mathmatical physics

1. NPTEL – Physics – Mathematical Physics - 1
Lecture 11
Orthogonal basis:
Gram Schmidt Orthogonalization (GSO)
Two 𝑛 − component vectors 𝑎⃗ and 𝑏⃗⃗ be orthogonal if < 𝑎⃗|𝑏⃗⃗ >= 0 . Let us
suppose we have n vectors 𝑣⃗1, 𝑣⃗2 … … … … … , 𝑣⃗𝑛 in 𝑅𝑛 which are mutually
orthogonal and all different from zero, such that,
< 𝑣⃗𝑖 |𝑣⃗𝑗 > =
0
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(11.1)
This set of vectors forms a basis for 𝑅𝑛. The proof follows immediately if we can show
that the set 𝑣⃗1 … … … … . . 𝑣⃗𝑛 is linearly independent since any set of linearly
independent vectors from 𝑅𝑛 forms a basis for 𝑅𝑛.
Consider the problem of finding the 𝜆𝑖 which will satisfy
𝜆1𝑣⃗1 + 𝜆2𝑣⃗2 + … … … … … … 𝜆𝑛𝑣⃗𝑛 = 0
Taking a scalar product with 𝑣⃗1 in (2)
𝜆1|𝑣1|2 + 𝜆2 < 𝑣⃗1|𝑣⃗2 > + … … … … 𝜆𝑛 < 𝑣⃗1|𝑣⃗𝑛 >=
0
Thus 𝜆1 = 0 as |𝑣1|2 ≠ 0
(11.2)
Similarly taking the scalar product with 𝑣⃗2, one can show, 𝜆2 = 0. Hence each of the 𝜆𝑖′𝑠
are zero. So the vectors are linearly independent and hence form a basis for 𝑅𝑛. Thus any
set of n mutually orthogonal nonzero vectors from 𝑅𝑛 yields a basis for 𝑅𝑛.
Let us divide each vector 𝑣⃗𝑖 by its length |𝑣⃗1|and write
𝑢⃗⃗ = |𝑣⃗ | ≠
0
𝑖 𝑖
𝑣
⃗⃗𝑖
|𝑣⃗⃗
|
𝑖
< 𝑢⃗⃗𝑖|𝑢⃗⃗𝑗 >= 𝛿𝑖𝑗
So the definition of orthonormal basis is a set of n mutually orthogonal
vectors of unit length from 𝑅𝑛 forms what is called as an orthonormal
basis for 𝐸𝑛.
Orthonormal bases are especially interesting because any vector 𝑥⃗ in 𝑅𝑛
can be expressed very easily as a linear combination of the orthonormal
basis vectors.
𝑥⃗ = 𝜆1𝑢⃗⃗1 + 𝜆2𝑢⃗⃗2 + … … … . 𝜆𝑛𝑢⃗⃗𝑛
where 𝜆𝑖 =< 𝑢⃗⃗𝑖|𝑥⃗ >

2. NPTEL – Physics – Mathematical Physics - 1
Any set of n given linearly independent vectors from 𝑅𝑛 can be converted into an
orthonormal basis by a procedure known as Gram Schmidt (GS) orthogonalisation
procedure. Let us suppose that 𝑎⃗1 … … … . 𝑎⃗𝑛 are n linearly independent vectors forming
𝑅𝑛. We select any vector from this set, for example, 𝑎⃗1. This vector defines a direction in
space, and we build the orthonormal set around it.
1
Let us define the vector of unit length 𝑢⃗⃗ as 𝑢⃗⃗ = 𝑎⃗⃗1
. To obtain a vector 𝑣⃗
orthogonal
1 1 |𝑎⃗⃗
|
2
to 𝑢⃗⃗1, subtract from 𝑎⃗2 a scalar multiple of 𝑢⃗⃗1, that is, 𝑣⃗2 which is expressed
as,
𝑣⃗2 = 𝑎⃗2 − 𝛼1𝑢⃗⃗1
and 𝛼1 is determined so that < 𝑢⃗⃗1|𝑣⃗2 > = 0
Thus, 𝛼1 =< 𝑢⃗⃗1|𝑎⃗2 >, 𝛼1 can be thought of as the component of 𝑎⃗2 along 𝑢⃗⃗1.
Thus,
𝑣⃗2 = 𝑎⃗2−< 𝑢1|𝑎2 > 𝑢1
A second unit vector orthogonal to 𝑢⃗⃗1 is defined by,
𝑢⃗⃗
=
2
𝑣
⃗⃗2
|𝑣⃗⃗
|
2
A vector orthogonal to 𝑢⃗⃗1 and 𝑢⃗⃗2 is found by subtracting from 𝑎⃗3, the vector
components of 𝑎⃗3 along 𝑢⃗⃗1 and 𝑢⃗⃗2.
𝑣⃗3 = 𝑎⃗3−< 𝑢⃗⃗1|𝑎⃗3 > 𝑢⃗⃗1− < 𝑢⃗⃗2|𝑢⃗⃗3 > 𝑢⃗⃗2
The vector 𝑣⃗3 is clearly orthogonal to 𝑢⃗⃗1 and 𝑢⃗⃗2 . Thus the third unit vector
which is
orthogonal to 𝑢⃗⃗ and 𝑢⃗⃗ is 𝑢⃗⃗ = 𝑣⃗⃗3
. The procedure is continued until an
orthonormal
1 2 3 |𝑣⃗⃗
|
3
basis is obtained. In general,
𝑣⃗𝑟 = 𝑎⃗𝑟 − ∑𝑟−1 < 𝑢⃗⃗𝑖 |𝑎⃗ > 𝑢⃗⃗ . Thus,
𝑢⃗⃗ =
𝑖=1 𝑟 𝑖
𝑟
𝑣⃗⃗
𝑟
|𝑣⃗⃗
𝑟|
Example
Using GS procedure, construct an orthonormal
basis form
𝑎⃗1 = [2,3,0], 𝑎⃗2 = [6,1,0] and 𝑎⃗3 = [0,2,4]. It can be seen that 𝑎⃗1, 𝑎⃗2 and 𝑎⃗3
are all linearly independent as they do not lie in a plane. Then if,
𝑢⃗⃗ = = [2,3,0] = [0.554,0.831,0]
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1
𝑎
⃗⃗1
|𝑎⃗⃗ | √13
1
1
𝑣⃗2 = 𝑎⃗2−< 𝑢1|𝑎⃗2 > 𝑢⃗⃗1
< 𝑢⃗⃗1|𝑎⃗2 > = 4.16 ; < 𝑢⃗⃗1|𝑎⃗2 > 𝑢⃗⃗1 =
[2.30, 3.45, 0].

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NPTEL – Physics – Mathematical Physics - 1
Thus, 𝑣⃗2 = [3.70, −2.45, 0]
𝑢⃗⃗
=
2
𝑣
⃗⃗2
|𝑣⃗⃗
|
2
Similarly carry on for 𝑢3 and so on.