Contact Interactions in Non-Relativistic Quantum Mechanics

Contact interactions in non-relativistic quantum mechanics

Bachelorarbeit der Philosophisch-naturwissenschaftlichen Fakult¨at der Universit¨atBern vorgelegt von

Philippe Widmer

2009

Betreut durch Prof. Dr. U.-J. Wiese Institut f¨urtheoretische Physik, Universit¨atBern Contents

1 Introduction 3

2 Theory of self-adjoint extensions of hermitian operators4 2.1 General boundary condition for contact interactions...... 4 2.2 Physical properties of general contact interactions...... 5 2.2.1 Transmission and reﬂection coeﬃcients...... 5 2.2.2 Number of bound states and their respective energies...... 6 2.2.3 Parity considerations...... 6 2.2.4 Number of physical solutions and parity of the wave function...... 7

3 The delta potential8 3.1 The standard approach...... 8 3.2 The momentum cut-oﬀ regularization...... 9 3.3 The dimensional regularization...... 11 3.4 Coeﬃcients of self-adjoint extension theory...... 13

4 The derivative of the delta-function 14 4.1 The problem...... 14 4.2 Point-splitting, renormalization and regularization...... 14 4.2.1 Sum of two delta-functions...... 14 4.2.2 The derivative of the delta-function...... 15 4.2.3 Regularization and renormalization...... 16 4.3 The momentum cut-oﬀ method...... 16 4.4 The dimensional regularization...... 18 4.5 Coeﬃcients of self-adjoint extension theory...... 19

5 The second derivative of the delta-function 20 5.1 The momentum cut-oﬀ regularization...... 20 5.2 The dimensional regularization...... 22 5.3 Coeﬃcients of self-adjoint extension theory...... 23

6 Conclusion 24

2 1 Introduction

Contact interactions are interactions between two particles that happen only when they meet at exactly the same point in space. One motivation behind studying them is that they may open the possibility to construct an interesting theory of relativistic quantum mechanics without the need to use quantum field theory. This is because in relativistic theories for every interaction between two particles there needs to be a casual connection between the two, either using a third mediating particle (which ensures that the interaction doesn’t happen faster than with the speed of light) or by being in the same place. The first possibility implies the creation and annihilation of particles which is not possible in quantum mechanics (there is no formalism which would allow this) and which would therefore require a quantum field theory. The other possibility leads to contact interactions. This explains why it might be useful to study these. Mathematically, in quantum mechanics this implies a potential which is zero everywhere except at one point. This leads to potentials consisting of a delta-function or its derivatives. Evaluating the latter leads to divergences which need to be handled with techniques known as regularization and renormalization in order to get finite results. These techniques are well known in quantum field theories and will be explained in this text. Another approach to solving these singular potentials is the use of the theory of self-adjoint extensions of hermitian operators which will be introduced in the next section at least so far as it is needed in further calculations. The goal of this thesis is to understand one dimensional contact interactions in non-relativistic quantum mechanics. As will be seen, this alone covers already quite a lot of material. Therefore the relativistic part of contact interactions is beyond the scope of this work.

3 2 Theory of self-adjoint extensions of hermitian operators 2.1 General boundary condition for contact interactions For studying contact interactions it is obviously a good idea to assume a potential which is zero except at a single point at which the value is undeﬁned. Without losing generality it can be assumed that this point is the origin at x = 0. To ensure that the Hamilton operator is still self-adjoint one needs to verify that the relation

hφ | Hψi = hHφ | ψi (1) still holds (though this is actually just a necessary condition, not a sufficient one as will be explained later). Calculating this explicitly results in   Z − 1 d2 Z ∞ 1 d2  hφ | ψi = lim  dx φ∗(x) − ψ(x) + dx φ∗(x) − ψ(x) . H  2 2  →0  −∞ 2m dx 2m dx  | {z } =:A The two integrals are basically the same. Therefore we first look at just one of them, namely the integral A: 1 Z ∞ dφ∗(x) dψ(x) dψ() A = lim dx + φ∗() 2m →0 dx dx dx Z ∞ 2 ∗ ∗ 1 d φ (x) ∗ dψ() dφ () = lim − dx 2 ψ(x) + φ () − ψ() , 2m →0 dx dx dx where both steps are achieved by partial integration using the fact that φ and ψ vanish at infinity. The other integral gives the same result up to the sign of the additional terms (because now the upper end of the integration interval is evaluated). Combining these calculations yields

hφ | Hψi = hHφ | ψi + B where 1 dψ(−) dφ∗(−) dψ() dφ∗() B := lim −φ∗(−) + ψ(−) + φ∗() − ψ() . 2m →0 dx dx dx dx In order to have a self-adjoint Hamilton operator, i.e. one which satisﬁes equation (1), the additional terms B need to cancel out. Therefore the relation dφ∗ dψ dφ∗ dψ lim ψ − φ∗ = lim ψ − φ∗ . (2) →0 →0 dx dx x=− dx dx x= between φ and ψ and their derivatives must hold. Satisfying this equation therefore ensures that the Hamilton operator is hermitian but it might still not be self-adjoint. To achieve this, the domain D(H) must equal the domain of the adjoint operator, i.e. † D(H) = D(H ).

dΨ(0) Inserting e.g. ψ(0) = dx = 0 in equation (2) would result in a self-adjoint operator but the domains wouldn’t be equal as there would be a constraint on ψ but not on φ. Therefore the boundary conditions have to be distributed equally on both ψ and φ. It is possible to show (though this mathematical prove is omitted here) that this is in the most general case satisﬁed for all wave functions which comply with the conditions ψ a b ψ = , (3) dψ c d dψ −

4 dψ where a, b, c, d ∈ C and dψ := dx . Obviously to have symmetric boundary conditions it is necessary to impose the same constraint on φ as well. As this text concerns itself only with rather speciﬁc contact interactions (namely those of non- relativistic one-dimensional quantum mechanics), it is still possible to simplify the above condition somewhat. Taking equation (2) and inserting the equations from (3) yields for x = − (omitting the limit)

dφ∗ψ − φ∗dψ = (c∗φ∗ + d∗dφ∗)(aψ + b dψ) − (a∗φ∗ + b∗dφ∗)(cψ + d dψ) =⇒ c∗a = c a∗, d∗b = d b∗, 1 = a d∗ − b∗c, 1 = a∗d − b c∗, where the second line follows from the first one through a comparison of the coefficients (as the equation has to be true for every allowed pair of wave functions ψ and φ). The first two conditions imply that a and c and d and b, respectively, got the same phase: Let z1 = r1 exp(iϕ1) and z2 = r2 exp(iϕ2). Then

∗ ∗ z1 z2 = z1z2 ⇔ r1r2 exp(i(−ϕ1 + ϕ2)) = r1r2 exp(i(ϕ1 − ϕ2)), which generally can only be satisﬁed if ϕ1 = ϕ2. Inserting this result into the other two equations and using a := ra exp(iϕ1), b := rb exp(iϕ2), c := rc exp(iϕ1) and d := rd exp(iϕ2) gives

1 = rard exp(i(−ϕ1 + ϕ2)) − rbrc exp(i(ϕ1 − ϕ2)) = rard exp(i(−ϕ1 + ϕ2)) − rbrc exp(i(ϕ1 − ϕ2))

(rard + rbrc) exp(i(−ϕ1 + ϕ2)) = (rard + rbrc) exp(i(ϕ1 − ϕ2)), which again implies ϕ1 = ϕ2. Additionally 1 = rard − rbrc must hold. Combining these results gives the general form of contact interactions in quantum mechanics,

ψ a b ψ = exp(iφ) , ad − bc = 1, (4) dψ c d dψ − with a, b, c, d, φ ∈ R. It is important to note that these are well-deﬁned and unique for every physical problem (i.e. there is just one set of constants per problem).

2.2 Physical properties of general contact interactions Contact interactions in one-dimensional quantum physics have two relevant physical properties which describe the process. First, there is the number, energy and parity of bound states and second, one wants to know the transmission and reflection coefficients T and R, respectively, whose norm squared is a measure of the probability of a particle being transmitted or reflected. In the next section we present the calculation of these properties from the general formula (4) above.

2.2.1 Transmission and reflection coefficients Assuming an energy E > 0 one can solve the eigenvalue problem everywhere except at x = 0. The differential equation to be solved is

1 d2 ψ(x) = − ψ(x) = Eψ(x). H 2m dx2 The solutions are all functions of the form

ψ(x) = A exp(±ikx) √ with k = 2mE. To calculate the transmission and reﬂection coeﬃcients, the ansatz ( exp(ikx) + R exp(−ikx) x < 0 ψ(x) = T exp(ikx) x > 0

5 is assumed. Inserting this into equation (4) yields

T a b 1 + R = exp(iφ) . ikT c d ik(1 − R)

Solving this for R and T results in

c + ik(d − a) + k2b 2ik R = ,T = exp(iφ) (5) k2b − c + ik(a + d) ik(a + d). + k2b − c

2.2.2 Number of bound states and their respective energies To calculate the bound states we assume an energy E < 0. This means having to solve the same diﬀerential equation as above (for x 6= 0 and therefore V (x) = 0), namely

1 d2 ψ(x) = − ψ(x) = Eψ(x). H 2m dx2

Solutions of this are

ψ(x) = A exp(−κ|x|) with κ = p2m(−E) > 0, as can be easily veriﬁed. Therefore κ is directly related to the energy and it suﬃces to solve the equations for κ. Inserting these solutions into equation (4) yields

ψ a b ψ = exp(iφ) dψ c d dψ − 1 a b 1 ⇒ = exp(iφ) . −κ c d κ

These are two linear equations. Dividing the second by the ﬁrst one results in c + dκ −κ = ⇒ bκ2 + (a + d)κ + c = 0. a + bκ Solving this quadratic equation leads to the solutions √ ( (a−d)2+4 − a+d ± b 6= 0 κ = 2b 2b (6) c − a+d b = 0.

2.2.3 Parity considerations Assuming parity symmetry is equivalent to having a wave function which has one of the following two forms:

ψ() = ψ(−) and dψ() = −dψ(−) (even function) ψ() = −ψ(−) and dψ() = dψ(−) (odd function)

For even functions this yields

ψ ψ a b ψ = = exp(iφ) dψ −dψ c d dψ − − a exp(iφ) − 1 b exp(iφ) ⇒ det = 0 c exp(iφ) d exp(iφ) + 1 ⇒ exp(iφ)(a − d) = 1 − exp(2iφ),

6 where the second line follows from the ﬁrst through the fact that one wants to have non-trivial solutions. Doing the same calculation for odd functions results in

exp(iφ)(a − d) = exp(2iφ) − 1.

To allow odd and even functions simultaneously both equations need to be fulﬁlled concurrently. This yields constraints on the parameters, namely

a = d, exp(iφ) = 1 (7) where the second equation means that generally φ can be assumed to be zero. Inserting this into the results for the bound states yields

( a 1 − b ± b b 6= 0 κ = c (8) − 2a b = 0. The transmission and reﬂection coeﬃcients simplify to k2b + c 2ik R = ,T = . (9) 2ika + k2b − c 2ika + k2b − c

2.2.4 Number of physical solutions and parity of the wave function In physical examples one always assumes parity symmetry, i.e. either an even or an odd wave function. Additionally the solution must be physically meaningful which implies κ > 0. From this one can calculate the parity of the wave function based on the coeﬃcients a, b, c and d. As before the cases b = 0 and b 6= 0 have to be distinguished. b = 0: In this case one can trivially insert the ansatz for the wave function (either odd or even) into equation (3). This results in the following two possibilities for κ:

c • Even wave function: κ = 1+a or equivalently a = 1. c > 0 to be physically meaningful. −c • Odd wave function: κ = 1−a or equivalently a = −1. c < 0 to be physically meaningful. Therefore either one (if κ > 0) or zero (if κ < 0) bound states exist. b 6= 0: Again inserting the ansatz yields:

c 1−a • Even wave function: κ = 1+a or equivalently κ = b . −c 1+a • Odd wave function: κ = 1−a or equivalently κ = b . The calculation of the number of bound states is a bit more complicated: Obviously if there exist two solutions with κ > 0 for both, then one of them needs to be even and the other odd. With no solution with κ > 0, there is of course also no physically meaningful solution. The most interesting case is the one with just one positive solution for κ: Then the parity of the physical solution depends on the actual values and signs of a, b and c. It’s rather easy to calculate this knowing the actual values for the constants, but doing it generally is rather cumbersome and doesn’t yield any new insights wherefore it is omitted here.

General number of solutions and their parity As calculated before, if there exist two physical solutions, then one of them is even and the other odd where generally either of them could be the ground state. With only one solution, it can be either odd (implying b 6= 0) or even (no constraint on b). Therefore in principle there could be a potential with an odd ground state even though one would normally assume it to be even (as it is for every standard example like e.g. the harmonic oscillator).

7 3 The delta potential

Solving the delta-function potential, V (x) = V0δ(x), doesn’t provide any major difficulties, es- pecially there are no divergences. Therefore it is perfect for introducing the different methods of solving the Schrödingerequation for such potentials. First we will discuss a more standard approach to the problem in order to be able to compare the different methods’ results.

3.1 The standard approach Inserting the potential into the Schr¨odingerequation yields 1 d2 ψ(x) = − + V δ(x) ψ(x) = Eψ(x). H 2m dx2 0 Integrating the whole equation from − to and taking the limit → 0 results in 1 Z d2 Z lim − dx 2 ψ(x) + V0 dx δ(x)ψ(x) = 0, →0 2m − dx − 1 dψ() 1 dψ(−) lim − + + V0ψ(0) = 0. →0 2m dx 2m dx This gives an easy to use boundary condition dψ() dψ(−) lim − = 2mV0ψ(0). (10) →0 dx dx Comparing this to equation (4) one sees that this equation already has the correct form.

Bound states For x 6= 0 there is again a vanishing potential. Therefore one can us the same solution for the wave function as before, namely

ψ(x) = A exp(−κ|x|) with κ = p2m(−E). Inserting this into the boundary condition (10) results in

lim (−2Aκ) = 2mAV0. →0 Solving this equation for the energy yields mV 2 E = − 0 . 2 This means that the delta-function allows exactly one bound state with the above energy (and an even wave function).

Transmission and reﬂection coeﬃcients Assuming the wave function ( exp(ikx) + R exp(−ikx) x < 0 ψ(x) = T exp(ikx) x > 0 and inserting it into the boundary condition yields

T ik − ik + Rik = 2mV0T. Using this and the continuity equation 1 + R = T one can solve this for R and T :

ik −1 mV −1 R = − 1 ,T = 1 − 0 . (11) mV0 ik

8 3.2 The momentum cut-off regularization This method tries to solve the problem in momentum space. The meaning of “cut-off” and “regularization” will be explained in the section about the derivative of the delta-function as these techniques are not needed for the simple delta-function. To switch to momentum space the Fourier transform is used. As a reminder, this operation and its inverse are defined as Z ∞ 1 Z ∞ ψe(p) = dx ψ(x) exp(−ipx) and ψ(x) = dp ψe(p) exp(ipx), −∞ 2π −∞ where ψe(x) is the Fourier transform of ψ(x). Infinite integral boundaries are omitted from here on.

Bound states Starting with the same Schr¨odingerequation as before one can transform it into momentum space by integrating the whole equation,

1 d2 − + V δ(x) ψ(x) = Eψ(x), 2m dx2 0 p2 ψe(p) + V0ψ(0) = Eψe(p), 2m 2 Z p V0 ψe(p) + dp ψe(p) = Eψe(p). 2m 2π | {z } :=A

This can be solved for ψe (as A is constant),

V0A 2mV0A 1 ψe(p) = − 2 = − p 2π p2 − 2mE 2π( 2m − E) Z Z 2mV0A 1 ⇒ A = dp ψe(p) = − dp . 2π p2 − 2mE

To ﬁnd bound states, one can set EB := −E > 0. Then the integral is easily solvable using the residue theorem from Complex Analysis: Z 1 Z π p p −1 √ dp 2 = dp((p + i 2mEB)(p − i 2mEB)) = . p + 2mEB 2mEB Inserting this into the equation for A results in 2mV A π A = − 0 √ . 2π 2mEB A drops out and then it can be solved for the energy, yielding

mV 2 E = 0 . (12) B 2 Obviously this is the same bound state energy as the one obtained with the previous approach. However, in contrast to the other method, the form of the wave function is yet unknown. This can be changed by calculating the inverse Fourier transform of the ψe function calculated above. 1 Z Z exp(ipx) ψ(x) = dp ψeexp(ipx) = −AmV0 dp 2 . 2π p + 2mEB This integral can be solved for either x < 0 or x > 0. In the former case exp(ipx) goes to zero for all complex numbers p with |p| = ∞ and Im p ≥ 0. This allows one again to use the residue

9 theorem with a curve which follows the real axis and then forms a semicircle of inﬁnite radius in the upper half-plane. This yields

ψ(x) = A exp(mV0x).

For x > 0 basically the same can be done. One just needs to integrate over a curve in the lower half-plane (as exp(ipx) vanishes there at inﬁnity). Using this results in a slightly diﬀerent wave function

ψ(x) = A exp(−mV0x).

Combining these results yields again the same function as in section 3.1,

ψ(x) = A exp(−mV0|x|). (13)

Transmission and reflection coefficients To get T and R the same ansatz is again made, ( exp(ikx) + R exp(−ikx) x < 0 ψ(x) = T exp(ikx) x > 0, with the continuity equation T = 1 + R. As a first step this wave function can be calculated in momentum space,

Z 0 Z ∞ ψe = dx (exp(i(k + p)x) + R exp(−i(k + p)x)) + dx(1 + R) exp(−i(k + p)x) −∞ 0 Z ∞ = dx exp(i(k + p)x) + φe = 2πδ(k + p) + φ,e −∞ where the unknown function φe was introduced. As calculated before, the Schr¨odingerequation in momentum space takes the form

2 p V0 ψe(p) + ψ(0) = Eψe(p). 2m 2π Inserting the ansatz yields

2 Z p V0 (2πδ(k + p) + φe(p)) + dp(2πδ(k + p) + φe(p)) = E(2πδ(k + p) + φe(p)). 2m 2π To get any meaningful solution, the delta-functions on both sides need to cancel out at the point k2 R p = k. Therefore E = 2m must hold. Deﬁning the constant A := dp φe and omitting the delta-function part one can solve this equation for φe,

(2π + A)V0 1 (2π + A)2mV0 1 φe(p) = − = − . 2π p2 − 2mE 2π p2 − k2 Again the constant A can be calculated, though this time a small trick is needed: The singularity of φe lies on the real axis. Therefore one needs to move it by i for a small which will be sent to zero at the end of the calculation. With this the residue theorem can again be used: Z Z (2π + A)2mV0 1 A = dp φe = − dp 2π p2 − k2 (2π + A)2mV Z 1 = lim − 0 dp →0 2π (p − k − i)(p + k + i) (2π + A)2mV 2πi (2π + A)2mV πi = lim − 0 = − 0 . →0 2π 2k + i 2π k

10 Solving this equation for A yields 2π A = . k i − 1 2mV0

With this equation φ(x), the inverse Fourier transform of φe(p), can now be calculated, Z −1 Z 1 mV0i V0 exp(ipx) φ(x) = dpφe(p) exp(ipx) = − 1 + dp 2 . 2π k 2π p 2m − E This is principally the same calculation as the inverse Fourier transform for the bound state above, i.e. one needs to use again the residue theorem. The result is

ki −1 φ(x) = − 1 exp(ik|x|). mV0 The next step is to get a relation between the coefficient of exp(ik|x|) in φ(x) and the constants R and T . This can be done by analyzing the definition of φe(p): Z 0 Z ∞ φe(p) = R dx exp(ipx) exp(−ikx) + R dx exp(ipx) exp(ikx) −∞ 0 Z ∞ = R dx exp(ik|x|) exp(ipx). (14) −∞ Therefore the coefficient of φ(x) is exactly R and one again gets

ik −1 mV −1 R = − 1 ,T = 1 − 0 . mV0 ik

3.3 The dimensional regularization The ﬁnal method used here is the so-called dimensional regularization. It is normally used in perturbative quantum ﬁeld theory. Unfortunately this method doesn’t always give correct results (at least possibly for non-perturbative problems like this). However other methods are used here as well and so there will be the possibility of comparing the results with those from other methods, including the previously described theory of self-adjoint extensions. If they all agree one can safely assume them to be correct. The approach starts with formulating the problem not in one but in D general dimensions and then solving the integrals in momentum space using Γ-functions. Finally the limit D → 1 can be taken.

Bound state As has been explained, the Schrödingerequation first has to be formulated in D dimensions and then transformed (with the D-dimensional Fourier transform) into momentum space (analogous to the momentum cut-off method), which yields 1 d2 − + V δ(x) ψ(x) = Eψ(x), 2m dx2 0 2 Z p D ψe(p) + V0 d x δ(x)ψ(x) = Eψe(p), 2m p2 ψe(p) + V0ψ(0) = Eψe(p). 2m

Solving this equation for ψe(p) (and setting EB := −E) leads to V ψ(0) ψ(p) = − 0 . e p2 2m + EB

11 Now one can insert this into the deﬁnition of ψ(0), Z Z 1 D 2mV0ψ(0) D 1 ψ(0) = D d p ψe(p) = − D d p 2 . (2π) (2π) p + 2mEB The integral can be solved using a trick which leads to the use of Γ-functions. As a reminder, these are deﬁned as Z ∞ Γ(x) = dt tx−1 exp(−t). 0 Additionally there are two important calculation rules, 1 √ Γ( ) = π and Γ(x + 1) = xΓ(x). 2

D π 2 With this, the above integral can be solved: First one deﬁnes SD−1 := D D, the surface of Γ( 2 +1) the (D − 1)-dimensional unit-sphere. With this the calculation can be continued,

Z Z ∞ D−1 2mV0 D 1 2mV0 p A := D d p 2 = D SD−1 dp 2 (2π) p + 2mEB (2π) 0 p + 2mEB Z ∞ Z ∞ 2mV0 D−1 2 = D SD−1 dp p dt exp(−t(p + 2mEB)) (2π) 0 0 D −1 Z ∞ Z ∞ 0 2 2mV0 exp(−2mEBt) 0 p 0 = D SD−1 dt dp exp(−p ), (2π) 0 2t 0 t where the substitution p0 = p2t was used. Using the Γ-function one can now evaluate the integrals,

D −1 Z ∞ Z ∞ 0 2 2mV0 exp(−2mEBt) 0 p 0 A = D SD−1 dt dp exp(−p ) (2π) 0 2t 0 t Z ∞ 2mV0 D D − 2 = D SD−1Γ( ) dt exp(−2mEBt)t 2(2π) 2 0 D − 2 V0 1 D D = D SD−1 Γ( ) Γ(1 − ). (15) 2(2π) EB 2mEB 2 2 With this one now has to solve the equation

ψ(0) = −ψ(0)A.

This can again be done for EB (as A depends on EB), the dimension can be set to D = 1 and the Γ functions can then be evaluated. This leads again to the by now well-known result

mV 2 E = 0 . (16) B 2 One can calculate the wave function with the inverse Fourier transform. However, having set the dimension to D = 1, this will obviously lead to the exact same equations and results as in the momentum cut-oﬀ method and the calculation is therefore omitted.

Transmission and reﬂection coeﬃcients Starting again with the usual ansatz, ( exp(ikx) + R exp(−ikx) x < 0 ψ(x) = T exp(ikx) x > 0,

12 which in momentum space transforms to

ψe = (2π)Dδ(k + p) + φe(p), and inserting this into the Schr¨odingerequation yields

2 Z p V0 ((2π)Dδ(k + p) + φe(p)) + dDp (2πδ(k + p) + φe(p)) = E((2π)Dδ(k + p) + φe(p)). 2m (2π)D

k2 Again E = 2m has been deﬁned to get rid of the delta-functions outside the integral. This leads to

2 p V0 φe(p) + A = Eφe(p) 2m (2π)D using the deﬁnition Z A := dDp φe(p).

The Schr¨odinger equation above can be solved for φe(p),

D ((2π) + A)2mV0 1 φe(p) = − . (2π)D p2 − k2

This can be inserted into the deﬁnition of A and solved for the constant (using the calculations of the integrals from the bound state calculation and setting D = 1) yielding 2π A = . k i − 1 2mV0 Using equation (14) one can then calculate R and T ,

ik −1 mV −1 R = − 1 ,T = 1 − 0 . mV0 ik

3.4 Coefficients of self-adjoint extension theory As expected, each of the three different approaches yields the same results. This actually strongly suggests that the methods are correct. Therefore one can now try to describe the delta-function potential within the framework of the previously derived theory which basically means to simply calculate the matrix coefficients. This is most easily done using the results from section 3.1, namely the parity of the wave function ψ() = ψ(−) and equation (10). The latter actually got the correct form already so the values of the constants a, b, c and d can be easily read off. This gives the relation

ψ 1 0 ψ = . dψ 2mV 1 dψ 0 − As can easily be veriﬁed with the equations from section2, these coeﬃcients lead to the same physical properties as the calculations above.

13 4 The derivative of the delta-function

Now a more complex example is presented: A potential consisting of the ﬁrst derivative of the 0 delta-function, V (x) = V0δ (x). As δ(x) is not actually a function but rather a distribution the derivative is only deﬁned for an integral over δ0(x), namely through partial integration, i.e. Z Z dx δ0(x)f(x) = − dx δ(x)f 0(x).

4.1 The problem The obvious way for trying to solve this is to use the same approach as before, i.e. integrating the Schr¨odingerequation over a small interval around x = 0. This leads to

Z 2 Z 1 d 0 lim − dx 2 ψ(x) + V0 dx δ (x)ψ(x) = 0, →0 2m − dx − 0 0 0 lim (ψ () − ψ (−)) = 2mV0ψ (0). →0 As can be easily seen, this equation is quite problematic: The derivative of ψ at x = 0 is not well deﬁned (unless the left side of the equation evaluates to 0), wherefore ψ0(0) doesn’t exist. This actually means that there is no way to solve this equation. One has to try another method.

4.2 Point-splitting, renormalization and regularization Another natural way is to use the so called diﬀerential quotient,

0 δ(x + ) − δ(x − ) V (x) = V0δ (x) = lim V0. →0 2

4.2.1 Sum of two delta-functions As one can therefore approximate δ0(x) by two delta-functions this general potential is being solved below, namely the potential V (x) = V0δ(x − ∆x) + V1δ(x + ∆x). The Schr¨odingerequation can again be integrated around ∆x and −∆x. This yields two diﬀerent boundary conditions,

0 lim (ψ(−∆x + ) − ψ(−∆x − )) = 2mV0ψ (−∆x) →0 and

0 lim (ψ(∆x + ) − ψ(∆x − )) = 2mV0ψ (∆x). →0

Bound state Again one needs to solve the Schr¨odingerequation everywhere except at x = −∆x and x = ∆x. This yields the solution  A exp(κx) x < −∆x  ψ(x) = B exp(κx) + C exp(−κx) |x| < ∆x (17) D exp(−κx) x > ∆x with κ = p2m(−E). The continuity of the wave function is again demanded which yields two additional equations,

A − B = C exp(2κ∆x) and D − C = B exp(2κ∆x).

14 Together with the two boundary conditions one can now eliminate the four constants and get an equation for κ. As this is a simple but tedious calculation, it is not repeated here and only the result is shown, m (κ + (V + V ))2 − m2(V + V )2 = m2V V (exp(−4κ∆x) − 1). (18) 2 0 1 0 1 0 1 This is not solved more explicitly for κ as the goal here is to calculate the derivative of the delta-function (as it will get quiet a lot easier when one inserts this special case).

Transmission and reﬂection coeﬃcients The ansatz now has the form  exp(ikx) + R exp(−ikx) x < −∆x  ψ(x) = A exp(ikx) + B exp(−ikx) |x| < ∆x T exp(ikx) x > ∆x.

Again there are two continuity equations,

exp(−ik∆x) + R exp(ik∆x) = A exp(−ik∆x) + B exp(ik∆x) and

A exp(ik∆x) + B exp(−ik∆x) = T exp(ik∆x).

These together with the boundary conditions can be solved for T and R. This is again somewhat tedious and will be omitted here. The results are

mV mV m2V V −1 T = 1 − 1 1 − 0 + 0 1 exp(4ik∆x) , (19) ik ik k2 mV R = T 1 (1 − exp(−4ik∆x)) exp(2ik∆x) + exp(−2ik∆x)(T − 1). (20) ik

4.2.2 The derivative of the delta-function As shown before, one can now approximate the derivative with two delta-functions using the following deﬁnitions:

0 V0 • V0 := 2∆x

0 V0 • V1 := −V0 = − 2∆x These deﬁnitions can then be inserted into equation (18), which yields

V 02 κ2 = − 0 (exp(−4κ∆x) − 1). (2∆x)2

As ∆x will shortly be sent to 0, the exponential function can be approximated for small x,

exp(x) ≈ 1 + x.

This results in V 02 κ = 0 . (21) ∆x Unfortunately, κ again diverges for ∆x → 0 and inﬁnite energy is not an allowed solution. However now there is a way out as will be shown in the next section.

15 4.2.3 Regularization and renormalization To solve this problem one can use techniques known from quantum field theory: Basically a physical quantity is replaced by its measured value and this value is kept constant. For example one can decide to assume a measured value for the energy of the bound state EB and then vary the potential strength V0 such that this fixed value results from calculating it with the Schrödinger equation. This redefinition of constants is called renormalization. The mathematical procedures how to do this (e.g. the momentum cut-off method or the dimensional regularization) are called regularizations. An important point to note here is that the result should be independent of the regularization. Therefore here at least two regularization per potential will be used to be able to verify this. One can now try to use this approach for the problem at hand: First of all, from equation (21) it can be seen that there exists only one energy eigenvalue (one solution for κ). Therefore it is reasonable to assume κ to be constant and to choose a suitable V0. This can be achieved by solving 0 0 the said equation for V0 and by inserting this result into the definition of V0 , i.e. one can redefine √ V 0 κ∆x r κ V := 0 = = . 0 2∆x 2∆x 2∆x

Obviously a formula for EB cannot be calculated now, it would need to be measured. However one can still calculate other physical quantities, namely the wave functions and the transmission and reflection coefficients using the redefined potential. The wave function was actually already calculated up to normalization factors in equation (17). The constants themselves can be calculated using the continuity equations, the boundary conditions, and the normalization equation R dx |ψ(x)|2 = 1, and then finally taking the limit ∆x → 0. Doing this yields √ ψ(x) = κ exp(−κ|x|).

Obviously this is again the same even function as the solution of the simple δ(x) potential. The coefficients can be calculated using equations (19) and (20) with the redefined potential. This yields ik −κ T = ,R = . (22) ik + κ ik + κ As can be easily shown by inserting the definition of κ into these equation, this is actually the exact same result as before for the attractive δ(x) potential!

4.3 The momentum cut-off method 0 Bound state One can now try to solve the potential V (x) = V0δ (x) with the momentum cut-off regularization. Again the Schrödinger equation is transformed into momentum space. This leads to 2 Z p 0 ψe(p) + V0 dx δ (x)ψ(x) exp(−ipx) = Eψe(p), 2m p2 Z dψ(x) ψe(p) + V0 dx δ(x)( exp(−ipx) − ipψ(x) exp(−ipx)) = Eψe(p), 2m dx p2 dψ(0) ( − E)ψe(p) + V0( − ipψ(0)) = 0, 2m dx V (ψ0(0) − ipψ(0)) ⇒ ψ(p) = − 0 . e p2 2m + EB

16 Here partial integration was used and the bound state energy EB := −E was defined. Now the constant ψ(0) can be calculated, Z Z Z 1 2mV0 0 1 p ψ(0) = dp ψe(p) = − ψ (0) dp 2 − iψ(0) dp 2 2π 2π p + 2mEB p + 2mEB mV = − 0 ψ0(0), κ where the second integral vanishes because the integrand is an odd function√ and the first one is calculated using the residue theorem. Finally, we defined again κ := 2mEB. The same can be 0 R p2 done for ψ (0), but here lies a problem: An integral of the type dp p2+A with some constant A. This is not solvable using the residue theorem (as the integrand doesn’t vanish at infinity). Therefore this needs to be regularized: One integrates not from minus infinity to infinity but instead from −Λ to Λ. Then the potential can be redefined as before and the results can be calculated with this new regularized potential. At the end one can then send Λ to infinity and the results should still remain finite. Doing this yields

Z Z Z 2 0 2mV0 0 p p ψ (0) = dp ipψe(p) = − iψ (0) dp 2 + ψ(0) dp 2 2π p + 2mEB p + 2mEB Z Λ 2 2mV0ψ(0) p + 2mEB − 2mEB = − dp 2 2π −Λ p + 2mEB Z 2mV0ψ(0) 1 2mV0ψ(0) = − 2Λ − 2mEB dp 2 = − (2Λ − κπi) . (23) 2π p + 2mEB 2π

One can now insert the calculated value for ψ(0) and solve this for V0, which yields the potential π κπ V 2 = ≈ , 0 2 2Λ 2m2Λ m ( κ − πi) where the last step assumes that Λ π. Again an equation for the energy can’t be calculated as it was intendedly left constant to be determined from another source. One can still calculate the wave function in coordinate space though

Z 0 2mV0 ψ (0) − ipψ(0) ψ(x) = dp 2 exp(ipx) 2π p + 2mEB Z Z 2mV0 0 exp(ipx) p exp(ipx) = ψ (0) dp 2 − iψ(0) dp 2 . 2π p + 2mEB p + 2mEB

0 Now there are two constants ψ(0) and ψ (0) which depend on V0 and a relation between them. Therefore one of them can be replaced by the other. As both integrals are finite, it remains to be seen which of the constants remain finite and unequal zero. If one replaces ψ(0) by ψ0(0) one gets a wave function ψ(x) = 0 (when the limit Λ → ∞ is being taken) which is not an acceptable solution. Replacing ψ0(0) by ψ(0), the constant before the first integral remains unequal zero and the other one vanishes. This leads to wave functions of the form (after normalization with the remaining constant) √ ψ(x) = κ exp(−κ|x|).

Transmission and reflection coefficients To calculate the coefficients the usual ansatz is again used, ( exp(ikx) + R exp(−ikx) x < 0 ψ(x) = T exp(ikx) x > 0,

17 which yields

ψe(p) = 2πδ(k + p) + φe(p) as calculated before. One can now again transform the Schr¨odingerequation into momentum space and then insert the ansatz, p2 dψ(0) ( − E)ψe(p) + V0( − ipψ(0)) = 0, 2m dx 2 Z p V0 ( − E)(2πδ(k − p) + φe(p)) + dp0 [ip0 − ip] ψe(p0) = 0. 2m 2π

k2 As before the deﬁnition E := 2m was used. Inserting the ansatz into the integral and simplifying the delta-functions gives   2 Z Z p V0  0 0 0 0 0  ( − E)φe(p) + 2π(k − p) + dp p φe(p ) −p dp φe(p ) = 0 2m 2π   | {z } | {z } =:B =:A

2miV0(2π(k − p) + B − Ap) ⇒ φe(p) = − . 2π(p2 − 2mE) Now one can again calculate a relation for A and B. This leads to the usual integrals and is easily solvable. The results are B miV (2π + A) A = mV (2π + ) and B = 0 (2Λ + πik). 0 k π Solving these equations for B yields

2 4πΛ + 4πΛmV0 + πik + 2π ikmV0 πikmV0 B = π k − . − 2ΛmV0 k miV0 R can now be calculated using equation (14), Z 1 mV0 R exp(ik|x|) = dp φe(p) exp(ipx) = (k + B) exp(ik|x|) 2π 2πk mV (k + B) κ ⇒ R = lim 0 = − Λ→∞ 2πk ik + κ ik ⇒ T = . ik + κ Fortunately this yields the same results as with the other regularization.

4.4 The dimensional regularization Obviously it would be a good idea to test the results with another regularization, the dimensional regularization, in order to show that it yields the same result. Unfortunately, this doesn’t seem to be possible for the δ0(x) potential: One would need to generalize the potential into more dimensions. The obvious way to do so seems to be using the gradient of the delta-function. However, this doesn’t work, as from this operation one gets a vector instead of a scalar and this can’t be inserted into the Schr¨odingerequation (which is a scalar equation). Furthermore, there doesn’t seem to be any other useful generalization. Because of this, just the two regularizations shown above are considered for this potential (which fortunately give the same results wherefore it is reasonable to conclude that it’s the correct one). For the potential consisting of the second derivative of the delta-function there exists a natural extension into D dimensions as will be shown in the next section.

18 4.5 Coefficients of self-adjoint extension theory As in the case of the δ(x) potential, again the same result was obtained with both regularizations, as expected. More surprisingly, it was exactly the same solution as before. There doesn’t seem to be any (measurable) difference between the attractive delta-function and its derivative. Using the theory of self-adjoint extensions one obviously gets again the same matrix coefficients. As a slightly different notation was used here (mainly because of the need to regularize and renormalize the problem), one gets a somewhat different representation of the element c. Inserting the results yields the relation

ψ 1 0 ψ = . dψ 2κ 1 dψ −

19 5 The second derivative of the delta-function

The last potential to be examined here is the second derivative of the delta-function, V (x) = 00 V0δ (x). Again, two diﬀerent regularizations will be used to test the results. First the momentum cut-oﬀ method is presented.

5.1 The momentum cut-off regularization Bound state As usual one first has to transform the Schrödingerequation into momentum space, yielding 2 Z p 00 ψe(p) + V0 dx δ (x)ψ(x) exp(−ipx) = Eψe(p), 2m 2 Z 2 p d ψ(x) dψ(x) 2 ψe(p) + V0 dx δ(x)( − 2ip − p ψ(x)) exp(−ipx) = Eψe(p), 2m dx2 dx 2 2 p d ψ(0) dψ(0) 2 ψe(p) + V0( − 2ip − p ψ(0)) = Eψe(p) 2m dx2 dx 2 d ψ(0) dψ(0) 2 2mV0( dx2 − 2ip dx − p ψ(0)) ⇒ ψe(p) = − 2 , p + 2mEB where partial integration was again used and EB := −E was defined. One now has three constants and therefore three equations to solve. As can easily be seen, all of them are linear. Therefore they can be written in matrix form:

2  2mV0 R p 2mV0 R 1  1 − dp 2 0 dp 2   2π p +2mEB 2π p +2mEB ψ(0) 2 4mV0 R p 0  0 1 + dp 2 0  ψ (0) = 0  2π p +2mEB    4 2 00 2mV0 R p 2mV0 R p ψ (0) dp 2 0 1 − dp 2 2π p +2mEB 2π p +2mEB This system has got one obvious solution, namely the trivial one ψ(0) = ψ0(0) = ψ00(0) = 0, which is not an acceptable solution. To allow other solutions one therefore needs to demand that the determinant of the matrix equals zero, det M = 0, if the matrix is denoted by M. This yields the equation h i (1 + 2I(2)) (1 − I(2))2 − I(0)I(4) = 0, where the deﬁnition Z x 2mV0 p I(x) = dp 2 2π p + 2mEB was used. Next these integrals need to be calculated. This is basically the same relatively straight- forward calculation that was done before, so nothing but the results is given here r m I(0) = V0, 2EB 2mV0Λ p I(2) = − 2mE mV , π B 0 2mV0 2 3 2 2 π I(4) = ( Λ − 4mEBΛ + 4m EB √ ). 2π 3 2mEB With this one can now try to solve the determinant equation. It is the product of two brackets where either of them could be zero, so they can be examined separately. The ﬁrst one reads

4mV0Λ p 1 + 2I(2) = 1 + − 2 2mE mV = 0 π B 0 1 2m 2 ⇒ EB = √ + √ Λ . 3 3 2m V0 2m π

20 Obviously the energy goes to infinity when Λ goes to infinity. As usual one wants therefore to choose the potential strength V0 such that the energy remains finite. However, as can be easily seen, there doesn’t exist any value for V0 with this property (as the rightmost term still remains infinite independent of the value of the potential strength). Therefore one needs to take a look at the other part of the determinant equation, (1 − I(2))2 − I(0)I(4) = 0 √ 2 √ p 2 2mπV0 3p p 2 2 (π − 2mV0Λ + 2mEBmV0π) = (Λ mEB − 6mE mEB + 3 2m EBπ). 3EB Unlike before, this time an ansatz for the potential strength is made. As can be seen be examining the above equation, the only possibility for it not to diverge for Λ → ∞ or alternatively be equal to zero on both sides is the ansatz λ V = , 0 Λ3/2 where λ is a freely chooseable proportionality constant. It actually takes the role of the (measurable) binding energy in previous calculations. Inserting this into the equation, solving it for EB, and taking the limit of Λ to infinity yields a relation between λ and the binding energy EB, 2m3λ4 E = . (24) B 9π2 This means that again there is just one bound state. It remains to calculate the wave function. Performing the integral and taking the limit for Λ is again the same as before, wherefore the calculation is omitted. The result of this is √ ψ(x) = κ exp(−κ|x|), √ with κ = 2mEB, which is again the same function as the one from the delta-function potential.

Transmission and reﬂection coeﬃcients With the usual ansatz ( exp(ikx) + R exp(−ikx) x < 0 ψ(x) = T exp(ikx) x > 0, which yields

ψe(p) = 2πδ(k + p) + φe(p), and inserting this into the Schr¨odingerequation gives 2 p 00 0 2 ψe(p) + V0 ψ (0) − 2ipψ (0) − p ψ(0) = Eψe(p) 2m 2 2 2mV0 2π(k − 2kp + p ) + A − 2Bp + Cp ⇒ φe(p) = , 2π p2 + 2mE

k2 where again E = 2m was used to get rid of the delta-function outside the integrals and the following constants were deﬁned Z Z Z A := dp p2φe(p),B := dp pφe(p),C := dp φe(p).

These constants define an inhomogeneous system of equations which can easily be solved. The integrals are again basically the same ones already solved in equation (23). Therefore the calculation is not repeated here. After taking the limit Λ → ∞, the reflection coefficient can again be calculated using equation (14). This yields the result 1 R = 3kπ . −i 2m2λ2 − 1

21 This got again the same form as the one from the delta-function potential. One can additionally replace λ by EB using equation (24) and the deﬁnition of κ. This yields the by now well-known results κ ik R = − ,T = . ik + κ ik + κ

5.2 The dimensional regularization Bound state Starting again by transforming the Schr¨odingerequation formulated in D dimensions into momentum space gives 2 Z p D ψe(p) + V0 d x ∆δ(x)ψ(x) exp(−ipx) = Eψe(p), 2m 2 Z p D ψe(p) − V0 d x ∇δ(x) · (∇ψ(x) − ipψ(x)) exp(−ipx) = Eψe(p), 2m 2 Z p D 2 ψe(p) + V0 d x δ(x)(∆ψ(x) − 2p · ∇ψ(x) − p ψ(x)) exp(−ipx) = Eψe(p), 2m 2 p 2 ψe(p) + V0(∆ψ(0) − 2p · ∇ψ(0) − p ψ(0)) = Eψe(p) 2m ∆ψ(0) − 2p · ∇ψ(0) − p2ψ(0) ⇒ ψe(p) = −2mV0 2 . p + 2mEB where EB := −E was deﬁned as usual. As before this gives a homogeneous system of linear equations which can be written down as a matrix equation,  ψ(0)  2 dψ(0)  2mV0 R p 2mV0 R 1    1 − (2π)D dp p2+2mE 0 (2π)D dp p2+2mE  dx1  B B  .   0 A 0   .  = 0,  4 2    2mV0 R p 2mV0 R p dψ(0) D dp 2 0 1 − D dp 2   (2π) p +2mEB (2π) p +2mEB  dxD  ∆ψ(0) where A is the D × D matrix

2  4mV0 R p1  1 + dp 2 0 0 2πD p +2mEB  .   0 .. 0  .  2  4mV0 R pD 0 0 1 + dp 2 2πD p +2mEB Again one has some integrals to solve. Fortunately all of those can be solved at once using the deﬁnition Z x 2mV0 p I(x, D) := D dp 2 . (2π) p + 2mEB As basically the same calculation was already done for the dimensional regularization of the delta- function potential (the only diﬀerence being the additional factor px which doesn’t fundamentally change the calculation), just the result will be stated here,

D+x − 2 V0 1 D + x D + x I(x, D) = D SD−1 Γ( ) Γ(1 − ). 2(2π) EB 2mEB 2 2 In order to have non trivial solutions the determinant of the matrix equation must be zero, leading to the equation 2 [1 + I(2,D)]D [(1 − I(2,D))2 + (−1)D+2I(0,D)I(4,D)] = 0. D

22 Inserting the values of the integrals and setting D = 1 yields the simplified equation √ √ 3 p 3 p (−1 + 2 2m V0 EB)(1 + 2 2m V0 EB) = 0 1 ⇒ EB = 2 3 . 8V0 m As can be seen from the first of the two previous equations, there is one solution for positive potentials and one for V0 < 0. The first one implies ∇ψ(0) = 0 and leads to the exact same calculations as before, resulting in an even wave function. Negative potentials however imply ∆ψ(0) = ψ(0) = 0. Doing the same calculation then results in an odd wave function. Combining these results, the wave function can be summarized as ( A exp(−κ|x|) V > 0 ψ(x) = 0 sign(x)A exp(−κ|x|) V0 < 0, √ where A is the normalization constant and κ = 2mEB.

Transmission and reﬂection coeﬃcients Using again the ansatz ( exp(ikx) + R exp(−ikx) x < 0 ψ(x) = T exp(ikx) x > 0 yields

ψe(p) = (2π)Dδ(k + p) + φe(p). This can again be inserted into the Schr¨odingerequation in momentum space

2 p 2 ψe(p) + V0 ∆ψ(0) − 2ip · ∇ψ(0) − p ψ(0) = Eψe(p) 2m D 2 2 P 2 2mV0 (2π) (k − 2k · p + p ) + A − 2 Bipi + Cp ⇒ φe(p) = , (2π)D p2 + 2mE

k2 where again E = 2m was used to get rid of the delta-functions outside the integrals. Additionally the following constants were deﬁned Z Z Z 2 A := dp p φe(p),Bi := dp piφe(p),C := dp φe(p).

These constants deﬁne an inhomogeneous system of equations which can easily be solved. The integrals are again basically the same as the ones already solved in equation (15). Therefore the calculation is omitted here. Setting D = 1 the equations can easily be solved for R. This gives a solution for R and T , κ ik R = − ,T = . ik + κ ik + κ

5.3 Coefficients of self-adjoint extension theory Apart from the different result for the wave function from the dimensional regularization (which we will ignore as explained in the next section), the calculations lead again to the same results from every regularization and it is even the same as for the delta-function potential. Inserting them into the equations for the matrix coefficients, one gets exactly the same solutions as for the first derivative of the delta-function, ψ 1 0 ψ = . dψ 2κ 1 dψ −

23 6 Conclusion

A summarization of the results is now presented. First, this is a list of all of them (omitting T = R + 1, a = d = 1, and b = 0, as they are all trivial or can easily be calculated from the ones below):

Potential Regularization EB Reg. |ψ(x)| Parity R c 2 −κ V0δ(x) Standard ∝ V0 V0 ∝ 1 A exp(−κ|x|) even ik+κ 2κ 2 −κ V0δ(x) Momentum cut-off ∝ V0 V0 ∝ 1 A exp(−κ|x|) even ik+κ 2κ 2 −κ V0δ(x) Dimensional reg. ∝ V0 - A exp(−κ|x|) even ik+κ 2κ 1 0 2 2 −κ V0δ (x) Standard ∝ V0 V0 ∝ (∆x) A exp(−κ|x|) even ik+κ 2κ 1 0 4 − 2 −κ V0δ (x) Momentum cut-off ∝ V0 V0 ∝ Λ A exp(−κ|x|) even ik+κ 2κ 3 00 4 − 2 −κ V0δ (x) Momentum cut-off ∝ V0 V0 ∝ Λ A exp(−κ|x|) even ik+κ 2κ 00 1 1 −κ V0δ (x) Dimensional reg. ∝ 2 - A exp(−κ|x|) both 2κ V0 ik+κ Obviously most of the results are the same for every regularization and even for every potential. While the latter is rather surprising, the former was actually expected, since the formalism of renormalizations and regularizations is well established in quantum field theory. The only things not quite in agreement with the rest are the binding energy and the parity of the wave function in the δ00(x) potential with the dimensional regularization: The former actually decreases with increasing potential strength V0 which is rather counter intuitive. The latter offers an odd ground state which was found nowhere else. Both irregularities are most probably due to the already mentioned problem where this regularization doesn’t always give correct results in a nonpertur- bative context. We can safely assume this, as other regularizations (not considered in this thesis) actually agree with the result of the momentum cut-off method. The only difference between the potentials does seem to be the changing “strength” of the regularization needed. Taking the momentum cut-off method as an example, one has an increasing dependence on Λ with each additional differentiation of the delta-function. The equivalence of the potentials actually prevents a deeper understanding what each parameter of the theory of self-adjoint extensions does. Unfortunately, these delta-functions seem to just grant access to one of them, namely the one called c. The others seem to be unreachable through delta-functions, at least up to the second derivative. The last point being mentioned here is that none of these results depend on the sign of the potential (except the one for the delta-function itself, which e.g. only as solutions for the binding energy in its attractive form). It doesn’t seem to matter whether one got an attractive or a repulsive derivative of a delta-function in the potential.

1 even for V0 > 0 and odd for V0 < 0

24 References

[1] Prof. Dr. G. Colangelo. Quantentheorie 1. Universit¨atBern, 2007.

[2] Prof. Dr. C. Greub. Mathematische und numerische Methoden der Physik I. Universit¨atBern, 2007. [3] Prof. Dr. C. Tretter. Komplexe Analysis. Universit¨atBern, 2008.