Lecture 6: Pictures
Read Sakurai and Napolitano Chapter 2.1 and 2.2
Both operators and state vectors are unobservable, only the inner products are related to
observable quantities. Under the requirement of keeping the inner products invariant, we can
use different pictures to formulate the time-evolution in quantum mechanics. They are related
to similar transformations.
1
Schrödinger picture
The time evolution is expressed as the evolution of state vectors. The canonical coordinates,
momenta, spin do not change with time.
|ΨS (t)i = T (t,t0 )|Ψs (t0 )i.
(1)
Assume that F s is an operator for the observable F in the Schrödinger picture, we have its
expectation value
F s = hΨs (t)|F s |Ψs (t)i.
(2)
If H s does not depend on t, then T (t, 0) = exp{− h̄i Ht}. If H s explicitly depend on t, the
expression of T is not so simple. According to ∂t∂ T (t, 0) = − h̄i H s (t)T (t, 0), we have
−i
T (t, 0) = 1 +
h̄
Z t
0
dt1 H s (t1 )T (t1 , 0),
(3)
and by iteration,
t
t1
−i t
−i
dt1 H s (t1 ) + ( )2 dt1
dt2 H s (t1 )H s (t2 )
h̄ 0
h̄
0
0
Z
Z tn−1
−i n t
dt1 ...
dtn H s (t1 )H s (t2 )...H s (tn ) + ...
+ ... + ( )
h̄
0
0
Z
Z t
+∞
1 −i n t
= ∑ ( )
dt1 ... dtn T H s (t1 )H s (t2 )...H s (tn )
n!
h̄
0
0
n=0
Z t
−i
= T exp{
H(t ´ )dt ´ }.
h̄ 0
Z
Z
Z
T (t, 0) = 1 +
(4)
T is the time-ordered operator. T is defined as
T H(t1 )H(t2 )...H(tn ) = ∑ θ(t p1 > t p2 > ... > t pn )H(t p1 )H(t p2 )...H(t pn ),
(5)
p
where p is a permutation of 1, 2, ..., n, and θ-function equals to 1 if the condition t p1 > t p2 >
... > t pn is satisfied and otherwise 0.
6-1
2
Heisenberg picture
We can also fix state vector stationary with time, say, set the state vector in Heisenberg picture
as the that of the Schrödinger one at t = 0, and let operator to evolve with time:
|ΨH i = |Ψs (0)i,
F̂ H (t) = T t (t, 0)F̂ s T (t, 0),
(6)
such that
hΨH |F H |ΨH i = hΨs |F s |Ψs i.
(7)
Actually, F s can also explicitly depend on time as F s (t).
Now let us derive the equation of motion of operators. From the Schrödinger equation in which that H s may explicitly depend on time, ih̄ ∂t∂ |Ψs (t)i = H s (t)|Ψs (t)i, we have
ih̄ ∂t∂ T (t, 0)|Ψs (0)i = H s (t)T (t, 0)|Ψs (0)i, thus we have
∂
T (t, 0) = H s (t)T (t, 0)
∂t
∂ †
−ih̄ T (t, 0) = T † (t, 0)H s (t).
∂t
ih̄
(8)
Then we have dtd F H (t) = ∂t∂ T t (t, 0)F s (t)T (t, 0)+T t (t, 0) ∂t∂ F s (t)T (t, 0)+T t (t, 0)F s (t) ∂t∂ T (t, 0),
and then
1
∂F s (t)
d H
F (t) = [F H (t), H H (t)] + T † (t, 0)
T (t, 0),
dt
ih̄
∂t
(9)
where H H (t) = T † (t, 0)H s (t)T (t, 0). In particular, for the canonical coordinate and momentum, we have
1 H
d H
q j (t) =
[q (t), H H (t)],
dt
ih̄ j
d H
1 H
p j (t) =
[p (t), H H (t)].
dt
ih̄ j
(10)
Example 1) Consider the Hamiltonian of an harmonic oscillator in the Schrödinger picture
HS =
pS,2 1
+ mω2 xS,2 .
2m 2
6-2
(11)
Then the operators in the Heisenberg picture xH and pH can be solved in the following way.
d H
1 H
1
x (t) =
[x (t), H H ] = eiHt [x, H S ]e−iHt ,
dt
ih̄
ih̄
d H
1 H
1
p (t) =
[p (t), H H ] = eiHt [p, H S ]e−iHt .
dt
ih̄
ih̄
(12)
Using the fact that
1
ih̄
[x, pS,2 ] = pS
2m
m
(13)
1
mω2 [pS , x2 ] = −ih̄mω2 xS ,
2
(14)
pH (t) d H
,
p (t) = −mω2 xH (t),
m
dt
(15)
[xS , H S ] =
and
[pS , H s ] =
we arrive at
d H
x (t) =
dt
the solution will be
pS
sin ωt,
mω
pH (t) = −mωxS sin ωt + pS cos ωt.
xH (t) = xS cos ωt +
(16)
Exercise 1 Please prove Eq. 16 for harmonic oscillators by directly using
i
S
i
i
S
S
i
S
xH (t) = e h̄ H t xS e− h̄ H t , pH (t) = e h̄ H t pS e− h̄ H t .
(17)
Hint: You may use the Baker-Hausdorff lemma in page 95 in Sakurai and Napolitano’s book.
This formula is proved in my notes in lecture 7.
3
Interaction picture
We decompose the Hamiltonian H s of the Schrödinger picture into the free part H0 and the
perturbative part V as
H s = H0 +V,
(18)
which H0 is independent of time; V may depend on time. We define that the state vector
evolves with time as
|ΨI (t)i = eiH0t/h̄ |ΨS (t)i = eiH0t/h̄ T (t, 0)|ΨS (0)i,
6-3
(19)
and correspondingly the operator
F I (t) = eiH0t/h̄ F s e−iH0t/h̄ .
(20)
In such a convention, we keep the inner product invariant
hΨIA (t)|F I (t)|ΨIB (t)i = hΨSA (t)|F S (t)|ΨSB (t)i
(21)
Now let us derive equation of motion, we have
1
d I
∂F s (t) −iH0t/h̄
F (t) = [F I (t), H0 ] + eiH0t/h̄
e
.
dt
ih̄
∂t
(22)
For state vector, we have
i
1
∂ I
|Ψ (t)i =
H0 eiH0t/h̄ |Ψs (t)i + eiH0t/h̄ H S |ΨS (t)i
∂t
h̄
ih̄
i
1
= eiH0t/h̄ (H0 − H s )e−iH0t/h̄ |ΨI (t)i = V I (t)|ΨI (t)i.
h̄
ih̄
(23)
From Eq. 23, we can derive the time-evolution operator U(t,t0 ) in the interaction picture as
|ΨI (t)i = U(t,t0 )|ΨI (t0 )i,
Z
i t ´ I ´
U(t,t0 ) = T exp{−
dt V (t )}.
h̄ t0
(24)
Exercise 2
1. Prove Eq. 22 of
d I
dt F (t).
2. Prove the above expression of U(t,t0 ).
Example Let us consider a perturbed Harmonic potential
H = H0 +V (x),
(25)
where H0 = p2 /2m + 12 mω2 x2 and V (x) is a small perturbation. In the interaction picture, we
have
i
i
p
xI (t) = e h̄ H0t xe− h̄ H0t = x cos ωt +
sin ωt,
mω
i
i
(26)
pI (t) = e h̄ H0t pe− h̄ H0t = −mωx sin ωt + p cos ωt,
and thus
V I (t) = V (xI (t)).
6-4
(27)
The time-evolution operator for the state vectors are
U(t,t0 ) = T exp{−
i
h̄
Z t
6-5
t0
dtV (xI (t))}.
(28)