Physical examples of operators

QUANTUM OPERATORS
later algebraic convenience:
A, eλB =
A,
∞
(λB)n
n!
n=0
=
∞
λn
[ A, B n ]
n!
n=0
=
∞
λn
nB n−1 [ A, B ] , using the earlier result,
n!
n=0
=
∞
λn
nB n−1 [ A, B ]
n!
n=1
=λ
∞
λm B m
[ A, B ] , writing m = n − 1,
m!
m=0
= λeλB [ A, B ] .
Now consider the derivative with respect to λ of the function
f(λ) = eλA eλB e−λ(A+B) .
In the following calculation we use the fact that the derivative of eλC is CeλC ; this is the
same as eλC C, since any two functions of the same operator commute. Differentiating the
three-factor product gives
df
= eλA AeλB e−λ(A+B) + eλA eλB Be−λ(A+B) + eλA eλB (−A − B)e−λ(A+B)
dλ
= eλA (eλB A + λeλB [ A, B ] )e−λ(A+B) + eλA eλB Be−λ(A+B)
− eλA eλB Ae−λ(A+B) − eλA eλB Be−λ(A+B)
= eλA λeλB [ A, B ] e−λ(A+B)
= λ [ A, B ] f(λ).
In the second line we have used the result obtained above to replace AeλB , and in the last
line have used the fact that [ A, B ] commutes with each of A and B, and hence with any
function of them.
Integrating this scalar differential equation with respect to λ and noting that f(0) = 1,
we obtain
1 2
ln f = 12 λ2 [ A, B ] ⇒ eλA eλB e−λ(A+B) = f(λ) = e 2 λ [ A,B ] .
Finally, post-multiplying both sides of the equation by eλ(A+B) and setting λ = 1 yields
1
eA eB = e 2 [ A,B ]+A+B . 19.2 Physical examples of operators
We now turn to considering some of the specific linear operators that play a
part in the description of physical systems. In particular, we will examine the
properties of some of those that appear in the quantum-mechanical description
of the physical world.
As stated earlier, the operators corresponding to physical observables are restricted to Hermitian operators (which have real eigenvalues) as this ensures the
reality of predicted values for experimentally measured quantities. The two basic
656
19.2 PHYSICAL EXAMPLES OF OPERATORS
quantum-mechanical operators are those corresponding to position r and momentum p. One prescription for making the transition from classical to quantum
mechanics is to express classical quantities in terms of these two variables in
Cartesian coordinates and then make the component by component substitutions
r → multiplicative operator r
and p → differential operator − i∇.
(19.22)
This generates the quantum operators corresponding to the classical quantities.
For the sake of completeness, we should add that if the classical quantity contains
a product of factors whose corresponding operators A and B do not commute,
then the operator 12 (AB + BA) is to be substituted for the product.
The substitutions (19.22) invoke operators that are closely connected with the
two that we considered at the start of the previous subsection, namely x and
∂/∂x. One, x, corresponds exactly to the x-component of the prescribed quantum
position operator; the other, however, has been multiplied by the imaginary
constant −i, where is the Planck constant divided by 2π. This has the (subtle)
effect of converting the differential operator into the x-component of an Hermitian
operator; this is easily verified using integration by parts to show that it satisfies
equation (17.16). Without the extra imaginary factor (which changes sign under
complex conjugation) the two sides of the equation differ by a minus sign.
Making the differential operator Hermitian does not change in any essential
way its commutation properties, and the commutation relation of the two basic
quantum operators reads
∂
[ px , x ] = −i , x = −i.
(19.23)
∂x
Corresponding results hold when x is replaced, in both operators, by y or z.
However, it should be noted that if different Cartesian coordinates appear in the
two operators then the operators commute, i.e.
[ px , y ] = [ px , z ] = py , x = py , z = [ pz , x ] = [ pz , y ] = 0.
(19.24)
As an illustration of the substitution rules, we now construct the Hamiltonian
(the quantum-mechanical energy operator) H for a particle of mass m moving
in a potential V (x, y, z) when it has one of its allowed energy values, i.e its
energy is En , where H|ψn = En |ψn . This latter equation when expressed in a
particular coordinate system is the Schrödinger equation for the particle. In terms
of position and momentum, the total classical energy of the particle is given by
E=
p2x + p2y + p2z
p2
+ V (x, y, z) =
+ V (x, y, z).
2m
2m
Substituting −i∂/∂x for px (and similarly for py and pz ) in the first term on the
657
QUANTUM OPERATORS
RHS gives
(−i)2 ∂ ∂
(−i)2 ∂ ∂
(−i)2 ∂ ∂
+
+
.
2m ∂x ∂x
2m ∂y ∂y
2m ∂z ∂z
The potential energy V , being a function of position only, becomes a purely
multiplicative operator, thus creating the full expression for the Hamiltonian,
2
2
∂
∂2
∂2
H =−
+
+
+ V (x, y, z),
2m ∂x2
∂y 2
∂z 2
and giving the corresponding Schrödinger equation as
2 ∂ 2 ψn
∂ 2 ψn
∂ 2 ψn
+
+
+ V (x, y, z)ψn = En ψn .
Hψn = −
2
2
2
2m
∂x
∂y
∂z
We are not so much concerned in this section with solving such differential
equations, but with the commutation properties of the operators from which they
are constructed. To this end, we now turn our attention to the topic of angular
momentum, the operators for which can be constructed in a straightforward
manner from the two basic sets.
19.2.1 Angular momentum operators
As required by the substitution rules, we start by expressing angular momentum
in terms of the classical quantities r and p, namely L = r × p with Cartesian
components
Lz = xpy − ypx ,
Lx = ypz − zpy ,
Ly = zpx − xpz .
Making the substitutions (19.22) yields as the corresponding quantum-mechanical
operators
∂
∂
−y
,
Lz = −i x
∂y
∂x
∂
∂
−z
,
(19.25)
Lx = −i y
∂z
∂y
∂
∂
−x
.
Ly = −i z
∂x
∂z
It should be noted that for xpy , say, x and ∂/∂y commute, and there is no
ambiguity about the way it is to be carried into its quantum form. Further, since
the operators corresponding to each of its factors commute and are Hermitian,
the operator corresponding to the product is Hermitian. This was shown directly
for matrices in exercise 8.7, and can be verified using equation (17.16).
The first question that arises is whether or not these three operators commute.
658
19.2 PHYSICAL EXAMPLES OF OPERATORS
Consider first
∂
∂
∂
∂
−z
−x
Lx Ly = −2 y
z
∂z
∂y
∂x
∂z
∂2
∂2
∂2
∂2
∂
2
+ yz
− yx 2 − z 2
+ zx
= − y
.
∂x
∂z∂x
∂z
∂y∂x
∂y∂z
Now consider
∂
∂
∂
∂
−x
−z
y
Ly Lx = −2 z
∂x
∂z
∂z
∂y
2
2
∂2
∂2
∂
∂
∂
− z2
− xy 2 + x
+ xz
= −2 zy
.
∂x∂z
∂x∂y
∂z
∂y
∂z∂y
These two expressions are not the same. The difference between them, i.e. the
commutator of Lx and Ly , is given by
∂
∂
Lx , Ly = Lx Ly − Ly Lx = 2 x
−y
= iLz .
∂y
∂x
(19.26)
This, and two similar results obtained by permutting x, y and z cyclically,
summarise the commutation relationships between the quantum operators corresponding to the three Cartesian components of angular momentum:
Lx , Ly = iLz ,
Ly , Lz = iLx ,
(19.27)
[ Lz , Lx ] = iLy .
As well as its separate components of angular momentum, the total angular
momentum associated with a particular state |ψ is a physical quantity of interest.
This is measured by the operator corresponding to the sum of squares of its
components,
L2 = L2x + L2y + L2z .
(19.28)
This is an Hermitian operator, as each term in it is the product of two Hermitian
operators that (trivially) commute. It might seem natural to want to ‘take the
square root’ of this operator, but such a process is undefined and we will not
pursue the matter.
We next show that, although no two of its components commute, the total
angular momentum operator does commute with each of its components. In the
proof we use some of the properties (19.17) to (19.20) and result (19.27). We begin
659
QUANTUM OPERATORS
with
L2 , Lz = L2x + L2y + L2z , Lz
= Lx [ Lx , Lz ] + [ Lx , Lz ] Lx
+ Ly Ly , Lz + Ly , Lz Ly + L2z , Lz
= Lx (−i)Ly + (−i)Ly Lx + Ly (i)Lx + (i)Lx Ly + 0
= 0.
Thus operators L2 and Lz commute and, continuing in the same way, it can be
shown that
2
(19.29)
L , Lx = L2 , Ly = L2 , Lz = 0.
Eigenvalues of the angular momentum operators
We will now use the commutation relations for L2 and its components to find
the eigenvalues of L2 and Lz , without reference to any specific wavefunction. In
other words, the eigenvalues of the operators follow from the structure of their
commutators. There is nothing particular about Lz , and Lx or Ly could equally
well have been chosen, though, in general, it is not possible to find states that are
simultaneously eigenstates of two or more of Lx , Ly and Lz .
To help with the calculation, it is convenient to define the two operators
U ≡ Lx + iLy
and D ≡ Lx − iLy .
These operators are not Hermitian; they are in fact Hermitian conjugates, in that
U † = D and D† = U, but they do not represent measurable physical quantities.
We first note their multiplication and commutation properties:
UD = (Lx + iLy )(Lx − iLy ) = L2x + L2y + i Ly , Lx
= L2 − L2z + Lz ,
DU = (Lx − iLy )(Lx + iLy ) = L2x + L2y − i Ly , Lx
= L2 − L2z − Lz ,
[ Lz , U ] = [ Lz , Lx ] + i Lz , Ly = iLy + Lx = U,
[ Lz , D ] = [ Lz , Lx ] − i Lz , Ly = iLy − Lx = −D.
(19.30)
(19.31)
(19.32)
(19.33)
In the same way as was shown for matrices, it can be demonstrated that if two
operators commute they have a common set of eigenstates. Since L2 and Lz
commute they possess such a set; let one of the set be |ψ with
L2 |ψ = a|ψ
and Lz |ψ = b|ψ.
Now consider the state |ψ = U|ψ and the actions of L2 and Lz upon it.
660
19.2 PHYSICAL EXAMPLES OF OPERATORS
Consider first L2 |ψ , recalling that L2 commutes with both Lx and Ly and hence
with U:
L2 |ψ = L2 U|ψ = UL2 |ψ = Ua|ψ = aU|ψ = a|ψ .
Thus, |ψ is also an eigenstate of L2 , corresponding to the same eigenvalue as
|ψ. Now consider the action of Lz :
Lz |ψ = Lz U|ψ
= (ULz + U)|ψ, using [ Lz , U ] = U,
= Ub|ψ + U|ψ
= (b + )U|ψ
= (b + )|ψ .
Thus, |ψ is also an eigenstate of Lz , but with eigenvalue b + .
In summary, the effect of U acting upon |ψ is to produce a new state that
has the same eigenvalue for L2 and is still an eigenstate of Lz , though with that
eigenvalue increased by . An exactly analogous calculation shows that the effect
of D acting upon |ψ is to produce another new state, one that also has the same
eigenvalue for L2 and is also still an eigenstate of Lz , though with the eigenvalue
decreased by in this case. For these reasons, U and D are usually known as
ladder operators.
It is clear that, by starting from any arbitrary eigenstate and repeatedly applying
either U or D, we could generate a series of eigenstates, all of which have the
eigenvalue a for L2 , but increment in their Lz eigenvalues by ±. However, we
also have the physical requirement that, for real values of the z-component, its
square cannot exceed the square of the total angular momentum, i.e. b2 ≤ a. Thus
b has a maximum value c that satisfies
c2 ≤ a but (c + )2 > a;
let the corresponding eigenstate be |ψu with Lz |ψu = c|ψu . Now it is still true
that
Lz U|ψu = (c + )U|ψu ,
and, to make this compatible with the physical constraint, we must have that
U|ψu is the zero ket vector | ∅ . Now, using result (19.31), we have
DU|ψu = (L2 − L2z − Lz )|ψu ,
⇒
0| ∅ = D| ∅ = (a2 − c2 − c)|ψu ,
⇒
a = c(c + ).
This gives the relationship between a and c. We now establish the possible forms
for c.
If we start with eigenstate |ψu , which has the highest eigenvalue c for Lz , and
661
QUANTUM OPERATORS
operate repeatedly on it with the (down) ladder operator D, we will generate a
state |ψd which, whilst still an eigenstate of L2 with eigenvalue a, has the lowest
physically possible value, d say, for the eigenvalue of Lz . If this happens after n
operations we will have that d = c − n and
Lz |ψd = (c − n)|ψd .
Arguing in the same way as previously that D|ψd must be an unphysical ket
vector, we conclude that
0| ∅ = U| ∅ = UD|ψd = (L2 − L2z + Lz )|ψd , using (19.30),
= [ a − (c − n)2 + (c − n) ]|ψd ⇒
a = (c − n)2 − (c − n).
Equating the two results for a gives
c2 + c = c2 − 2cn + n2 2 − c + n2 ,
2c(n + 1) = n(n + 1),
c = 12 n.
Since n is necessarily integral, c is an integer multiple of 12 . This result is valid
irrespective of which eigenstate |ψ we started with, though the actual value of
the integer n depends on |ψu and hence upon |ψ.
Denoting 12 n by we can say that the possible eigenvalues of the operator
Lz , and hence the possible results of a measurement of the z-component of the
angular momentum of a system, are given by
,
( − 1),
( − 2),
... ,
−.
The value of a for all 2 + 1 of the corresponding states,
|ψu ,
D|ψu ,
D2 |ψu ,
... ,
D2 |ψu ,
is ( + 1)2 .
The similarity of form between this eigenvalue and that appearing in Legendre’s
equation is not an accident. It is intimately connected with the facts (i) that L2
is a measure of the rotational kinetic energy of a particle in a system centred
on the origin, and (ii) that in spherical polar coordinates L2 has the same form
as the angle-dependent part of ∇2 , which, as we have seen, is itself proportional
to the quantum-mechanical kinetic energy operator. Legendre’s equation and
the associated Legendre equation arise naturally when ∇2 ψ = f(r) is solved in
spherical polar coordinates using the method of separation of variables discussed
in chapter 21.
The derivation of the eigenvalues ( + 1)2 and m, with − ≤ m ≤ , depends
only on the commutation relationships between the corresponding operators. Any
662
19.2 PHYSICAL EXAMPLES OF OPERATORS
other set of four operators with the same commutation structure would result
in the same eigenvalue spectrum. In fact, quantum mechanically, orbital angular
momentum is restricted to cases in which n is even and so is an integer; this
is in accord with the requirement placed on if solutions to ∇2 ψ = f(r) that are
finite on the polar axis are to be obtained. The non-classical notion of internal
angular momentum (spin) for a particle provides a set of operators that are able
to take both integral and half-integral multiples of as their eigenvalues.
We have already seen that, for a state |, m that has a z-component of
angular momentum m, the state U|, m is one with its z-component of angular
momentum equal to (m + 1). But the new state ket vector so produced is not
necessarily normalised so as to make , m + 1 | , m + 1 = 1. We will conclude
this discussion of angular momentum by calculating the coefficients µm and νm in
the equations
U|, m = µm |, m + 1
and D|, m = νm |, m − 1
on the basis that , r | , r = 1 for all and r.
To do so, we consider the inner product I = , m |DU| , m, evaluated in two
different ways. We have already noted that U and D are Hermitian conjugates
and so I can be written as
I = , m |U † U| , m = µ∗m , m | , mµm = |µm |2 .
But, using equation (19.31), it can also be expressed as
I = , m |L2 − L2z − Lz | , m
= , m |( + 1)2 − m2 2 − m2 | , m
= [ ( + 1)2 − m2 2 − m2 ] , m | , m
= [ ( + 1) − m(m + 1) ] 2 .
Thus we are required to have
|µm |2 = [ ( + 1) − m(m + 1) ] 2 ,
but can choose that all µm are real and non-negative (recall that |m| ≤ ). A
similar calculation can be used to calculate νm . The results are summarised in the
equations
( + 1) − m(m + 1) | , m + 1,
D| , m = ( + 1) − m(m − 1) | , m − 1.
U| , m =
It can easily be checked that U|, = | ∅ = D|, −.
663
(19.34)
(19.35)
QUANTUM OPERATORS
19.2.2 Uncertainty principles
The next topic we explore is the quantitative consequences of a non-zero commutator for two quantum (Hermitian) operators that correspond to physical
variables.
As previously noted, the expectation value in a state |ψ of the physical quantity
A corresponding to the operator A is E[A] = ψ| A |ψ. Any one measurement of
A can only yield one of the eigenvalues of A. But if repeated measurements could
be made on a large number of identical systems, a discrete or continuous range
of values would be obtained. It is a natural extension of normal data analysis
to measure the uncertainty in the value of A by the observed variance in the
measured values of A, denoted by (∆A)2 and calculated as the average value of
(A − E[A])2 . The expected value of this variance for the state |ψ is given by
ψ |(A − E[A] )2 |ψ.
We now give a mathematical proof that there is a theoretical lower limit for
the product of the uncertainties in any two physical quantities, and we start by
proving a result similar to the Schwarz inequality. Let |u and |v be any two state
vectors and let λ be any real scalar. Then consider the vector |w = |u + λ|v and,
in particular, note that
0 ≤ w | w = u | u + λ(u | v + v | u) + λ2 v | v.
This is a quadratic inequality in λ and therefore the quadratic equation formed
by equating the RHS to zero must have no real roots. The coefficient of λ is
(u | v + v | u) = 2 Re u | v and its square is thus ≥ 0. The condition for no real
roots of the quadratic is therefore
0 ≤ (u | v + v | u)2 ≤ 4u | u v | v.
(19.36)
This result will now be applied to state vectors constructed from |ψ, the state
vector of the particular system for which we wish to establish a relationship between the uncertainties in the two physical variables corresponding to (Hermitian)
operators A and B. We take
|u = (A − E[A]) | ψ
and
|v = i(B − E[B]) | ψ.
(19.37)
Then
u | u = ψ |(A − E[A] )2 |ψ = (∆A)2 ,
v | v = ψ |(B − E[B] )2 |ψ = (∆B)2 .
Further,
u | v = ψ | (A − E[A])i(B − E[B]) | ψ
= iψ | AB | ψ − iE[A]ψ | B | ψ − iE[B]ψ | A | ψ + iE[A]E[B]ψ | ψ
= iψ | AB | ψ − iE[A]E[B].
664
19.2 PHYSICAL EXAMPLES OF OPERATORS
In the second line, we have moved expectation values, which are purely numbers,
out of the inner products and used the normalisation condition ψ|ψ = 1.
Similarly
v | u = −iψ | BA | ψ + iE[A]E[B].
Adding these two results gives
u | v + v | u = iψ | AB − BA |ψ,
and substitution into (19.36) yields
0 ≤ (iψ | AB − BA |ψ)2 ≤ 4(∆A)2 (∆B)2
At first sight, the middle term of this inequality might appear to be negative, but
this is not so. Since A and B are Hermitian, AB −BA is anti-Hermitian, as is easily
demonstrated. Since i is also anti-Hermitian, the quantity in the parentheses in
the middle term is real and its square non-negative. Rearranging the equation and
expressing it in terms of the commutator of A and B gives the generalised form
of the Uncertainty Principle. For any particular state |ψ of a system, this provides
the quantitative relationship between the minimum value that the product of the
uncertainties in A and B can have and the expectation value, in that state, of
their commutator,
(∆A)2 (∆B)2 ≥ 14 | ψ | [ A, B ] | ψ |2 .
(19.38)
Immediate observations include the following:
(i) If A and B commute there is no absolute restriction on the accuracy with
which the corresponding physical quantities may be known. That is not to
say that ∆A and ∆B will always be zero, only that they may be.
(ii) If the commutator of A and B is a constant, k = 0, then the RHS of
equation (19.38) is necessarily equal to 14 |k|2 , whatever the form of |ψ,
and it is not possible to have ∆A = ∆B = 0.
(iii) Since the RHS depends upon |ψ, it is possible, even for two operators
that do not commute, for the lower limit of (∆A)2 (∆B)2 to be zero. This
will occur if the commutator [ A, B ] is itself an operator whose expectation
value in the particular state |ψ happens to be zero.
To illustrate the third of these, we might consider the components of angular
momentum discussed in the previous subsection. There, in equation (19.27), we
found that the commutator of the operators corresponding to the x- and ycomponents of angular momentum is non-zero; in fact, it has the value iLz .
This means that if the state |ψ of a system happened to be such that ψ|Lz |ψ = 0,
as it would if, for example, it were the eigenstate of Lz , |ψ = |, 0, then there
would be no fundamental reason why the physical values of both Lx and Ly
should not be known exactly. Indeed, if the state were spherically symmetric, and
665
QUANTUM OPERATORS
hence formally an eigenstate of L2 with = 0, all three components of angular
momentum could be (and are) known to be zero.
Working in one dimension, show that the minimum value of the product ∆px × ∆x for
a particle is 12 . Find the form of the wavefunction that attains this minimum value for a
particle whose expectation values for position and momentum are x̄ and p̄, respectively.
We have already seen, in (19.23) that the commutator of px and x is −i, a constant.
Therefore, irrespective of the actual form of |ψ, the RHS of (19.38) is 14 2 (see observation
(ii) above). Thus, since all quantities are positive, taking the square roots of both sides of
the equation shows directly that
∆px × ∆x ≥ 12 .
Returning to the derivation of the Uncertainty Principle, we see that the inequality becomes
an equality only when
(u | v + v | u)2 = 4u | u v | v.
The RHS of this equality has the value 4||u||2 ||v||2 and so, by virtue of Schwarz’s inequality,
we have
4u2 v2 = (u | v + v | u)2
≤ (|u | v| + |v | u|)2
≤ (u v|| + v u)2
= 4u2 v2 .
Since the LHS is less than or equal to something that has the same value as itself, all of
the inequalities are, in fact, equalities. Thus u|v = u v, showing that |u and |v are
parallel vectors, i.e. |u = µ|v for some scalar µ.
We now transform this condition into a constraint that the wavefunction ψ = ψ(x)
must satisfy. Recalling the definitions (19.37) of |u and |v in terms of |ψ, we have
d
−i
− p̄ ψ = µi(x − x̄)ψ,
dx
1
dψ
+ [ µ(x − x̄) − ip̄ ]ψ = 0.
dx
ip̄x
µ(x − x̄)2
, giving
−
The IF for this equation is exp
2
d
µ(x − x̄)2
ip̄x
ψ exp
= 0,
−
dx
2
which, in turn, leads to
µ(x − x̄)2
ip̄x
ψ(x) = A exp −
exp
.
2
From this it is apparent that the minimum uncertainty product ∆px × ∆x is obtained when
the probability density |ψ(x)|2 has the form of a Gaussian distribution centred on x̄. The
value of µ is not fixed by this consideration and it could be anything (positive); a large
value for µ would yield a small value for ∆x but a correspondingly large one for ∆px . 666
19.2 PHYSICAL EXAMPLES OF OPERATORS
19.2.3 Annihilation and creation operators
As a final illustration of the use of operator methods in physics we consider their
application to the quantum mechanics of a simple harmonic oscillator (s.h.o.).
Although we will start with the conventional description of a one-dimensional
oscillator, using its position and momentum, we will recast the description in
terms of two operators and their commutator and show that many important
conclusions can be reached from studying these alone.
The Hamiltonian for a particle of mass m with momentum p moving in a
one-dimensional parabolic potential V (x) = 12 kx2 is
p2
p2
1
1
+ kx2 =
+ mω 2 x2 ,
2m 2
2m 2
H=
where its classical frequency of oscillation ω is given by ω 2 = k/m. We recall that
the corresponding operators, p and x, do not commute and that [ p, x ] = −i.
In analogy with the ladder operators used when discussing angular momentum,
we define two new operators:
A≡
ip
mω
x+ √
2
2mω
and A† ≡
ip
mω
.
x− √
2
2mω
(19.39)
†
Since both x and p are Hermitian, A and A are Hermitian conjugates, though
neither is Hermitian and they do not represent physical quantities that can be
measured.
Now consider the two products A† A and AA† :
mω 2 ipx ixp
p2
H
i
x −
+
+
=
− [ p, x ] =
2
2
2
2mω
ω
2
p2
H
i
mω 2 ipx ixp
x +
−
+
=
+ [ p, x ] =
AA† =
2
2
2
2mω
ω
2
From these it follows that
A† A =
and that
H = 12 ω(A† A + AA† )
A, A = .
[ H, A ] =
=
Similarly,
(19.40)
†
Further,
H
− ,
ω
2
H
+ .
ω
2
†
H, A
=
1
(19.41)
†
†
2 ω(A A + AA ), A
† 1
†
†
2ω A 0 + A , A A + A A , A
1
2 ω(−A − A) = −ωA.
†
= ωA
+ 0 A†
(19.42)
(19.43).
Before we apply these relationships to the question of the energy spectrum of
the s.h.o., we need to prove one further result. This is that if B is an Hermitian
operator then ψ | B 2 | ψ ≥ 0 for any |ψ. The proof, which involves introducing
667
QUANTUM OPERATORS
an arbitrary complete set of orthonormal base states |φi and using equation
(19.11), is as follows:
ψ | B 2 | ψ = ψ | B × 1 × B | ψ ψ | B |φi φi | B |ψ
=
=
=
i
∗
ψ | B |φi φi | B |ψ∗
i
∗
ψ | B |φi ψ | B † |φi i
=
ψ | B |φi ψ | B |φi ∗ ,
since B is Hermitian,
i
=
| ψ | B |φi |2 ≥ 0.
i
We note, for future reference, that the Hamiltonian H for the s.h.o. is the sum of
two terms each of this form and therefore conclude that ψ|H|ψ ≥ 0 for all |ψ.
The energy spectrum of the simple harmonic oscillator
Let the normalised ket vector |n (or |En ) denote the nth energy state of the s.h.o.
with energy En . Then it must be an eigenstate of the (Hermitian) Hamiltonian H
and satisfy
H|n = En |n with m|n = δmn .
Now consider the state A|n and the effect of H upon it:
HA|n = AH|n − ωA|n,
using (19.42),
= AEn |n − ωA|n
= (En − ω)A|n.
Thus A|n is an eigenstate of H corresponding to energy En − ω and must be
some multiple of the normalised ket vector |En − ω, i.e.
A| En ≡ A|n = cn |En − ω,
where cn is not necessarily of unit modulus. Clearly, A is an operator that
generates a new state that is lower in energy by ω; it can thus be compared to
the operator D, which has a similar effect in the context of the z-component of
angular momentum. Because it possesses the property of reducing the energy of
the state by ω, which, as we will see, is one quantum of excitation energy for the
oscillator, the operator A is called an annihilation operator. Repeated application
of A, m times say, will produce a state whose energy is mω lower than that of
the original:
Am |En = cn cn−1 · · · cn−m+1 |En − mω.
668
(19.44)
19.2 PHYSICAL EXAMPLES OF OPERATORS
In a similar way it can be shown that A† parallels the operator U of our angular
momentum discussion and creates an additional quantum of energy each time it
is applied:
(A† )m |En = dn dn+1 · · · dn+m−1 |En + mω.
(19.45)
It is therefore known as a creation operator.
As noted earlier, the expectation value of the oscillator’s energy operator
ψ|H|ψ must be non-negative, and therefore it must have a lowest value. Let
this be E0 , with corresponding eigenstate |0. Since the energy-lowering property
of A applies to any eigenstate of H, in order to avoid a contradiction we must
have that A|0 = | ∅ . It then follows from (19.40) that
H|0 = 12 ω(A† A + AA† )|0
= 12 ωA† A|0 + 12 ω(A† A + )|0,
=0+0+
using (19.41),
1
2 ω|0.
(19.46)
This shows that the commutator structure of the operators and the form of the
Hamiltonian imply that the lowest energy (its ground-state energy) is 12 ω; this
is a result that has been derived without explicit reference to the corresponding
wavefunction. This non-zero lowest value for the energy, known as the zero-point
energy of the oscillator, and the discrete values for the allowed energy states
are quantum-mechanical in origin; classically such an oscillator could have any
non-negative energy, including zero.
Working back from this result, we see that the energy levels of the s.h.o. are
1
3
5
1
2 ω, 2 ω, 2 ω, . . . , (m + 2 )ω, . . . , and that the corresponding (unnormalised)
ket vectors can be written as
|0,
A† |0,
(A† )2 |0,
... ,
(A† )m |0,
... .
This notation, and elaborations of it, are often used in the quantum treatment of
classical fields such as the electromagnetic field. Thus, as the reader should verify,
A(A† )3 A2 A† A(A† )4 |0 is a state with energy 92 ω, whilst A(A† )3 A5 A† A(A† )4 |0 is
not a physical state at all.
The normalisation of the eigenstates
In order to make quantitative calculations using the previous results we need to
establish the values of the cn and dn that appear in equations (19.44) and (19.45).
To do this, we first establish the operator recurrence relation
Am (A† )m = Am−1 (A† )m A + mAm−1 (A† )m−1 .
669
(19.47)
QUANTUM OPERATORS
The proof, which makes repeated use of A, A† = , is as follows:
Am (A† )m = Am−1 AA† (A† )m−1
= Am−1 (A† A + )(A† )m−1
= Am−1 A† A(A† )m−1 + Am−1 (A† )m−1
= Am−1 A† (A† A + )(A† )m−2 + Am−1 (A† )m−1
= Am−1 (A† )2 A(A† )m−2 + Am−1 A† (A† )m−2 + Am−1 (A† )m−1
= Am−1 (A† )2 (A† A + )(A† )m−3 + 2Am−1 (A† )m−1
..
.
= Am−1 (A† )m A + mAm−1 (A† )m−1 .
Now we take the expectation values in the ground state |0 of both sides of
this operator equation and note that the first term on the RHS is zero since it
contains the term A|0. The non-vanishing terms are
0 | Am (A† )m | 0 = m0 | Am−1 (A† )m−1 | 0.
The LHS is the square of the norm of (A† )m |0, and, from equation (19.45), it is
equal to
|d0 |2 |d1 |2 · · · |dm−1 |2 0 | 0.
Similarly, the RHS is equal to
m |d0 |2 |d1 |2 · · · |dm−2 |2 0 | 0.
√
It follows that |dm−1 |2 = m and, taking all coefficients as real, dm = (m + 1).
Thus the correctly normalised state of energy (n + 12 ), obtained by repeated
application of A† to the ground state, is given by
|n =
(A† )n
|0.
(n! n )1/2
To evaluate the cn , we note that, from the commutator of A and A† ,
A, A† |n = AA† |n − A† A|n
|n = (n + 1) A |n + 1 − cn A† |n − 1
√
= (n + 1) cn+1 |n − cn n |n,
√
= (n + 1) cn+1 − cn n,
√
which has the obvious solution cn = n. To summarise:
√
cn = n
and
dn = (n + 1).
(19.48)
(19.49)
We end this chapter with another worked example. This one illustrates how
the operator formalism that we have developed can be used to obtain results
670