The Dirac Delta Function

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The Dirac delta function

August 9, 2015

1 Introduction

The Dirac delta function δ(x) is characterised by

δ(x) = 0 for x 6= 0 (1) ( Z b 1 whenever 0 ∈ (a, b) δ(x) dx = (2) a 0 otherwise

The delta function is a misnomer, in that it is not really a function R → R: no function R → R can have this property. (What would its value at 0 be?) Technically, it is a generalised function or distribution, and a proper treatment of it would require a foray into distribution theory. But there are several ways of thinking about the delta function intuitively.

2 As a symbol

The simplest way to think of the delta function is just as a symbol, giving it meaning only when it appears inside an integral, deﬁning it by ( Z b f(0) whenever 0 ∈ (a, b) f(x)δ(x) dx = (3) a 0 otherwise for a smooth function f. This follows from Equation 2, noting that the integrand is zero whenever x is away from the origin, and that by continuity of f we have f(x) approximately constant near x = 0, so that

Z b Z f(x)δ(x) dx = f(x)δ(x) dx (4) a − Z = (f(0) + O())δ(x) dx (5) − Z = f(0) (1 + O())δ(x) dx (6) − = f(0) + O() (7) for any > 0, and taking the limit → 0. This is called the sampling property of the Dirac delta function. Note that this is the continuous analogue of the result δijaj = ai for the Kronecker delta δij. We can also evaluate integrals which contain δ(g(x)) by making the substitution y = g(x) to obtain an integral containint δ(y). In particular, we have

Z b f(x)δ(x − c) dx = f(c) (8) a

1 whenever c ∈ (a, b), and Z b 1 f(x)δ(kx) dx = f(0). (9) a k These ideas are explored in more detail in the Part IB Methods course.

3 As a limit of increasingly spiky functions

A more sophisticated way to deﬁne the value of the integral

Z b I = f(x)δ(x) dx (10) a is to think of it as the limit of integrals of the form

Z b In = f(x)Kn(x) dx (11) a where the functions Kn are integrable and satisfy

Kn(x) → 0 as n → ∞ for all x 6= 0 (12) Z ∞ Kn(x) dx = 1. (13) ∞ (where the convergence is uniform outside (−, )).

Equations 12 and 13 mean that the graphs of the Kn’s are increasingly tall and thin, becoming a narrow spike at x = 0. Equation 13 means that the spikes all have the same area. For example, we could take ( n/2 for |x| < n Kn(x) = (14) 0 for |x| > n

This motivates the deﬁnition of the delta function δ(x) as the pointwise limit of the sequence Kn(x). (Note that this relies on ‘integral of limit equals limit of integral’; see Analysis II for a more rigorous treatment.)

4 As a probability distribution

We can give a probabilistic interpretation to this. Let Kn be the probability density function of the normal 2 2 distribution, with mean 0 and variance σn = 1/n , so that n 1 2 2 Kn(x) = √ exp − n x (15) 2π 2

For each n, let Xn be a random variable with probability density function Kn. Since the variances are decreasing as n increases, the Xn’s are increasingly likely to take values near 0. In the limit n → ∞, we have σn → 0: which means that the random variable may take no value other than 0. Thus, the delta function is the probability density function of a ‘random’ variable that has mean 0 and variance 0: it can take no value other than 0. 0). Of course, we can think of things like δ(x − c) and 1 2 (δ(x) + δ(x − 1)) in a similar way. Instead of probability distributions, we could think of distributions of some other quantity, such as mass density or charge density. A point mass or point charge can then be represented using a delta function (or its equivalent in two or three dimensions). See Vector Calculus.

2 5 As an impulse force

There is a mechanical interpretation of the Dirac delta function as an impulse force. Consider a particle with mass m moving in one dimension with position x(t). If a force F (t) is applied on the particle, then Newton’s second law states that mx¨(t) = F (t) (16) Deﬁning the momentum of the particle as p(t) = mx˙(t), we have

p˙(t) = F (t) (17) wherefore Z t2 0 0 I = p(t2) − p(t1) = F (t ) dt : (18) t1 the change in momentum (the impulse) is equal to the integral of an applied force. Suppose that a constant force f is applied for 0 < t < T and is otherwise zero. For t1 < 0 < t2, we then have

I = fT. (19)

The impulse is unchanged if f is doubled and T is halved. We can repeat this process. We think of the delta function as an infinite force which is applied over an infinitesimal time interval, such that the change in momentum that it causes is finite. Such a force is called an impulse force, and can be used to model collisions. For example, suppose a free particle moves along the positive x-axis and collides elastically with a wall at x = 0 at time t = 0. For t < 0 the particle is moving with velocityx ˙ = −v and for t > 0 the particle is moving with velocityx ˙ = v, where v > 0. In other words,

x˙(t) = v(2H(t) − 1) (20) where H is the Heaviside step function. The acceleration of the particle is inﬁnite at t = 0 and otherwise zero. Indeed, x¨(t) = 2vδ(t) (21) This illustrates the fact that the integral of the delta function is the Heaviside step function. The displacement of the particle is given by ( −vt t < 0 x(t) = (22) vt t > 0 = v|t| (23) illustrating that x(t) is continuous at t = 0.

6 Solving diﬀerential equations involving the delta function

The above example illustrates how to solve linear diﬀerential equations which have the delta function as a forcing term, such as Equation 21. Consider the equation

x¨(t) + p(t)x ˙(t) + q(t)x(t) = δ(t) (24)

Assume that p(t) and q(t) are not singular at or near t = 0.

Suppose we are given initial conditions at some t0 < 0. We want to ﬁnd x(t). Recall that δ(t) = 0 for t 6= 0. We therefore have the two domains t < 0 and t > 0 on which the RHS of (24) is zero, i.e.

x¨(t) + p(t)x ˙(t) + q(t)x(t) = 0 for t 6= 0 (25)

3 This is a homogeneous linear equation, and it admits two linearly independent solutions, x1 and x2, which we ﬁnd either in closed form or by using a series solution. Then the general solution of (25) is

x(t) = Ax1(t) + Bx2(t), (26) a linear combination of the two solutions. The values of A and B will be diﬀerent for t < 0 and t > 0, so that ( A x (t) + B x (t) for x > 0 x(t) = + 1 + 2 (27) A−x1(t) + B−x2(t) for x < 0 and we want to ﬁnd the four unknowns A+, B+, A− and B−. Now think back to the interpretation of §5, in which the impulsive force instantly changes the momentum of the particle, but does not change its position. Mathematically, this means that x(t) must be continuous butx ˙(t) jumps. The requirement that x(t) must be continuous gives one relationship between the four unknowns, namely A+x1(0) + B+x2(0) = A−x1(0) + B−x2(0). (28)

To work out how muchx ˙(t) must jump, integrate Equation 24 across a small interval (−, ): Z Z x¨(t) + p(t)x ˙(t) + q(t)x(t) dt = δ(t) dt. (29) − − The RHS is equal to 1. The ﬁrst term on the LHS integrates to the jump ofx ˙ across t = 0:

[x ˙]− =x ˙() − x˙(−). (30)

The second and third terms vanish, because p, q and x are continuous and non-singular at or near t = 0. Thus, Equation 29 gives us a second relationship: A+x˙ 1(0) + B+x˙ 2(0) − A−x˙ 1(0) + B−x˙ 2(0) = 1 (31)

The initial conditions give two more relationships between the unknowns. In total, we have four equations for four unknowns, which may be solved.