Winter 2013
Chem 356: Introductory Quantum Mechanics
Chapter 2 – Intro to Math Techniques for Quantum Mechanics ............................................................... 22
Intro to differential equations ................................................................................................................ 22
Boundary Conditions............................................................................................................................... 25
Partial differential equations and separation of variables ..................................................................... 25
Introduction to Statistics......................................................................................................................... 30
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
Intro to differential equations
Function y  y ( x) is to satisfy a differential equation
d2 y
dy
 5  6 y  0 x
2
dx
dx
(1)
For this ‘type’ of Differential equation (more later), try solution y  e x
Then
dy
  e x
dx
d2 y
  2 e x
2
dx
dn y
  n e x
n
dx
Substitute into differential equation (1)
 2 e x  5 e x  6e x  0
x
Or
 2  5  6  0
(  3)(  2)  0
Solutions:
But:
  3,   2
e3 x :
9e3 x  15e3 x  6e3 x  0
e2 x :
4e3 x  10e3 x  6e3 x  0
any linear combination of solutions is also solution
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Winter 2013
Chem 356: Introductory Quantum Mechanics
y ( x)  c1e3 x  c2 e 2 x
9c1e3 x  4c2 e 2 x
15c1e3 x  10c2 e 2 x
6c1e3 x  6c2 e 2 x
0
0
Let us try another one
d2 y
y0
dx 2
x
 2 e x  e x  0
  i
2 1  0
General Solution:
ci e  c2e ix
ix
Alternative way to write:
eix  cos x  i sin x
eix  cos x  i sin( x)  cos x  i sin x
y( x)  (c 1  c2 ) cos x  i(c1  c2 )sin x
 d1 cos x  d2 sin x
define
d1  c1  c2 , d2  i(c1  c2 )
→ choose d1 , d 2 ‘real’
Verify:
d2
d
cos x  ( sin x)   cos x
2
dx
dx
d2
d
sin x  (cos x)   sin x
2
dx
dx
d2 y
 y  0 , as expected
dx 2
Type of solutions
e  x , e   x , cos  x , sin  x ,  real,  >0
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Chem 356: Introductory Quantum Mechanics
When does this work?
0  cB y  c1
dy
d2 y
d3y
 c2 2  c3 3  .....
dx
dx
dx
(1) Constant coefficients in front of y and its derivatives
d2 y
 x2 y  0
2
dx
Linear in function y
→ not:
(2)
d2 y
dy
y
0
→ not:
2
dx
dx
(3)
Homogeneous equation
→ not: c2
2 y
y
 c1
 c3 y  0
2
x
x
For inhomogeneous differential equation:
→ Find particular solution y  P( x) add to this the general solution of inhomogeneous
equation.
For more complicated differential equations (ie. Not homogeneous DE with constant
coefficients) solutions are often hard to find



Many tricks of the trade
Use symbolic math program (it knows many of the tricks)
Numerical approaches (often work very easily → picture of solution)
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Winter 2013
Chem 356: Introductory Quantum Mechanics
Boundary Conditions
Let us consider our original differential equation.
d2 y
dy
5  y  0
2
dx
dx
y ( x)  c1e3 x  c2 e 2 x
Now impose further conditions. Eg:
y (0)  0
c1  c2  0
 dy 
  1
 dx  x 0
3c1  2c2  1
c2  c1
3c1  2c1  1
y ( x )  e3 x  e 2 x
c2  1
c1  1
statistics DE and boundary conditions
Solution is completely specified if one supplies as many conditions as one has free
coefficients c1 , c2 , …. in the solution
→ always linear set of equations
So recipe is very simple
Try y( x)  e x and work it out!
Partial differential equations and separation of variables
Consider problem of vibrating string (eg. guitar, violin)
We want to describe the amplitude u ( x, t )
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Winter 2013
Chem 356: Introductory Quantum Mechanics
2u
1 2u
(
x
,
t
)

( x, t )
x 2
v 2 t 2
v : Velocity of wave propagation in string, related to spring constant, (as sound of the
DE
string)
Boundary Condition:
u (0, t )  u (a, t )  0
t
 u 
u ( x, t ) : function of 2 variables → use partial derivatives  
 x t
In math we typically do not write is kept constant
(compare thermodynamics)
How to solve PDE (partial differential equation)?
Try solution u ( x, t )  X ( x)T (t )
Simple product of a function of x and a function of t
Boundary Condition: X (0)  X (a)  0
Substitute trial function into PDE
d 2 X ( x)
1 d 2T
T
(
t
)

X ( x)
dx 2
v 2 dt 2
Divide both sides by X ( x)T (t ) :
1 d 2 X ( x ) 1 1 d 2T
 2
X ( x) dx 2
v T (t ) dt 2
only depends on x only depends on t
Like f ( x )  g (t ) should be true for all x ,
t
f ( x)  g (t0 )
→
f ( x ) is constant
 g (t1 )
→
f ( x ) is constant, should be same
g (t )  f ( x0 )
→
g (t ) is constant
 f ( x1 )
f ( x)  g (t )
x, t
Can only be true if both functions are constant! Ie. The same constant
Let us call this constant, the separation constant  k 2 (for later simplicity)
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Chem 356: Introductory Quantum Mechanics
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d2X
 k 2 X ( x)
2
dx
X (0)  X (a)  0
1 d 2T
 k 2T (t )
2
2
v dt
no boundary conditions
Now we can use techniques discussed before ( k is constant)
Try X ( x)  e x
 2 e x  k 2 e x
  ik ,   ik
X ( x)  c1eikx  c2 e  ikx
Note k could be imaginary im
k 2  (im)2   m2
0
However we know that the string will oscillate, and hence can anticipate e ikx , with k real
Using what we did before
c1eikx  c2 eikx  (c1  c2 ) cos kx  i(c1  c2 )sin kx
 d1 cos kx  d2 sin kx
This is general solution. Now consider boundary conditions.
x 0:
d1 cos k 0  d2 sin k 0
d11  d2 0  0
x a:
d2  0
d1  0
d 2 sin(ka)  0
(flat string possibility)
When is sin( ka )  0 ?
ka  n
or
sin( ka )  0
x  0,  , 2 , 3 ...
k
 n x 
General solution: X ( x)  d n sin 

 a 
n
This is solution for x at particular value for kn 
a
n
a
kn 2  
n 2 2
a2
Now consider corresponding solution for T (t )
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Winter 2013
Chem 356: Introductory Quantum Mechanics
1 d 2T
 n 
 
 T (t )
2
2
v dt
 a 
2
d 2T
  wn 2T (t )
2
dt
wn 
n v
a
Similar equation as before:
T (t )  c1 sin wn (t )  c2 cos( wn t )
If we combine this with X we get
 n x   n vt 
 n x 
 n vt 
u ( x, t )  An sin 
 sin 
  Bn sin 
 cos 

 a   a 
 a 
 a 
This is a solution for any value of An , Bn and any value of n  0, 1, 2, 3
Most general solution:
u( x, t ) 
 n x 
sin 
  An sin( wn t )  Bn cos( wn t ) 
 a 
n 0,1,2,3...

You can verify that this indeed satisfies PDE
How can we interpret this?
sin
n x
:
a
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Winter 2013
Chem 356: Introductory Quantum Mechanics
n x
  n x 
:
sin 
   sin
a
 a 
Same solution, restrict n  0,1, 2,3
( n  0 , nothing extra)
So a string vibrates as a linear combination of modes, each of the modes oscillates in time at a
different frequency.
n x
n v
sin
wn 
a
a
 nw0
All multiples of fundamental frequency
This defines the pitch of the sound
v
a
 w0
w0
The other modes are called over tones
Meaning of coefficients An , Bn ?
t 0
 n v 
Bn cos 

 a 
 n x 
sin 

 a 
The initial shape of function
du
n v 

 An  cos

dt
a 

the initial velocity of the string.
Different instruments, guitars, violins, cello
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Winter 2013
Chem 356: Introductory Quantum Mechanics
 different An , Bn
How you attack the string determines the initial shape/velocity
 compare Chinese zither: hit the snare in different spots or twang the string
None of this affects the pitch  w0 the general harmonic
Period sin( wt )  sin w(t  T )
sin( wt  2 )  sin( wt )
2  wt
T
2
w
;
w
2
T
Introduction to Statistics
We will see that quantum mechanics is essentially a statistical theory. We can predict
the results and their distribution from a large number of repeated experiments only. We cannot
predict (even in principle) the outcome of an individual experiment.
Let us therefore talk about statistics using a simple example: the dice
If you throw the dice, each throw will yield the result 1, 2, 3, 4, 5 or 6.
If you throw the dice many times,
say 6000 times, we might get
1
2
3
4
5
6
Total
1010
980
995
1025
1030
960
6000
For a fair dice, each number has equal chance, and so we say the probability to throw
1
for example a ‘3’ is . This is reflected by the actual numbers we got in the example.
6
ni
1000 1

  Pi
6000 6000 6
1
In the limit that we throw a very large (infinite) times, we get closer and closer to
6
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Chem 356: Introductory Quantum Mechanics
Winter 2013
ni
only has meaning for many repeated experiments
Ntotal
Pi 
P 1
i
We might call ai the actual outcome of experiment, here ai = 1, 2, 3, 4, 5, 6
The average is given by
1
Ntot
n a
 Pa
i i
i
i
i
i
For dice:
1
21
1
(1  2  3  4  5  6) 
3
6
6
2
The average value does not need to be a possible outcome of an individual throw.
We are also interested in the variance of the results. We call the average A or A .
(both are used)
Then the variance is given by:
 A2   Pi (ai  A)2
0
i
Let us take a dice with 5 sides to make the numbers easier
ai  1, 2, 3, 4, 5
Pi 
,
1
5
A 3
1
5
1
 [4  1  0  1  4]  2
5
 A2  [(1  3) 2  (2  3) 2  (3  3) 2  (4  3) 2  (5  3) 2 ]
Standard Deviation  A  2
I can write the variance differently as
 P (a
i
i
A) 2   Pi (ai 2  2ai A  A 2 )
i
2
2
  Pa
i i  2 A  Pa
i i  A  Pi
i
i
i
 A2  2 AA  AA
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Winter 2013
Chem 356: Introductory Quantum Mechanics
 A2  A
  A2
2
Let us check for the 5 face dice:
1
A2  (12  22  32  42  52 )
5
1
 (1  4  9  16  25)
5
1
 (55)  11
5
A
2
 33  9
A2  A
2
 2   A2
(as before)
Of course: We proved this is true mathematically!
This concludes (for now) discussion of discrete statistics.
Now consider the case of a continuous distribution. For example a density distribution.
 ( x)dx  dm
 the mass between x and x  dx
(in 1 dimension)

  ( x)dx  M
total mass

b
Also
  ( x)dx 
mass between points a and b
a
If we normalize
1
 ( x) dx
M
probability to find a fraction of the total mass between x and x  dx
P ( x) dx 
We can define average position

x 
 xP( x)dx


x
2

 x P( x)dx
2

 x 2  x2  x
2
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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Chem 356: Introductory Quantum Mechanics
Winter 2013
Example: take a box between 0 and a
1
a
0 xa
0
elsewhere
P( x) 
Uniform

a
1
x
 P( x)dx  0 a dx  a  0  1
1)
a
“normalized”

a
1
x2 
a

2)  xP ( x)dx   xdx 

a
2a  0 2

0
a

a
1 2
x3 
1
 a2
3)  x P ( x)dx   x dx 

a
3a  0 3

0
a
2
2
4)  x  x
2
 x
2
2
1
a2
1 
 a2   a  
3
12
2 
 0 Always
More complicated distributions are possible of course
Famous is the Gaussian distribution
 x2
P( x)  ce 2 a
‘ a ’ parameter (will be  x )
2
c : normalization constant

 P( x)  1
c

Then
x 
x
2

1
a 2

 xe
1
2 a 2
 x2
2 a2
0

1
a 2

xe
2
 x2
2 a2
dx
Odd function f ( x)   f ( x)

a 2 2 a 2


 2  a2
2
a 2
 x  a as advertised
1
(this was the reason to define the Gaussian like this)
See book for discussion integrals
Chapter 2 – Intro to Math Techniques for Quantum Mechanics
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