R. I. Badran Complete set of orthogonal basis Mathematical Physics

Complete set of orthogonal functions

Discrete set of vectors:

  ˆ ˆ ˆ ˆ ˆ ˆ The two vectors A  AX i A y j  Az k and B  BX i B y j  Bz k are   3 orthogonal if A B  0 or  Ai Bi  0 . i1

Continuous set of functions on an interval (a, b):

a) The two continuous functions A(x) and B (x) are orthogonal b on the interval (a, b) if A(x)B(x)dx  0. a

b) The two complex functions A (x) and B (x) are orthogonal on b the interval (a, b) if A (x)B(x)dx  0 , where A*(x) is the a complex conjugate of A (x).

c) For a whole set of functions An (x) (where n= 1, 2, 3,) and on the interval (a, b) b  0 if m  n An (x)Am (x)dx  a  , const.t  0 if m  n

An (x) is called a set of orthogonal functions.

Examples:

 0 m  n sin nx sin mxdx  i)   ,  m  n  0 where sin nx is a set of orthogonal functions on the interval (-, ).

Similarly  0 m  n cos nx cos mxdx  if    if m  n  0 R. I. Badran Complete set of orthogonal basis Mathematical Physics

 ii) sin nx cos mxdx  0 for any n and m   0 m  n (einx ) eimxdx  if iii)   2 if m  n  0 1 vi) P (x)Pm (x)dx  0 unless.  m 1

[Try to prove this; also solve problems (2, 5) of section 6]. 1 v) P (x)  (any polynomial of degree <  ) dx = 0.   1 [Solve problems (6 of section 6) & (4, 5 and 6 of section 7)].

 ˆ ˆ ˆ Conclusion: The vector A  AX i A y j  Az k in 3-D is expanded in terms of unit vectors iˆ, ˆj and . kˆ Then and form complete set of orthogonal basis vectors. Thus any vector (or function) can be expanded in terms of a set of orthogonal basis vectors (or functions).

Normalization of functions:   The scalar product A A  A2 gives the square of the length of a vector), where the length of a vector is also called the norm of a vector.

What is the norm of a unit vector?

How can we find the norm of a function A (x) on the interval (a, b)?

b b 2 Answer:  A (x)A(x)dx   A(x) dx  N 2  N is called the norm of the a a function over the interval (a, b).

Can we get a normalized function? A(x) Answer: Yes, we just take as a normalized function where N N-1 is called the normalization factor. R. I. Badran Complete set of orthogonal basis Mathematical Physics

Exercise: a) What is the norm of the function sin nx on (0, )?  2  Answer: Firstly, we find  sin nx dx  and then the norm is  . 0 2 2

b) Find the normalized function. 2 Answer: The normalized function is sin nx . [How can you be  so sure of that?!

Orthonormal set of vectors:

The unit vectors iˆ, ˆj and kˆ are a set of orthonormal vectors because they are orthogonal to each other and each has a norm equal 1.

Note: Any set of normalized and orthogonal functions is called 2 orthonormal (e.g. sin nx is an orthonormal set of basis  functions on (0 ,).

Any function (or a vector in vector space) like f (x) can be expanded on (0, ) in a Fourier sine series as 2 f (x)   Bn sin nx n  Here f(x) is considered a vector (or a function) with components

Bn while are the basis vectors.

[Note: In Q. M. a physical system is expressed as either a state function or a state vector].

Normalization of the Legendre polynomials:

What is the norm of P (x) on the interval (-1, 1)? R. I. Badran Complete set of orthogonal basis Mathematical Physics

1 2 [P (x)]2 dx  Answer: The norm of P (x) is.   1 2 1

Proof: Use the identity xP(x)  P1(x)  P (x) and multiply both sides by and integrate to get

1 1 1 2  [P (x)] dx  xP (x)P(x)dx  P (x)P1(x)dx 1 1 1

The last integral is zero (See problem 4 of section 7). Use the method of integration by part to find the 1st integral on R. H. S of the equation as:

1 1 1 2 xP (x)P(x)dx  1 [P (x)] dx 1 2 1 [Hint: to reach this answer you may need the identity  ( P (1)  (1) ](See problem 2, section 2). Substitute this result back into the previous equation to get:

1 2  [P (x)]2 dx    Q. E. D. 1 2 1

Orthonormal set of Legendre functions:

2 1 a. The functions P (x) form an orthonormal set of functions 2  n (-1, 1).

1 2 1 2m 1 0 if   m b. P (x) Pm (x)dx   m , where  is 1  m 2 2 1 if   m called Kronecker delta.

Reminder for suggested problems: Solve problems (6.2, 6.5, 6.6), 7.4, (8.1, 8.2, 8.5)