Download MATH3385/5385. Quantum Mechanics. Handout # 5: Eigenstates of

MATH3385/5385. Quantum Mechanics. Handout # 5: Eigenstates of the
Hamiltonian and the One-dimensional Stationary Schrödinger Equation
Time-independent potentials
In the case of a structureless particle subject to a potential V (r, t) (possibly time-dependent!)
we have seen in Handout # 3 that the Hamiltonian can be represented by the differential
operator:
2
b = − ~ ∇2 + V (r, t)
H
2m
(i.e. from the replacement of the momentum vector p by the operator −i~∇ in the classical
Hamiltonian function H(r, p; t)). If now the potential V is actually time-independent, i.e.
V (r, t) = V (r) only depending on the position vector r, then we can look for solutions of
the Schrödinger (SE) equation in the separated form:
ψ(r, t) = χ(t)φ(r)
which inserted in the SE leads to
i~
1 dχ
1b
= Hφ
=E
χ dt
φ
where the l.h.s. of the equation depends only on t, whereas the r.h.s. depends only on r.
Thus, both sides must be equal to a constant E, and we can thus separate the equations
into
b
Hφ
= Eφ
dχ
= Eχ
i~
dt
the time − independent SE
(1a)
(1b)
The second equation (1b) can be readily solved to give:
χ(t) = e−iEt/~χ(0)
leading to a simple time-evolution1 . The first equation (1a), i.e. the stationary or timeindependent Schrödinger equation, which for a single structureless particle reads:
−
~2 2
∇ φ + V (r)φ = Eφ
2m
(2)
has the form of an eigenvalue problem –with eigenvalue E, the energy– of the differenb Solving this eigenvalue (or “spectral”) problem under certain boundary
tial operator H.
conditions will lead to a possible set of values that E can take, i.e. the eigenvalues of the
1
Obviously, the separated solution ψ = χφ is not the general solution of the time-dependent Schrödinger
equation, but only a particular solution: see Handout # 4.
1
operator under the boundary condtions, and this set of values is called the spectrum2 . From
now on we will concentrate, in given problems, in developing the techniques to calculate
the spectrum of the Hamiltonian: once we know this and the corresponding eigenfunctions
φ = φE (r), the quantum mechanical problem is practically solved. All physical quantities
(such as expectation values of observables) can then in principle be calculated.
b † = H,
b the eigenfunctions for different
Since the Hamiltonian operator is Hermitian H
eigenvalues are orthogonal w.r.t. the inner product:
Z
(φE , φE ′ ) = dr φ∗E (r)φE ′ (r) = 0 provided E 6= E ′ .
Recalling further that if φE (r) is a solution of the eigenvalue problem (1a), e.g. (2),
with eigenvalue E, then so is any constant multiple cφE (r). We can encounter then the
following situations:
• the spectrum is discrete: the eigenvalues of the stationary SE take on a discontinuous
set of values En , n = 1, 2, . . . . In this case the corresponding eigenfunctions φn (r) =
φEn (r) can be normalised, i.e.
Z
2
k φE k = dr |φE (r)|2 = 1
The corresponding physical states of the system are called bound states.
• the spectrum is continuous: the eigenvalues can take on a continuous range of values,
say in an interval E0 < E < E1 . In that case the eigenvalues cannot be normalised,
but instead we can choose c to have an orthogonality relation of the form
Z
dr φ∗E (r)φE ′ (r) = δ(E − E ′ )
The corresponding physical states of the system are called scattering states. The
completeness relation in this case can be written as:
Z E1
φE (r)φ∗E (r ′ ) dE = δ(r − r ′ ) .
E0
• the spectrum is partly discrete and partly continuous. In this mixed situation there
are in the system bound states as well as scattering states.
It can happen that we have different (i.e. independent) eigenfunctions for one and the
same eigenvalue E: in that case we say that the corresponding “energy level” is degenerate.
The corresponding eigenspace is then no longer one-dimensional as would be the case if the
eigenvalue were nondegenerate (i.e. when the eigenfunction is unique up to a multiplicative
constant). In the degnerate case we need additional “quantum numbers” to characterise
the eigenstate uniquely. As we have noted in Handout # 4, this means that we need
to look for other physical observables whose operators commute with the Hamiltonian
b and consequently for which we can then impose an additional eigenvalue
operator H,
problem consistent with the one for the Hamiltonian.
2
The name, which has become customary in mathematics, is clearly inspired by physics: it is the set of
values of E that corresponds to the measured spectra of atoms in experiments.
2
Example 1: particle in an infinite potential well
As a first example of a system with a discrete spectrum, i.e. physically representing bound
states, we consider the case of a one-dimensional particle in an infinite potential well given
by:
0
,
−a ≤ x ≤ a
V (x) =
∞
otherwise
In view of the fact that outside the interval −a ≤ x ≤ a the potential is infinite (and
thus we have a potential “barrier” inhibiting the particle from being found anywhere else
than within the mentioned interval) we solve the eigenvalue problem by imposing:
−
~2 d2 φ
= Eφ
2m dx2
φ(x) = 0
,
−a≤x≤a
, x ≤ −a or x ≥ a
implying that the particle has zero probability of being found outside the mentioned interval. So, concentrating on the problem within the interval [−a, a], we note that we have
a simple linear homogenous differential equation with constant coefficients to solve. We
distinguish two cases: the case that E > 0 and the one when E ≤ 0.
E>0
In this case the stationary Schrödinger equation takes the form:
d2 φ
+ κ2 φ = 0
dx2
φ(x) = Aeiκx + Be−iκx ,
⇒
(3)
√
with κ = 2mE/~ (which is real in this case). imposing continuity of the eigenfunction
we have that at the boundaries of the interval, at x = ±a the wave function vanishes:
Aeiκa + Be−iκa = 0
φ(±a) = 0
⇒
Ae−iκa + Beiκa = 0
This coupled set of equations has only a nontrivial solution for the coefficients A and B if
the following condition holds:
e2iκa − e−2iκa = 2i sin(κa) = 0
⇒
2κa = nπ
(n ∈ N) .
Thus, we find the following energy levels:
E = En =
~2 n2 π 2
2m 4a2
associated with the following eigenfunctions:
2iAn sin( nπx
2a )
φn (x) =
2An cos( nπx
2a )
,
,
(4)
n even
n odd
The coefficients An are determined by the normalisation of the eigenfunctions:
Z a
Z a
nπx 2
nπx 2
) dx = 1
resp.
4|An |2
) dx = 1
cos2 (
4|An |2
sin2 (
2a
2a
−a
−a
3
(5)
leading in both cases to |An |2 = 1/(4a) . Thus, we can take for the normalised eigenfunctions:
( 1
√ sin( nπx )
,
n even
2a
a
(6)
φn (x) =
√1 cos( nπx )
,
n odd
2a
a
Exercise:
Check that these eigenfunctions are orthogonal, i.e. that we have:
Z a
φ∗n (x)φm (x) dx = 0
, n 6= m .
−a
E≤0
In this case κ is purely imaginary, and we have instead to introduce the real quantity:
√
−2mE
k≡
~
in which case the stationary Schrödinger equation takes the form:
d2 φ
− k2 φ = 0
dx2
φ(x) = αekx + βe−kx .
⇒
(7)
Again imposing continuity of the eigenfunction at the boundaries of the interval, the wave
function vanishes:
αeka + βe−ka = 0
φ(±a) = 0
⇒
αe−ka + βeka = 0
which only leads to a nontrivial solution for α, β if
e4ka = 1
⇒
k=0.
Thus, in this case only the eigenvalue E = 0 is allowed, from which we conclude that
there are no negative eigenstates! Finally, we have also to reject the eigenvalue E = 0,
because it is easy to see that if k = 0 the above boundary conditions only lead to the case
α = β = 0 which would lead to the trivial function φ(x) = 0 which is to be rejected as an
eigenfunction.
Thus, we can draw the following conclusions about the spectrum for the infinite potential
well:
Eigenvalues:
Eigenfunctions:
En =
~2 n2 π 2
2m 4a2
φn (x) =









,
√1
a
√1
a
(n positive integer)
sin( nπx
2a )
nπx
cos( 2a )
0
4
,
n even
,
n odd
if :
−a < x < a
otherwise
Remark: Since we have only a discrete spectrum in this example, we say that the
particle can only exist in bound states, meaning that the particle is confined to such a
discrete energy state, unless it gets a “kick” of enough energy to jump to the next state.
In Example 1 we have investigated a siuation where the potential is unbounded, namely
V = ∞ at the boundaries of the interval [−a, a]. In that case the derivatives of the solution
φ at the boundaries of the infinite potential well do not play a role. In the examples that
follow now, V (x) is discontinuous but bounded in a neighborhood of the discontinuity. In
that case we have to consider also the boundary values for the derivative φ′ (x). In fact,
from the stationary SE we have:
Z ǫ
Z ǫ 2
dφ
V (x) − E
dφ
d φ
φ(x) dx = 0 as ǫ → 0 ,
dx =
−
=
2
2
dx x=ǫ
dx x=−ǫ
−ǫ ~ /(2m)
−ǫ dx
whenever V (x) is bounded around x = 0 and since φ(x) in the last integrand is continuous,
implying continuity of φ′ (x) at x = 0.
Example 2: Finite one-dimensional well
In example 1, we have seen a situation where the spectrum is discrete leading to bound
states only, whereas in example 2 we have seen the other extreme situation, where the
spectrum is fully continuous leading to scattering states characterised by the reflection
and transmission coefficients. In the last example of this handout we consider a mixture
of both situations, namely the case in which the spectrum consists of both a discrete part
as well as a continuous part. However, the two parts of the spectrum have to be dealt
with separately. We consider again a one-dimensional particle, now in a finite potential
well given by:
0
|x| > a
V (x) =
−V0
−a ≤ x ≤ a
with fixed V0 > 0. Again the analysis of this case proceeds in different steps according to
whether E ≥ V0 or 0 ≤ E < V0 , (it can easily be shown that the value E < −V0 does not
lead to nontrivial physical solutions.
−V0 ≤ E < 0
In this case it is convenient to introduce the wave numbers:
p
√
,
~k = 2m(E + V0 ) ,
~κ = −2mE
pertinent to the regimes |x| > a respectively −a < x < a. The stat. SE in these regimes
read:
φ′′ = −k2 φ
′′
φ
|x| < a ,
2
= κ φ
|x| > a ,
5
and these are readily solved in the various areas as follows

,
x < −a
 αeκx
ikx
−ikx
Ae + Be
,
−a < x < a
φ(x) =

βe−κx
,
x>a
(8)
where we have already taken into account the requirement that the solution should be
bounded as x → ±∞. Imposing φ(x) and φ′ (x) to be continuous at x = ±a leads to the
following matching conditions:
αe−κa
= Ae−ika + Beika
continuity at x = −a :
−κa
καe
= ik Ae−ika − Beika
βe−κa
= Aeika + Be−ika
continuity at x = a :
−κa
−κβe
= ik Aeika − Be−ika
Eliminating α from the first set of equations and β from the second we obtain two different
expressions for the fraction B/A leading to the relation:
ik + κ 2ika 2
B
ik + κ 2ika ik − κ −2ika
=1,
=
e
=
e
⇒
e
A
ik − κ
ik + κ
ik − κ
and thus
ik + κ 2ika
B
=
e
= ±1 ,
(9)
A
ik − κ
where depending on whether we have the + sign or the − sign we will have either one of
the following relations:
(+)
k tan(ka) = κ
resp.
(−) k cot(ka) = −κ ,
(10)
which follow after some algebra. These relations between k and κ actually determine
the allowed values of E (i.e. the eigenvalues of the SE), although these equations being
transcendental we cannot give explicit algebraic expressions for the solutions for the eigenvalues E. However, as we shall show below, there is a nice countable discrete of solutions
of these equations which we can visualise through a graphical proceduere. Then, having
obtained numerical values for the eigenvalues En , and hence also of the corresponding wave
numbers kn and κn , the corresponding eigenfunctions are immediately obtained from the
relations (8) and (9) to yield:
2A cos(k x)
,
(+)
n
n
ikn x
−ikn x
(11)
=
±e
φn = An e
2iAn sin(kn x)
,
(−)
where An may be determined by normalisation (one can absorb the i in the second row
in the An if necessary). Thus, the solutions going with the (+) are even functions, whilst
the ones determined by the (−) equation of the spectrum are odd3 .
To analyse the equations (10) for the spectrum further we note that from the definition
of k and κ we get:
~2 k2 + ~2 κ2 = 2m(E + V0 ) − 2mE = 2mV0
3
⇒
(κa)2 + ξ 2 = γ 2 ≡
2mV0 2
a ,
~2
Note that in this problem all the difficulty stems from the determination of the spectrum through the
equations (10), not so much from the eigenfunctions themselves!
6
where we have introduced as new variable ξ = ka. Using the relations for the spectrum,
we get then:
tan2 ξ
ξ = γ| cos ξ|
,
(+)
2
2
ξ 1+
=γ
⇒
2
cot ξ
ξ = γ| sin ξ|
,
(−)
where (as a consequence of (10) noting that κ/k is positive) the equations for ξ should be
solved subject to the additional conditions that:
(+)
tan(ξ) > 0
,
(−)
tan(ξ) < 0 .
The solutions can be graphically found from the intersection points of the graphs (in the
(ξ, η)-plane) of the line η = ξ/γ and the function η = | cos ξ| respectively η = | sin ξ|, as
indicated in the figure.
1
0.8
0.6
0.4
0.2
0
2
4
6
8
10
12
x
Although we cannot give analytic expressions for the intersection points indicated in
the graph, it is clear that for given value of V0 (which through γ −1 determines the slope
of the line in the figure) we will have a finite number of discrete values for ξ and hence
for E. If γ is large enough and, hence, the slope is very small (indeed the slope is of the
order of ~!) then approximately the lowest eigenvalues correspond to
ξ ≈ nπ/2
⇒
En ≈ −V0 +
(using the definitions of ξ respectively k).
7
π 2 ~2 n2
8ma2
E>0
In this case we will have, as in Example 2, only scattering states that can be characterised through the reflection and transmission coefficients R and T . The solution can be
analysed in a similar manner as the next example. We will, therefore not treat this case
separately here.
Example 3. The quantum harmonic oscillator
The quantum mechanical system of the one-dimensional harmonic oscillator is one of the
simplest nontrivial quantum models. In the classical model the potential is given by
V (x) = 12 kx2 leading to the linear equations of motion
mẍ + kx = 0
i.e. the Newton equation for a particle on a spring (following Hooke’s law). Obviously, the
classical equation is easily solved in trigonometric funcions, taking k = mω 2 . The model
is of key importance, since it represents the basic first step in studying the quantum
mechanics of systems of vibrating particles. In principle any smooth potential V (x) in the
neighborhood of a local minimum x0 can be approximated by a harmonic oscillator, in
view of the Taylor expansion:
1
V (x) = V (x0 ) + V ′′ (x0 )(x − x0 )2 + . . .
2
so any one-dimensional potential behaves approximately as a harmonic oscillators for
low enough energy values. We will study the model here in detail , because it connects
quantum mechanics with another interesting area of mathematics: that of special functions
and orthogonal polynomials.
a) Eigenvalue Problem for the harmonic Oscillator
The classical Hamiltonian of the one-dimensional harmonic oscillator reads
H(x, p) =
p2
1
+ mω 2 x2
2m 2
which is quantised by taking for the quantum Hamiltonian the form:
2
b = pb + 1 mω 2 x
b2
H
2m 2
(12)
leading to the eigenvalue problem (i.e. the stationary Schrödinger equation):
−
~2 d2 φ 1
+ mω 2 x2 φ = Eφ .
2m dx2
2
To analyse this equation it is convenient to go to some dimensionless variables:
r
~
x = λξ
,
λ=
mω
8
(13)
(14)
and rewrite the eigenfunctions as:
1 2
1
φ(x) = √ e− 2 ξ H(ξ) ,
λ
(15)
where the new function H(ξ) should not be confused with the notation for the Hamiltonian!
(It is for historical reasons that we use the symbol H for these functions). Implementing
the change of variables we are led to the equation :
2E
dH
d2 H
+
−1 H =0 ,
(16)
− 2ξ
dξ 2
dξ
~ω
which is a well-known differential equation called the Hermite equation. Note that this is a
linear homogeneous second order equation, but the coefficients are not constant (otherwise
we could readily solve the equation). If the original eigenfunctions are normalised it is
easy to check that through this change of variables the new functions H(ξ) are normalised
as well:
Z ∞
Z ∞
2
2
(17)
|H(ξ)|2 e−ξ dξ = 1 .
|φ(x)| dx = 1
⇒
−∞
−∞
It turns out (cf. Appendix D) that the solutions of the Hermite equation (16) that are of
physical interest are only those for which the coefficient in the last term takes on integer
values, i.e. from which:
2E
− 1 = 2n
,
n∈N,
(18)
~ω
in which case remarkably the solutions are actually polynomials: the so-called Hermite
polynomials. Thus, associated with the values of the energy given by (18) we have polynomial Hn (ξ) of order n solving the equation:
Hn′′ − 2ξHn′ + 2nHn = 0 ,
(19)
in which the primes denote differentiation w.r.t. the variable ξ.
To summarise our findings from the above investigation of the the quantum eigenvalue
problem associated with the harmonic oscillator:
eigenvalues:
E = En = (n + 12 )~ω
, n = 0, 1, 2, . . .
x 2
(
λ)
φ(x) = φn (x) = (2ne n!√πλ)1/2 Hn (x/λ)
−1
2
eigenfunctions:
q
~
, Hn (ξ) are the standard Hermite polynomials
where λ = mω
Thus, we have for the quantum harmonic oscillator a discrete spectrum, where the
eigenvalues En represent the (non-degenerate) energy-levels. Note that the spacing between two adjacent levels is always the same (i.e. they are equidistant), namely given by
quantum ~ω, the lowest energy-level being given by E0 = 21 ~ω. The latter corresponds
to the so-called ground state of the model, and the fact that we have a non-zero energy
for this ground state is referred to as zero-point energy. We mention also that the occurrence of discrete eigenvalues of the form given above is in agreement with the basic
9
assumption made in 1905 by A. Einstein to explain the behaviour of specific heat as a
function of temperature in crystals: the vibrational modes of atoms in the crystal can be
effectively modelled by quantum harmonic oscillators. The normalised eigenstates (given
above, where the Hermite polynomials Hn are described in the next section) form an orthonormal basis of the Hilbert space of the model. The completeness relation follows from
the following property of the Hermite polynomials:
∞
X
1 2
2
Hn (ξ)Hn (η)
√
= δ(ξ − η)e 2 (ξ +η )
n
2
n!
π
n=0
(see Appendix D).
10
(20)
0.3
0.5
0.2
Figure 2:
0.15
0.15
4
2
x
0
-2
0.2
Figure 1:
0.3
0.25
0.3
0.2
Figure 3:
11
0.2
-4
4
2
x
0
-2
-4
0.25
0.35
0.3
0.4
0.1
0.05
0.1
4
2
x
0
-2
-4
4
2
x
0
-2
-4
0.4
0.1
0.05
0.1
A plot of the probability densities |φn (x)|2 for the first few energy levels is given in
Figures 1-4.
0.35
∗ b) Operator Solution
Once we have established the eigenfunctions of the Hamiltonian and its spectrum for the
harmonic oscillator we can, in principle, calculate all physically interesting quantities, in
particular expectation values and matrix elements of operators. However, in practice,
these calculations can become technically difficult requiring the evaluation of infinite sums
(namely sums over all eigenvalues). In order to tackle this problem we will now reformulate
the quantum harmonic oscillator in terms of some new operators which allows us to deal
with these calculations in an algebraic way.
Let us introduce the following operators (called ladder operators, or lowering resp.
raising operators for reasons that will become clear):
r
r
1
1
mω
mω
†
ib
p
, b
a =
ib
p ,
(21)
x
b+
x
b−
b
a=
2~
mω
2~
mω
which are eachother’s Hermitian conjugate in view of the hermiticity of x
b and pb: x
b† = x
b,
†
pb = pb . The commutation relations for these operators are easily checked:
[x
b , pb ] = i~b
1
[b
a, b
a† ] = b
1.
⇒
(22)
We can now write the Hamiltonian in the following suggestive way
1
b = 1 pb2 + 1 mω 2 x
H
b2 = ~ω(b
n+ )
2m
2
2
,
where n
b=b
a† b
a
(23)
where the operator n
b is referred to as the number operator. This operator has the following
commutation relations with b
a and b
a† :
[n
b, b
a] = [b
a†b
a, b
a] = [b
a† , b
a ]b
a = −b
a
[n
b, b
a† ] = [ b
a† b
a, b
a† ] = b
a† [ b
a, b
a† ] = b
a†
which can also be written as:
n
bb
a=b
a (b
n−b
1)
,
More generally, for k integer, we have:
or equivalently
Exercise:
n
bb
ak = b
ak (b
n − kb
1)
[n
b, b
ak ] = −kb
ak
n
bb
a† = b
a† (b
n+b
1)
n
b (b
a† )k = (b
a† )k (b
n + kb
1)
,
[n
b , (b
a† )k ] = k(b
a† )k .
,
(24)
(25)
(26)
Verify the above commutation relations.
Now, we make the assumption (justified by the conclusions of section 1) that there is
a distinguished nonzero quantum state, which we denote by φ0 (x) , called the vacuum
state, which is annihilated by the operator b
a, i.e.
b
aφ0 = 0
normalised as
12
kφ0 k = 1 .
(27)
(Note: this vacuum state is not the zero vector in the Hilbert space). In fact, the vacuum
state can be obtained by solving (27) as a differential equation. From (21) we have
mω ~ dφ0
1
ib
p φ0 =
+ xφ0 = 0 ⇒ φ0 (x) = C exp
x2
x
b+
mω
mω dx
2~
which corresponds to the exponential prefactor in (15). Determining C by normalisation
subsequently leads to the solution φ0 given in part (a). From the algebra given above, we
then have that
b 0 = 1 ~ωφ0 ,
n
bφ0 = 0
⇒
Hφ
2
using (23). Thus, the vacuum state is an eigenstate (with eigenvalue 0) of the number
operator n
b, and hence also an eigenstate of the Hamiltonian with eigenvalue (energy)
E0 = 21 ~ω.
To obtain the other eigenstates of the Hamiltonian we make the observation that from
(25) we have
a† )k (b
n + kb
1)φ0 = k(b
a† )k φ0 ,
n
b(b
a† )k φ0 = (b
implying that the state (b
a† )k φ0 is an eigenstate of n
b with integer eigenvalue k, and
hence from (23) it is an eigenstate of the Hamiltonian associated with the energy level
Ek = (k + 21 )~ω . What remains to be done is to normalise the eigenstate by including a
multiplicative factor. Setting now
φk (x) = αk (b
a† )k φ0 (x) ,
to be this normalised eigenstate, i.e. imposing: kφk k = 1 , we can easily find this
multiplicative factor αm . In fact, from the commutation relations (22) it is not difficult to
demonstrate the following identity:
[b
ak , (b
a† )k ] φ0 = k!φ0 ,
(e.g. by induction), and then we can assert that:
1 = (φk , φk ) = |αk |2 ((b
a† )k φ0 , (b
a† )k φ0 ) = |αk |2 (φ0 , b
ak (b
a† )k φ0 )
= (φ0 , [ b
ak , (b
a† )k ] φ0 ) = k! |αk |2 (φ0 , φ0 ) = k! |αk |2 ,
√
and thus αk = 1/ k! up to a complex phase factor which we can neglect. Thus, we find
that the normalised eigenstates of the quantum harmonic oscillator are given by:
1
φk (x) = √ (b
a† )k φ0 (x) .
k!
(28)
Using the commutation relations
[b
a, (b
a† )k ] = k(b
a† )k−1
,
[b
a† , b
ak ] = −kb
ak−1 ,
one can obtain recursion formulae following from the action of b
a and b
a† on the eigenstates
as follows:
1
1
b
aφk = √ b
a(b
a† )k φ0 = √ [ b
a, (b
a† )k ] φ0
k!
k!
p
k (k − 1)!
1
† k−1
√
a ) φ0 =
φk−1
= √ k(b
k!
k!
13
whilst
1
a† )k+1 φ0 =
b
a φk = √ (b
k!
Thus, we obtain the following formulae:
†
p
(k + 1)!
√
φk+1 .
k!
√
b
aφk = √k φk−1
b
a† φk = k + 1 φk+1
Thus, it has now become clear why we call b
a† and b
a a raising- resp. lowering operator:
they are the operators that (mathematically) bring you from one energy state into the
next higher (resp. lower) energy state!
One merit of the operator description is that it facilitates the calculation of “physical”
quantities such as expectation values. For this we need to consider the matrix elements of
the operators b
a and b
a† , which are easily calculated from the above formulae:
√
√
(φk , b
aφl ) =
l(φk , φl−1 ) = l δk,l−1
(29a)
√
√
†
l + 1(φk , φl+1 ) = l + 1 δk,l+1
(29b)
(φk , b
a φl ) =
and from these we can calculate the matrix elements of x
b and pb by inverting the relations
(21), i.e.
r
r
~
mω~ †
†
x
b=
(b
a+b
a)
,
pb = i
(b
a −b
a) .
(30)
2mω
2
We can actually present the matrix elements Xkl ≡ hk|b
x|li and Pkl ≡ hk|b
p|li (k, l =
0, 1, 2, . . . ) of the position and momentum operators as semi-infinite matrices as follows:
√
√



1
0
0
.
.
.
1
0
0
0
0
−
√
√
√
√



r
r
1 0
2 0 ... 
1 √0
− 2
0
√
~ 
mω~ 
 0 √2 0 √3 . . . 
 0
2
0
−
3
X=
 , P =i


√
√


2mω  0
2
0
0
3 0 ... 
3
0

 0
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
.
(31)
Expectation values of the observables x and p and its powers can now be simply calculated
by using combinations of these semi-infinite matrices, which multiply according to the rule:
∞
X
Xkj Pjl
(k, l = 0, 1, 2, . . . ) .
(XP )kl =
j=0
Example: To calculate the expectation values hxi and hpi in the eigenstate |ki we need
only to find the corresponding diagonal elements of the matrices X resp. P :
hxi = Xkk = 0
whilst the spreading is calculated from
~
hx2 i = (X 2 )kk = (2k + 1)
2mω
and
and
14
hpi = Pkk = 0
hp2 i = (P 2 )kk = (2k + 1)
mω~
.
2
...
...
...
...
..
.




 .

