Ch 4. Using Quantum
Mechanics on Simple Systems
- Discussion of constrained and not constrained
particle motion
ex) free particle, In 2-D or 3-D boxes, In vice versa
- Continuous energy spectrum of Q.M free particle
- Discrete energy spectrum and preferred position of
Q.M particles in the box (Quantized energy levels)
MS310 Quantum Physical Chemistry
4.1. The free particle
Free particle : no forces
Classical 1-dimension, no forces :
d2x
F  ma  m 2  0
dt
Solution : x = x0 + v0t
x0 ,v0 : initial condition, constants of integration
Explicit value : must be known initial condition
How about the free particle in Q.M?
Time-independent Schrödinger Equation in 1-dimension is
 2 d 2 ( x )

 V ( x ) ( x )  E ( x )
2
2m dx
MS310 Quantum Physical Chemistry
constant V(x) : can choose the reference V(x)=0 (absolute
potential reference doesn’t exist)
d 2 ( x )
2m
→ reduced to
  2 E ( x )
2
dx

p  2mE , k 
Use these notations
2

Solution is given by   ( x )  A e  i
  ( x )  A e

p
 2mE /  2

( 2 mE /  2 ) x
 i ( 2 mE /  2 ) x
 A e ikx
 A e  ikx
 ( x )    ( x )    ( x )  A e ikx  A e  ikx
Obtain Ψ(x,t) : multiply each e-i(E/ℏ)t or equivalently e-iωt (E = ℏω)
 2k 2
Eigenvalue E 
: not quantized(all energy allowed : k is
2m
continuous variable)
MS310 Quantum Physical Chemistry
Plane wave cannot be localized.
→ cannot speak about the position of particle.
Then, what about probability of finding a particle?
→ also cannot calculate (wave function cannot be normalized
in interval -∞ < x < ∞)
However, if x is ‘restricted’ to the interval –L ≤ x ≤ L then
P ( x )dx 
 * ( x ) ( x )dx
L
*

 ( x ) ( x )dx
L

A A e ikxe  ikxdx
L
A A  e ikxe  ikxdx

dx
2L
L
P(x) : independent of x → no information about position
MS310 Quantum Physical Chemistry
What about the momentum of particle?
 

 ( x )   i A e ikx  kA e ikx  k  ( x )
x
x


ˆp x  ( x )   i   ( x )   i
A e  ikx  kA e  ikx  k  ( x )
x
x
pˆ x  ( x )   i
ψ+(x) : state of momentum + ℏk(positive direction)
ψ-(x) : state of momentum –ℏk(negative direction)
MS310 Quantum Physical Chemistry
4.2 The particle in a One-dimensional box
1-dimensional box : particle in the range 0<x<a only
impenetrable : infinite potential
V(x) = 0 for 0 < x < a
= ∞ for x ≥ a , x ≤ 0
MS310 Quantum Physical Chemistry
Schrödinger Equation is changed by
d 2 ( x ) 2m
 2 [V ( x )  E ] ( x )
dx 2

If ψ(x) ≠ 0 outside the box, then value of ψ’’(x) becomes
infinite because value of V(x) is infinity outside the box.
However, 2nd derivative exists and well-behaved
→ ψ(x) must be 0 outside the box
boundary condition : ψ(0) = ψ(a) =0
Inside the box : same as the free particle
We can write the solution by the sin and cos.
 ( x )  A e ikx  A e  ikx
 A (cos kx  i sin kx )  A (cos kx  i sin kx )
 ( A  A ) cos kx  i ( A  A ) sin kx
 A sin kx  B cos kx
MS310 Quantum Physical Chemistry
From the consideration of boundary conditions
 ( 0)   ( a )  0
 ( 0)  0  B  0
 ( x )  A sin kx
 (a )  A sin ka  0
 ka  n , n  0, 1, 2, 3, 
n
k
a
 n ( x )  A sin
n
x
a
(notice : n)
Normalization

1    n ( x ) n ( x )dx  A 2  sin 2
a
0

 A2 
a
0
n x
dx
a
1
2nx 
1 2
2 1 
1

cos
dx

A
x

aA




2
a 
 2 0 2
a
2
2
nx
A
, n ( x ) 
sin(
)
a
a
a
Energy of the particle
2
2
2

d

 n 
Hˆ  n ( x )  

(
x
)


  n ( x )  E n n ( x )
n
2
2m dx
2m  a 
2
 2  n 
n 2 h2
En 

 
2m  a 
8ma 2
2
‘Quantization’ arises by the boundary condition
Particle is ‘quantized’, n : quantum number
Ground state : n=1
However, energy of n=1 is not zero : zero point energy(ZPE)
2   
h2

  
2m  a 
8ma 2
2
E n 1
particle in a box : ‘stationary’ wave(not a traveling wave)
Also, n increase → # of node increase → wave vector k increase
because
2 p
k
  2mE /  2
 
Finally, what about a classical limit?
→ same as result of C.M(same probability in everywhere)
MS310 Quantum Physical Chemistry
graph of ψn(x) and ψn*(x)ψn(x)
MS310 Quantum Physical Chemistry
Graph of ψn2(x) / [ψ12(x)]max
n increase : large energy
Lower resolution : cannot precise measure → near to C.M
Result of Q.M ‘approach’ to the C.M when classical limit
MS310 Quantum Physical Chemistry
4.3 Two- and Three- dimensional boxes
Boundary condition : similar to 1-dimensional box
V(x, y, z) = 0 for 0 < x < a, 0 < y < b, 0 < z < c
= ∞ otherwise
Inside the box, Schrödinger Equation is given by
2  2
2
2

( 2  2  2 ) ( x , y, z )  E ( x , y, z )
2m x
y
z
Solving by separation of variable
And equation is changed by
2
d2
d2
d2

(Y ( y ) Z ( z ) 2 X ( x )  X ( x ) Z ( z ) 2 Y ( y )  X ( x )Y ( y ) 2 Z ( z ))
2m
dx
dy
dz
 EX ( x )Y ( y ) Z ( z )
MS310 Quantum Physical Chemistry
Divide both side by X(x)Y(y)Z(z)
2
1 d 2 X ( x)
1 d 2Y ( y )
1 d 2 Z (z)

(


) E
2
2
2
2m X ( x ) dx
Y ( y ) dy
Z ( z ) dz
E : independent to coordinate → E = Ex + Ey + Ez and original
equation(PDE) reduced to three ODEs.
2 d 2 X ( x)
 2 d 2Y ( y )
2 d 2 Z (z)

 E x X ( x ),
 E yY ( y ),
 Ez Z ( z )
2
2
2
2m dx
2m dy
2m dz
Solution of each equation is already given.
n yy
n z
nxx
 nx n y nz ( x , y , z )  N sin
sin
sin z
a
b
c
2
h2 nx2 n y nz2
And energy is given by E  E x  E y  E z 
( 2  2  2)
8m a
b
c
MS310 Quantum Physical Chemistry
Normalization
n yy
b
c
n z
n xx
2
1     n d  N  sin
dx  sin
dy  sin 2 z dz
0
0
0
a
b
c
2n yy 
a 1
b1
c 1
2nzx 
2n xx 
2


 N   1  cos
dx
1

cos
dx
1

cos

dz
 0 


0 2
0
a 
2
b 
2
c 


n
2
a
1  1
 N 2  x 
 2 0  2
a
b
2
c
1
 1 
y   z   abcN 2
0  2 0 8
n yy
nxx
8
8
n z
N 
, nx n y nz ( x, y, z ) 
sin
sin
sin z
abc
abc
a
b
c
(nx , n y , nz : 1,2,3...)
MS310 Quantum Physical Chemistry
If total energy is sum of independent terms
→ wave function is product of corresponding functions
Solution has a three quantum numbers : nx, ny, nz
→ more than one state may have a same energy
: energy level is degenerate and # of state is degeneracy
ex) if a=b=c, energy of (2,1,1), (1,2,1), and (1,1,2) is same.
h2
3h 2
2
2
2
E1,1, 2 
(1  1  2 ) 
2
8ma
4ma 2
in this case, state (2,1,1), (1,2,1) and (1,1,2) is degenerate and
degeneracy of the level is 3.
2-dimensional box problem : similar to 3-dimensional problem
(end-of-chapter problem)
MS310 Quantum Physical Chemistry
4. Using the postulate to understand the particle in the box and
vice versa
Postulate 1 : The state of a quantum mechanical system is
completely specified by a wave function Ψ(x,t). The probability
that a particle will be found at time t in a spatial interval of
width dx centered at x0 given by Ψ*(x0,t)Ψ(x0,t)dx.
We see the postulates of Q.M using the particle in a box.
Ex) 4.2
ψ(x) = c sin (πx/a) + d sin (2πx/a)
a. Is ψ(x) an acceptable wave function of particle in a box?
b. Is ψ(x) an eigenfunction of the total energy operator Ĥ?
c. Is ψ(x) normalized?
MS310 Quantum Physical Chemistry
Sol)
a. Yes.
ψ(x) = c sin (πx/a) + d sin (2πx/a) satisfies the boundary
condition, ψ(0) = ψ(a) = 0 and well-behaved function.
Therefore, ψ(x) is acceptable wave function.
b. No.
2
2
2 2

d

x
2

x


x
2x
Hˆ  ( x )  
(
c
sin

d
sin
)

(
c
sin

4
d
sin
)
2
2
2m dx
a
a
2ma
a
a
Result of Ĥψ(x) is not ψ(x) multiplied by constants. Therefore,
ψ(x) is not a eigenfunction of the total energy operator.
c. No
a
 | c sin
0
x
a
2
 d sin
2x
x
2x
*
*
  | c | sin
dx   | d | sin
dx   (cd  c d ) sin sin
dx
a
a
a
a
0
0
0
a
2
2
x
2x
| dx
a
a
a
2
2
MS310 Quantum Physical Chemistry
Third integral becomes zero because of orthogonality.
a
2
|
c
|
 sin
0
2 x
a
a
dx   | d |2 sin 2
0
2x
dx
a
1 a
a
2 1 a
| c | [ ]0  | d | [ ]0  (| c |2  | d |2 )
2
2
2
2
Therefore, ψ(x) is not normalized. However, the function
2
x
2x
[c sin  d sin
]
a
a
a
is normalized when |c|2+|d|2=1
Superposition state depends on time. Why?
2  iE1t / 
x
2x
 iE2 t / 
( x , t ) 
[ce
sin  de
sin
]   ( x ) f (t )
a
a
a
Therefore, this state doesn’t describe the stationary state.
MS310 Quantum Physical Chemistry
Then, what about a probability of particle in the interval?
Ex) 4.3
probability of ground-state particle in the central third?
Sol) ground state :
2a / 3
2
P    ( x ) 1 ( x )dx 
a
a/3
*
1
 1 ( x) 
2a / 3

a/3
sin 2
2
x
sin
a
a
x
2 a a
4
2
dx  [ 
(sin
 sin )]  0.609
a
a 6 4
3
3
Probability of finding a particle in central third is 60.9%.
However, we cannot obtain this result by one individual
measurement. We can only predict the result of large
number of experiment(60.9%).
MS310 Quantum Physical Chemistry
We can understand the two postulates together.
Postulate 3: In any single measurement of the observable that
corresponds to the operator Â, the only values that will ever be
measured are the eigenvalues of that operator.
Postulate 4 : If the system is in a state described by the wave
function Ψ(x,t), and the value of the observable a is measured
once each on many identically prepared systems, the average
value(also called expectation value) of all of those
measurement is given by

*

 ( x , t ) Aˆ ( x, t )dx
 a   
  ( x , t )( x , t )dx
*

MS310 Quantum Physical Chemistry
1) wave function is an eigenfunction of operator.
→ all measurement gives same value and it is average value
Ex) ground state of particle in a box
 d
  
Hˆ  1 ( x )  

(
x
)

   1 ( x )  E1 1 ( x )
1
2
2m dx
2m  a 
2
2
2
2
  
h2
E1 
  
2m  a 
8ma 2
2
2
2) wave function is not an eigenfunction of operator.
→ each measurement gives different value
Ex) normalized superposition state
 ( x) 
2
x
2x
[c sin  d sin
], | c |2  | d |2  1
a
a
a
MS310 Quantum Physical Chemistry


2
2

d
 E    ( x ) Hˆ  ( x )dx    ( x )[ 
 V ( x )] ( x )dx
2
2m dx


*
*
2
x
2x
2 d 2
x
2x
*
*
  (c sin  d sin
)[
]( c sin  d sin
)
2
a0
a
a
2m dx
a
a
a
2
x
2x
 [| c |2 E1  sin 2
dx  | d |2 E 2  sin 2
dx ]
a
a
a
0
0
a
a
2
x
2x
x
2x
 [c *dE 2  sin sin
dx  cd * E1  sin sin
dx ]
a
a
a
a
a
0
0
a
a
Last two integrals are zero by orthogonality and final result is
2
x
2x
 E  [| c |2 E1  sin 2
dx  | d |2 E 2  sin 2
dx ] | c |2 E1  | d 2 | E 2
a
a
a
0
0
a
where
a
n 2 h2
En 
8ma 2
MS310 Quantum Physical Chemistry
However, result of individual experiment is only E1 or E2 by the
postulate 3. How can represent the result?
→ by the postulate 4, result of the large number of individual
experiment, probability of E1 is |c|2 and probability of E2 is |d|2,
and the ‘average value’ of energy <E> = |c|2E1| + |d|2E2.
More generally, we can think about this case
Ψ(x) = cΨ1(x) + dΨ2(x) + 0(Ψ3(x) + Ψ4(x) + …)
All coefficient except Ψ1(x) and Ψ2(x) is zero. Therefore, no other
energy is measured except the E1 and E2.
MS310 Quantum Physical Chemistry
Now, consider the momentum.
d
ˆ
p


i

We know x
dx and can calculate the average value of
Momentum of nth state.
2
n x
sin
a
a
 ( x) 
2
n x
d
nx
 p    ( x ) pˆ  ( x )dx   sin
[ i (sin
)]dx
a
a
dx
a
0
0
a
a
*
 2 i n 
n x
nx
 2 i n 

sin
cos
dx

(sin 2 n  sin 2 0)  0
2
2

a
a
a
a
0
a
In the Q.M, momentum of particle : cannot be zero
(energy E = p2 / 2m cannot be zero in Q.M)
→ average of two superposition state is zero!
We can rewrite the wave function by complex form.
e ix  e  ix
(use the sin x 
)
2i
MS310 Quantum Physical Chemistry
2
nx
2 e inx / a  e  inx / a
 ( x) 
sin

(
)
a
a
a
2i
d
n inx / a
pˆ e inx / a   i e inx / a 
e
dx
a
d  inx / a
n  inx / a
 inx / a
pˆ e
  i e

e
dx
a
In the case of momentum, two probability of positive momentum
and negative momentum is same. Therefore, the average value
seems to zero.
MS310 Quantum Physical Chemistry
Ex) 4.4
Particle in the ground state.
a. Is wave function the eigenfunction of position operator?
b. calculate the average value of the position <x>.
Sol)  ( x )  2 sin x
a
a
2
x
2
x
a. position operator : xˆ  x, x ( x ) 
x sin
c
sin
 c ( x )
a
a
a
a
Therefore, wave function is not an eigenfunction of position operator.
b. expectation value is calculated by postulate 4.
2
x
x
2
x
 x    ( x ) xˆ  ( x )dx   sin
x sin dx   x sin 2
dx
a
a
a
a
a
0
0
0
a
a
a
*
2x
2x
x sin
2
2
2
x
2
a
a
a
a
a
a 
a ]  [( 
[ 

0
)

]

0
 2

4
a 4 8 2
8 2
2
8( )
4
a
a
2
cos
Average value of particle is half, the expected position.
MS310 Quantum Physical Chemistry
Summary
- The motion of particle which is not constrained shows
continuous energy spectrum however, the particle in a box
has a discrete energy spectrum.
- The state of a quantum mechanical system is completely
specified by a wave function Ψ(x,t). The probability that a
particle will be found at time t in a spatial interval of width dx
centered at x0 given by Ψ*(x0,t)Ψ(x0,t)dx.
- In any single measurement of the observable that
corresponds to the operator Â, the only values that will ever
be measured are the eigenvalues of that operator.
MS310 Quantum Physical Chemistry
If the system is in a state described by the wave function
Ψ(x,t), and the value of the observable a is measured once
each on many identically prepared systems, the average
value(also called expectation value) of all of those
measurement is given by

*

 ( x , t ) Aˆ ( x, t )dx
 a   
  ( x , t )( x , t )dx
*

MS310 Quantum Physical Chemistry