Dimensional Regularization and Asymptotic Freedom of the δ-function Potential in Relativistic Quantum Mechanics

Transcription

1 Diensional Regularization and Asyptotic Freedo of the δ-function Potential in Relativistic Quantu Mechanics Bachelor Thesis at the Institute for Theoretical Physics, University of Bern Written by Igor Jurosevic and supervised by Prof. Dr. Uwe-Jens Wiese Septeber 04

2 Abstract This bachelor thesis is divided into three chapters. Chapter treats a non-relativistically oving particle in a δ-potential well. In a first attept, the setting is investigated in position and oentu space. The second chapter treats the sae potential, but this tie the particle is allowed to ove relativistically. The appearing divergences are treated by applying a ethod known as diensional regularization. In the treatent of the so-called scattering states, the running coupling constant λ, b is studied together with the corresponding β function and the asyptotic freedo of the λ, b coupling is shown. Since the potential can be regarded as caused by a super heavy ass sitting in the origin, the whole setting can be thought of as a relativistic scattering experient with a contact interaction. Since such contact interactions are not excluded by Leutwyler s classical non-interaction theore, the above described setting could be a hint to a quantu loop hole in the theore. Chapter 3 gives a sall recapitulation of the conclusions and rearks ade over the whole thesis.

3 Contents Preface Non-relativistic particle in the δ-potential well 3. xploration in position space using the particle physicist s approach The attractive case The bound state The scattering states The repulsive case Considering the scattering states by using the spatial syetry of the Dirac δ-potential. 0. xploration of the attractive δ-potential in oentu space The bound state The scattering states Orthogonality in oentu space Relativistic particle in the δ-potential well. The bound state Diensional regularisation and renoralization of the δ-potential and construction of the wave function The non-relativistic liit, λ b, ɛ, and the wave function The scattering states Regularisation, renoralization, and construction of the wave function The running coupling constant λ, b, asyptotic freedo, and the β-function Transission and reflection coefficients Orthogonality in oentu space Recapitulation, Rearks, and Conclusions 4 3. Chapter Chapter

4 Preface In order to create an easier access for physics students to topics such as diensional regularization and renoralization and to cross-check soe results obtained through the theory of self-adjoint extensions of the pseudo-differential operator H = p +, M. H. Al-Hashii, A. M. Shalaby, and U. - J. Wiese wrote a paper entitled Asyptotic Freedo, Diensional Transutation, and an Infra-red Conforal Fixed Point for the δ-function Potential in -diensional Relativistic Quantu Mechanics. Soe probles investigated in the paper are also treated in this thesis. Retrospectively we can say, that the paper constituted a kind of road ap for this thesis. We wish to ephasize the word retrospectively for the following reason: In order to give us an idea of the scientific work of a theoretical physicist and to keep us surrounded by the fog of exploration and ystery, our supervisor only gave us hints on the tasks we should perfor and kept the above entioned paper in secret. Therefore, we only gained knowledge of the existence of the above entioned paper after we had perfored our own coputations and thoughts about these topics.

5 Chapter Non-relativistic particle in the δ-potential well. xploration in position space using the particle physicist s approach.. The attractive case... The bound state In this chapter we will begin with the study of the Dirac δ [5] potential well. To understand a particle s behaviour in this well, we have to solve the one-diensional Schrödinger equation. In order to do so, we assue that the particle is oving in a non-relativistic way i.e. the particle s kinetic energy is sall copared to its rest ass energy. The potential has the following for, where λ is in R V x = λδx.. There are two possible types of such wells. The attractive ones, where λ < 0 and the repulsive ones, where λ > 0. The case λ = 0 deands a special treatent. We will consider it in ore detail in a few pages. As we can see, the potential doesn t depend on tie and therefore we are allowed to use the separation of variables approach for solving the Schrödinger equation. In this case the tie-dependence just affects a phase factor to the stationary part. Therefore, we shall focus on the explicit solution of the stationary equation d ψx + λδxψx = ψx.. dx For now we will consider an attractive potential, avoid the point x = 0 and focus on the sections where x 0. Further we shall define λ R + and attach the sign which corresponds to one of the two possible potential types i.e. a negative sign for the attractive case and a positive one for the repulsive case. We look for solutions, whose energy binds the particle to the well <0. The ter bind ay sound strange, but if one thinks of the δ-potential as a sequence of ever deepening and narrowing finite box potentials, where a particle ay indeed be inside the box, the ter becoes self-explanatory. For x < 0, < 0 d ψx = ψx..3 dx This differential equation can be solved by using an exponential ansatz ψ x = Ae kx + Be kx, k =..4 Since the unitarity principle deands noralizable solutions i.e. the probability to find the particle soewhere has to be, we have to abandon the second ter in the su. Otherwise, the function will blow up as x goes 3

6 to and we will fail at noralizing it. For x > 0, < 0 the situation is analogous ψ + x = Ce kx + De kx, k =..5 Again the unitarity principle forces us to drop a ter. This tie it has to be the first ter, or the function will blow up, as x goes to and this would, again, screw up the noralization of ψx. The solution is now of the for { Ae kx, x < 0 ψx = De kx, x > 0..6 So far the solution is a function with a discontinuity at x = 0. In order to patch it up at x = 0, we need conditions how to do this. Since we solved a second order differential equation we have also two degrees of freedo which we can fix with appropriate boundary conditions. The standard set of boundary conditions for probles like ours is. ψx continuous.. dψx dx continuous alost everywhere. The continuity of ψx can be achieved easily li ψx = li ψ x = li x 0 x 0 x Aekx = A,.7 0 li ψx = li ψ +x = li x + 0 x + 0 x De kx = D So if we want ψx to be continuous at x = 0, we have to ipose the condition A = D. Figure.: The for of ψx. The spike at x = 0 indicates that the derivative of ψx at x = 0 can t be continuous. When one looks at the plot in Figure., one can see that the continuity of dψx dx will, at least for x = 0, be violated. Not being continuous, doesn t ean, that there is no way to characterize the wave function s behaviour, as we now shall see: Consider the Hailtonian with x ] ɛ, ɛ[: d Hψx = ψx λδxψx = ψx..9 dx Unleashing an integral on this relation, iplies 4

7 ɛ ɛ Hψxdx = ɛ ɛ d dx ψxdx + ɛ ɛ ɛ λδxψxdx = ψxdx..0 ɛ With the integral on the Hailtonian we obtain two siplifications. First we get an expression with the derivative of ψx in it. Second, the part of the su containing the δ function strictly speaking, it is a distribution, but physicist s language conventions label it as a function becoes treatable in the sense, that it is no longer infinite! ɛ Hψxdx = ɛ d ɛ dx ψx ɛ λψ0 = ψxdx.. ɛ The equation yields a few ore things for harvesting. To bring in this harvest, we shall take a closer look at the first ter on the left side and the ter on the right side of the equation siultaneously ɛ ɛ d dx ψx = d ɛ dx ψɛ d ψ ɛ,. dx ɛ 0 ɛ ψxdx = ψxdx + ψxdx..3 ɛ 0 ɛ Considering the liit ɛ 0 yields li d ψx = ɛ 0 li ɛ 0 dx ψ ɛ d 0 ɛ Ae kx dx + Ae kx dx ɛ 0 li ɛ 0 dx ψɛ = li ɛ 0 = li ɛ 0 Ake kx 0 ɛ Ake kx ɛ 0 By inserting equations.4 and.5 into equation., we get Ake kɛ Ake kɛ = Ak,.4 = 0..5 Ak λψ0 = 0 Ak = λψ0..6 With the coputation above, we deterined the behaviour of the derivative of ψ. Although it is clear that integrating over a function with a finite nuber of discontinuities will deliver a soother function integrating has soothing properties, it is still very fascinating, that the singularity at x = 0 disturbs the continuity of the derivative only in such a way, that the corresponding function receives just a kink at this point. If we recall that ψ0 is the value of the wave function at the contact point x = 0, which we deterined to be A, we can siplify further k = λ..7 Again we have to reeber soething. While using the exponential ansatz for solving the Schrödinger equation we defined k, or the wave nuber, in dependence of. Therefore, we are now able to deterine its for, which is k = = λ 4 = λ..8 So the not yet noralized solution to the stationary Schrödinger equation looks the following way Ae λ x, x < 0 ψx = A, x = 0..9 Ae λ x, x > 0 It is possible to write it ore copactly. Then, it reads ψx = Ae λ x..0 5

8 Since we would like to have a physically eaningful description, we have to noralize the solution. A noralized solution eans, that the probability, which is the integral of the odulus of ψ squared, for finding the particle soewhere on the x-axis is one. Foralized, the noralization looks the following way, where ψx is the coplex conjugate of ψx. ψx dx = ψx ψxdx = Once ore we will consider both ters siultaneously 0 ψx ψxdx + 0 ψx ψxdx =.. Hence we obtain 0 0 A e λ x dx = λ A e λ x 0 A e λ x dx = λ A e λ x 0 = A λ,. = A λ..3 Therefore we arrive at A λ + = A λ λ = A = λ..4 λ A = A = ±..5 By convention we take the positive value as A. So the solution of the Schrödinger equation, or of the eigenvalue proble, takes the following for ψx = λ e λ x, = λ..6 Suarizing the thoughts of the last few pages we can say that we looked at a very special potential known as the Dirac δ-potential. We investigated its properties in the sense that we considered it in the stationary Schrödinger equation, and there we asked whether it provides bound states. Indeed, we found one bound state, which we then noralized. Figure. provides exaples with concrete values of and λ. Figure.: xaples of properly noralized bound state wave functions. 6

9 ... The scattering states Now we shall set our attention on an attractive potential, where the energy of the particle is greater than zero i.e. > 0, λ < 0. The procedure is analogous to the solution of the bound case. Again we avoid the point x = 0 and look at the two sectors, where x < 0 respectively x > 0. For x < 0, > 0 the Schrödinger equation reads d Hψx = d ψx = ψx ψx = ψx..7 dx dx This tie there is a sall coplication. The factor in front of ψx is negative. Such differential equations can be solved by using a coplex exponential ansatz ψ x = Ae ikx + Be ikx, k =..8 There is one big proble with this procedure: Since the solution represents cosine and sine which are oscillating functions, without any daping, noralization, in the usual sense, can not be obtained. Nevertheless, it can still be of great use, if one is just interested in the behaviour of the syste. The solution is describing a free particle and therefore there is a physical otivation for not discarding it. Furtherore, one can construct noralizable wave packages out of a bunch of given k s or wave nubers for the procedure take for instance [6], if desired. So if we constrain ourselves to the study of the behaviour of the syste, we can continue with our work. For x > 0, > 0 we obtain This tie our basic solution reads ψx + = Ce ikx + De ikx, k = ψx =..9 { Ae ikx + Be ikx, x < 0 Ce ikx + De ikx, x > Again ψ has a discontinuity at x = 0. We shall use our standard set of boundary conditions to patch up the function at x = 0. First, we investigate the continuity of ψx li ψx = li ψ x = li x 0 x 0 x Aeikx + Be ikx = A + B,.3 0 li ψx = li ψ +x = li x + 0 x + 0 x Ceikx+ De ikx = C + D If we deand continuity at x = 0, we have to ipose A + B = C + D. Now we shall focus on the derivative of ψ. The sae procedure as in the bound state case yields the following expression iλ ikc D A + B = λa + B C D A + B = A + B = A + B..33 ik k With further siplifications, we can get an expression in which we can deterine how C and D depend on A and B C D = iλ k A + A + iλ k B B = A + iλ k B iλ k..34 To aintain soe degree of siplicity in our notation, we define β = λ k. This alters equation.34 a bit C D = A + iβ B iβ..35 Up to this point we perhaps used algebra and standard procedures for solving differential equations. The reader has already noticed that we have five paraeters A, B, C, D and β and not enough inforation to fix the all. In order to circuvent this proble we will now take a ore physical approach to the proble. Let s iagine a 7

10 Figure.3: In order to reduce the nuber of free paraeters, we iagine a scattering experient, where the particle is fired fro the left on to the well. A, B, C represent the corresponding aplitudes of our ansatz. scattering experient, where we fire a particle fro the left on to the potential. As we already entioned, our ansatz represents unbound particles. The ters with a positive sign in their exponent are oving to the right, and the ters with the negative sign in the exponent are oving to the left. Since we fire fro the left as shown in Figure.3, there will be no incoing particles fro the right side. Therefore, the aplitude D can be set to zero. What reains is the incoing particle which can be reflected at the edges of the δ-peak. Further, the particle could also just pass through the well. Beware of thinking of the last stateent as the tunnel-effect! The particle has a bigger energy than foreseen by the potential! Tunnelling describes the fact, that a particle with a saller energy than the axial well-energy still ay be found outside the well. In contrast to this, our particle has enough energy to pass the potential without bending to it. Foralizing the stateents above delivers C D = A + iβ B iβ D=0 C = A + iβ B iβ..36 If we recall the continuity condition for ψx, we can express equation.36 still a bit differently, C = A + iβ B iβ A + B = A + iβ B iβ..37 Now, we are able to deterine B or A in dependence of A respectively B. Of course we are copletely free in our choice which paraeter shall be expressed in dependence of which. In our scattering set up we chose to fire the particle fro the left side on to the well and therefore it appears soehow natural to set A as the free paraeter. The dependency is obtained as follows A+B = A+iβ B iβ B +B iβ = A+iβ A B = A iβ iβ B = A iβ iβ..38 Reebering the fact that C = A + B delivers an expression for C C = A + B = A + A iβ iβ A Aiβ + Aiβ iβ = A iβ..39 We anaged to reduce the nuber of free paraeters fro 5 to, which is the best we can do. We could, however, create the already entioned wave package and so fix A as a function of k, the energy of the particle. In our way of treatent we have to chose the incident aplitude A and the energy of the particle. With the coputations above, we are even able to calculate the transission and reflection coefficients for the scattering process, which depend on only one paraeter, as we shall see now: The transission coefficient is defined as the odulus of the quotient of the passed aplitude and the incident aplitude squared. Forally this is T = C A = C A = iβ 8 = + β..40

11 The definition of the reflection coefficient is siilarly to the one for the transission R = B A = B A = iβ iβ = β + β..4 If we reeber that β = λ k, we can see better how energy, ass, and the depth λ of the well influence the reflection and transission behaviour of the well. Before we proceed further, we shall, as proised on page 3, investigate the case λ = 0. To have a forally clear and fir stateent, we will anticipate a procedure, which will be used and introduced properly in the next section the reader ay please forgive the disruption of the reading flow. If we use a Fourier transforation on the Schrödinger equation with our potential plugged in, we get the expression given below k ψk + λδxψxe ikx dx = ψk..4 valuating the integral in.4 generates k ψk λψ0 = ψk..43 Since λ = 0, the second ter vanishes and we obtain the description of an object which corresponds to a planar wave or an unbound particle as will be shown in the next sub-chapter. Therefore, the case λ = 0 is the first unbound or scattering state... The repulsive case As already entioned, a repulsive potential is defined with the positive sign of λ. We assue that our particle oves, again, in a non relativistic anner and we avoid the point x = 0 and look at the regions x < 0 and x > 0. Now there is a big difference between the attractive and the repulsive potential. Since the repulsive potential is greater than zero, there can not exist any bound states. In order to see why this fact holds, we recall the generic stationary Schrödinger equation Rewriting it a bit, leads us to d ψx + V xψx = ψx..44 dx d ψx = V x ψx..45 dx < V in iplies, that the sign of the second derivative of ψ can never change. Moreover, the sign is also always the sae as that of ψ itself. If we recall what the second derivative of a function eans i.e. the way a function is being bent, we can see, that we will have trouble to noralize ψ. It is, in any case, bent away fro the x-axis, which is equivalent to a function without daping. Therefore, the inial energy has to be bigger than V in, if we want noralizable physical solutions. Now we shall take a look at the Schrödinger equation for x < 0, > 0. It reads d ψx = ψx,.46 dx which is the sae expression as for the scattering states in the attractive case. Since further processing would just produce a copy of the scattering solutions, we will not think the repulsive potential through. One final aazing reark has to be ade, though. Since both cases scattering states in the attractive case and all solutions to the repulsive potential lead to plane waves, we can construct for both scenarios transission and reflection coefficients. Since the procedure for those coefficients does not change and they depend only on λ, there would be no difference in the results! Therefore we can say that, as far as the transission and reflection probabilities are concerned, we can not distinguish a well fro a barrier! 9

12 ..3 Considering the scattering states by using the spatial syetry of the Dirac δ-potential On the last few pages we solved the proble by eliinating paraeters through introducing a physically eaningful experiental setting. As we now shall see, there is a way to solve the proble without taking recourse to an experiental setting. Before we start, lets again take a look at the potential. As it can be seen in Figure.4, the potential is spatially syetric around the peak. This allows us to consider even and odd functions as solutions [6]. Those solutions we shall figure out on the next few pages. Since we have the above entioned parity syetry, it is sufficient to consider one side of the well. The opposite side is then fixed autoatically by irroring the obtained solution properly. Figure.4: If one sets up the reference frae in such a way that the δ peak is at the origin, one gets a spatially syetric setting around 0. This allows the construction of solutions which respect the parity syetry. We will now take a look on the right hand side of the well and consider the Schrödinger equation there. With the sae assuptions as in the previous sub-sections and for x > 0 it reads d ψx = ψx..47 dx This equation can be solved, if we try a sinx ansatz. ψx = A sin..48 Of course, the reader ay take the ansatz and check by differentiating two ties that it solves the Schrödinger equation. To patch together the solution for both regions we will reflect the solution. To reflect it properly we have to recall that sinx is an odd function. One now ay ask what odd or even ay ean for functions. Foralized, behind odd and even are the following definitions: f is odd : f x = fx..49 f is even : f x = fx..50 With the previous definition we can now reflect the solution. Since x < 0 provides already a sign, we have to attach another to it so the odd property is copensated. After this, we can attach the regular irroring sign to the solution. The process cobined with the usual boundary conditions i.e. continuity of ψ and continuity alost everywhere of ψ yields 0

13 A sin kx, x < 0, ψx = 0, x = 0, A sinkx, x > 0..5 Now we will focus on the derivative of ψ. To deterine its behaviour, we will use the sae trick as in the previous sub-sections. ɛ ɛ ɛ ɛ ψ xdx + λδxψxdx = ψxdx..5 ɛ ɛ The right-hand side and the second ter of the equation are zero. The first ter deands a closer look ψ x ɛ ɛ = Ak cos kɛ Ak coskɛ = In the line above we exploited the fact, that cosx is an even function and so the sign of the arguent does not atter at all. Therefore also the first ter becoes zero and hence the derivative of ψ at x = 0 is zero. With the knowledge we gathered above, we can say that odd functions do not sense the δ-peak. Let us recall the Schrödinger equation for x > 0 again d ψx = ψx..54 dx On the previous pages we used a sinx, which fulfills the equation. However, as the reader can convince hiself, also a cosx solves the equation. We will now try the following ansatz ψx = B coskx + φ, x > The constant phase φ can be justified the following heuristic way. The cosine has a axiu at x = 0. If we put there an obstacle e.g. a repulsive δ peak, it would be wrong if the odulus of ψx = 0 squared, would be one still. After all, there is a force at x = 0 that pushes the particle aside. Reflecting our ansatz properly delivers B cos kx φ, x < 0, ψx = B cosφ, x = 0,.56 B coskx + φ, x > 0. Once ore, we use our trick to look at the derivative of ψ ɛ ɛ ɛ xdx + ɛ ψ λδxψxdx = ψxdx..57 ɛ ɛ Coputation of the three ters yields k sinkɛ + φ k sinkɛ + φ + λ cosφ = The first ter can be siplified further ɛ 0 generates k sinkɛ + φ = k sinkɛ + φ..59 Inserting the result.60 into equation.58 we get k li ɛ 0 sinkɛ + φ = k sinφ..60 k sinφ = λ cosφ..6

14 If we push further towards an expression for φ we can take the φ-containing factors on one side and the other things on the other. In addition, if we recall that sinx cosx = tanx [, p. 5] we obtain a nice and siple expression tanφ = λ k..6 Unleashing the arctan on equation.6, we obtain an expression in which φ is explicit φ = arctan λ k = arctan λ k.63 Until now, we spent not a single word on the properties of the potential. Since the property of repulsiveness attractiveness is hidden in the sign of λ, we have now to consider it. According to equation.63, a repulsive potential produces a negative φ, while an attractive one generates a positive φ. a Attractive case b Repulsive case Figure.5: The influence of the sign of λ on the phase φ and the behaviour of the wavefunction around x = 0. If we take a look at Figure.5, we can see that even in the attractive case the probability to find the particle at x = 0 is only a local axiu despite the fact, that there sits an attracting force. The probability to find it at soe points away fro the well is higher. The right picture also shows very ipressively how the repulsive potential pushes the wave function away fro x = 0. One ay ask now, what the difference between the approach of the previous sub-sections and the actual procedure is. In the previous sub-sections we had to use an experiental setting in order to fix or eliinate free paraeters. In the actual approach we just generated two fundaental solutions without any recourse to an experiental setting, which is not only useful for a atheatical purist. If we recall the uler relation e ix = cosx + i sinx[, p. 3], we are able to construct the experiental approach fro the present one.. xploration of the attractive δ-potential in oentu space The last section was dedicated to the exploration of our proble in position space. In order to prepare ourselves for the consideration of the ore sophisticated relativistic proble, we shall now explore the sae proble as in the previous section. However, this tie we will explore it in oentu space. Since oentu and position space are linked by a Fourier transforation [5], we will first check how the Schrödinger equation behaves under such a transforation d ψx + V xψx = ψx,.64 dx where V x is a generic potential. Applying a Fourier transforation to the expression above, yields the equation d ψx + V xψx e ikx dx = ψxe ikx dx..65 dx Since the Fourier transforation is a linear ap, we can siplify equation.65 to d dx ψxe ikx dx + V xψxe ikx dx = ψxe ikx dx..66

15 xploiting the property of the Fourier transforation, that derivatives of the variable, which is being transfored, turn into factors of the new variable, equation.66 becoes k ψk + V xψxe ikx dx = ψk..67 Further siplifications are not possible, until one knows the properties of the potential V x. With the insights we gained above, we will now focus on the sae proble as in the previous section. All assuptions and conventions we ade there shall now be used once ore... The bound state Since the Dirac δ-potential is a concrete one, we can plug it into equation.67 and see what we get k ψk λδxψxe ikx dx = ψk..68 The second ter contains a δ-function and therefore we can copute it by using the coputation rules [5] for the δ-function. λδxψxe ikx dx = λψ0 = c b..69 With the considerations above c b is the bound case constant, equation.68 can be written as k ψk + cb = ψk..70 The last result we gained is the Schrödinger equation we have to solve in oentu space, if we are interested in the wave function. xecuting soe algebraic anipulations in.70 delivers an expression for the wave function in oentu space c b = ψk..7 k This ay soehow be a bit surprising. After all, in position space we had to solve a second order differential equation. However, we ust not forget that thanks to the properties of the Fourier transforation we transfored the derivatives into factors! Let us now figure out the energy eigenvalue for the bound state. For this task we shall again use a well known property of the Fourier transforation ψ0 = Recalling the expression for c b, we get ψke ik0 dk = ψkdk = c b dk..7 k Further processing delivers a useful relation: ψ0 = = λ k λψ0 dk..73 k.74 + dk. This relation is known as the gap equation, which we shall now investigate further. Since we are considering the bound state case, the energy is saller than zero and therefore we write the additional sign in front of explicitly in order to prevent confusion about the square root of negative values Making the substitution u = further λ + dk = λ k.75 k + dk. k, du dk = brings us to a well known integral, which we can process 3

16 λ Recalling the relation of equation.74, we get u + du = λ arctanu = λ..76 = λ..77 Fro here on we obtain an expression for, which states = λ =: b..78 This is the sae eigenvalue that we obtained in the previous section. To suarize, we have a wave function and an eigenvalue, which take the for.. The scattering states ψk = c b, k b = λ..79 As in the exploration in position space, we shall now look at states with positive energy or scattering states. Since the parity approach was an eleentary one, we would like to use it also in oentu space. If we want to keep an analogous procedure to that one in position space, we will hit on a proble: In order to keep up copatibility between the solutions in position and oentu space, we have to use the position space functions equivalents in oentu space. More concretely: We require Fourier transfors of functions, which are not exactly Fourier transforable sine and cosine are not L [4] functions. However, there is a way to handle this issue. Before we do it, we shall conduct a sall exercise, which will hopefully clarify the procedure on the next few pages and iniize confusion. Although the sine function is not a eber of the L space, we can still apply the Fourier transforation to it and look at what we get sink 0 xe ikx dx..80 This expression is not very encouraging, but if we reeber once ore the uler relation, we can rewrite the sine and insert its new for into equation.80: i e ik0x e ik0x e ikx dx = e ik0 kx dx e ik0+k dx..8 i The right-hand side of the equation above corresponds to the Fourier transforation of the constant function fx =, which is by definition the Dirac δ [5]. e ik k0x dx e ik0+k dx = i i δk k 0 δk + k 0..8 xecuting the inverse Fourier transforation yields i δk k 0 δk + k 0 e ikx dk = δk k 0 e ikx dk δk + k 0 e ikx dk,.83 4i which after evaluating the integrals becoes 4i eik0x e ik0x = 4i i sink 0x = sink 0x..84 4

17 Hence, if we want a proper Fourier transforation of a sine the cosine is exained analogously, we have to attach soe factors. With the exercise in ind, we can now write down the Fourier transfors of sine and cosine Fsink 0 x = i δk k 0 δk + k 0,.85 Fcosk 0 x = δk k 0 + δk + k 0,.86 where the F stands for the Fourier transforation. With the above insights, we now write down an ansatz for the odd solution of the transfored Schrödinger equation Inserting this ansatz in the transfored Schrödinger equation yields ψ odd p = δp k δp + k..87 i p i δp k δp + k + c s = δp k δp + k..88 i Recalling that c s where the s stands for scattering is λψ0 or reebering that odd solutions do not sense the δ-peak, we can further siplify the expression above p i δp k δp + k = δp k δp + k..89 i If we wish our ansatz to solve the transfored equation, we have to identify with k. If we do so, we have the odd solution to the transfored Schrödinger equation. Now we ake an ansatz for the even solution ψ even p = δp k + δp + k + φp..90 Inserting it in the transfored Schrödinger equation we get an expression which is a bit ore coplicated than in the odd case p δp k + δp + k + φp + c s = δp k + δp + k + φp..9 This tie, there is no reason to set c s to zero, since the corresponding ansatz in position space is not zero in general. While the su of the δ-functions would provide the cosine in position space, the ter which contains φp is a puzzle, to which we will now give further thoughts. If we recall a not correct but pragatic definition of the Dirac δ i.e. everywhere zero, except at the origin we can exploit the fact that it is zero for values of p which are not equal ±k. If we look at those values, expression.9 becoes uch sipler to work with. The δ ters vanish and what survives, after further eleentary algebraic operations, looks the following way Fro this equation, we can derive an expression for φp, which states that p φp + c s = φp..9 c s φp =..93 p Suarizing the last few lines, we now have the solution to the Fourier transfored Schrödinger equation. This solution takes the for ψ even c s p = δp k + δp + k p If we want to check the copatibility of our ansatz, we need the inverse Fourier transfor of φp. Foralized, this eans 5

18 φx = c s e ipx dp = p c s k p e ipx dp..95 Recalling how it is possible to express φ0 in ters of its Fourier transfor φp, we can gain an expression for c s which depends just on φ c s = λψ0 = λ ψpe ip0 dp = λ δp k + δp + k + φpdp..96 Picking the right-hand side of.96 and further processing yields c s = λ + φpdp = λ + λφ0..97 The reader ay now ask hiself, why we derived such an expression. After all, we already have one which works well. The expression above is copletely independent of any pre-knowledge of things in position space and therefore we are able to develop a solution in oentu space and copare it with the one we constructed in position space. So this path presents a very good cross check opportunity. We will keep the alternative description of c s in ind but now we shall return to the for of φx. As seen in expression.95, we have to solve the following integral c s k p e ipx dp = c s k p eipx dp = c s k p eipx dp..98 Perhaps the reader has already noticed, that despite its siple appearance, the integrand has very sharp teeth. It contains two singular points at p = ±k over which we ust integrate. If we can circuvent the, we ay even be able to fix the in a further step. Indeed, there is a way to avoid these points. Since the integrand is a quotient of two polynoials, it has an analytic continuation to C, which looks the sae as the original function. Therefore, we shall consider the integrand as a ap fro C to C and try to integrate it there. Since we now have one ore diension of spatial freedo, we can ove around the critical points. As for every integration in C, we have to patch together an integration path. Figure.6 provides an illustration of the path we will take. Figure.6: The integration path we shall take orientation: counter-clockwise in order to evade the singular points and to deterine φx. A way to paraetrize the path above is given on the next page. 6

19 γ : [0, k] C, γ t = t ɛ, γ =..99 γ : [, 0] C, γ t = k + ɛe it, γ = iɛe it..00 γ 3 : [k, R] C, γ 3 t = t + ɛ, γ 3 =..0 γ 4 : [0, ] C, γ 4 t = Re it, γ 4 = ire it..0 γ 5 : [ R, k] C, γ 5 t = t ɛ, γ 5 =..03 γ 6 : [, 0] C, γ 6 t = k + ɛe it, γ 6 = iɛe it..04 γ 7 : [ k, 0] C, γ 7 t = t + ɛ, γ 7 =..05 Therefore, the integral takes the for γ + e itz k z dz = 0 k 0 e it ɛx k t ɛ dt + e ireit x k Re it ireit dt + k R 0 e ik+ɛeit x k k + ɛe it iɛeit dt + e it ɛx k t ɛ dt + 0 R k e it+ɛx k t + ɛ dt+ e i k+ɛeit x k k + ɛe it iɛeit dt + 0 k e it+ɛx k t + ɛ dt..06 Since the function is holorophic on the doain we defined with the closed integration path, writing the ters in a different order, considering the liit ɛ 0 and using the residue theore [3] provides R R e itx 0 k dt+ li t ɛ 0 e ik+ɛeit x 0 k k + ɛe it iɛeit dt + e i k+ɛeit x k k + ɛe it iɛeit dt + 0 e ireit x k Re it ireit dt = The first ter is the integral we wish to deterine, while the fourth ter will vanish, when we send R to infinity the reader ay convince hiself quite easily that the integrand is bounded and will vanish. Ters two and three arise fro the paths we used to circuvent the critical points. In the liit ɛ 0 we have to take a closer look at the. Let us investigate ter nuber two first li ɛ 0 0 Further siplifying yields e ik+ɛeit x k k + ɛe it iɛeit dt li ɛ 0 0 = li ɛ 0 0 e ik+ɛeit x 0 k ɛe it idt = An analogous procedure siplifies the third ter to 0 li ɛ 0 e i k+ɛeit x k k + ɛe it iɛeit dt e ik+ɛeit x k k + kɛe it + ɛe it iɛeit dt..08 ie ikx i dt = k k eikx..09 = i k e ikx..0 The second ter is valid for x > 0 and the third holds for x < 0. Since the third ter contains an additional sign in the exponent, we have to take the odulus of x in order to patch together the correct behaviour i k eikx i k e ikx = i k i sink x = sink x.. k Hence, φx has the following for φx = c s sink x.. k even With the insights we gathered above, we are now able to write down the inverse Fourier transfor of ψ. 7

20 ψ even x = c s δk p + δk p + p e ipx dp = coskx + c s sink x..3 k Reebering expression.97 we can further siplify the above equation, which yields coskx + c s λ sink x = coskx + k sink x..4 k Rearkably the factor in front of the sine is nothing else than tanφ see equation.6 and do not forget the sign convention for the attractive potential! Therefore, we can write.4 still a bit differently Multiplying the last equation with cosφ yields coskx + λ sinφ sink x = coskx sink x..5 k cosφ cosφcoskx sinφ sink x = coskx cosφ sinφ sink x..6 cosφ Invoking the relation cosα±β = cosα cosβ sinα sinβsee[, p. 54] would allow further siplifications. In order to prevent a ess with signs after all, there is a odulus to be considered, we shall look at x > 0 first. So equation 6 becoes The sae relation applied for x < 0 yields coskx cosφ sinφ sinkx = coskx + φ..7 coskx cosφ sinφ sin kx = coskx cosφ + sinφ sinkx = coskx φ = = cos kx + φ = cos kx + φ..8 quations.7 and.8 can be written copactly as coskx cosφ sinφ sink x = cosk x + φ,.9 which is nothing else than the ansatz we used to solve the Schrödinger equation in position space! Therefore, if we add the iraculous factor cosφ to the ansatz in oentu space, we achieve full copatibility of the approaches in oentu and position space..3 Orthogonality in oentu space Before we progress any further, we have to check if the acquired solutions in oentu space fulfill orthogonality after all we consider the to be eigenstates of a physical and therefore self-adjoint syste. Therefore, we shall dedicate this section to the checking of orthogonality of the solutions we gathered in the previous section. First, we will check if the wave function of the bound state is orthogonal to the unbound states. For the odd wave function this looks like ψ odd ψ b = i δp k δp + k Further siplifications lead us to the following expression ψ odd ψ b = c b i b k b k c b b p dp..0 = 0.. Therefore, the odd scattering solutions are orthogonal to the bound state. For the even solutions the sae procedure as above, yields 8

21 ψ even ψ b = δp k + δp + k + c s p c b b p dp.. This tie, further eleentary processing generates an expression, which looks a bit ore coplicated than in the odd case ψ even ψ b = c b + c s c b dp..3 b k p b p The second ter on the right-hand side deands further treatent. Since we have a product of two rational polynoials we can conduct a partial fraction expansion. The corresponding coputation is not coplicated but tiresoe and bulky we encourage the reader though, still to control the result for hiself. Therefore, here we provide just the result. c s p After the partial fraction expansion we have 4 c s c cb 4 c b dp = c sc b b p α p β p = 4 c s c b 4 β α p b α p dp p 4 4 dp..4 β p dp..5 If we recall the integral of equation.98 and its result. we can use the fact that the first ter is nothing else than φ0, which is 0. Hence, equation.5 gets reduced to 4 c s c b 4 β α β p dp = 4 c s c b 4 β α dp,.6 p + β where we absorbed the sign into the negative β which is now the positive β. valuating the integral analogously to.75, the expression above becoes 4 c s c b 4 β α Siplifying and reconstituting α, β, β yields p + β dp = 4 c s c b 4 β α β..7 Further processing delivers 4 c s c b 4 β α β = 4 c s c b 4 b b..8 4 c s c b 4 b = b c s c b b b..9 Now our knowledge of c s c b coes into play. Reebering their for and recalling.6 in order to express b, allows further refining of expression.9. c s c b b = λ λ λ b b The expression on the right of.30 is nothing else than λ = λ λ b..30 Plugging this into equation.3 results in c b b..3 c b b + c b b =

22 At this point, we exploited the fact that = k. Therefore, also the even solution is orthogonal to the bound state. As next we shall check the orthogonality between two different odd, respectively even, wave functions. Taking the scalar product between two different i.e. solutions have not the sae energy or wave nuber odd wave functions looks the following way. ψ odd ψ odd = i δp k δp + k i δp k δp + k dp..33 xpanding and siplifying the product leads to ψ odd ψ odd = δp k δp k δp k δp+k δp+k δp k +δp+k δp+k dp..34 Recalling how a δ-function acts on another function allows us to rewrite.34 as ψ odd ψ odd = δk k δk + k δk + k + δk k,.35 which is nothing else, than ψ odd ψ odd = δk k δk + k..36 The expression above is zero as long as k k. Therefore, odd wave functions are orthogonal to each other. Now we take a look at even functions ψ even ψ even = δp k + δp + k + φ δp k + δp + k + φ dp..37 Again, further processing delivers a not as siple expression as in the odd case ψ even ψ even = δk k + δk + k + φ k + φ k + φ p φ pdp..38 Inserting the explicit for for φ i p see equation.93, eliinating the first two ters analogous to.36, the above expression gets orphed in ψ even ψ even = 0 + c + c c c + dp..39 p p Since the for of c si only depends on λ, the ters two and three can be considered to be antisyetric and therefore cancel each other out. What has survived to this point is the integral, which deands additional thoughts. Once ore, the path over a partial fraction expansion provides the key to the coputation of it. The ters generated by the procedure can be further treated if one recalls how φ i 0 can be coputed. Since this special value of φ i is zero see equation., all ters vanish and the orthogonality of two even wave functions has been shown. What reains to be exained is the orthogonality of an even to an odd wave function. Fortunately in this case we can argue the following way: The product of an even and an odd function reains odd. An odd function integrated over a syetric interval in our case the interval is ], [ is zero and therefore orthogonality holds. As we already entioned, position space and oentu space are connected through a Fourier transforation. Since the Fourier transforation belongs to the faily of unitary aps, it preserves the previously coputed scalar product between the eigenstates. Therefore we can further say that our syste reains orthogonal also in position space. 0

23 Chapter Relativistic particle in the δ-potential well The last chapter was dedicated to the Dirac δ potential for a particle, which oved in a non-relativistic anner. This chapter is devoted to the sae potential, but the particle now has the freedo to ove relativistically. While the procedure in chapter one was quite straightforward, the sall change has, as we shall see, grave ipacts on the whole scenario. Where in the previous chapter our greatest challenge was the treatent of two poles, here we well hit on divergences, whose treatent will deand the usage of a ost sophisticated ethod known as diensional regularisation. We will also revert to coputer progras in order to solve soe, otherwise difficult to evaluate integrals! Fortunately our playing ground the last chapter provides a solid foundation on which we can build chapter two.. The bound state.. Diensional regularisation and renoralization of the δ-potential and construction of the wave function As first action, we shall forulate the Hailtonian for the new proble. Since the particle now oves in a relativistic way and the potential reains an attractive Dirac δ well, we can add it to the Hailtonian of a free and relativistically oving particle. All things patched together lead us to the exprssion c 4 + p c ψp λδxψxe ipx dp = ψp.. In order to keep a siple notation we shall set = c =. For the sae reason, we just keep in ind that we consider the attractive case and drop therefore also the sign in front of λ. With these conventions, which will be applied until the end of this thesis, equation. takes the for + p ψp + λψ0 = b ψp.. The observant reader ay now ask herself, why we forulated the proble directly in oentu space. Besides the fact that the relation between relativistic oentu and energy uses explicitly the oentu quantity, there is a subtle atheatical reason which akes quite an ipact. If we want to express oentu in position space, we have to use the corresponding position space operator. It reads i x. Inserting it into equation. yields x ψp + λψ0 = b ψp..3 Now, if we want to proceed in position space, we ust first understand what the square root of a partial derivative eans and how it can be handled. Although the task is not copletely hopeless it relies on atheatics we, unfortunately, do not have at our disposal. Thus, we circuvent the issue by forulating the proble directly in oentu space. In analogy to the bound non-relativistic case, we will forulate again a gap equation. take equation. and after soe algebraic anipulations we obtain For this task, we

24 λψ0 ψ b p = b + p..4 A siilar procedure to the non-relativistic case generates the following relation, which is the gap equation we sought after λ = b dp..5 + p As we can see, the integral contains now a square root. Since the integral is logarithically divergent once ore, we encourage the reader to convince hiself of the latter stateent, the odification through the square root destroys any hope to solve it. As it looks now, we hit a dead end. Fortunately, there are still ways to handle the proble. In order to blow in a breach in the obstacle above, we shall add zero to equation.5. λ + 0 = b + p + p + dp dp..6 p + The anner in which we added zero to the above equation ay look a bit weird. The full reach of the action will becoe clear, when we process.6 further. For now let us just say, that we are trying to separate the probleatic part fro the rest of the integral and that the way above provides a successful procedure for this. Since the left side is quite trivial, we shall focus on the processing of the right side. Taking the first and second ter together generates b b p + p + dp p + dp..7 While the first ter of the expression above sells soehow like an arctanx, the second ter now contains the bare divergence. Although we can sell the arctanx in the first ter its solution would consue uch tie. The reader ay do the coputation by hand, we however, fed it to a coputation progra we used Matheatica. The progra spit out the following result b b λ = b arctan + b b dp..8 p + Since the bound state energy has to be saller than the rest ass of the particle otherwise the particle would behave like an unbound one, we can define a new paraeter x = b, which fulfills the condition x <. While the latter constraint would allow the consideration of negative x which iplies also states that we call strong bound states with b < 0 the focus of this thesis is only on the bound states with 0 < x <. The above condition allows the use of the relation [, p. 86] x arctan = arcsinx..9 x Therefore,.8 can be rewritten in a ore copact anner as λ = b b + arcsin b dp..0 p + So far we have characterized the first ter of equation.7. As already entioned, the second ter contains the divergence, which we shall now investigate further. In order to do so, we first take the integral to n = D diensions i.e. looking at the integral in R n=d. There, it reads D R D p + dd p.. Since the integrand depends only on the euclidean nor of p squared, we can siplify it further by switching to spherical coordinates.

25 D D S D p D p + dd p = 0 p + dp = S D D 0 p D dp.. p + S D is nothing else than the volue of the D -diensional sphere. So far, the integral looks not very helpful, since for any diension D it reains divergent. Just ignoring the divergence is not an option after all it is a part of the total energy of the particle. A rude but viable way would be the introduction of an upper bound for the oentu, the particle is allowed to take. This ethod is often referred to as oentu cut-off. It is very siple, but in general its application does not preserve all inforations about the syste [7]. At this point, we introduce the ethod known as diensional regularisation [7]. Instead of continuing the integral just to additional real diensions, it allows the diension, by introduction of a paraeter, to becoe any even coplex nuber. Included in this are also diensions between zero and one. In this range the procedure exploits one peculiarity of soe divergent integrals. Those integrals can, however, be paraetrized in such a way that their integration yields a result which contains the Γ function and the paraeter as its arguent. Figure. shows the behaviour of the Γ function in a certain range. Figure.: The behaviour of the Γ function for x [ 3, 3]. As long as the arguent stays between inus one and zero, the Γ function and therefore also the integral which it represents, stays finite. It blows up to infinity when the introduced paraeter hits the value, which corresponds to the original diension of the integral. In a next step often through soe power expansion in ters of the paraeter it is possible to forulate an expression which allows direct access to the paraeter. Suarized, the procedure gives a way to represent the divergence with a paraeter, the so called regulator and therefore control the divergence s behaviour directly. Moreover, as long as the divergence is switched off, it is still possible to exaine the physics, where otherwise we could only see! So uch for the idea. Now we shall apply the ideas described above to our situation. We will not try to paraetrize. anually. We fed the integral to our progra, which gave the response below S D p D D 0 p + dp = D D D Γ D, 0 < ReD <,.3 In the above expression a ass ter D appeared. We shall give a heuristic explanation for this appearance. If we perfor the substitution y = p, we ust also adjust the ter p D in the integral above. This yields the ter D. Therefore we can say that the ass ter has to appear for diensional copatibility reasons. Interestingly the coupling λ is a diensionless nuber only in one spatial diension. If we want to keep it that way, we have to attach a ter D to the coupling. All further coputations were carried out with the diensionless coupling. Now let us focus again on the processing of.3. As long as D <, the Γ function reains finite and everything is finite. For D = it blows up to infinity, exactly as our original integral. At this point we shall introduce the regulator, ɛ := D which will do the trick, since it is zero when D = and therefore it generates the full scale divergence, according to the behaviour of.. Now we have to find a 3

26 way to access the regulator directly. Since we are interested in the behaviour of the Γ function around 0, we will conduct a Laurent expansion. The result, which once ore was generated by the coputation progra, is given below. D D D Γ Laurentexp. = ɛ + γ + ln + ln + Oɛ..4 The right-hand side of the equation above states that the integral diverges with one over ɛ as ɛ goes to zero respectively D goes to one. Plugging it into.0 yields λɛ, b = b b + arcsin + γ ln ln + Oɛ..5 b ɛ In analogy to the non-relativistic setting,.5 can be considered as the solution to the gap equation. But there are big differences! While in chapter one λ paraetrized the bound state energy, here we have no chance to do it this way, since λ would just blow up. We choose a bound energy b and fix the regulator s behaviour in this way. This procedure is known as renoralization. We could also say that the regulator s behaviour is set in relation to b and therefore renoralized to it. We characterize the depth or coupling strength with ɛ and b, which has now becoe a paraeter the opposite way of the non-relativistic setting! Instead of an eigenvalue we have gained a paraetrized coupling strength λ b, ɛ. As far as the wave function is concerned, we can write down a not yet noralized wave function. ψ b p = A b p +,.6 where A is the noralization constant. Noralizing the wave function leads to A b A p + b dp = A p + b dp =..7 p + Once ore Matheatica helped us solving the above expression. A b b b + b + arcsin =..8 3 Further algebraic processing yields A = b 3 b b + + arcsin..9 b Hence, the noralized oentu space wave function has the for ψ b p = b 3 b b + + arcsin b b p +..0 As entioned above,.0 is the representation in oentu space. Of course, it would be very interesting to see its for in position space. Therefore, we will now apply the inverse Fourier transfor on expression.0 ψ b x = b 3 b b + + arcsin b e ipx b dp.. p + In order to solve the integral we perfor an analytic continuation to C the whole procedure is siilar to the one in chapter. Fortunately, also square-root functions have a continuation which looks the sae way as the function in R. There is one proble with square roots in the coplex plain. In order to keep the bijective, 4