Quantum Mechanics for Scientists and Engineers. David Miller

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Quantum Mechanics for Scientists and Engineers David Miller

Background mathematics 5 Sum, factorial and product notations

Summation notation If we want to add a set of numbers a 1, a 2, a 3, and a 4, we can write S a1a2 a3 a4 or we can use summation notation If the range of j is obvious S a j Here j is an index 4 S a a a j j j j1 j1,2,3,4 j1,...,4 j plural of index indexes or indices

a1a2 a3 a4 Example arithmetic series For a set of numbers spaced by a constant amount, an arithmetic progression or sequence i.e., where the nth number is an a1 n1 d e.g., the numbers 4, 7, 10, and 13 i.e., a1 4, a2 7, a3 10 and a4 13 with, therefore d 3 and m 4 terms in the progression gives the series (sum of the terms) 4 7 10 13 34

a1a2 a3 a4 Example arithmetic series For a set of numbers spaced by a constant amount, an arithmetic progression or sequence i.e., where the nth number is an a1 n1 d e.g., the numbers 4, 7, 10, and 13 i.e., a1 4, a2 7, a3 10 and a4 13 with, therefore d 3 and m 4 terms in the progression gives the series (sum of the terms) 4 m aj a1 j 1d j1 j1

Example arithmetic series For a set of numbers spaced by a constant amount, an arithmetic progression or sequence i.e., where the nth number is an a1 n1 d e.g., the numbers 4, 7, 10, and 13 i.e., a1 4, a2 7, a3 10 and a4 13 with, therefore d 3 and m 4 terms in the progression gives the series (sum of the terms) a1a2 a3 a4 m 1 1 j1 a j d m a a 1 4 2

Example arithmetic series For a set of numbers spaced by a constant amount, an arithmetic progression or sequence i.e., where the nth number is an a1 n1 d e.g., the numbers 4, 7, 10, and 13 i.e., a1 4, a2 7, a3 10 and a4 13 with, therefore d 3 and m 4 terms in the progression gives the series (sum of the terms) m a1 am 413 a 1a2 a3 a4 a1 j1d m 4 34 2 2 j1

a1a2 a3 a4 Example geometric series With a constant ratio between successive terms, a geometric progression or sequence n 1 i.e., where the nth term is an ar 1 e.g., the numbers 3, 6, 12, and 24 i.e., a1 3, a2 6, a3 12 and a4 24 with, therefore r 2 and m 4 terms in the progression gives the series (sum of the terms) 3 6 12 24 45

a1a2 a3 a4 Example geometric series With a constant ratio between successive terms, a geometric progression or sequence n 1 i.e., where the nth term is an ar 1 e.g., the numbers 3, 6, 12, and 24 i.e., a1 3, a2 6, a3 12 and a4 24 with, therefore r 2 and m 4 terms in the progression gives the series (sum of the terms) 4 a m j j1 j1 j 1 ar 1

Example geometric series With a constant ratio between successive terms, a geometric progression or sequence n 1 i.e., where the nth term is an ar 1 e.g., the numbers 3, 6, 12, and 24 i.e., a1 3, a2 6, a3 12 and a4 24 with, therefore r 2 and m 4 terms in the progression gives the series (sum of the terms) 4 m m j1 1 r a1a2 a3 a4aj ar 1 a1 1 r j1 j1

Example geometric series With a constant ratio between successive terms, a geometric progression or sequence n 1 i.e., where the nth term is an ar 1 e.g., the numbers 3, 6, 12, and 24 i.e., a1 3, a2 6, a3 12 and a4 24 with, therefore r 2 and m 4 terms in the progression gives the series (sum of the terms) 4 m m j1 1r 116 a1a2 a3 a4aj ar 1 a1 3 315 45 1r 12 j1 j1

Summation over multiple indexes We can extend the summation notation Suppose we have two lists of numbers a 1, a 2, a 3, a 4 and b 1, b 2, b 3 and we want to add up all the products R ab 1 1ab 1 2ab 1 3 ab 2 1ab 2 2ab 2 3 ab 3 1ab 3 2ab 3 3 abab ab 4 1 4 2 4 3

Summation over multiple indexes Then we can write R ab ab ab 1 1 1 2 1 3 abab ab 2 1 2 2 2 3 ab ab ab 3 1 3 2 3 3 abab ab 4 1 4 2 4 3 4 3 ab j k j1 k1 and, because order of addition does not matter in various equivalent notations 4 3 3 4 R ab ab ab ab j k j k j k jk, j k j1 k1 k1 j1 j, k

Factorial notation Quite often, we need a convenient way of writing the product of successive integers e.g., 1 234 We can write this as 1 2344! called four factorial and using the exclamation point! The notation is obvious for most other cases Note, though, that we choose 0! 1

Product notation Generally, when we want to write the product of various successive terms a a a a 1 2 3 4 by analogy with the summation notation we can use the product notation a1a2a3a 4 a For example, for all integers n 1 n n! p p1 4 j1 j

Background mathematics 5 Power series

Analytic functions and power series For a very broad class of the functions in physics we presume they are analytic (except possibly at some singularities ) i.e., at least for some range of values of the The ellipsis means we omit writing some terms explicitly argument x near some point x o the function f x can be arbitrarily well approximated by a power series i.e., with x o =0for simplicity we have 2 3 f x ao a1xa2x a3x which may be an infinite series

Analytic functions and power series For a very broad class of the functions in physics we presume they are analytic (except possibly at some singularities ) i.e., at least for some range of values of the argument x near some point x o the function f x can be arbitrarily well approximated by a power series i.e., generally 2 3 o 1 o 2 o 3 o f x a a xx a xx a xx

Maclaurin series Taking the simplest case of x o =0 first 2 3 f x ao a1xa2x a3x Obviously at x =0 f 0 ao so, trivially, ao f Now, df 2 fx a12a2x3a3x dx so df f 0 a1 dx 0 0

Maclaurin series Continuing 2 d f f x 2a 2 2 32a3x2! a2 3! a3x dx so 2 d f f 0 2! a 2 2 dx 0 and 3 d f f 0 3! a 3 3 dx and so on 0

f f x x o Maclaurin and Taylor series Continuing gives the Maclaurin series 2 2 n n xdf x d f x d f f x f 0 2 n 1! dx 2! dx n! dx Repeating the same procedure around x x o with 2 3 f x a a xx a xx a xx gives the Taylor series 0 0 0 o 1 o 2 o 3 o 2 2 xx xx xx n n o df o d f o d f 2 n 1! dx x 2! dx n! dx x x o o o

Example power series expansions 1 1 1 x x x 2 3 x 2 3 x x expx1x 2! 3! 2 4 x x cos x 1 2! 4! sin x 3 5 x x x 3! 5!

Power series for approximations Maclaurin and Taylor series allow approximations for small ranges of the argument about a point Examples for small x Approximations to first order in x 1/ 1x 1x exp x 1 x sin x x 1x 1 x /2 tan x Note lowest order dependence on x for cos x is second order 2 cos x 1 x / 2 ln 1 x x x