Lecture 23. Representation of the Dirac delta function in other coordinate systems

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1 Lectue 23 Repesentation of the Diac delta function in othe coodinate systems In a geneal sense, one can wite, ( ) = (x x ) (y y ) (z z ) = (u u ) (v v ) (w w ) J Whee J epesents the Jacobian of the tansfomation. a) Cylindical Coodinate System The volume element in given by, dv =ρ dρ d dz J(ρ,, z) = x ρ y ρ z ρ x y z x z x =ρ cos y y =ρ sin z z z = z z = cos sin sin cos The deteminant is J which is cos 2 + sin 2 = Thus ( ) = ( ) ( ) (z z ) Also, f(x, y, z ) = f (x, y, z)δ(x x ) δ(y y )δ(z z )dxdydz Joint initiative of IITs and IISc Funded by MHRD Page 8 of 5

2 = f (,, z) δ( )δ( )δ(z z ) d d dz = f(,, z ) b) Spheical pola coodinate system Following the definition as befoe, ( ) = ( ) ( ) ( ) 2 sin and f(x, y, z ) = f(,, z ). An impotant elation in Electodynamics Let us fist state the elation, 2 ( ) = 4πδ() We shall pove the above elation now. Using 2 = 2 (2 ) 2 ( ) = fo all >. But as the above identity does not stand as the opeato itself in not defined at =. (because of the facto 2). To know the behavio at =, conside Gauss's divegence theoem. v. Adv = A s. ds Suppose A = ( ) = 2 Joint initiative of IITs and IISc Funded by MHRD Page 9 of 5

3 In ode to evaluate the divegence of A at the oigin ( = ) Conside a sphee of adius R suounding the oigin. On the suface, A has a constant value R 2. Integating ove the spheical suface as shown in figue, A. ds = R 2 (. ) = 4 π θ= 2π Φ= The answe we have got is independent of R. Thus putting it in the divegence theoem,. ( )dv = A. ds = 4π R 2 sinθdθdφ 2 ( )dv = 4π The above esult is tue even in the limit R Using the integal popety of the - function δ()dv = Joint initiative of IITs and IISc Funded by MHRD Page of 5

4 Thus, 2 ( ) = 4 () In a geneal sense, we can wite 2 ( ) = 4 ( ) Applications to Physical Poblems As deived ealie, Whee = 2 ( ) = 4πδ() () The electostatic potential is of the function Thus, multiplying Eq. () with q 4πε Φ () = q 4πε 2 q ( 4πε ) = 4πq δ() = q δ() = 4πε ε ε Thus we ecove the Laplace s equation. Completeness condition of Special functions in tems of - function in quantum mechanics, the wavefunctions fo a hamonic oscillato wavefunctions ae given by, n (x)=a n H n (x) e x2 2 n =,,2.. Joint initiative of IITs and IISc Funded by MHRD Page of 5

5 Coesponding to an enegy spectum given by E n = (n + 2 ) h is the fequency and H n (x) epesents a complete set of othonomal functions in the domain < x <. H n 's ae called the Hemite polynomials and A N is the nomalization constant, A n = π 2 2 n n! The othogonality of the wave function is epesented by, m (x) n (x)dx = mn Since n (x) ae assumed to fom a complete set of functions, we can expand any well behaved function (x) as (x) = c n n n (x) We multiply above by m (x) and integated to obtain, m (x) (x)dx = n c n m (x) n (x)dx = n c n δ nm = c m Thus, substituting fo c n (x) = n [ n (x ) (x )dx ] = n (x) Whee the pimed summation is used as a dummy vaiable. Intechanging the summation and integation, (x) = dx (x )[ n n (x) n (x )] Joint initiative of IITs and IISc Funded by MHRD Page 2 of 5

6 Since the wavefunction foms an othonomal set, n n (x ) n (x) = (x x ): completeness condition. Thus plugging in the fom fo n (x) in tems of the Hemite polynomials, e (x2 +x 2 )/2 π n= H 2 n n! n Similaly fo the Legende polynomials, P n (x) (x)h n (x ) = δ (x x ) Fo - < x, x < (2n + )P 2 n= n (x ) = δ(x x ) x, x Similaly fo sinusoidal functions, f(x) = Asinkx = Asin nπx L ; x L A 2 sin nπx n sin nπx L L δ(x x ) Fo x, x L. Evaluate the integal, 6 (3x 2 2x )δ(x 3)dx 2 Tutoial Solution: Using f(x)δ(x a)dx = f(a) Hee f(a) = f(x = 3) = 27 6 = 2 2. Show that x d δ(x) = δ(x) dx Solution: f(x) [x d dx δ(x)] = xf(x)δ(x) Whee f(x) is an abitay function. The fist tem on the RHS is zeo as d dx (xf(x))δ(x)dx Joint initiative of IITs and IISc Funded by MHRD Page 3 of 5

7 δ(x) = at x = ± Also d dx (xf(x)) = f(x) + xf (x) Thus LHS = (f(x) + x df dx ) δ(x)dx = f() = f() = f(x)δ(x)dx Thus, x d dx δ(x) = δ(x) Poved. 3. Show that the deivative of a -function is a -function. A -function is defined by, θ(x) = fo x > = fo x Solution: Poceeding as in the pevious poblem, f(x) dθ dx dx = f(x)θ(x) df θ(x)dx dx = f() df ddx dx = f() f() + f() = f() = f(x)δ(x)dx Thus, dθ dx = δ(x) 4. Pove that δ(αx) = δ(x) whee α is a constant α Joint initiative of IITs and IISc Funded by MHRD Page 4 of 5

8 Solution: f(x)δ(αx)dx Changing vaiable fom x p = αx x = p α and dx = α dp If α is positive then the integation uns fom α to +α. With α as negative, x = α implies p = α and vice vesa. Thus the limits ae intechanged fo negative α that entails a negative sign. f(x)δ(αx)dx = ± f ( p dp ) δ(p) α α = ± α f() = α f() The poof follows in the same manne at poblems (2) and (3). 5. Evaluate the integal I = e (. v 2) dv Whee v is a sphee of adius R. ˆ Solution: I = e 4πδ 3 ( ) dv v = 4πe = 4π Joint initiative of IITs and IISc Funded by MHRD Page 5 of 5