Download 14 The Postulates of Quantum mechanics

14
The Postulates of Quantum mechanics
• Postulate 1: The state of a system is described completely in terms
of a state vector Ψ(r, t), which is quadratically integrable.
• Postulate 2: To every physically observable there exist a linear Hermitian operator.
• Postulate 3: In any measurement of the observable associated with
operator Â, the only values that will ever be observed are the eigenvalues ai , which satisfy the eigenvalue equation Âgi = ai gi .
• Postulate 4: The eigenfunctions of operators corresponding to observable forms a complete set.
• Postulate 5: If a system is in a state described by a normalized wave
function Ψ, then the average
! ∞ ∗value of the observable corresponding to
 is given by < A >= −∞ Ψ ÂΨdτ .
• Postulate 6:The wavefunction or state function of a system evolves in
time according to the time-dependent Schrdinger equation ĤΨ(r, t) =
.
i! ∂Ψ
∂t
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14.1
Measurement and superposition of states
Postulate 4 allows us to expand any state functions in terms of the complete
eigenfunctions of an operator
"
Ψ(q, t) =
ci (t)gi (q)
(551)
i
where Ψ is normalized and q is the 3n coordinates of the system. The eigenfunctions obey the following eigenvalue equations for the hermitian operator
B
B̂gi (q) = bi gi (q)
(552)
Using the normalization condition we have
# ""
# "
"
∗ ∗
c∗i ck gi∗ gk dτ
1=
ci g i
ck gk dτ =
i
i
k
(553)
k
Assuming that we can interchange summation and integrations we get
#
""
∗
ci ck gi∗ gk dτ = 1
(554)
i
k
We know that the eigenfunctions are orthogonal since B̂ is hermitian thus
"
""
c∗i ck δik =
|ci|2 = 1
(555)
i
i
k
and we see that the sum of the square of the coefficients of a complete set
adds to one. Now lets consider the importance of this for measurements. We
are interested in determining the average value of the operator B̂ as
< B >=< Ψ|B̂|Ψ >
"
"
ck g k >
=<
ci gi |B̂|
i
=
=
c∗i ck
< gi |B̂|gk >
(558)
c∗i ck bk < gi |gk >
(559)
""
(560)
k
""
i
(557)
k
""
i
(556)
k
=
i
c∗i ck bk δik
k
=
"
i
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|ci |2 bi
(561)
and the expectation values is the weighted sum of the eigenvalues. The
square of the expansion coefficients are thus the probability of measuring the
eigenvalues. We find the coefficients by project the eigenfunction onto the
state functions as
#
"#
"
∗
< gj |Ψ >= gj Ψdτ =
gj∗ gi dτ =
ci δij = cj
(562)
i
j
and we call the integral the probability amplitude. Thus, we see two things
The first is that the state functions is given by a superposition of the eigenfunctions of the operator and we find the expectation valued by summing the
square of the amplitudes. This also means that once we performed a measurement of the system it is collapsed to to an eigenfunctions of the system.
This means that in quantum mechanics the act of performing a measurement
perturbs the system.
15
The time-evolution of the wave function
Consider a hamiltonian that is time-independent. This means that the state
function obeys
ĤΨ = EΨ
(563)
and the time-dependent Schrödinger equations becomes
i!
∂Ψ
= EΨ
∂t
(564)
which can be integrated to give
Ψ(q, t) = exp(−iEt/hbar)ψ(q)
(565)
Therefore, if the hamiltonian is time-independent there exist stationary states
Ψn = exp(−iEn t/!)ψn (q)
(566)
where ψ)n is a solution to the time-independent Schrödinger equation. We
can now expand any function in terms of these eigenfunctions as
"
"
Ψ=
cn Ψ n =
cn exp(−iEn t/!)ψn (q)
(567)
n
n
67
where cn are time-independent. This function is not an eigenfunction of the
hamiltonian, and thus do not have a definite energy. To find the coefficients
at time t0 we evaluate the projection
"
"
< ψj (q)|Ψ(q, t0 ) >=
cn exp(−iEn t/!) < ψj |ψn >=
cn exp(−iEn t/!)δjn = cj exp(−iEj t/!)
n
n
(568)
Thus, the general solution to the time-dependent Schrödinger equation is
given by
"
Ψ=
< ψj (q)|Ψ(q, t0) > exp(−iEn (t − t0 )/!)ψj (q)
(569)
j
if the hamiltonian is independent of time.
15.1
Example
As an example consider a state which consist of a superposition of the two
lowest eigenfunctions of the particle in a 1D box
%
&
$
1
2
Ψ(x, 0) = 2/A √ sin(2πx/a) + √ sin(πx/z)
(570)
5
5
√
√
Our expansion coefficients are then c1 = 2/ 5 and c2 = 1 5. and the state
of the system at t > 0 is given by
%
&
$
1
2
Ψ(x, t) = 2/A √ exp(−iω2 t) sin(2πx/a) + √ exp(−iω1 t) sin(πx/a)
5
5
(571)
where !ωn = En . The probability of measuring is then
√
2
P (E1 ) = |c1 |2 = [ √ exp(−iω1 t)]∗ [f rac2 5 exp(−iω1 t)] = 4/5
5
(572)
and is thus constant in time. What about the expectation value
1
< E >t>0 = (< exp(−iω2 t)φ2 | +2 < exp(−iω1 t)φ1 |)
' 5
(
×
Ĥ| exp(−iω2 t)φ2 | + 2Ĥ| exp(−iω1 t)φ1 >
1'
=
< φ2 |Ĥ|ψ2 > +4 < φ1 |Ĥ|φ1 >
5
68
(573)
(574)
(575)
+
=
(
2 exp(−i(ω1 + ω2 )t)[< φ1 |Ĥ|φ2 > + < φ2 |Ĥ|φ1 >] (576)
1
(E2 + 4E1 ) =< E >t=0
5
and also the expectation value of the energy is constant in time.
69
(577)