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```Lecture 4
Partial differentiation
(c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has
been developed and made available online by work supported jointly by University of Illinois, the
National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus
Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and
conclusions or recommendations expressed in this material are those of the author(s) and do not
necessarily reflect the views of the sponsoring agencies.
Partial differentiation



In quantum chemistry, we often deal with a
function of more than just one variables, for
example, f(x,y,z).
When considering the derivatives of a multivariable function, we must be aware of with
respect to which variable we are
differentiating the function.
 f  is used to indicate the derivative is
  with respect to z, and x and y are
 z  x, y held fixed in the differentiation.
Partial differentiation


∂ (round d) indicates the partial
differentiation.
Consider a function of space (x) and time (t)
variables, f(x, t). Let the space variable also
depend on time x = x(t). In this case, a partial
derivative of f with respect to t is different
from the exact derivative because
df ( x, t )  f   f  dx  f 
    
 
dt
 t  x  x t dt  t  x
Partial differentiation

Other than that, partial differentiation follows
essentially the same rules as usual
differentiation.
æ dy ö æ dx ö
1
çè dx ÷ø = çè dy ÷ø is true
Is this true?
 y   x 
   
 x   y 
YES and NO - this is only true if the variables
held fixed are identical in the left- and righthand sides.
-1

Partial differentiation


Consider the change in function f(x,y,z)
caused by an increase in x (y and z held
fixed) and then in y (x and z held fixed).
The result would be the same if we increase
y first and then x.
 f
 f

xy yx
2
2
Partial differentiation
 f
 f

xy yx
2
2
Partial differentiation
f ( x, y)  ax y  by
3
 f 
2
   3ax y
 x  y
 f
2
 3ax
yx
2
2
 f 
3
   ax  2by
 y  x
 f
2
 3ax
xy
2
Time-dependent
Schrödinger equation


We use partial derivatives for operators.
For example, the energy operator is,

E  i 
 t  x , y , z
We do not differentiate x, y, z dependent part
of the wave function by t (see the simple
wave in the previous lecture)
Time-dependent
Schrödinger equation

The time-dependent Schrödinger equation is:
  
ˆ
H  i

 t  x, y , z
The Schrödinger equation

For one-dimension, it is
2
2
é
ù
d
ˆ
ˆ
HY = ê + V (x)ú Y = EY
2
ë 2m dx
û

The kinetic energy operator comes from the
classical to quantum conversion of the
momentum operator
d
p  i
dx
The Schrödinger equation

In three-dimension, we have three Cartesian
components of a momentum:



px  i ; p y  i ; p z  i
x
y
z


Accordingly, the momentum operator is a
vector operator:





p  ( px , p y , pz )    i ,i ,i   i
x
y
z 

 (“del”) is a vector  / x,  / y,  / z 
The Schrödinger equation

Kinetic energy in classical mechanics:
2
p

2m
p x2  p y2  p z2
2m
 
2
2
2
p

p

p

p

p
(The vector inner product is
x
y
z )

In quantum mechanics:
2
2
2





 
2
 2  2  2 

 
2m
2m  x y z 
2
2
Ñ2 ( Del squared) = D ( Laplacian)
The Schrödinger equation

The Schrödinger equation in three
dimensions is,
2
2
2
2
é
ù
æ
ö ˆ
¶
¶
¶
ˆ
HY = ê + 2 + 2 ÷ + V(x, y, z)ú Y = EY
2
ç
ë 2m è ¶x ¶y ¶z ø
û
The Schrödinger equation
in spherical coordinates

coordinates (x, y, z), it is
sometimes more
convenient to use
spherical coordinates
(r, θ, φ)
x  r sin  cos
y  r sin  sin 
z  r cos
The Schrödinger equation

The kinetic energy operator can be written in
two ways – Cartesian or spherical.
 2 2

H  
  V ( x, y, z )   E
 2m




  2 2 2
x y z
2
2
2
2
2

2  1
2
  2
 2
r
r r r
 1 2
1 
 

sin 
 2

2
sin  
 
 sin  
Homework Challenge #2

Derive the spherical-coordinate expression of
2 (the green panel) using the equations in

blue and orange panels.
x  r sin  cos
2
2
2



2
y  r sin  sin 
  2 2 2
x y z
z  r cos
2

2  1
2
  2
 2
r
r r r
 1 2
1 
 

sin 
 2

2
sin  
 
 sin  
Summary


We use partial derivatives to define quantummechanical operators.
In using partial derivatives, we must be aware
of which variables are being held fixed.
```