ −e ⎞⎟

⎟⎟ S ⋅ B .
H = −µ ⋅ B = ⎜⎜⎜
⎜⎝ mec ⎟⎟⎠
(3.23)
We consider a static uniform magnetic field in the z-direction, i.e., B = Bẑ , and H is written as
⎛ eB ⎞⎟

⎟⎟S .
H = −µ ⋅ B = −⎜⎜⎜
⎜⎝ mec ⎟⎟⎠ z
(3.24)
H and Sz commute and share the common eigenkets ± :
Sz ± = ±
with eigenenergies E ± = ∓

± ; H ± = E± ± ,
2
(3.25)
e"B
eB
. We can define ω ≡| E + − E− | / = −
, so that H = ωSz ,
2mec
mec
and the time-evolution operator
U(t) = e −iHt/ = e
−iωSzt/
.
(3.26)
At time t = 0 , ψ(0) = c+ + +c1 − , applying Eq. (3.26), the state ket evolves as
ψ(t) = c+e −iωt/2 + +c−e iωt/2 − .
(3.27)
Specifically, when c+ = c− = 1 / 2 , ψ(0) = x + , while c+ = −c− = 1 / 2 ⇒ ψ(0) = x − ,
where x ± are eigenkets for operator Sx with eigenvalues of ± / 2 , respectively. Eqs. (3.17)
and (3.18) suggest that here the mixing angle θ = −π / 4 for x ± .
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If the system starts with ψ(0) = x + , the possibility of finding it in the same state at time t:
| x + ψ(t) |2 = cos2(ωt / 2) ,
(3.28)
while in state x − the probability is
| x − ψ(t) |2 = sin 2(ωt / 2) ,
(3.29)
and the expectation value of Sx is
Sx =



cos2(ωt / 2)− sin 2(ωt / 2) = cos(ωt) .
2
2
2
(3.30)
This is called spin procession. Question: (1) For this initial state, will we observe procession for
Sy ? or Sz ? (2) If the initial state ψ(0) = + , will we observe procession for Sx , Sy or Sz ?
Experimental verification: the observation of the precession of muon spin by G. W. Bennett
et al., Phys. Rev. D 73, 072003 (2006). This experiment measured the deviation of gS from 2.
Muons were injected in the storage ring so that their spins will precess in lock step with their
momentum vector if gS = 2 . Consequently, observation of their precession measures gS − 2 directly, in good agreement with the theoretical value derived from the quantum field theory.
Muon Storage Ring
Muon Spin Precession
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3.3 Heisenberg Picture
Under unitary transformation (e.g., time-evolution, space translation, rotation, etc.), the inner
product of two state vectors α β is invariant:
α β → α U †U β = α β ,
(3.31)
However, the matrix elements for a (Hermitian) operator A might change
α A β → α U †AU β ,
(3.32)
except for [A,U ] = 0 .
In order to study the properties of a physical system under unitary transformation, one can
take two approaches:
Approach 1 (Schrödinger picture): the state vector is changed, α →U α , with operators unchanged.
†
Approach 2 (Heisenberg picture): the operator is changed, A →U AU , with state vectors unchanged.
α(H ) = α(S )(0)
(3.33)
A(H )(t) =U †(t)A(S )U(t); and A(H )(0) = A(S )
(3.34)
The expectation value of A obviously should not change:
α(S )(t) A(S ) α(S )(t) = α(H ) A(H ) α(H ) ,
(3.35)
which suggests that these two pictures describe the same physics (using distinct mathematic formalisms).
The Heisenberg picture is better suited to bringing out fundamental features, such as symmetries and
conservation laws, and it is indispensible in systems with many degrees of freedom (QFT, statistical physics). Here we briefly discuss the dynamics in Heisenberg picture. In contrast to the Schrödinger picture in
which the dynamics involves time-evolution state vectors, the dynamics in Heisenberg picture is represented by the time-evolution of operators, similar to classical physics where the physical quantities
change in time:
i
⎛ ∂U † (S )
dA(H )
∂U ⎞⎟⎟
= i ⎜⎜⎜
A U +U †A(S )
⎟.
⎜⎝ ∂t
dt
∂t ⎟⎠
(3.36)
Using Eq. (3.3) and assuming a time-independent Hamiltonian H (in Schrödinger picture), so that
U(t) = e −iHt/ and
H (H ) =U †H (S )U = H (S ); thus H (H ) = H (S ) = H ,
42
(3.37)
we derived the Heisenberg equation of motion
i
dA(H )
= [A(H ),H ] .
dt
(3.38)
This has exactly the same form of the classical equation of motion in Poisson bracket:
dA
= {A,H }PB ,
dt
(3.39)
considering the correspondence between classical Poisson bracket and the quantum commutator
Eq. (1.169).
Another interesting analogue is the Ehrenfest theorem
m
d2 r
=
dt 2
d p
dt
= − ∇V(r) ,
(3.40)
to Newton’s second law
m
d 2r dp
=
= −∇V(r) = F .
dt
dt 2
Proof: in Heisenberg picture (omitting
i
(H )
(3.41)
in the following),
dp
= [p,H ] = −i∇V(r) ,
dt
(3.42)
dr
p
= [r,H ] = i .
dt
m
(3.43)
i
Combining Eq. (3.42) and (3.43), we find that
m
d 2r dp
=
= −∇V(r) .
dt
dt 2
(3.44)
[Note that Eqs. (3.41) and (3.44) have exactly the same format; however they are very different!]
Taking expectation value, we thus proved the Ehrenfest theorem [Eq. (3.40)].
Eq. (3.38) also suggests that if [A,H ] = 0 , i.e., A commutes with H , then A(H ) is a constant
of motion, corresponding to a conservation law. For example, a free particle’s momentum,
dp(t)
1
p2
= [p,
]= 0.
dt
i 2m
(3.45)
the angular momentum of a particle in a central potential,
dL(t)
1
= [L,H ] = 0 .
dt
i
43
(3.46)
3.4 Time Development of 1D Simple Harmonic Oscillator
In this class, the default picture is Schrödinger; but in the following we use the Heisenberg picture, and we suppress the superscript
(H )
. It is easy to show the dynamics of a quantum system in
Heisenberg picture, as already seen clearly – the close correspondence between classical dynamical equations for observables and quantum mechanical dynamic equations for operators in the
Heisenberg picture.
The Heisenberg equation of motion [Eq. (3.38)] for position ( X ) and momentum ( P ) operators of a one-dimensional SHO are
dX
1
P
= ⎡⎢X,H ⎤⎥ = .
⎣
⎦
dt
i
m
(3.47)
dP
1
= ⎡⎢P,H ⎤⎥ = −mω 2X ,
⎦
dt
i ⎣
(3.48)
You might find that this set of equations are exactly the same in format as that in classical mechanics, therefore the solutions look exactly the same as well:
X(t) = X(0)cos(ωt) +
P(0)
sin(ωt) ,
mω
P(t) = −mωX(0)sin(ωt) + P(0)cos(ωt) ,
(3.49)
(3.50)
for the initial conditions of X(0) and P(0) .
Alternatively, we can evaluate X(t) directly:
X(t) =U †(t)X(0)U(t) = e iHt/ X(0)e −iHt/ ,
using the Baker-Hausdorff lemma, (Homework 07, Problem 1)
1
1
e ABe −A = B +[A,B]+ [A,[A,B]]+ [A,[A,[A,B]]]+ 
2!
3!
(3.51)
(3.52)
We obtain
2
⎛ it ⎞
1 ⎛ it ⎞
X(t) = X(0) + ⎜⎜⎜ ⎟⎟⎟ ⎡⎢H, X(0)⎤⎥ + ⎜⎜⎜ ⎟⎟⎟
⎦ 2! ⎝  ⎟⎠
⎝  ⎟⎠ ⎣
⎡H, ⎡H, X(0)⎤ ⎤ + 
⎢⎣ ⎣⎢
⎦⎥ ⎥⎦
(3.53)
Then we need to evaluate commutators [H, X(0)] , [H,[H, X(0)]] , …
i
P(0) ,
m
(3.54)
[H,P(0)] = imω 2X(0)
(3.55)
[H, X(0)] = −
44
P(0) (ωt)2
(ωt)3 P(0)
−
X(0)−
+!
mω
2!
3! mω
P(0)
= X(0)cos(ωt) +
sin(ωt)
mω
X(t) = X(0) + ωt
(3.56)
Similarly, we can prove Eq. (3.50). Next, we find the ladder operators a(t) and a †(t) . Using the
fundamental commutators [a,a † ] = 1 and H = ω(a †a + 1 / 2) , as a practice please prove that
a(t) = e iHt/a(0)e −iHt/ = a(0)e −iωt ,
(3.57)
a †(t) = e iHt/a †(0)e −iHt/ = a †(0)e iωt .
(3.58)
Question: Do Eqs. (3.50) and (3.51) suggest that the expectation values of X and P (which
are measurable) for energy eigenstates n oscillate in time?
n X(0) n = 0 and n P(0) n = 0 ⇒ n X(t) n = 0 and n P(t) n = 0 . These actually
are quite obvious in the Schrödinger picture. Then how to construct a quantum state to show the
characteristic dynamics of SHO, i.e., the classical-like oscillation?
One example is the coherent state defined as the eigenstate of the lowering operator
a λ =λ λ
(3.59)
In homework (HW#6 problem 3c) you already have proved that a coherent state can be obtained
by applying the translation operator e
−iPx 0 /
to the ground state 0 :
λ =e
Then
X(t) = λ X(t) λ = 0 e
iP(0)x 0 /
−iPx 0 /
0 .
⎡
⎤
⎢X(0)cos(ωt) + P(0) sin(ωt)⎥ e −iP(0)x 0 / 0 .
⎢
⎥
mω
⎣
⎦
(3.60)
(3.61)
Because
e
iP(0)x 0 /
X(0)e
−iP(0)x 0 /
e
= X(0) +
iP(0)x 0 /
P(0)e
ix 0

[X(0),P(0)] = X(0)− x 0 ,
−iP(0)x 0 /
= P(0) ,
(3.62)
(3.63)
the expectation value of X(t) oscillates classically:
X(t) = λ X(t) λ = −x 0 cos(ωt) .
(3.64)
P(t) = λ P(t) λ = mωx 0 sin(ωt) .
(3.65)
Similarly,
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