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Lecture 15 Phys 3750 D M Riffe -1- 2/18/2009 The Dirac Delta Function Overview and Motivation: The Dirac delta function is a concept that is useful throughout physics. For example, the charge density associated with a point charge can be represented using the delta function. As we will see when we discuss Fourier transforms (next lecture), the delta function naturally arises in that setting. Key Mathematics: The Dirac delta function! I. Introduction The basic equation associated with the Dirac delta function ()xδ is ()() ()0fdxxfx =∫∞∞−δ, (1) where ()xf is any function that is continuous at 0=x . Equation (1) should seem strange: we have an integral that only depends upon the value of the function ()xf at 0=x . Because an integral is "the area under the curve," we expect its value to not depend only upon one particular value of x. Indeed, there is no function ()xδ that satisfies Eq. (1). However, there is another kind of mathematical object, known as a generalized function (or distribution), that can be defined that satisfies Eq. (1). A generalized function can be defined as the limit of a sequence of functions. Let's see how this works in the case of ()xδ. Let's start with the normalized Gaussian functions ()2nxnenxg−=π. (2) Here 21σ=n , where σ is the standard Gaussian width parameter. These functions are normalized in the sense that their integrals equal 1, ()1=∫∞∞−dxxgn (3) for any value of n (>0). Let's now consider the sequence of functions for K,3,2,1=n ,Lecture 15 Phys 3750 D M Riffe -2- 2/18/2009 ()211xexg−=π, ()2222xexg−=π, … , ()2100100100xexg−=π, … (4) What does this sequence of functions look like? We can summarize this sequence as follows. As n increases (a) ()0ng becomes larger; (b) ()0≠xgn eventually becomes smaller; (c) the width of the center peak becomes smaller; (d) but ()1=∫∞∞−dxxgn remains constant. The following figure plots some of the functions in this sequence. 2 10 1 205101520n = 1n = 4n = 16n = 64n = 256n = 1n = 4n = 16n = 64n = 256xgn(x) Let's now ask ourselves, what does the ∞=n limit of this sequence look like? Based on (a) through (d) above we would (perhaps simplistically) say (a) ()∞=∞0g ; (b) ()00 =≠xgn; (c) the width of the center peak equals zero;Lecture 15 Phys 3750 D M Riffe -3- 2/18/2009 (d) but ()1=∫∞∞−∞dxxg . Note that (a) and (b) are not compatible with (d) if )(xg∞is a function in the standard sense, because for a function (a) and (b) would imply that the integral of )(xg∞ is zero. So how should we think of this sequence of functions, then? Well, the sequence is only really useful if it appears as part of an integral, as in, for example, () () ()∫∫∞∞−−∞→∞∞−∞→= dxxfedxxfxgnxnnnn2limlimπ. (5) Let's calculate the integral, and then the limit in Eq. (5). The following figure should help with the calculation. 2 10 1 2051015g256(x)f(x)g256(x)f(x)xgn(x), f(x) As n get large, ()2nxnnexg−=π becomes narrower such that it only has weight very close to 0=x . Thus, as far as the integral is concerned, for large enough n only ()xf at 0=x is important. We can thus replace ()xf by ()0f in the integral, which gives usLecture 15 Phys 3750 D M Riffe -4- 2/18/2009 () () () ()01lim0lim0lim22ffdxefdxxfennxnnnxnn===∞→∞∞−−∞→∞∞−−∞→∫∫ππ (6) Thus, the integral on the lhs of Eq. (1) is really shorthand for the integral on the lhs of Eq. (6), That is, the Dirac delta function is defined via the equation () () ()∫∫∞∞−−∞→∞∞−= dxxfedxxfxnxnn2limπδ (7) Now often (as physicists) we often get lazy and write ()2limnxnnex−∞→=πδ, (8) but this is simply shorthand for Eq. (7). Eq. (8) really has no meaning unless the function 2nxne−π appears inside an integral and the limit ∞→nlim appears outside the same integral. However, after you get used to working with the delta function, you will rarely need to even think about the limit that is used to define it. One other thing to note. This particular sequence of functions ()2nxnnexg−=π that we have used here is not unique. There are infinitely many sequences that can be used to define the delta function. For example, we could also have defined ()xδ via ()()xnxxnsin1limπδ∞→= . (9) The sequence of functions ()()xnxπsin is illustrated in the figure at the top of the next page. Notice that the key features of both of these two difference sequences are expressed by (a) – (d) at the top of page 5. II. Delta Function Properties There are a number of properties of the delta function that are worth committing to memory. They include the following, ()() ()xfdxxfxx′=′−∫∞∞−δ, (10)Lecture 15 Phys 3750 D M Riffe -5- 2/18/2009 () () ()0fdxxfx′−=′∫∞∞−δ (11) () ()xaaxδδ= (12) The proof of Eq. (10) is relatively straightforward. Let's change the integration variable to xxy′−= , dxdy = , which gives ()() ()()∫∫∞∞−∞∞−′+=′− dyxyfydxxfxxδδ. (13) Then using Eq. (1), we see that Eq. (10) is simply equal to ()xf′. QED. Let's also prove Eq. (12). We do this in two steps, for 0>a and then for 0<a . (i) First, we assume that 0>a. Then 4 20240510152025n = 1n = 4n = 8n = 16n = 32n = 1n = 4n = 8n = 16n = 32xsin(nx)/(pi*x)Lecture 15 Phys 3750 D M Riffe -6- 2/18/2009 ()()∫∫∞∞−∞∞−= dxaxdxaxδδ (14) Changing integration variable axy =, adxdy =, this last equation becomes () ()∫∫∞∞−∞∞−= dyyadxaxδδ (15) and changing variables back to x via yx=, dydx= gives () ()∫∫∞∞−∞∞−= dxxadxaxδδ (16) and so for 0>a we have ()()xaaxδδ=. (ii) We now assume 0<a . Then we have ()()∫∫∞∞−∞∞−−= dxaxdxaxδδ (17) Changing integration variable axy −= , adxdy −= , this last equation becomes () ()()()∫∫∫∫∞∞−∞∞−−∞∞∞∞−==−=dxxadyyadyyadxaxδδδδ (18) and so for 0<a we also have ()()xaaxδδ= . QED. We leave the proof of Eq. (11) as an exercise.Lecture 15 Phys 3750 D M Riffe -7- 2/18/2009 III. Fourier Series and the Delta Function Recall the complex Fourier series representation of a function ()xf defined on LxL ≤≤− , ()∑∞−∞==nLxninecxfπ, (19a) ()∫−−=LLLxinndxexfLcπ21. (19b) Let's now substitute nc from


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USU PHYS 3750 - The Dirac Delta Function

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