Operator Formalism II
7
7.1
Completeness
To each observable corresponds a Hermitian operator, say Q̂. There exists
a set of eigenfunctions ψn, each corresponding to an eigenvalue qm (may
be more eigenfunctions ψn than eigenvalues qm, not vice versa. )
Completeness: Any physical state ψ can be expanded in this set,
ψ=
X
cnψn
(1)
cn = (ψn, ψ)
(2)
n
If the ψn are normalized,
(Analogy: in 3-space, any vector of unit length may be written v̂ = v1î +
v2ĵ + v3k̂, with v1 = î · v̂, v2 = ĵ · v̂, and v3 = k̂ · v̂.)
Example: Free particle momentum eigenstates in periodic universe:
1
ψp(r) = √ 3 eip·r/h̄,
L
with pα = 2πnαh̄/L
(3)
(4)
Note (ψp, ψp0 ) = δpp0 , i.e. 1 or 0.
Any state ψ(r) may be represented as
ψ(r) =
with
X
cpψp(r)
X cp
√ eip·r/h̄
=
p
L3
p
(5)
(6)
d3r −ip·r/h̄
cp = (ψp, ψ) = L3 √ 3 e
ψ(r),
(7)
L
nothing more than a definition of a 3D Fourier series! Note the cp are
dimensionless.
Z
1
Example II: Momentum eigenstates with L → ∞ (Continuous spectrum).
If we let L → ∞ the eigenfunctions (3) no longer well defined. How do we
normalize plane wave states in real (infinite) universe? From (4), difference between allowed momenta ∆p = 2πh̄/L also → ∞ =⇒ momentum
eigenvalues continuous. Instead of requiring (ψp, ψp0 ) = δpp0 (Kronecker
δ-fctn.), normalize with Dirac δ-fctn.:
(ψp, ψp0 ) = δ(p − p0)
Choose normalization
ψp(r) =
(8)
1
eip·r/h̄
3/2
(2πh̄)
(9)
Verify Eq. (8). (See def. (5.37) of δ-fctn., use p = h̄k).
Then if we expand ψ in terms of these fctns:
ψ(r) =
Z
3
d p c(p) ψp(r) =
c(p) = (ψp, ψ) =
Z
Z
d3p
c(p)eip·r/h̄,
3/2
(2πh̄)
d3r
ψ(r)e−ip·r/h̄
3/2
(2πh̄)
(10)
(11)
just Fourier integral transform!
N.B. If ψ is superposition of energy eigenstates, ψ =
into S.-eqn. ih̄ ∂ψ
∂t = Hψ to see time evolution:
P
n cn ψn ,
∂ψ
∂X
X dcn
cnψn = ih̄
= ih̄
ψn
n
n dt
∂t
∂t
X
X
X
= H cnψn =
cn(Hψn) = cnEnψn
ih̄
n
n
can plug
(12)
n
multiply by ψm, take inner product of lhs and rhs, get
dcn
ih̄
= Encn
dt
2
(13)
soln: cn = dne−iEnt/h̄, or
ψ(r, t) =
X
n
dnψn(r)e−iEnt/h̄
(14)
where dn is const. to be determined by initial conditions. Note that ψn(r)
itself is a stationary state—not a function of time. It is the form of the
energy eigenstate with eigenvalue En at t = 0.
7.2
Measurement
This is just a summary of some results we’ve written down before. Want
to measure observable, represented by Hermitian operator Q̂ which has
complete set of eigenfunctions ψn (assume nondegenerate for now) corresponding to eigenvalues qn.
• If we measure Q in state ψ probability that measurement yields the
result qn is
Pn = |(ψn, ψ)|2 = |cn|2
if ψ =
P
(15)
n cn ψn .
Can check that prob. is normalized
P
n Pn
= 1!
• Expectation values. Average result of many measurements is
< Q̂ >≡
X
n
qnPn = (ψ, Q̂ψ)
(16)
pf:
X
n
Pn =
X
n
(ψ, ψn)qn(ψn, ψ)
= (ψ,
X
n
= (ψ, Q̂
qnψn(ψn, ψ))
X
n
ψn(ψn, ψ))
= (ψ, Q̂ψ).
since ψ =
P
n ψn (ψn , ψ).
3
(17)
• Pure states. If ψ = ψn before the measurement, result is always qn.
• Simultaneous eigenstates. If [P̂ , Q̂]=0, can find complete set of simultaneous eigenfctns ψpq , such that P̂ ψpq = pψpq and Q̂ψ = qψpq . If
[P̂ , Q̂] 6= 0, and system is in state of definite P , must be mixture of
different eigenvalues qn, i.e. measurement of Q is uncertain.
7.3
Parity
Definition: Parity operator Π̂ defined by
Π̂ψ(r) = ψ(−r)
(18)
Easy to check that (φ, Π̂ψ) = (Π̂φ, ψ) so Π̂ is Hermitian and an observable!
Eigenfunctions and Eigenvalues
Obviously Π̂2 = 1, since −(−r) = r. Then if ψ = const.(6= 3.14159265...!)
is the eigenvalue of parity, Π̂ψ = πψ, must have π 2 = 1, and since Π̂ is
Hermitian, π must be real =⇒ π = ±1.
Therefore classify eigenfunctions of parity as π = ±1, i.e. even or odd under spatial inversion. But energy eigenstates may (e.g, symmetric square
well) or may not be eigenstates of parity. Need to find out under what
circumstances may expect parity eigenstates.
Commutation relations:
1. [Π, p̂] = −2p̂Π̂, or Π̂p̂ = −p̂Π̂ (check!).
2. [Π̂, p̂2] = 0 (check!)
√ 2
3. Central potential, V = V (r), with r = x + y 2 + z 2: [Π̂, V (r)] = 0,
so with previous relation find [Π̂, H] = 0.
For the real H of the universe, [Π̂, H] 6= 0. While parity good symmetry
for strong and electromagnetic interactions, violated in weak interaction
decays.
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If [Π̂, H] = 0, what then?
1. If
∂ψ
= Hψ,
∂t
(19)
∂ Π̂ψ
= H Π̂ψ
∂t
(20)
ih̄
then
ih̄
so if ψ is allowed state of system, mirror image state also possible
state.
2. Can find complete set of eigenstates of Π̂ and H, so every energy
eigenstate can be assigned a definite parity.
3. Parity is conserved, i.e. π associated with eigenfctn ψ doesn’t flip
with time. • Note how symmetry yields a conservation law!
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