Unformatted text preview: Chapter 14 The Quantum Mechanical Postulates
Questions on Concepts
Q14.1
Is it correct to say that all the information about a quantum mechanical system is in its
wavefunction?
Yes. All information about a quantum mechanical systems is contained in its
wavefunction. In other words, as long as the wavefunction is given, anything such as
the eiegenvalues of an operator can be obtained. In many cases, one cannot find the
exact wavefunction, instead only approximate wavefunction is found; in other cases,
even approximate wavefunction is not possible to find and only an approximate
electronic density functional is to be calculated.
Q14.2
Why is a wavefunction normalized?
The total probability of finding a particle in all possible states must be 1. (The
particle/system has to be in some state)
Q14.3
What requirements should be satisfied for a function to be a legitimate quantum
mechanical wavefunction?
The wavefunction is for a physical system, hence some practical restrictions apply: (i)
the wavefunction must be continuous (the probability cannot make sudden changes);
(ii) the wavefucntion must be singlevalued (one particle can take two or more
positions in QM, but cannot take two or more different probabilities for being in a
certain state); (iii) The wavefunction must be finite so that the total probability must
be normalized to be 1. (The wavefunction cannot be infinite nor being zero
everywhere).
Q14.4
Why are Hermitian operators used for quantum mechanical observables?
The eigenvalues of a Hermitian operator are real. (All measurement results of a
physical quantity must be real).
Q14.5
Is it correct to say that a quantum mechanical operator can have many, even infinite,
different eigenvalues?
Yes. All eigenvalues of a (Hermitian) operator form the ‘spectrum’ of the operator.
The spectrum may contain finite or infinite number of eigenvalues, continuous
number or discrete number of eigenvalues.
Q14.6
Is it correct to say that the eigenfunctions belonging to different eigenvalues must be
orthogonal to each other? Yes. From the definition of eigen equation, we have
ˆ
ˆ
Oφm = λmφm ; Oφn = λnφn
ˆ
φ *Oφ = λ φ *φ
n m mnm *
ˆ
φ Oφn = λnφmφn
*
m According to the definition of Hermitian operator, the right hand side of the above
twp equations must be equal to each other
ˆ
∫ dxφ Oφ
*
n m *ˆ
= ∫ dxφmOφn *
*
from which we immediately have (λm − λn ) ∫ dxφmφn = 0 ⇒ ∫ dxφmφn = 0. Q14.7
Is it correct to say that a quantum mechanical system can only be in one of its
eigenstates?
No. The system may be in a superposition state of two or more eigenstates: ∫ dxφ φ *
mn = δ mn . Q14.8
Explain what is an orthonormal and complete set.
A set of functions that satisfies the following conditions ∫ dxφ φ *
mn = δ mn . is called a set of orthonormal functions. If any function can be expanded in terms of a set of functions: ψ = ∑ bnφn , bn = ∫ dxψ *φn , then the set { φn } n is called a complete set.
Q14.9
Is it to say that a single time measurement on a physical quantity (quantum
mechanical observable) can only generate one of its eigenvalues?
Yes. That’s exactly the content of the measurement postulate.
Q14.10
Is it right that the average of measurement results of a quantum mechanical
observable can be very different from its eigenvalues?
Yes. The average of many eigenvalues can be very different from any of the
eigenvalues:
*ˆ
*
ˆ
< O >= ∫ dxψ *Oψ = ∫ dx ∑ cmφmO ∑ cnφn
m n **
**
*
ˆ*
= ∫ dx ∑ cmφm ∑ cn Oφn = ∫ dx ∑ cmφm ∑ cn λnφn
m n m n *
**
= ∑ cm cn λn ∫ dxφmφn Q14.11 m,n *
*
= ∑ cm cn λnδ mn = ∑ cn cn λn = ∑  cn 2 λn
m,n n n In a doubleslit experiment, the interference pattern is found on the screen behind the
slits. This means the particle passing through the doubleslit was in the superposition
states of two slits. Is this statement correct?
Yes. It is correct. The interference is a consequence, or a manifestation, of the
superposition state of the particle passing through the double slit.
Q14.12
Write down the wavefunction of an electron in a hydrogen atom which is in the
superposition of 1s and 2px.
The general superposition state of 1s and 2px is: ψ = aψ 1s + bψ 2 p , a 2 +  b 2 = 1.
x Q14.13
Is it reasonable to say that the Schroedinger equation in quantum mechanics plays the
same role as Newton’s second law does in classical mechanics?
Yes. Both equations predict the future states of a system based on present conditions.
In Newton’s equation, the reason of change is force; in Schroedinger’s equation, the
reason of change is the energy operator (Hamiltonian).
Q14.14
_____________________________________________________________________ Problems
P14.1
Are the following functions qualified to be a quantum mechanical wavefunction?
(1)sin π x (x = [−2, 2]) (2) cos8π x (x = [−2, 2]) (4) e−2 x ( x = [0, ∞)) (5) ei 2π x (x = (−∞, ∞)) (6)
2 (3) ei 4π x ( x = [−1,1])
1
1+ x (x = (−∞, ∞)) One can readily check and find that the functions (1,2,3,4,6) are continuous,
singlevalued and finite but not zero everywhere (normalizable), but function (5) is
not normalizable.
P14.2
Normalize the following wavefunctions:
(1)sin 3x (2) cos 2 x (3) eiax (a ∈ R) (4) e−bx (b is positive, real, x = [0, ∞)) (5) eiax (a ∈ R) (6)
2 1
1+ x 2 (x = (−∞, ∞)) a (1)sin 3x ⇒ N 2 ∫ dx sin 2 3x = N6 (a − sin 6a ) = 1 ⇒ N =
2 6
a −sin 6 a 0 a (2) cos 2 x ⇒ N 2 ∫ dx cos 2 2 x = N4 (sin 4a − a) = 1 ⇒ N =
2 4
sin 4 a − a 0 a (3) eiax (a ∈ R) ⇒ N 2 ∫ dxe −iax eiax =N 2 (a − 1) = 1 ⇒ N = 1
a −1 0 (4) e−bx (b is positive, real, x = [0, ∞))
a N
⇒ N 2 ∫ dxe−2bx = −2b (e−2ba − 1) = 1 ⇒ N =
2 2b
1− e−2 ba 0 a (5) e iax 2 (a ∈ R) ⇒ N 2 ∫ dx =N 2 (a − 1) = 1 ⇒ N = 1
a −1 0 ∞ (6) 1
1+ x 2 (x = (−∞, ∞)) ⇒ N 2 ∫ −∞ 1
(1+ x 2 )2 dx =N 2 = 1 ⇒ N = 1. P14.3
Check what in the following functions are the eigenfunction of momentum operator
ˆ
px = −i ∂∂x ：
(1)sin x (3) eiax (a ∈ R) (2) cos x (4) ebx (b ∈ R) (5) eiax (a ∈ R) (6)
2 ˆ
(1) px sin x = −i ∂ (sin x )
∂x 1
1+ x = −i cos x ≠ λ sin x (2) Similar to (1).
ˆ
(3) px eiax = −i
ˆ
(4) px ebx = −i ∂ ( ebx )
∂x = aeiax ⇒ λ = a ∂ ( eiax )
∂x ˆ
(5) px eiax = −i
2 ˆ1
(6) px 1+ x = −i = −i bebx ⇒ λ = −i b (the eigenvalue is not real). 2 ∂ ( eiax )
∂x
1
∂ ( 1+ x )
∂x = a (2 x)eiax ≠ λ eiax =i 2 2 1
≠ λ 1+ x 1
(1+ x ) 2 P14.4
Check what in the following functions are the eigenfunction of the kinetic energy
ˆ
operator of a particle moving in a onedimensional space, Ek = − 2 m ∂∂x ：
2 2 2 (1)sin x (3) eiax (a ∈ R) (2) cos x (4) ebx (b ∈ R) (5) eiax (a ∈ R) (6)
2 ˆ
(1) Ek (sin x) = − 2 m ∂ (sin x ) =
∂x
2 2 2 ˆ
(2) Ek (cos x) = − 2 m ∂ (cos x ) =
∂x
2 2 2 2 1
1+ x sin x
2m 2 ,λ = cos x
2m ,λ = 2 2m 2 2m ˆ
(3) Ek (eiax ) = − 2 m ∂ ∂(xe ) =
2 2 iax ae
2m ˆ
(4) Ek (ebx ) = − 2 m ∂ ∂(xe ) = −
2 2 bx 2 ˆ
(5) Ek (eiax ) = − 2 m ∂
2 2 2 2 ( eiax )
∂x 2 ,λ = 2 2 iax 2 ,λ = − 2 2 bx be
2m 2 =− 22
a
2m 22
b
2m ( −4 a 2 x 2 + 2 ax ) eiax
2m 2 (negative kinetic energy!)
≠ λ eiax 2 1
ˆ1
(6) Ek ( 1+ x ) = − 2 m ∂ ∂(x ) = − m (1+ x ) ≠ λ 1+ x
2 2 1
1+ x
2 2 3 P14.5
Check what in the following functions are the eigenfunction of the kinetic energy
ˆ
operator of a particle moving in a onedimensional space, Ek = − 2 m ∂∂x ：
2 2 2 (1)sin x (3) eiax (a ∈ R) (2) cos x (4) ebx (b ∈ R) (5) eiax (a ∈ R) (6)
2 1
1+ x The same as P14.4.
P14.6
An electron in a molecule is in the following state:
Ψ = ψ 1s + ψ 2 p x
where ψ 2 s ,ψ 2 px are normalized. (a) Is the wavefunction Ψ normalized? If not, normalize it.
(b) What is the probability of finding the electron in the state ψ 2 px ? (a) Check if is normalized: ∫ dxΨ Ψ = ∫ dx(ψ
* 1s + ψ 2 px )* (ψ 1s + ψ 2 px ) *
*
= ∫ dx(ψ 1*sψ 1s + 2 pxψ 1s +ψ 2 pxψ 1*s + ψ 2 pxψ 2 px ) = 1 + 0 + 0 + 1 = 2.
Therefore, the normalized wavefuntion is
Ψ= 1
2 (ψ 1s + ψ 2 px ) (b) Since Ψ = 1
2 (ψ 1s +ψ 2 px ) , the probability to find the system in ψ 2 px is   =1.
2 12
2 (這是 promotion + hybridization 的最簡單例子，其中 1s 電子先提升然後與 2p
電子做 sp 混成)
P14.7
An electron in a molecule is in the following state:
Ψ = ψ 2 s + ψ 2 p x − ψ 2 p y − ψ 2 pz
where ψ 2 s ,ψ 2 px ,ψ 2 py ,ψ 2 pz are normalized. (a) Is the wavefunction Ψ normalized? If not, normalize it.
(b) What is the probability of finding the electron in the state ψ 2 px ?
（a） ∫ dxΨ Ψ = ∫ dx(ψ
* 2s + ψ 2 px −ψ 2 p y −ψ 2 pz )* (ψ 2 s + ψ 2 px −ψ 2 p y −ψ 2 pz ) *
*
*
*
*
*
*
*
*
= ∫ dx(ψ 2 sψ 2 s +ψ 2 sψ 2 px −ψ 2 sψ 2 py −ψ 2 sψ 2 pz +ψ 2 pxψ 2 s + ψ 2 pxψ 2 px −ψ 2 pxψ 2 p y −ψ 2 pxψ 2 pz
*
*
*
*
*
*
*
*
−ψ 2 p yψ 2 s −ψ 2 p yψ 2 px + ψ 2 p yψ 2 p y + ψ 2 p yψ 2 pz −ψ 2 pzψ 2 s −ψ 2 pzψ 2 px + ψ 2 pzψ 2 p y + ψ 2 pzψ 2 pz ) = 1 + 0 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 0 + 1 = 4. The normalization factor is therefore 1/2 and the normalized wavefunction is
Ψ = 1 (ψ 2 s + ψ 2 px −ψ 2 p y −ψ 2 pz )
2 （b） Since Ψ = 1 (ψ 2 s +ψ 2 px −ψ 2 p y −ψ 2 pz ) , the probability to find the system in
2 ψ 2 p is  1 2 = 1 .
2
4
x (這是 promotion + hybridization 的著名例子，其中 2s 電子先提升然後與 2p 電
子做 sp3 混成)
P14.8
An electron in a molecule is in the following state:
Ψ = 0.5ψ 2 s + 0.5ψ 2 px + 0.5ψ 2 p y + 0.5ψ 2 pz
where ψ 2 s ,ψ 2 px ,ψ 2 py ,ψ 2 pz are normalized. (a) Is the wavefunction Ψ normalized? If not, normalize it.
(b) What is the probability of finding the electron in the state ψ 2 px ?
(a) It is normalized, as shown in P14.7.
(b) The same as P14.7.
P14.9
Find the expectation values of the momentum operator px = −i
ˆ
states: ∂
∂x of the following (1)sin 2 x (3) eiax (a ∈ R) (2) cos 5 x (4) ebx (b ∈ R) (5) eiax (a ∈ R) (6)
2 1
1+ x The expectation value of an operator px = −i
ˆ ∫ dxψ * ( −i ∂
∂x in the state ψ is ∂
∂x )ψ . Therefore, b b (1) ∫ dx sin 2 x(−i ∫ dx sin 2 x cos 2 x = ) sin 2 x = −2i ∂
∂x a b (2) ∫ dx cos 5 x(−i ∂
∂x ) cos 5 x = 5i a b ∂
∂x )eiax = a a 1
2 i (sin 2 5b − sin 2 5a ) ∫ dxe − iax iax e = a (b − a ) a c c ∂
∂x a )ebx = −i b ∫ dxebx ebx = − i2 (e 2bc − e 2ba )
a b b (5) ∫ dxe −iax (−i
2 ∂
∂x )eiax = a
2 a ∫ dxe − iax 2 2 xeiax = a (b 2 − a 2 )
2 a b 1
(6) ∫ dx 1+ x (−i ∫ dx cos 5 x sin 5 x =
a b (4) ∫ dxebx (−i i (sin 2 2a − sin 2 2b) a b (3) ∫ dxe −iax (−i 1
2 b ∂
∂x 1
) 1+ x = −i a ∫ dx −1
(1+ x )3 = i [ (1+1a )3 − (1+1b )3 ] a P14.10
ˆ
Find the expectation values of the operator of kinetic energy Ek = − 2 m ∂∂x of the
2 2 2 following states:
(1)sin 2 x (3) eiax (a ∈ R) (2) cos 5 x (4) ebx (b ∈ R) (5) eiax (a ∈ R) (6)
2 1
1+ x b (1) ∫ dx sin 2 x(− b 2 ∂2
∂x 2 ) sin 2 x = 4 a 2 ∫ dx sin 2 x sin 2 x = 2 2 (cos 4b − cos 4a ) a The rest should be the same as 14.9.
P14.11
A particle is prepared in the following state:
a sin x + b cos x (a, b ≠ 0, a, b ∈ R) What are the possible measurement results if one measures its momentum ( px = −i
ˆ
What is the probability for each measurement result?
The eigenfunction of the momentum operator px = −i
ˆ ∂
∂x is Ae−ipx x / ∂
∂x )? a sin x + b cos x = ∑ c px Ae−ipx x /
px ∞ c px = ∫ dxAe Aa
2i = Aa
2i ∞ (a sin x + b cos x) = −∞
∞ = − ipx x / ∫ dxe − ip x x / − ip x x / (a e ix − e− ix
2i +b e ix + e− ix
2 ) −∞ (eix − e− ix ) + ∞
Ab
2 −∞ ∫ dxe − ipx x / (eix + e− ix ) −∞ [δ ( x − 1) − δ ( x + 1)] +
p = ( Aa +
2i ∫ dxAe p Ab
2 )δ ( px − 1) + (− Aa +
2i Ab
2 p
[δ ( x − 1) + δ ( x + 1)] Ab
2 p )δ ( px + 1) Therefore, the measurement of momentum may yield two possible results:
px = ± with probabilities of  Aa
2i +  and  − Aa +
2i Ab 2
2 Ab 2
2  ,respectively. P14.12
A particle is prepared in the following state:
a sin x + b cos x (a, b ≠ 0, a, b ∈ R) What are the possible measurement results if one measures its kinetic
ˆ
energy( Ek = − 2 m ∂∂x )? What is the probability for each measurement result?
2 2 2 Similar to 14.11, one needs to decompose the state a sin x + b cos x into the ˆ
eigenfunctions of the kinetic energy operator Ek = − 2 m ∂∂x .
2 2 2 P14.13
A particle is prepared in the following state:
a sin( x + y ) + b cos(2 x − 3 y ) (a, b ≠ 0, a, b ∈ R) Find the expectation value (average) of the angular momentum operator
lˆz = −i ( x ∂∂y − y ∂∂x ) hd ∫ ∫ dxdy[a sin( x + y) + b cos(2 x − 3 y ) lˆz a sin( x + y ) + b cos(2 x − 3 y )] gc hd = −i ∫ ∫ dxdy[a sin( x + y) + b cos(2 x − 3 y) lˆz a sin( x + y) + b cos(2 x − 3 y)] gc
hd = −i ∫ ∫ dx[a sin( x + y) + b cos(2 x − 3 y )] ( x ∂∂y − y ∂∂x )[a sin( x + y ) + b cos(2 x − 3 y )] gc 下略。 P14.14
A particle is prepared in the following state:
a sin( x + y ) + b cos(2 x − 3 y ) (a, b ≠ 0, a, b ∈ R) What are the possible measurement results if one measures its angular momentum lˆz = −i ( x ∂∂y − y ∂∂x ) ? What is the probability for each measurement result? Hint: The eigenfunctions of lˆz = −i ( x ∂∂y − y ∂∂x ) are 1
2π eimφ ,where m are integers, then using the transformation between polar and Cartesian coordinates.
P14.15
A particle is prepared in the following state:
a sin( x + y )ei 2t / + b cos(2 x − 3 y )ei 2t / (a, b ≠ 0, a, b ∈ R) . Is the particle in a stationary state?
Yes, The state can be written as [a sin( x + y ) + b cos(2 x − 3 y )]ei 2t / . where the spatial and temporal variables are separated.
P14.16
A particle is prepared in the following state:
a sin( x + y )ei 2t / + b cos(2 x − 3 y )e −i 2t / (a, b ≠ 0, a, b ∈ R) . Is the particle in a stationary state?
No. The wavefunction cannot be separated as the product of two functions where one
contains spatial variables only and the other contains temporal variables only. In fact,
the state is a superposition of two stationary states: one with energy 2 and other with
energy 2. ...
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