Delta functions and all that1

We can define the step function (sometimes called the Heaviside step function), θ(x), by

θ(x) = 0 for x < 0 (1) = 1 for x > 0 . (2)

It is strictly not defined for x = 0 but physicists like to define it to be symmetric, θ(0) = 1/2. One can think of this discontinuous function as a limit:

1 + tanh(x/a) θ(x) = Lim + ; a→0 2 this is easy to check if one remembers that the hyperbolic tangent assumes the values ±1 as x → ±∞ respectively. a goes to zero from the positive direction. This limiting procedure is consistent with θ(0) = 1/2.

The step function is useful when something is switched on abruptly, say a voltage which is 0 for t < 0 and a constant above. In physical situations the step function is a convenient mathematical approximation.

The (Dirac) delta function is strictly not a function but is a so-called distribution or a generalized function. The theory behind this was worked out by Laurent Schwartz who received the Fields Medal, the mathematics equivalent of the Nobel Prize. We will use the delta function carefully, confident that the rigorous aspects have been checked.

The delta function, δ(x − a), is defined by the property Z c2 dx f(x) δ(x − a) = f(a) if a ∈ (c1, c2) (3) c1 = 0 otherwise (4) for any suitably continuous function.2 Note that if a does not lie in the range of integration then the argument of the delta function does not vanish and therefore

1For a rigorous yet clear exposition see M. J. Lighthill, ‘Introduction to Fourier Analysis and Generalized Functions. 2From the point of view of physics, this definition implies that the dimensions(units) of the delta function are the inverse of the dimensions of its argument.

1 it is always zero. In words, the delta function, when multiplied by a function and integrated, picks out the value of the function at the point where the argument of the delta function vanishes inside the range of the integral. If viewed naively as a function, the delta function x is zero when its argument is nonzero, i.e, when x 6= a and equal to ∞ when x = a. We can think of the delta function as the derivative of the step function: d δ(x) = θ(x) ; (5) dx clearly the derivative vanishes for x 6= 0 and because of the finite jump within an infinitesimal interval the derivative is infinite at x = 0.

The delta function can also be viewed as a limit of a sequence of functions or the limit of a function; for example, : 1 ² δ(x) = lim (6) ²→0 π x2 + ²2 1 x2 − 2 = limσ→0 √ e 2σ (7) 2πσ and other such representations. Check in each case that when x 6= 0 taking the spec- ified limit yields 0 and when x = 0 the limit yields infinity.

How are the constants in the above limits chosen ? From the fundamental property of the delta function we have Z ∞ dx δ(x) = 1 −∞ and the constants in the limits above are chosen to obey this integral constraint. For example, (changing variables from x to y = x/σ in the integral below) we have

Z ∞ 2 Z ∞ 1 − x 1 −y2/2 dx f(x) √ e 2σ2 = √ dy f(σy) e −∞ 2πσ 2π −∞ and upon expanding f(σy) for small σ as f(0) + O(σ) we find that all the terms proportional to powers of σ vanish and we are left behind with f(0) which is precisely the definition of a delta function. We also note that 1 sin νx lim = δ(x) . (8) ν→∞ π x From the study of Fourier integrals we find another of the more useful representa- tions of the delta function: Z ∞ dk 0 δ(x − x0) = eik(x−x ) . (9) −∞ 2π

2 Equivalently, since the variables in question are dummy variables,

Z ∞ dx ei(k−k0)x = 2π δ(k − k0) . (10) −∞ The material presented may be used in this course.

I list below some properties of the delta function which are useful. All the results in which a delta function appears without being integrated over should be viewed as being multiplied (on both sides of the equation) by a a good function f(x) being integrated over a range of x, say (−∞, +∞).

The delta function is even : δ(x) = δ(−x) , (11) For a real constant c, we have 1 δ( c(x − a) ) = δ(x − a) . (12) | c |

We also have, as can be shown with a bit of work, X 0 δ( f(x) ) = [ δ(x − xi)/ | f (xi) | ] (13) i

0 0 where f (x) = df/dx, f (xi) = df/dx|x=xi , and the {xi} are the roots of the equation f(x) = 0 and the sum is over all the roots. For example, 1 δ(x2 − a2) = [ δ(x − a) + δ(x + a)] 2a for a > 0.

One can (blithely) define the derivative of the delta function, δ0(x), even though it is not a function! Z c2 0 df dx δ (x) f(x) = − |x=0 if 0 ∈ (c1, c2) (14) c1 dx i.e., the derivative of the delta function picks out the value of the derivative of the function at the point where the argument of the delta function vanishes apart from a negative sign. This is easily verified by using integration by parts.

3 Exercise: Generalize the result to the nth derivative of the delta function by re- peated integration by parts.

Z ∞ Answer: dx δ(n)(x) f(x) = (−1)n f (n)(0) −∞ where the superscript n denotes the nth derivative. Another obvious result is

f(x) δ(x − a) = f(a) δ(x − a) .

One can generalize delta functions defined in one dimension above to two and three dimensions in Cartesian coordinates easily and in non-Cartesian coordinates with some care and work. The Jacobian appears naturally as is shown below. In two dimensions let δ(~r − ~r0) = δ(x − x0) δ(y − y0) = D(r, θ) δ(r − r0) δ(θ − θ0) in polar coordinates. Note that D has dimensions of 1/L. We have Z d2r g(~r) δ(~r − ~r0) = g(~r0) = g(r0, θ0) Ω ZZ = rdr dθ D(r, θ) δ(r − r0) δ(θ − θ0) g(r, θ) (15) and for this to be true for all g we have D = 1/r, i.e., it is the inverse of the Jacobian.

In three dimensions we have 1 1 δ(~r − ~r0) = δ(r −r0) δ(θ −θ0) δ(φ−φ0) = δ(r −r0) δ(cos θ − cos θ0) δ(φ−φ0) r2 sin θ r2 where we have used δ(θ − θ0) δ(cos θ − cos θ0) = sin θ observing that in spherical polar coordinates sin θ is non-negative. For cylindrical coordinates 1 δ(~r − ~r0) = δ(ρ − ρ0) δ(φ − φ0) δ(z − z0) . ρ

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