Review of Quantum Mechanics

and the same expansion with r ? rr in for the right bar obtaining

¸
(O) =
¸
=

¸
d3r
¸
d3r


d3rr(?|r)(r|O|rr)(rr|?)

d3rr??(r, t)O(r, rr)?(rr, t) . (3.40)

To see that this is the same as the position representation version of (3.2) is a little trickier. We substitute the identity expansion (3.34) in for the left bar

Hermitian operators in quantum mechanics often turn out to be local, and hence
O(r, rr) = O(r, r)?(r ? rr) . (3.41)
This fact reduces (3.40) to (3.2) as desired. More generally we can interpret
O?(r, t) on the right-hand side of (3.2) as
¸

O?(r, t) =

d3rrO(r, rr)?(rr, t) .


Similarly substituting the expansion (3.37) into (3.39), we ?nd the ex- pectation value (3.24), where the operator matrix elements written earlier as (3.16) can be written simply as
Omn = (m|O|n) . (3.42)

Problem 3.2 shows that this equals (3.16). It is also useful to express the operator O directly in terms of the basis set. This then reads as
O = . Onm|n)(m| . (3.43)
n,m

Finally, we note that the state vector version of the Schro¨dinger equation (3.5) is




where

?
ik ?t |?) = H|?) , (3.44)

pˆ2
H = 2m + V (ˆr) . (3.45)

This reduces to the Schr¨odinger equation (3.5) in the coordinate representa- tion.


Coordinate and Momentum Representations

Equation (3.36) introduced the coordinate representation wave function
?(r, t) = (r|?). One can in a similar fashion introduce the momentum repre-
sentation wave function
?(p, t) = (p|?) . (3.46)


In order to discuss the connection between the coordinate and momentum representations, we restrict ourselves to a one-dimensional situation where
pˆ ? pˆ and xˆ ? xˆ, with [xˆ, pˆ] = ik. We proceed by introducing the translation,
or shift operator
S(?) = exp(?i?pˆ/k) . (3.47)
S(?) is unitary, with S†(?) = S?1(?) = S(??), and it has a number of interesting properties. For instance, using the commutation relations
[xˆ, f (pˆ)] = ikf r(pˆ) , (3.48)
[pˆ, g(xˆ)] = ?ikgr(xˆ) , (3.49)

which are proven in Prob. 3.19, one ?nds readily that
[xˆ, S(?)] = ?S(?) , (3.50)

so that xˆS(?) = S(?)[xˆ + ?]. Since the ket |x) is an eigenstate of the operator xˆ with xˆ|x) = x|x), we have therefore that xˆS(?)|x) = S(?)(x + ?)|x) = (x + ?)S(?)|x). In other words, the state S(?)|x) is also an eigenstate of xˆ, but with eigenvalue (x + ?): the action of the operator S(?) on the ket |x) is to transform it into a new eigenvector of xˆ with eigenvalue shifted by the arbitrary amount ?,
S(?)|x) = |x + ?) . (3.51)

This also proves that the spectrum of xˆ is continuous. Finally, we note that the coordinate representation wave function corresponding to the ket S(?)|?)
is
(x|S(?)|?) = (x ? ?|?) = ?(x ? ?) . (3.52)

The translation operator permits one to easily determine the action of the momentum operator pˆ in the coordinate representation. Considering a small shift ? such that S(??) = exp(i?pˆ/k) c 1 + i ? pˆ + O(?2), we have





so that

?
(x|S(??)|?) = ?(x) + i


(x|pˆ|?) + O(?2) = ?(x + ?) , (3.53)

(x|pˆ|?) =

k lim i ??0

?(x+ ?) ? ?(x)
?

= k d
i dx

?(x) . (3.54)

The action of pˆ in the coordinate representation is therefore

d
pˆ ? ?ik dx . (3.55)

A similar derivation shows that in the momentum representation, the action of xˆ is
d
xˆ ? ik dp . (3.56)


Armed with this knowledge, it is quite easy to obtain the form of the Schr¨odinger equation in the momentum representation from its coordinate
representation of (3.6). This requires evaluating (p|V (xˆ)|?(t)). Introducing the identity ¸ dp |p)(p| = 1, we have

¸
(p|V (xˆ)|?(t)) =

¸
dpr(p|V (xˆ)|pr)(pr|?(t)) =


dprV (p ? pr)?(pr, t) , (3.57)


where V (p ? pr) are of course the matrix elements of the potential V in
the momentum representation. They can be obtained from the coordinate
representation by noting that

¸
(p|V (x)|pr) =


dx(p|x)(x|V (xˆ)|x)(x|pr) . (3.58)


Since (x|pr) may be interpreted as the coordinate representation wave func- tion ?(x) associated with the state vector |pr) we have from (3.54)
d?(x)



so that

(x|pˆ|p) = ?ik

dx ,

?(x) =

. ? ik . d?(x)

p dx
or
1
(x|p) = ? exp(ipx/k) , (3.59) 2?k
where we have used the plane wave normalization appropriate for the one- dimensional situation at hand. Introducing this result into (3.58) yields

1 ¸
V (p) = ?
2?k


dx e


?ipx/k


V (x) , (3.60)

that is, V (p) is the Fourier transform of V (x). With this result, the momen- tum representation of the Schro¨dinger equation is therefore

ik d?(p) =
dt

. p2 .
2m

1 ¸
?(p) +
(2?k)n/2


dnp V (p ? p )?(p ) (3.61)

r r r
Here, we have extended the one-dimensional result to n dimensions in a straightforward way. Note that in contrast to the coordinate representation, where the potential energy term is usually local and the kinetic energy term is not, the kinetic energy term is now local, but the potential energy term is not.


Schr¨odinger, Heisenberg and Interaction Pictures

The problem we are normally interested in quantum optics is to determine the expectation values (3.39) of observables at the time t. To do this we typ-
ically start with a system in a well de?ned state at an earlier time and follow the development up to the time t using the Schro¨dinger equation (3.44). It is
possible to follow this evolution in three general ways and in many combina-
tions thereof. The one we use primarily in the ?rst part of this book is called the Schro¨dinger picture and puts all of the time dependence in the state vec- tor. The interaction picture puts only the interaction-energy time dependence into the state vector, putting the unperturbed energy dependence into the operators. The Heisenberg picture puts all of the time dependence into the operators, leaving the state vector stationary in time. In the remainder of this section, we review the way in which these three pictures are tied together.
The Schr¨odinger equation (3.44) can be formally integrated to give
|?(t)) = U (t)|?(0)) , (3.62)
where the evolution operator U (t) for a time-independent Hamiltonian is given by
U (t) = exp(?iHt/k) . (3.63)
Substituting (3.62) into the expectation value (3.39) we obtain
(O) = (?(0)|U †(t)O(0)U (t)|?(0)) . (3.64)

We can also ?nd this same value if we can determine the time dependent operator.
O(t) = U †(t)O(0)U (t) . (3.65)
As Heisenberg ?rst showed, it is possible to follow the time evolution of the quantum mechanical operators. In fact we can obtain their equations of motion by di?erentiating (3.65). This gives


d dU †

?O dU



i.e.,

dt O(t) =

dt OU + U † ?t U + U †O dt ,

d i

?O

dt O(t) = k [H, O] + U

U , (3.66)
?t

where

[H, O] ? HO ? OH (3.67)

is the commutator of H with O. In deriving (3.66), we have used the fact that H commutes with U , which follows from the fact that U is a function of H only. The ?O/?t accounts for any explicit time dependence of the Schro¨dinger operator O.


In general when the system evolution is determined by integrating equa- tions of motion for the observable operators, we say the Heisenberg picture is being used. When the evolution is determined by integrating the Schro¨dinger equation, we say that the Schro¨dinger picture is being used. In either case, (3.64) shows that we get the same answers. You ask, why use one picture instead of the other? The answer is simply, use the picture that makes your life easier. Typically the insights obtained with one di?er somewhat from the other, but you get the same answer with either. Traditionally the Schro¨dinger picture is the ?rst one taught to students and many people feel more com- fortable with it. Much of this book is carried out in the Schro¨dinger picture.
On the other hand, the Heisenberg picture is a “natural” picture in the sense that the observables (electric ?elds, dipole moment, etc.) are time-
dependent, exactly as in classical physics. As a result, their equations of motion usually have the same form as in the classical case, although they are operator equations, which modi?es the way one can integrate and use them.
Another aspect is that in the Schro¨dinger picture, one has to ?nd |?(t))(or its generalization the density operator ?) before computing the desired expec- tation values. Since |?(t)) contains all possible knowledge about the system,
you have to solve the complete problem, which may be more than you need.
In many cases, you only want to know one or a few observables of the system. The Heisenberg picture allows you to concentrate on precisely those observ- ables, and with some luck, you may not have to solve the whole problem to get the desired answers.
In discussing (3.13, 3.22), we hinted at another way of following the time
dependence, namely, we put only the time dependence due to the interaction energy into the Cn(t), while the time dependence of the total Hamiltonian is contained in the cn(t). The state vector of (3.37) is the Schr¨odinger-picture
state vector, while the state vector
|?I (t)) = . Cn(t)|n) (3.68)
n

is said to be the interaction-picture state vector. The thought behind using the interaction picture is to take advantage of the fact that we often face situations where we already know the solutions of the problem in the absence of the interaction.
More formally, to eliminate the known part of the problem, we substitute the state vector



where

|?S (t)) = U0(t)|?I (t)) , (3.69)

U0(t) = exp(?iH0t/k) (3.70)

into the Schro¨dinger equation (3.44). We include the subscript S in (3.69) to remind ourselves that |?S(t)) is the Schr¨odinger-picture state vector. We
?nd

d i
dt |?I (t)) = ? k VI (t)|?I (t)) , (3.71)
where we have de?ned the interaction-picture interaction energy
VI (t) = U †(t)VS U0(t) (3.72)
and put a subscript S on the RHS to remind ourselves that VS is in the Schr¨odinger picture. From (3.69, 3.39), we also immediately ?nd that the
ex-ceptation value of an operator O in the interaction picture is given by
(O(t)) = (?I (t)|OI (t)|?I (t)) , (3.73)


where


OI (t) = U †(t)OS U (t)0 . (3.74)

Note that since we know the solution of the unperturbed problem, OI (t) is
already known. Comparing the equation of motion (3.71) achieved with the
original Schro¨dinger equation (3.44), we see that we have achieved our goal, namely, that we have eliminated the part of the problem whose solution we al- ready knew. We see in Chap. 4 that the interaction picture (or more precisely, an interaction picture) is particularly helpful in visualizing the response of a two-level atom to light.