Operator Formalism I
6
6.1
Momentum operator
In eqs. 5/(47-48) found
Z
Z
2
dx|ψ(x)| = dp|f (p)|2
(1)
suggesting that if |ψ(x)|2 is the probability density for finding
the particle in spatial range [x, x + dx], then |f (p)|2 is prob.
density for finding particle’s momentum in range [p, p + dp].
Plane wave ψp = eipx/h̄ is state of definite momentum. Define
momentum operator
∂
(2)
p̂ = −ih̄ ,
∂x
verify p̂ψp = pψp, note difference between operator p̂ and
eigenvalue p. Compare with Hψ = Eψ for state of definite
energy.
6.2
Expectation values
or, how the probabilistic nature of ψ affects physically measurable quantities.
Consider statistical ensemble of systems with 1 particle described by ψ. Prob. particle in d3r at r is dP = |ψ(r, t)|2d3r.
1
Calculating averages
Calculate avg. value by summing over all possible values of
something, weighted with prob. of each. For example, mean
value of particle’s position:
Z
< r >= ψ ∗rψ d3r
(3)
if ψ normalized. Called expectation value of r because it is at
< r > you would expect to find particle, on avg. Expectation
val. of any operator  is just
Z
< Â >= ψ ∗Â(r)ψ d3r
(4)
Avg. value of momentum (need to use momentum prob. dens.
|f (p)|2:
<p> =
=
Z
Z
d3p f (p)∗pf (p)
(5)
d3pd3r1d3r2 ∗
−ip·(r2 −r1 )/h̄
ψ
(r
)ψ(r
)pe
(6)
1
2
(2πh̄)3
Simplify by writing
pe−ip·(r2−r1)/h̄ = +ih̄∇2e−ip·(r2−r1)/h̄
(7)
where ∇2 grad wrt r2.
Integrate by parts in 3D: note for 2 functions g(r), h(r) which
fall off suff. rapidly at |r| → ∞,
Z
Z
3
d rg∇h = − d3r(∇g)h
(8)
since the surface term gh|∞ vanishes. So rewrite (6) as
d3pd3r1d3r2 ∗
−ip·(r2 −r1 )/h̄
ψ
(r
)(−ih̄∇
)ψ(r
)e
(9)
< p >=
1
2
2
(2πh̄)3
Z
2
Now perform int. over p, gives (2πh̄)3δ(r2 − r1), =⇒
Z
< p > = d3rψ ∗(r)p̂ψ(r),
p̂ ≡ −ih̄∇
(10)
(11)
looks just like eqn. (3) for position operator. Have found
a way to represent physical observable momentum in posiR
tion representation. Can show further < p2 >= d3rψ ∗p̂2ψ,
R
in general < f (p) >= d3rψ ∗f (p̂)ψ, provided f (p) can be
expanded in a power series. (check!).
6.3
Inner products
Generalize representation of momentum, position, energy as
operators on function ψ to any physical observable. Formally, operator Ô maps ψ onto some new function ψ 0, and is
linear:
Ô(αψ + βψ 0) = αÔψ + β Ôψ 0
(12)
3
ψ and ψ 0 are elements of a general linear function space
(like a vector space). We define a norm and scalar product on
this space just as for ordinary vectors.
Definition. Inner product of two functions ψ, φ:
Z
(ψ, φ) ≡ d3rψ ∗(r)φ(r)
(13)
has following properties:
(ψ, αφ + βχ) = α(ψ, φ) + β(ψ, χ)
(αψ, φ) = α∗(ψ, φ)
(ψ, φ)∗ = (φ, ψ)
(14)
(15)
(16)
Notation allows us to collect previous results in succinct way:
normalization condition 1
expectation values: < r >
<p>
<H>
6.4
=
=
=
=
..
(ψ, ψ)
(ψ, rψ)
(ψ, p̂ψ)
(ψ, Hψ)
(17)
(18)
(19)
(20)
Hermitian adjoint
Q: under what conditions can we expect general linear operator Ô to represent a physical observable? First must demand
expectation value (ψ, Ôψ) must be real.
Definition. Adjoint Ô† of operator Ô obeys
(Ô†ψ, φ) = (ψ, Ôφ)
(21)
4
Examples:
1. α a const. =⇒ α† = α∗
2. Ô =
∂
∂x :
∂ψ ∗ 3
ψ
dr = −
φd r
∂x
∂x
∂
∂
so
( )† = −
∂x
∂x
Z
∗ ∂φ 3
Z
(22)
(23)
3. (αÔ)† = α∗Ô†
†
∂
∂
∂
then p̂† = ih̄ ∂x
= −ih̄ ∂x
= p̂. p̂ is said
so if p̂ = −ih̄ ∂x
to be self-adjoint or Hermitian.
4. (P̂ + Ô)† = P̂ † + Ô†
5. (P̂ Ô)† = Ô†P̂ †
6. If (ψ, Ôψ) = (Ôψ, ψ)∀ψ, then (ψ, Ôφ) = (Ôψ, φ) so Ô
is self-adjoint. Check this!
7. From (16) get
(ψ, Ôφ)∗ = (Ôφ, ψ)
= (φ, Ô†ψ)
(24)
(25)
(ψ, Ôψ)∗ = (ψ, Ô†ψ)
(26)
so in particular
so that if Ô is self-adjoint, Ô† = Ô, then (ψ, Ôψ)real . All
observables wil be represented by self-adjoint operators.
5
6.5
Eigenvalues and eigenfunctions
Let Q̂ be a linear operator, and ψ an eigenfunction of Q̂ with
eigenvalue q, i.e. Q̂ψ = qψ. If Q̂ is Hermitian, Q̂ = Q̂†
then several properties follow:
1. All eigenvalues q are real! Start with
(Q̂ψ, ψ) = (ψ, Q̂ψ)
Q̂ψ = qψ,
(27)
(28)
(qψ, ψ) = (ψ, qψ)
q ∗(ψ, ψ) = (ψ, ψ)q =⇒
(29)
(30)
then
q∗ = q
(31)
2. Eigenfunctions belonging to different eigenvalues are orthogonal. If
Q̂ψ1
Q̂ψ2
then
(Q̂ψ1, ψ2)
q1(ψ1, ψ2)
= q1ψ1
= q2ψ2,
(32)
(33)
= (ψ1, Q̂ψ)
= q2(ψ1, ψ2),
(34)
(35)
so if q1 6= q2, then (ψ1, ψ2) = 0 and the two fctns said to
be orthogonal.
6
3. Linearly independent eigenfunctions belonging
to the same eigenvalue q can be made orthogonal. (Example: recall we studied eigenvalues of 3D SHO
in problem set, showed degeneracy—number of eigenfunctions corresponding to given eigenvalue– was N 2/2+3N/2+
1. )
Functions ψ1...ψm are linearly ind. if only soln. to Pm
n=1 ψn =
0 is cn = 0. Suppose
Q̂ψn = qψn,
n = 1, . . . m.
(36)
It might be that (ψm, ψn) 6= δmn for this choice of the ψn,
but can always find new set of φn obeying Q̂φn = qφn for
which (φm, φn) = δmn using Gram-Schmidt orthonormalization (as in linear algebra!):
Let φ1
φ2
with α chosen s.t.
(φ1, φ2)
let φ3
and require
(φ1, φ3)
= ψ1
= ψ2 + αψ1
(37)
(38)
= (ψ1, ψ2) + α(ψ1, ψ1) = 0 (39)
= ψ3 + βφ2 + γψ1
(40)
= (φ2, φ3) = 0.
(41)
Last line is two equations determining 2 unknowns β and
γ. And so on. . .
4. If P̂ and Q̂ commute, [P̂ , Q̂] = 0, can find set of
simultaneous eigenfunctions of P̂ and Q̂.
7
Simple case: suppose every eigenfctn. of P̂ corresponds to
different eigenvalue (nondegenerate). For all such eigenfctns.,
P̂ ψ = pψ
so
Q̂(P̂ ψ) = Q̂(pψ)
if [P̂ , Q̂] = 0 ↓
↓
P̂ (Q̂ψ) = p(Q̂ψ).
(42)
(43)
(44)
So Q̂ψ is eigenfctn. of P̂ with eigenvalue p. But there
is only one such lin. ind. fctn ψ by assumption, so must
have Q̂ψ ∝ ψ, means ψ also eigenfctn. of Q̂!
If 2 or more eigenfctns {ψ}n correspond to same p, more
complicated construction necessary to find set {ψn} which
are simultaneous eigenfctns. of P̂ and Q̂. Not terribly enlightening (see Peebles for details) but be aware it can
always be done. Point: if operators P̂ and Q̂ commute,
observable quantities p and q can be determined with arbitrary precision simultaneously! Not true in general,
e.g. [p̂i, xj ] = −ih̄.
Example: SHO wavefunctions Use wave functions we’re somewhat familiar with, 1D SHO energy eigenfctns given by
1
x − 21 xx 2
√ 1/2 Hn( )e 0
ψn(x) = n
(45)
(2 n!x0 π)
x0
s
where x0 =
h̄
mω
is the length scale we have encountered
8
before, and the Hn are specific polynomials which arise in the
solution of the S.-eqn. for higher n. H0 = 1 and H1 = 2x.
1. Check: energy eigenfunctions belonging to different eigenvalues are orthogonal.
(ψ1, ψ0) = 0
but (ψ0, ψ0) = 1, etc.
(46)
2. Check: are these momentum eigenfunctions too?
∂
− 12 xx 2
(47)
p̂ψ0 = −ih̄ ψ0 ∼ xe 0
∂x
so no, it’s not simultaneously an eigenfunction of momentum. We don’t expect this, since
p̂2
1
[H, p̂] = [
+ mω 2x2, p̂] = mω 2[x2, p̂] 6= 0
2m 2
(48)
3. Linearly independent eigenfunctions belonging to the same
eigenvalue q can be made orthogonal. The 1D SHO wavefunctions aren’t degenerate =⇒ can’t discuss this case.
9
But 3D SHO homework problem showed energies are
EN = h̄ω(n1 + n2 + n3 + 3/2), N = n1 + n2 + n3 (49)
and eigenfunctions are
ψn1,n2,n3 = ψn1 (x)ψn2 (y)ψn3 (z)
(50)
i.e., one for each possible combination n1, n2, n3. Consider
ψ100, ψ010, and ψ001. Each has N = 1, corresponding to
an energy E1 = 5h̄ω/2.
Physically they correspond to oscillator with one quantum
of excitation energy vibrating in x, y, or z directions, respectively. For isotropic potential, energies must be equal
(degenerate), so degeneracy is 3 for the eigenvalue E1.
They are already orthogonal, e.g.
Z
d3rψ1∗(x)ψ0(x)ψ0∗(y)ψ1(y)|ψ0(z)|2
(51)
= (ψ1, ψ0)1D (ψ0, ψ1)1D (ψ0, ψ0)1D = 0
(ψ100, ψ010) =
because the 1st two factors vanish.
10