The Sokhotski-Plemelj Formula

Transcription

1 hysics 24 Winter 207 The Sokhotski-lemelj Formula. The Sokhotski-lemelj formula The Sokhotski-lemelj formula is a relation between the following generalize functions (also calle istributions), ±iǫ = iπ(), () where ǫ > 0 is an infinitesimal quantity. This ientity formally makes sense only when first multiplie by a function f() that is smooth an non-singular in a neighborhoo of the origin, an then integrate over a range of containing the origin. We shall also assume that f() 0 sufficiently fast as ± in orer that integrals evaluate over the entire real line are convergent. Moreover, all surface terms at ± that arise when integrating by parts are assume to vanish. To establish eq. (), we shall prove that f() ±iǫ = f() where the auchy principal value integral is efine as: f() 0 { f() + iπf(0), (2) f(), (3) assuming f() is regular in a neighborhoo of the real ais an vanishes as. In these notes, I will provie three ifferent erivations of eq. (2). The first erivation is a mathematically non-rigorous proof of eq. (2), which shoul at least provie some insight into the origin of this result. A more rigorous erivation starts with a contour integral in the comple plane, f(z)z. z By efining appropriately, we will obtain two ifferent epressions for this integral. Setting the two resulting epressions equal yiels eq. (2) with the upper sign. omple conjugating this result yiels eq. (2) with the lower sign. Finally, an elegant thir proof makes irect use of the theory of istributions. Finally, a useful check is to consier the Fourier transform of eq. (), as iscusse in the Appeni to these notes. Note that eq. () can be generalize as follows, 0 ±iε = 0 iπ( 0 ), where { f() 0 f() f() The erivations presente below are easily moifie to encompass the above generalization.

2 2. A non-rigorous erivation of the Sokhotski-lemelj formula We begin with the ientity, ±iǫ = iǫ 2 +ǫ, 2 where ǫ is a positive infinitesimal quantity. Thus, for any smooth function that is non-singular in a neighborhoo of the origin, f() ±iǫ = f() f() iǫ 2 +ǫ 2 2 +ǫ. (4) 2 The first integral on the right ha sie of eq. (4), f() f() f() f() = ǫ 2 2 +ǫ 2 2 +ǫ 2 2 +ǫ. (5) 2 In the first two integrals on the right han sie of eq. (5), it is safe to take the it ǫ 0. In the thir integral on the right han sie of eq. (5), if is small enough, then we can approimate f() f(0) for values of <. Hence, eq. (5) yiels, However, { f() f() = + 2 +ǫ ǫ 2 = 0, f() +f(0) 2 +ǫ 2. (6) since the integran is an o function of that is being integrate symmetrically about the origin, an { f() f() f() +, 0 efines the principal value integral. Hence, eq. (6) yiels f() 2 +ǫ 2 = f(). (7) Net, we consier the secon integral on the right han sie of eq. (4). Since ǫ is an infinitesimal quantity, the only significant contribution from ǫ f() 2 +ǫ 2 can come from the integration region where 0, where the integran behaves like ǫ 2. Thus, we can again approimate f() f(0), in which case we obtain f() ǫ 2 +ǫ ǫf(0) = πf(0), (8) 2 2 +ǫ2 2

3 where we have mae use of 2 +ǫ = 2 ǫ tan (/ǫ) = π ǫ. Using the results of eqs. (7) an (8), we see that eq. (4) yiels, which establishes eq. (2). f() ±iǫ = f() iπf(0), (9) 2. A more rigorous erivation of the Sokhotski-lemelj formula We consier the following path of integration in the comple plane, enote by, shown below. Imz Rez That is, is the contour along the real ais from to, followe by a semicircular path (of raius ), followe by the contour along the real ais from to. The infinitesimal quantity is assume to be positive. Then f() = f() + f(), (0) where the principal value integral is efine in eq. (3). In the it of 0, we can approimate f() f(0) in the last integral on the right han sie of eq. (0). Noting that the contour can be parameterize as = e iθ for 0 θ π, we en up with Hence, f() 0 0 = f(0) 0 π ie iθ θ = iπf(0). eiθ f() = f() iπf(0). () We can also evaluate the left han sie of eq. () by eforming the contour to a contour that consists of a straight line that runs from + iε to + iε, where ε is a positive infinitesimal (of the same orer of magnitue as ). Assuming that f() has no 3

4 singularities in an infinitesimal neighborhoo aroun the real ais, we are free to eform the contour into without changing the value of the integral. It follows that f() +iε = f() = f(y +iε) y, (2) y +iε +iε where in the last step we have mae a change of the integration variable. Since ε is infinitesimal, we can approimate f(y+iε) f(y). Thus, after relabeling the integration variable y as, eq. (2) yiels f() = Inserting this result back into eq. () yiels f() +iε = f(). (3) +iε f() iπf(0). (4) Eq. (4) is also vali if f() is replace by f (). We can then take the comple conjugate of the resulting equation. The en result is 2 in agreement with eq. (9). f() ±iε = f() iπf(0), 3. An elegant erivation of the Sokhotski-lemelj formula Starting from the efinition of the auchy principal value given in eq. (3), we integrate by parts to obtain f() = f()ln f ()ln = f( ǫ)lnǫ f ()ln, f() = f()ln f ()ln = f(ǫ)lnǫ f ()ln, where f () f/ an we have assume that f() 0 sufficiently fast as ± so that the surface terms at ± vanish. Hence, { f() [f( ) f() ] = ln f ()ln f ()ln. 0 (5) More precisely, we can epan f(y+iε) in a Taylor series about ε = 0 to obtain f(y+iε) = f(y)+o(ε). At the en of the calculation, we may take ε 0, in which case the O(ε) terms vanish. 2 Alternatively, we can repeat the above erivation where the contour is replace by a semicircle of raius in the lower half comple plane, which yiels eq. () with i replace by i. Finally, after eforming the contour of integration to a new contour that consists of a straight line that runs from iε to iε, one obtains eq. (3) with i replace by i. 4

5 Since f() is ifferentiable an well behave, we can efine g() 0 f (t)t = f() f(0) which implies that g() is smooth an non-singular an, f() = f(0)+g(). (6) Inserting eq. (6) back into eq. (5) then yiels { f() [ 2g()ln = f ()ln 0 = f ()ln. f ()ln Note that ln is integrable at = 0, so that the last integral is well-efine. Finally, we integrate by parts an rop the surface terms at ± (uner the usual assumption that f () 0 sufficiently fast as ). The en result is f() That is, we have proven that as istributions, = f() ln. ln =. (7) We can employ eq. (7) to provie a very elegant erivation of eq. (). We begin with the efinition of the principal value of the comple logarithm, Ln z = ln z +iarg z, where arg z is the principal value of the argument (or phase) of the comple number z, with the convention that π < arg z π. In particular, for real an a positive infinitesimal ǫ, Ln(±iǫ) = ln ±iπθ( ), (8) ǫ 0 where Θ() is the Heavisie step function. Differentiating eq. (8) with respect to immeiately yiels the Sokhotski-lemelj formula, 3 where we have use eq. (7) an ±iǫ = iπ(), (9) Θ( ) = Θ() = (). 3 The erivative of the comple logarithm is Ln z/z = /z for z 0. 5

6 Appeni: Fourier transforms of istributions Eqs. (7) an (9), which we repeat below ln =, (20) ±iε = iπ(), (2) are only meaningful when multiplie by a test function f() an integrate over a region of the real line that inclues the point = 0. In the theory of tempere istributions, test functions must be infinitely ifferentiable an vanish at ± faster than any inverse power of. learly, e ik oes not satisfy this requirement for a test function. Nevertheless, one can efine Fourier transforms of tempere istributions by using the well known property of the Fourier transform, f(k)g(k)k = f(k) g(k), (22) where f(k) f()e ik. If f() is a tempere istribution an g() is a test function, then if follows that g() eists an is well efine. The Fourier transform of f(), enote by g(k), is efine via eq. (22). One can now check the valiity of eqs. (20) an (2) by computing their Fourier transforms. To compute the Fourier transform of eq. (20), we make use of the property of Fourier transforms that Hence, f() eik = ik f(k). ln eik = ik ln e ik. (23) The calculation of the right-han sie of eq. (23) is rather involve, since it only eists in the sense of istributions. One can show that 4 [ ln e ik = π f ] k +2γ(k), (24) where γ is the Euler-Mascheroni constant, an the istribution f(/ k ) is efine as f(k)f k k f(k) f(k) f(0) f(k) k + k + k, (25) k k k for any vali test function f(k). 4 See, e.g., Ram. Kanwal, Generalize Functions: Theory an Applications, Thir eition (Birkhäuser, Boston, 2004) pp an pp There are two typographical errors on these pages. In eq. (6.4.33), /u shoul be / u an in the last term in eq. (6.4.57), i(u i0) shoul be +i(u i0). Eq. (24) is a consequence of the correcte eq. (6.4.57). 6

7 Inserting the result of eq. (24) into eq. (23) an using k(k) = 0 an 5 ( k f ) = k = sgn(k), (26) k k the en result is given by, Net, we consier e ik = ln eik = iπsgn(k). (27) cos(k) sin(k) +i. (28) Since cos(k)/ is an o function of (i.e., it changes sign uner ), it immeiately follows from the efinition of the auchy principle value that cos(k) = 0. (29) Net, we observe that 0 sin(k)/ = k; that is, sin(k)/ is regular at = 0. Thus, sin(k) = sin(k) = sgn(k) siny y y = πsgn(k). (30) Note that the symbol has no effect on the integral given by eq. (30), since the integran is regular at = 0. The factor of sgn(k) arises after changing the integration variable, y = k. When k < 0, the integration its must be reverse, which then leas to the etra sign. Inserting eqs. (29) an (30) into eq. (28) then yiels, e ik = iπsgn(k). (3) In light of eqs. (27) an (3), we have verifie that the Fourier transform of eq. (20) is satisfie. Likewise, we can verify that the Fourier transform of eq. (2) is satisfie. The following result is require, e ik = 2πiΘ( k), (32) ±iε which was erive in Solution Set in hysics 25. Then, using eq. (3) an employing the two ientities, sgn(k) = Θ(k) Θ( k) an = Θ(k) + Θ( k), it follows that the Fourier transform of eq. (2) is 2πiΘ( k) = iπ [ Θ(k) Θ( k) ] iπ [ Θ(k)+Θ( k) ]. (33) It is a simple matter to check that eq. (33) is satisfie for either choice of sign. Since the Fourier transform of a tempere istribution an its inverse Fourier transform are unique, one can conclue that if the Fourier transforms of eqs. (20) an (2) are satisfie, then eqs. (20) an (2) are vali ientities. Thus, the Fourier transform technique ehibite in this Appeni provies a fourth inepenent erivation of the Sokhotski-lemelj formula. 5 When we multiply f(/ k ) by k, the singularity at k = 0 is cancele an the prescription inicate by eq. (25) is no longer require. Noting that k/ k is equal to sign of k for k 0, we en up with eq. (26). 7