Instantons and large N

Transcription

1 Preprint typeset in JHEP style - PAPER VERSION Instantons and large N An introduction to non-perturbative methods in QFT Marcos Mariño Département de Physique Théorique et Section de Mathématiques, Université de Genève, Genève, CH-1211 Switzerland marcos.marino@unige.ch Abstract: Book on advanced QFT.

2 Contents 1. Instantons in Quantum Mechanics Introduction QM as a one-dimensional field theory Unstable vacua in Quantum Mechanics Path integral around an instanton in QM Calculation of functional determinants I: solvable models Lifetimes in unstable vacua from instantons Instantons in the double well Instantons in periodic potentials Bibliographical notes Unstable vacua in QFT Instantons in scalar QFT The fate of the false vacuum Bibliographical notes Large order behavior and Borel summability Introduction Asymptotic expansions and Borel resummation Large order behaviour and Borel transforms The quartic anharmonic oscillator Instantons and large order behavior in quantum theories Stable vacua Unstable vacua Complex instantons Cancellation of nonperturbative ambiguities Bibliographical notes Nonperturbative aspects of Yang Mills theories Introduction Basics of Yang Mills theories Topological charge and θ vacua Instantons in Yang Mills theory Bibliographical notes Instantons and fermions Fermions and chiral symmetry in QCD The U(1) problem and the axial anomaly Bibliographical notes 99 1

3 6. Sigma models at large N The O(N) non-linear sigma model The P N 1 sigma model The model and its instantons The effective action at large N Topological susceptibility at large N Bibliographical notes The 1/N expansion in QCD Fatgraphs Large N rules for correlation functions QCD spectroscopy at large N: mesons and glueballs θ-dependence at large N The U(1) problem at large N. Witten Veneziano formula Bibliographical notes Solvable toy models at large N: matrix quantum mechanics The planar approximation in matrix quantum mechanics Some examples Instantons at large N Analyticity in the 1/N expansion Large N instantons Large N instantons in matrix quantum mechanics 141 A. Harmonic analysis on S B. Effective action for large N sigma models Instantons in Quantum Mechanics 1.1 Introduction In this Chapter we start our study of non-perturbative effects by looking at the simplest example, i.e. Quantum Mechanics, which can be regarded as a quantum field theory in one dimension. instantons, i.e. to nontrivial solutions to the classical equations of motion. If g is the coupling constant, these effects have the dependence e A/g. (1.1) Notice that this is small if g is small, but on the other hand it is completely invisible in perturbation theory, since it displays an essential singularity at g = 0. Instanton effects are responsible of one of the most important quantum-mechanical effect: tunneling through a potential barrier. This effect changes qualitatively the structure of the quantum vacuum. In a potential with a perturbative ground state degeneracy, like the one shown on the l.h.s. of Fig. 1, tunneling effects lift the degeneracy. There a single ground state, 2

4 Figure 1: Two quantum-mechanical potentials where instanton effects change qualitatively our understanding of the vacuum structure. and the energy difference between the ground state and the first excited state is an instanton effect of the form (1.1), E 1 (g) E 0 (g) e A/g. (1.2) In a potential with a metastable vacuum, like the one shown in the r.h.s. of Fig. 1, the perturbative vacuum obtained by small quantum fluctuations around this metastable vacuum will eventually decay. This means that the ground state energy has a small imaginary part, E 0 (g) = ReE 0 (g) + iim E 0 (g), ImE 0 (g) e A/g (1.3) which also has the dependence on g typical of an instanton effect. 1.2 QM as a one-dimensional field theory In this Chapter we will focus on quantum-mechanical models in one dimension, where instanton methods can be developed in complete detail. We will consider systems with a Hamiltonian of the form, H = 1 2 p2 + W(q), (1.4) where W(q) is the potential, and we will set = 1. If this system supports bound states, one basic question to ask is what is the energy of the ground state. This can be of course addressed by elementary methods, like stationary perturbation theory, but we want to formulate the problem in the language of path integrals, so that the intuition gained in this way can be applied to quantum field theories. The ground state energy of the quantum mechanical system described by (1.4) can be extracted from the small temperature behavior of the thermal partition function, Indeed, if we have a discrete spectrum with energies the thermal partition function can be written as Z(β) = tr e βh. (1.5) E 0 < E 1 < E 2 <, (1.6) Z(β) = e βen, (1.7) n=0 3

5 τ τ 2m e m τ τ g Figure 2: Feynman rules for the quantum mechanical quartic oscillator. therefore 1 E 0 = lim log Z(β). (1.8) β β On the other hand, the thermal partition function admits a path integral representation in terms of the Euclidean theory, in which we perform a Wick rotation to imaginary time t it, (1.9) and, because of the trace in (1.5), we have to consider periodic trajectories q(t) in imaginary time, q( β/2) = q(β/2), (1.10) where β is the period of the motion. After Wick rotation, the path integral involves the Euclidean action S(q), β/2 [ ] 1 S(q) = dt 2 ( q(t))2 + W(q(t)) (1.11) β/2 The thermal path integral is then given by Z(β) = D[q(t)]e S(q), (1.12) where the integration is performed over periodic trajectories. We note that the Euclidean action can be regarded as an action in Lagrangian mechanics, β/2 [ ] 1 S(q) = dt 2 ( q(t))2 V (q), (1.13) where the potential is β/2 V (q) = W(q), (1.14) i.e. it is the inverted potential of the original problem. It is possible to compute the ground state energy by using Feynman diagrams. We will assume that the potential W(q) is of the form W(q) = m2 2 q2 + W int (q) (1.15) 4

6 1 2a 2b 3a 3b 3c 3d Figure 3: Feynman diagrams contributing to the ground state energy of the quartic oscillator up to order g 3. where W int (q) is the interaction term. Then, the path integral defining Z can be computed in standard Feynman perturbation theory by expanding in W int (q). We will actually work in the limit in which β, since in this limit many features are simpler, like for example the form of the propagator. In this limit, the free energy will be given by β times a β-independent constant, as follows from (1.8). In order to extract the ground state energy we have to take into account the following 1. Since we have to consider F(β) = log Z(β), only connected bubble diagrams contribute. 2. The standard Feynman rules in position space will lead to n integrations, where n is the number of vertices in the diagram. One of these integrations just gives as an overall factor the volume, of spacetime i.e. the factor β that we just mentioned. Therefore, in order to extract E(g) we can just perform n 1 integrations over R. For β the propagator of this one-dimensional field theory is simply dp q(τ)q(τ e ip(τ τ ) e m τ τ ) = 2π p 2 = + m2 2m. (1.16) For a theory with a quartic interaction (i.e. the anharmonic, quartic oscillator) W int (q) = g 4 q4 (1.17) the Feynman rules are illustrated in Fig. 2 (an extra factor 4 n has to be introduced at the end, where n is the number of vertices, due to our normalization of the interaction). One can use these rules to compute the perturbation series of the ground energy of the quartic oscillator. We will set m = 1 in our calculations. At leading order in g, we find, Z(β) = Z G (β) ( 1 g dτ q(τ)q(τ)q(τ)q(τ) + ). (1.18) 5

7 Here, Z G (β) is the Euclidean partition function of the theory with the unperturbed Hamiltonian H = 1 2 p q2, (1.19) which is nothing but the thermal partition function of a harmonic oscillator with normalized frequency ω = 1, ( ) β Z G (β) = 4sinh 2. (1.20) 2 The subscript G indicates that, from the point of view of the path integral, this is a Gaussian theory. The correlation function in (1.18) can be easily evaluated, at large β, by using (1.16). There are three possible contractions of two pairs, and since they lead to coincident arguments, each of them gives q(τ)q(τ) q(τ)q(τ) = 1 4. (1.21) As usual, the contractions leading to this result can be represented by a Feynman diagram with a single quartic vertex. This is the first diagram shown in Fig. 3. It is straightforward to push the calculation to order g 3. The relevant Feynman diagrams are shown in Fig. 3, and the corresponding symmetry factors are given in table 1. diagram 1 2a 2b 3a 3b 3c 3d symmetry factor Table 1: Symmetry factors of the Feynman diagrams in Fig. 3. These numbers can be checked by taking into account that the total symmetry factor for connected diagrams with n quartic vertices is given by 1 n! (x4 ) n (c), (1.22) where (x 4 ) n = dxe x2 /2 x 4n dxe x2 /2 (1.23) is the Gaussian average. By Wick s theorem, this counts all possible pairings among n fourvertices, and we have to take the connected piece. Since we find One finds, for example, x 2k = (2k 1)!! = (2k)! 2 k k! 1 n! (x4 ) n (4n 1)!! = = (4n)! n! 4 n n!(2n)! x 4 (c) = 3, 1 2! (x4 ) 2 (c) = 1 2 ( (x4 ) 2 x 4 2 ) = 48. (1.24) (1.25) (1.26) 6

8 We can now compute the first corrections to the ground state energy. Putting together the Feynman integrals with the symmetry factors, we find, for the different diagrams, 1 : 1 4 2a : b : a : 3b : 3c : 3d : e 2 τ dτ = 36 16, e 4 τ dτ = , e τ 1 τ 2 τ 1 τ 2 dτ 1 dτ 2 = , e 2 τ 1 2 τ 2 2 τ 1 τ 2 dτ 1 dτ 2 = , e τ 1 τ 2 τ 1 3 τ 2 dτ 1 dτ 2 = , e 2 τ 1 τ 2 2 τ 2 dτ 1 dτ 2 = (1.27) This gives, E(g) = ( g 4) 21 ( g ) ( g ) O(g 4 ). (1.28) In Chapter 4 we will be interested in understanding this series in detail, and in particular we will look at the behavior of its coefficients at high order. The method of Feynman diagrams, although it emphasizes the parallelism with field theory, is not the most efficient one in order to generate the perturbative series for the ground state. In order to do that, it is better to write down the Schrödinger equation ( 1 d 2 ) 2 dx 2 + x2 2 + gx4 ψ(x) = E(g)ψ(x). (1.29) 4 We know that, for g = 0, the solution to this equation is the ground state of the harmonic oscillator, which is just the Gaussian e x2 /2. We will then write down an ansatz for the solution of the form ( ψ(x) = e x2 /2 g ) n Bn (x), B 0 (x) = 1. (1.30) 4 Plugging this ansatz into the above equation, and writing the energy as n=0 E(g) = ( g n a n (1.31) 4) n=1 we find the following recursive equation for the B n (x) and the a n : xb n(x) 1 2 B n(x) + x 4 B n 1 (x) = n a n p B p (x). (1.32) p=0 7

9 To solve this recursion, we further write 2i B i (x) = x 2j ( 1) i B i,j. (1.33) j=1 By looking at the term of degree zero in (1.32), we find that The coefficients B i,j satisfy the recursion relation a n = ( 1) n+1 B n,1. (1.34) i 1 2jB i,j = (j + 1)(2j + 1)B i,j+1 + B i 1,j 2 B i p,1 B p,j. (1.35) This recursion can be easily solved to high orders, and one finds for the very first coefficients, a 1 = 3 4, a 2 = 21 8, a 3 = , a 4 = , (1.36) in agreement with the Feynman diagram calculation (1.28). 1.3 Unstable vacua in Quantum Mechanics In order to begin our exploration of non-perturbative effects in quantum theory, we have to keep in mind that most quantities of interest will have both perturbative and non-perturbative contributions. For small coupling, perturbative contributions are typically dominant. In order to understand the idiosyncrasies of non-perturbative effects it is then convenient to focus on quantities which vanish in perturbation theory. p=1 Figure 4: Unstable minima in one-dimensional quantum-mechanical potentials. A situation where non-perturbative effects dominate the physics is the case of unstable minima in Quantum Mechanics. Let us consider a one-dimensional potential W(q) which has a relative minimum at the origin q = 0. Near this minimum, the potential is of the form W(q) 1 2 q2 + O(g). (1.37) 8

10 where g is a coupling constant which gives the strenght of the anharmonicity. Examples of such situation are the cubic potential W(q) = 1 2 q2 gq 3, (1.38) which is depicted in Fig. 4 (left), and the inverted quartic potential W(q) = q2 2 + g 4 q4, g = λ, λ > 0, (1.39) which is shown in the right hand side of Fig. 4. It is clear that these potentials do not admit bound states, since a particle trapped near the minimum of the potential at q = 0 will eventually decay by tunneling through the barrier. However, this is a priori not detected by doing conventional. stationary perturbation theory in the coupling constant g or λ: in both cases, one finds an infinite power series for the energy of, say, the ground state. In particular, for the inverted quartic oscillator, this series is simply (1.28) after setting g = λ: E(λ) = ( ) λ ( ) λ 2. (1.40) 4 This is then a situation where perturbation theory is unable to describe the essential physics of the problem. What are the interesting quantities that can be computed in unstable potentials like the ones shown in (4)? It turns out that, although they do not admit bound states, they admit resonant states. Resonant states can be defined by considering a scattering-like problem in the above potentials, and requiring that the amplitude of the incoming wavefunction vanish. This imposes boundary conditions (the so-called Gamow Siegert boundary conditions) which select a discrete set of energies in the Schrödinger equation. These energies turn out to be complex, with a negative imaginary part, and we can write them as E = ReE i Γ, Γ > 0. (1.41) 2 The interpretation of resonant states is well-known. Since the standard time evolution is given by e iet = e itre E e Γt/2, (1.42) they correspond to unstable states with a lifetime given by τ = 1/Γ. Resonant energies can be calculated in many ways, and they typically involve a procedure of analytic continuation which makes it possible to uncover the complex values for the energy. For example, one can use the technique of complex scaling, which is based on analytically continuing the Hamiltonian to a complex-valued function by means of a dilatation operation q e iθ q. (1.43) Here θ is a parameter which can be real or complex. It turns out that, when θ is larger than a certain threshold value, the rotated Hamiltonian has eigenfunctions which are square-integrable. The corresponding eigenvalues are the complex resonant energies one is looking for, and they are independent of the value of θ, provided θ is larger than the threshold value. Another possibility is to make an analytic continuation in the value of the coupling constant. For example, in the case of the quartic potential, one can first consider positive values of the coupling constant g > 0. For these values of g the potential admits bound states, whose energies can be computed. Then 9

11 one can make an analytic continuation to negative values of g. It turns out that, when doing this continuation, the energy levels of the potential for develop an imaginary part which corresponds to the resonant energies. In potentials of the form (1.37), the real part of the resonant energy has a small g expansion which is precisely the result obtained by stationary perturbation theory. In particular, each resonant state can be regarded as a perturbed level of the harmonic oscillator, since as the coupling constant goes to zero one must have ReE N + 1, g 0. (1.44) 2 The imaginary part of the energy can be estimated with WKB methods and it is exponentially small as the coupling constant goes to zero. For example, for the ground state of the quartic potential (i.e. the state which has N = 0 in (1.44)) one finds where Γ e A/λ, (1.45) q+ A = 2 p(q) dq, (1.46) 0 and q + is the turning point of the potential, satisfying W(q + ) = 0, and p(q + ) = 2W(q). (1.47) In the previous section we have seen that one can calculate the ground state energy of a quantum-mechanical system by looking at the path-integral representation of the thermal free energy. It is natural to ask how the imaginary part of the resonant energy can be computed from the path-integral point of view. One of the advantages of the path integral/instanton method is that it can be easily generalized to field theory, as we will eventually do. g z g z Figure 5: We can analytically continue the integral (1.48) to negative values of g by rotating the integration contour for z. This can be done in two ways: we can rotate g clockwise, and the integration contour counterclockwise, or the other way around. 10

12 In order to understand how complex energies appear in the path-integral formalism, we will consider a toy model for the quartic potential. As we just mentioned, one way of understanding the appearance of complex energies in this problem is to start with a stable potential with g > 0 and then move in the complex plane of the coupling constant until we reach the line g < 0. To understand the behavior of the path integral under such analytic continuation, we will perform it first in an ordinary integral, which can be regarded as the reduction of the anharmonic oscillator from one dimension to zero dimensions. This integral is defined as It is well defined as long as I(g) = 1 + dz e z2 /2 gz 4 /4. (1.48) 2π Re(g) > 0, (1.49) but we would like to define it for more general, complex values of g, in particular we would like to define it for negative values of g. This can be done by analytic continuation: we rotate the contour of integration for the z variable, so that Re (gz 4 ) > 0 (1.50) and the integral is still convergent. Equivalently, we give a phase to z in such a way that Arg z = 1 Arg g. (1.51) 4 Obviously, this analytic continuation of the integral is no longer real. In order to define the integral for negative g, we should rotate g towards the negative real axis. But it is clear that this can be done in two different ways: clockwise or counterclockwise, as shown in Fig. 5. The integration contour for z rotates correspondingly. Since the resulting integration contours are complex conjugate to each other, the two integrals defined in this way are also complex conjugate. For g g + i0, one has for z the integration contour C + : Arg z = π 4, (1.52) while for g g i0, one has C : Arg z = π 4, (1.53) see Fig. 6. This means that one can indeed obtain an analytic continuation of the integral I(g) to negative g, but the resulting function will have a branch cut along the negative real axis. The discontinuity across the cut is given by I(g + i0) I(g i0) = 2iIm I(g) = 1 2π C + C dz e z2 /2 gz 4 /4. (1.54) The discontinuity (1.54) can be computed by saddle-point methods. The saddle points of the integral occur at z = 0 or Therefore we have two nontrivial saddlepoints S 1,2 z + gz 3 = 0 z 2 = 1 g. (1.55) z 1,2 = ±e i(π/2+φg/2) g 1 2 (1.56) 11

13 g C + C z g + i0 g i0 Figure 6: The integration contours C ± correspond to the negative values of g = g ± i0. C + C S 1 S 2 Figure 7: The complex plane for the saddle-point calculation of (1.48). Here, C + and C are the rotated contours one needs to consider for g < 0. Their sum may be evaluated by the contribution of the saddle point at the origin. Their difference is evaluated by the contribution of the sub leading saddle points, here denoted as S 1 and S 2. where φ g is the phase of g. For g < 0, they are on the real axis, see Fig. 7x. The steepest descent trajectories passing through these points are determined by the condition Imf(z) = Im f(z i ), f(z) = z2 2 + g 4 z4. (1.57) For g < 0 these are hyperbolae x 2 y 2 = 1 g (1.58) passing through the saddlepoints S 1,2 at x = ± g 1 2, y = 0, see Fig. 7. From this figure it is also clear that the contour C + C appearing in (1.54) can be deformed into the sum of the steepest descent trajectories passing through S 1,2, therefore the imaginary part in (1.54) is given by ImI(g) 1 ( ) 1 exp, g 0. (1.59) 2 4g 12

14 The overall factor in (1.59) is obtained by doing the Gaussian integrations around the two saddles and adding up the results. Since the integral (1.48) is divergent for g < 0, the resulting complex function can not be analytic at g = 0. One consequence of this lack of analyticity is that the formal power series expansion around g = 0, where a k = ( 4) k 2π I(g) = a k g k, (1.60) k=0 dz z4k k! e z2 /2 = ( 4) k (4k 1)!!, (1.61) k! has zero radius of convergence. Its asymptotic behavior at large k is obtained immediately from Stirling s formula a k ( 4) k k!. (1.62) This factorial divergence is in fact a generic feature of perturbative series in quantum theory, as we will see. The moral of the simple analysis in the previous subsection is that, for negative g, the integral I(g) picks an imaginary part which is given by the contribution of the nontrivial saddlepoints. By analogy with this integral, we expect that the path integral of the quantum anharmonic oscillator will have the same behavior. Therefore, we expect an imaginary part in the thermal partition function, and we expect this imaginary part to be exponentially suppressed at small λ, just as in (1.59). The free energy then reads, when expanded formally in ImZ, F(β) = 1 β log Z = 1 β log(re Z) i β Therefore, at leading order in the exponentially suppresed factor we have and ImF(β) 1 β ImZ Re Z +, (1.63) Im Z Re Z, (1.64) 1 ImZ Im E(g) = lim Im F(β) lim β β β ReZ. (1.65) Furthermore, as in (1.54), we expect that the quantity discz( λ) = Z( λ + iǫ) Z( λ iǫ) = 2iIm Z( λ) (1.66) is given by the sum of the contributions of the nontrivial saddle-points of the path integral (1.12). We will now calculate these contributions. 1.4 Path integral around an instanton in QM We will now consider quantum-mechanical potentials W(q) which have a relative minimum at q = 0. Near this minimum, the potential is of the form (1.37). We will also assume that the potential is unstable, as in (8). We will study the imaginary part of the thermal free energy, Z(β), at large β, in order to extract the imaginary part of the ground state energy. To do that, we use the Euclidean path integral (1.12). The non-trivial saddle points of this path integral are time-dependent, periodic solutions of the EOM for the inverted potential, q c (t) + V (q c ) = 0. (1.67) 13

15 V (x) = W(x) E = 0 W(x) q q + Figure 8: A general unstable potential W(x) and the associated inverted potential V (x). A periodic solution with negative energy moves between the turning points q ±. The trajectory with zero energy, relevant for extracting the imaginary part of the ground state energy, is also shown. Examples of such nontrivial, periodic saddle points are oscillations around the local minima of V (q), as shown in Fig. 8. The period of such an oscillation between the turning points q and q + is given by q+ β = 2 q dq 2(E V (q)). (1.68) These trajectories satisfy in addition the energy conservation constraint and the action along such a trajectory is given by where W(E) = 1 2 q2 + V (q) = E(β), (1.69) S c S(q c (t)) = W(E) Eβ, (1.70) β/2 β/2 q+ dt( q c (t)) 2 = 2 p(q)dq. (1.71) q Notice that the period (1.68) varies between β = (corresponding to E = 0 in Fig. 8) and a minimum critical value β c corresponding to the minimum q 0 of the potential. This value can be computed as follows. Near the bottom of the inverted potential one has V (q) = V ω2 (q q 0 ) 2 + (1.72) where Let us define ǫ by the equation ω 2 = V (q 0 ). (1.73) E = V ω2 ǫ 2. (1.74) Then, at leading order in ǫ, the turning points can be approximated by q ± q 0 ± ǫ, (1.75) 14

16 q c (t) ( 2 2 λ)1 ( 2 λ)1 2 V (q) = 1 2 q2 + λ 4 q4 Figure 9: The inverted potential relevant for instanton calculus in the quartic case. The instanton or bounce configuration q c (t) leaves the origin at t =, reaches the zero (2/λ) 1 2 at t = t0, and comes back to the origin at t = +. We then find, q0 +ǫ β c = lim 2 ǫ 0 q 0 ǫ dq ω 2 (ǫ 2 (q q 0 ) 2 ) = 2 ǫ ω ǫ dζ ǫ 2 ζ 2 = 2π ω. (1.76) For β < β c there are no instanton trajectories. In terms of a thermal partition function, this means that for sufficiently high temperatures the instanton degenerates to a solution q(t) = q 0 staying at the top of the barrier. The decay mechanism above the temperature T c = 1/β c is just due to thermal excitations over the top of the barrier. Example 1.1. In the example of the quartic anharmonic oscillator, the inverted potential is V (q) = 1 2 q2 + λ 4 q4, (1.77) see Fig. 9. We can change variables q λ 1/2 q, E E/λ to set λ = 1, and the equation of motion reads q(t) + q(t) q 3 (t) = 0, (1.78) The inverted potential has minima at q ± 1, and zeros at q = ± 2. We will focus on the region q 0, the results in the region q < 0 foloows by the symmetry q q of the problem. We can now find a solution to the EOM (1.78), with energy 1/4 E 0, and turning points at q ± = 1 ± 1 + 4E, (1.79) in terms of Jacobi elliptic functions: q t 0 c (t) = q + dn (u m), (1.80) where u = q+ 2 (t t 0 ), m = 1 q2 q+ 2. (1.81) 15

17 Figure 10: The solution (1.84) with t 0 = 0. The period can be computed from (1.68), and reads β = 2 ( ) 2 m 1/2 2 K(m), (1.82) 2 where K(m) is the complete elliptic integral of the first kind. In terms of m, the energy reads E = 1 m (2 m) 2. (1.83) The value m = 0 corresponds to a particle with minimal energy E = 1/4, sitting at the bottom of the V (q), with period β c = 2/π. Since the frequency of the oscillations around the bottom is ω = 2, this is in accord with the result (1.76). The value m = 1 corresponds to a particle with energy E = 0 and infinite period β. Notice that the solution (1.80) has a parameter t 0, which corresponds to the initial point of the trajectory. Since t 0 is defined modulo the period of the motion, we can choose t 0 [ β/2,β/2]. In the limit m 0, corresponding to E = 0 and an infinite period, the solution simplifies. Since the Jacobi function dn(u m) becomes sech(u) when m 1, we find q t 0 c (t) = 2 cosh(t t 0 ). (1.84) This trajectory starts at the origin in the infinite past, arrives to the turning point q + at t = t 0, and returns to the origin in the infinite future, i.e. An example of (1.84) is shown at Fig. 10. Let us now return to the general case and expand the action around q c (t). We find, after writing q(t) = q c (t) + r(t) (1.85) that, at quadratic order in the fluctuations, S(q) S c + 1 dt 1 dt 2 r(t 1 )M(t 1,t 2 )r(t 2 ) (1.86) 2 where M is given by M(t 1,t 2 ) = δ 2 [ ( ) S d 2 δq(t 1 )δq(t 2 ) = V (q c (t 1 ))] δ(t 1 t 2 ). (1.87) q(t)=qc(t) dt 1 16

18 This distribution can be regarded as the kernel of an integral operator M, which acts on functions as (Mψ) (t) = M(t,t )ψ(t )dt, (1.88) and it is given explicitly by ( ) d 2 M = V (q c (t)). (1.89) dt In the quadratic (or one-loop) approximation, the path integral around the configuration q c (t) is then given by [ Dq(t)e S(q) e S(qc) Dr(t) exp 1 ] dt 1 dt 2 r(t 1 )M(t 1,t 2 )r(t 2 ). (1.90) 2 Since we are integrating over periodic configurations, the boundary conditions for r(t) are r( β/2) = r(β/2), ṙ( β/2) = ṙ(β/2). (1.91) Note that all possible values of the endpoints for r(t) are allowed, since we have to integrate over all possible periodic trajectories, and in particular over all possible endpoints. We now have to perform the Gaussian integration over r(t). In order to do this, we consider a complete set q n be orthonormal eigenfunctions of M, labeled by n = 0, 1,, and satisfying periodic boundary conditions appropriate for (1.90), Mq n = λ n q n, (1.92) q n ( β/2) = q n (β/2), q n ( β/2) = q n (β/2). (1.93) The eigenvalue problem can be written explicitly as ] [ d2 dt 2 V (q c (t)) q n (t) = λ n q n (t), n 0, (1.94) and orthonormality means that β/2 β/2 dt q n (t)q m (t) = δ nm. (1.95) Here, we have assumed that the spectrum is discrete. As we will see, in many cases this is not the case, but the formalism we are developing can be easily modified to account for a continuous spectrum. We now expand the fluctuations as r(t) = n 0c n q n (t). (1.96) This can be regarded as a change of variables from the set of paths r(t) to the coefficients c n. The measure for r(t) is then defined as the normalized Gaussian measure for the c n s, up to an overall normalization constant N which is indepedent of the potential: Dr(t) = N n 0 dc n 2π, (1.97) 17

19 and we find where [ Dr(t) exp 1 2 = N n 0 ] dt 1 dt 2 r(t 1 )M(t 1,t 2 )r(t 2 ) dc P n e 1 2 n 0 λnc2 n = N (detm) 1/2, 2π (1.98) detm = n 0λ n. (1.99) This derivation has been purely formal, and in order to perform the calculation of the determinant of M we have to take into account many subtleties Firs of all, if we take a further derivative w.r.t. t in (1.67) we find d 2 dt 2 q c(t) + V (q c (t)) q c (t) = 0. (1.100) Since q c (t) is periodic, q c (t) is periodic as well and the boundary conditions (1.91) are satisfied. Therefore, q c (t) is a zero mode of M, i.e. an eigenfunction with zero eigenvalue. It is also a normalizable function, therefore it must be (up to normalization) one of the eigenfunctions q n (t) of M, say q 1 (t). We will see in a moment that this eigenfunction is the first excited state of the spectrum, and the eigenfunction of the ground state will be denoted by q 0 (t). The normalized zero mode is q 1 (t) = 1 q c q c(t), (1.101) where the norm is given by q c 2 = β/2 β/2 dt ( q c (t)) 2 = W(E), (1.102) and we used (1.71). Note that, in the limit of large β, relevant to extracting the ground state energy, the trajectory q c (t) has E = 0, and we find W(E) = S c, β. (1.103) The origin of the zero mode q c (t) can be explained by time translation invariance. As we made clear in the examples in (??) and (1.84), the solution q c (t) depends on an arbitrary initial time t 0, and we should write it rather as q t 0 c (t). This trajectory solves the equations of motion for all values of the parameter t 0, δs δq(t) q(t)=q t 0 c (t) = 0. (1.104) This is a general fact: when we solve for a nontrivial saddle point we find in general a family of solutions. The parameters for such family are called moduli or collective coordinates. In the case at hand, we have a single modulus, namely the initial time t 0. If we now take a further derivative of (1.104) w.r.t. t 0, we obtain δ 2 S dt 2 δq(t 1 )δq(t 2 ) q(t)=q t 0 c (t) δq t 0 c (t 2 ) δt 0 = 0. (1.105) 18

20 The second functional derivative of S is M(t 1,t 2 ), and δq t 0 c (t 2 ) δt 0 = q t 0 c (t 2 ). (1.106) We conclude that dt 2 M(t 1,t 2 ) q t 0 c (t 2 ) = 0, (1.107) therefore q t 0 c (t) is a zero mode of M. Equivalently, we can consider the EOM (1.67), which is solved by q t 0 c (t) for any t 0, and take a derivative w.r.t. t 0. In this way one obtains again (1.100). Let us now address the issue raised by the existence of a zero mode, i.e. the fact that λ 1 = 0. Naively this leads to a vanishing result for the functional determinant (1.99), and a diverging result for (1.98). This is due to the fact that the mode c 1 has no damping factor in the Gaussian integral, since λ 1 = 0, and the infinite answer comes from the integration over c 1. We can now isolate this divergence as ( dc P n dc1 e 1 2 n 0 λnc2 n = )(det M) 1/2 (1.108) 2π 2π n where det M = n 1λ n (1.109) is the determinant of the operator M once the zero mode has been removed. However, the integration over c 1 should be treated more carefully, since this variable stands really for the collective coordinate t 0. To see this, notice that an arbitrary, periodic function of t can be expanded in two equivalent ways, either as in (1.96) or as q t 0 c (t) + n 1c n q n (t) (1.110) where t 0 now varies and parametrizes a direction in the space of path configurations. If we change c 1 in (1.96) we obtain q 1 (t)δc 1 = 1 q c qt 0 c (t)δc 1 (1.111) while varying t 0 in (1.110) gives q t 0 c (t)δt 0. (1.112) Both variations are proportional, therefore (1.110) parametrizes the same fluctuations as (1.96). The Jacobian of the change of variables from c 1 to t 0 can be easily computed by comparing both variations, J = δc 1 = qc = (W(E)) 1/2 (1.113) δt 0 Therefore, the integration over c 1 gives 1 2π dc 1 = J β/2 2π β/2 where we have used that the moduli space for t 0 is [ β/2,β/2]. dt 0 = β (W(E))1/2 2π, (1.114) 19

21 To summarize: instantons come in families parametrized by collective coordinates or moduli. This leads to zero modes in the quadratic operators that are obtained by looking at fluctuations around a fixed solution. The integration over these zero modes has to be translated into an integration over collective coordinates, and removes the apparent divergences associated to the zero modes in a naif treatment of the path integral. The second important property of the operator M is that it has one, and only one, negative mode. To see this, we note that it has an eigenfunction q c (t) with zero eigenvalue. But q c (t) changes sign at one of the turning points, and there should be an eigenfunction with a lower eigenvalue, which has to be negative. In the limit β, this can be also established by regarding (1.89) as a one-dimensional Schrödinger operator. The spectrum of such an operator has the well-known property that the ground state has no nodes, the first excited state has one node, etc. The function q c (t) has one node, so it is the first excited state of M and the ground state must have negative energy. This is the negative mode of M. Example 1.2. Let us consider again the quartic oscillator in the limit β and with λ = 1. The operator M is given in this case by M = d2 dt cosh 2 (t t 0 ). (1.115) Using translation invariance we can just set t 0 = 0 to study the spectrum. It is easy to see that Mψ(t) = 3ψ(t), ψ(t) = which is the single negative mode of this operator, and q 0 (t) ψ(t). 1 cosh 2 (t), (1.116) Since λ 0 < 0, in calculating the Gaussian integral (1.98) we have to make sense of the Gaussian integral with the wrong sign involving the mode c 0, dc0 2π e 1 2 λ 0 c 2 0. (1.117) This is done by analytic continuation: we rotate the integration contour of c 0 an angle of π/2, so that the resulting integral is done along the imaginary axis and is convergent. The result of the integration will be ±i λ 0 1/2, depending on whether we make the rotation clockwise or counterclockwise. Therefore, the final answer for (1.98) is imaginary, and there is sign ambiguity due to the analytic continuation. Equivalently, since M has one and only one negative eigenvalue, det M is negative, and in extracting its square root we will obtain an imaginary result. The sign ambiguity corresponds to the choice of branch cut of the square root. We now put everything together, and obtain Sc β (W(E))1/2 2iIm Z Ne (det M) 1/2. (1.118) 2π Note that, as we just discussed, the r.h.s. is imaginary due to the negative mode of M, which is consistent with the fact that we are computing the imaginary part of the partition function. Finally, we should fix the normalization of the path integral measure, N. In order to do this, it is convenient to use the (unperturbed) harmonic oscillator with ω = 1 as a reference point. Its 20

22 thermal partition function is given in (1.20). A path integral evaluation of this partition function along the lines of what we have done gives the well-known result where We then find, Z G (β) = N(detM 0 ) 1/2, (1.119) ( ) d 2 M 0 = + 1. (1.120) dt Im Z(β) 1 ( det ) 1 2i Z M 2 β (W(E)) 1/2 G(β) e Sc. (1.121) detm 0 2π This is the one-loop approximation to the full result. Since we have assumed that our potential is a perturbed quadratic potential, the real part of Z(β) is given, at one-loop, by Re Z Z G (β). (1.122) Therefore, after we take the limit β, or E 0, we find the following one-loop result for the imaginary part of the ground state energy, ( ) 1 Im E S1/2 c 2 det M 2 e S c. (1.123) 2π detm 0 Here we used the fact that, for β, W(E) becomes S c, as we pointed out in (1.103). The choice of branch cut for the square root in (1.121) corresponds to the choice of contour rotation in (1.117). The formula (1.123) gives the imaginary part of the ground state energy for general unstable potentials obtained by perturbing a quadratic potential. In the above formula it is understood that one considers the operator M in the limit of zero energy, which for example for the quartic oscillator is given by (1.115). Although we have discussed two of the subtleties appearing in the computation of the functional determinant, we still have to make sense of the infinite product appearing in (1.109). The eigenvalues of a Schrödinger operator of the form (1.89), on an interval [ β/2,β/2], grow like n as ( ) n 2 λ n, n 1, (1.124) β and the infinite product has to be regularized in an appropriate way. There are various ways to do this. One possibility is to consider quotients of determinants of different operators, as we have done in (1.123). In many cases, the divergent parts of the determinants cancel against each other, and one is left with a finite piece. Another possibility is to use a regularization prescription which removes the divergences and leads to a finite, physically meaningful finite piece. A particularly useful regularization is the zeta function regularization, which we explain in detail in Appendix??. We will now calculate the relevant determinants in our quantum-mechanical problem by using these two methods. 1.5 Calculation of functional determinants I: solvable models We will first consider a very direct approach to the computation of the determinant of M, starting from a computation of their spectrum. As is well-known from Quantum Mechanics, the spectrum of Schrödinger operators can be only computed exactly in some special solvable 21

23 models. Therefore, we have to consider potentials V (x) such that the operator appearing in (1.89) belongs to the class of solvable Schrödinger operators. It turns out that the operators one finds in the case of the quartic oscillator in the limit β, is precisely of this type, and belongs to a general family of operators called Pöschl Teller operators. They are labelled by two parameters l, m, and they have the form M l,m = d2 dt 2 + l(l + 1) m2 cosh 2 (t). (1.125) They can be regarded as Schrödinger operators in an inverted cosh squared potential, also called Pöschl Teller potential. Remarkably, the spectrum of these operators can be determined exactly. This is due to a factorization property first studied by Schrödinger, and which can be substantially clarified in the light of supersymmetric quantum mechanics, as we will do in Chapter 6. Let us introduce the operators It is immediate to compute that A l = d dt + ltanh t, A l = d + ltanh t. (1.126) dt A l A l = M l,m + l 2 m 2, A l A l = M l 1,m + l 2 m 2. (1.127) Notice that for l = 0 we recover the free particle. Also, we can obtain the ground state for the full family of potentials just by solving This is a first order ODE with solution The ground state energy for the operator M l,m is simply A l ψ (l) 0 (t) = 0. (1.128) ψ (l) 0 (t) 1 cosh l (t). (1.129) E (0) l,m = m2 l 2. (1.130) The properties above also make possible to calculate the excited states. To do this, notice that if ψ (l 1) (t) is an eigenfunction of M l 1,m with eigenvalue µ l 1, then ψ (l) (t) = A l ψ(l 1) (t) (1.131) is an eigenfunction of M l,m with the same eigenvalue. Indeed, ( M l,m ψ (l) (t) = A l A l + m 2 l 2) A l ψ(l 1) (t) = A ( l Ml 1,m l 2 + m 2) ψ (l 1) (t) + ( m 2 l 2) ψ (l) (t) = µ l 1 ψ (l) (t) (1.132) We can then construct the spectrum of M l,m by starting with the free particle l = 0 and applying the operators A l. For l = 0, the eigenfunctions are just plane waves (scattering states) e ikt (1.133) 22

24 A 1 A 2 ψ (0) 1 1 cosh(t) ψ (0) 2 1 cosh 2 (t) l = 1 l = 1 l = 2 Figure 11: The recursive solution of the spectrum of the Pöschl Teller potential. with energies Applying A 1 E l,m (k) = k 2 + m 2. (1.134) we obtain the scattering states of the l = 1 potential ψ (k) 1 (t) = A 1 e ikt (1.135) 1 + k 2 2π appropriately normalized. On top of that, we have the ground state (1.129) with l = 1, ψ (0) 1 1 cosh(t). (1.136) To go to l = 2, we apply A 2 to these states, and we obtain the scattering states a bound state and the new ground state ψ (k) 2 (t) = A 2 A 1 e ikt, (1.137) k k 2 2π 2 (t) 1 A 2cosh(t) ψ (1) (1.138) ψ (0) 2 (t) 1 cosh 2 (t). (1.139) Proceeding in this way we obtain the full spectrum of the l-th potential. It consists of scattering states ψ (k) A l l (t) = l 2 + k A 1 e ikt, (1.140) k 2 2π with energy E l,m = k 2 + m 2, (1.141) 23

25 and l bound states ψ (j 1) l (t) A l 1 A l j+1 cosh l j+1, j = 1,,l, (1.142) (t) with energy E (j) l,m = m2 (l j + 1) 2. (1.143) Since the spectrum of the operator M l,m is known, we should be able to compute its determinant. However, as we have explained above, this requires a regularization. We will do that by considering the quotient of the determinant of M l,m by a reference determinant. As in (1.121), it is convenient to take as our reference operator the harmonic oscillator operator M 0,m. From the point of view of Pöschl Teller potentials, this corresponds to the free particle. However, since we are working in the limit of infinite volume β, there is a part of the spectrum which is continuum. In our considerations in the previous section we have assumed that the spectrum is discrete, which is actually the case for finite β, so we have to be precise about what we mean by the determinant in this more general case. If an operator M has a discrete spectrum {λ n } and a continuum spectrum λ(k), the determinant should be understood as log detm = log(λ n ) + dk ρ(k) log (λ(k)), (1.144) n where ρ(k) is the density of states for the continuum part. This is easily determined by introducing an IR regulator and putting the system in a box. In the case of the Pöschl Teller potential, the calculation of ρ(k) goes as follows. A scattering state will experience phase shifts. Indeed, as t ±, we have and the asymptotic form of the scattering states will be ψ (k) l (t) A l d dt ± l (1.145) l ( ik ± j) e ikt, t ±. (1.146) j=1 Therefore the phase shifts, defined by [ ( ψ (k) l (t) exp i kt ± θ(k) )] 2 (1.147) are given by θ(k) 2 l ( ) k = tan 1 + π j 2 j=1 (1.148) The quantization condition once we put these scattering states in a box of length β is just ikβ + iθ(k) = 2πin (1.149) and the density of states is ρ(k) = dn dk = ρ free(k) + ρ θ (k), (1.150) 24

26 where In our case ρ free (k) = β 2π, ρ θ(k) = 1 2π θ (k). (1.151) ρ θ (k) = 1 π l j=1 j k 2 + j2. (1.152) Let us now compute log det M l,m by using these results. In the expression (1.144), the sum over λ n is over non-zero, discrete eigenvalues, and it is finite. The integration over the continuum part is divergent due to the contribution of ρ free (k). However, if we subtract log detm 0,m, this part cancels and we find Since ( det ) M l,m log = detm 0,m the end result is = 1 j l, j m 1 j l, j m log(m 2 j 2 ) + dk ρ θ (k)log(k 2 + m 2 ) log(m 2 j 2 ) 1 π l j=1 dk k 2 + j 2 log(k2 + m 2 ) = 2π j dk j k 2 + j 2 log(k2 + m 2 ). (1.153) log(j + m), (1.154) det M l,m 1 j l,j m = (m2 j 2 ) detm 0,m 1 j l (m +, (1.155) j)2 for (l,m) (1,1). In the case l = m = 1, the numerator should be taken as 1, since there is no contribution from discrete states, and one has 1.6 Lifetimes in unstable vacua from instantons det M 1,1 detm 0,1 = 1 4. (1.156) We now have all the necessary technical ingredients to calculate the imaginary part of the ground state energy in unstable vacua, given by (1.123). We can calculate the determinant of the relevant operators by using either the special results for solvable potentials, or the general expression (??). However, this expression was obtained for general β (or, equivalently, E), and to extract the ground state energy we are interested in the limit β, i.e. in the orbits of zero energy. This can be done as follows. First of all, we normalize our potential in such a way that the unstable minimum is at the origin. Periodic orbits of zero energy go from the unstable minimum q = 0 (which is also one of the turning points) to the other turning point q +. Clearly, as E 0, β. Let us consider the derivative of E w.r.t. β. In the large β limit, we write (1.68) as q+ β = 2 q dx 1 [ ] 1 2(E V (x)) [x 2 + 2E] 1 2 [x 2 + 2E] 1 2. (1.157) For E small, we have q 2E + O(E). (1.158) 25

27 The last integral gives ( 2log x + ) q + x 2 + 2E = log q+ 2 log( E/2) + O(E), E 0. (1.159) q The first two terms in (1.157) have a smooth limit at E 0, and they give q+ ( 1 2 dx 1 ). (1.160) 2W(x) x It follows that, as β, [ E(β) 2q+ 2 exp 2 therefore 0 q+ [ E q+ ( β 2q2 + exp 2 dx 0 0 ( 1 dx 1 )] e β, (1.161) 2W(x) x 1 1 2W(x) x )] e β, (1.162) which is manifestly positive, as anticipated above. We can now use this limiting expression in (??). Note that, due to the exponential dependence on β, det M diverges exponentially when β. However, after dividing by the determinant of the Gaussian operator M 0, we obtain a finite expression in this limit, given by det M = S c detm 0 2q+ 2 exp [ 2 q+ 0 dx ( 1 2W(x) 1 x )]. (1.163) Plugging this expression in (1.123) we finally obtain a general formula for the width of an unstable level in quantum mechanics (at one-loop): Im E ± 1 [ q+ ( 2 π q 1 + exp dx 1 )] e S(qc). (1.164) 2W(x) x 0 It is an interesting exercise to evaluate (1.163) for concrete potentials and check that indeed it agrees with (1.155). Example 1.3. Anharmonic oscillator. The action of the instanton is given by The exponent in (1.163) is q+ 0 The determinant is then given by 2/λ S c = 2 x 1 λ2 x2 dx = 2( 2 λx 2) 3/2 3 2λ 0 ( 1 dx 1 ) 2/λ 2 2 λx 2 = dx 2W(x) x x 2 λx 2 0 2/λ ( ) 2/λ = log 2 2 λx = log = 4 3λ. (1.165) (1.166) det M detm 0 = (1.167) 26

28 This can be also calculated by using Pöschl Teller operators: if we compare (1.115) to (1.125) we see that it is given by M 2,1, and formula (1.155) gives the same result as (1.167)). Using finally (1.123) we derive an explicit formula for the imaginary part of the ground state energy, Im E π 2 λ 2 e 4 3λ = 4 2πλ e 4 3λ. (1.168) Example 1.4. Cubic oscillator. Let us now study the lifetime of a particle in the ground state of the cubic potential The turning points are q = 0 and The instanton solution is and the operator M reads The action of the instanton is S c = 2 W(x) = 1 2 x2 gx 3. (1.169) q + = 1 2g. (1.170) 1 q c (t) = 2g cosh 2 ( ) (1.171) t 2 M = d2 dt cosh 2 ( ). (1.172) t 1/(2g) The nontrivial integral involved in the one-loop fluctuation is and we find 1/(2g) Therefore, the general formula (1.123) gives, 0 Im E 0 (g) 1 2π (x 2 2gx 3 ) 1 2dx = 2 15g 2. (1.173) dx x x 2 2gx 3 x x 2 2gx 3 = log 4, (1.174) det M detm 0 = (1.175) 1 2g 4 e 2/(15g2) = 1 πg 2 e 2/(15g2). (1.176) This agrees with the result obtained with the WKB method. The result (1.175) can be also obtained with the help of the appropriate Pöschl Teller operator. After rescaling t 2t we find that M = 1 4 M 3,2. (1.177) The behavior of the determinant of an operator under rescaling requires a careful analysis. After normalizing we find that det ( ) M 1 N 3,2 N 0,2 det M 3,2 = (1.178) detm 0 4 detm 0,2 where N 3,2 N 0,2 is the number of non-zero modes of M 3,2 minus the number of modes of M 0,2. This can be computed in general by mimicking the procedure in (1.153). Notice that M l,m has 27

29 j 1 discrete non-zero modes for m l, plus a continuum. To calculate the difference between the zero modes in the continuum for M l,m and M 0,m we can use again the spectral density. We find, N l,m N 0,m = j 1 + dk ρ(k) = j 1 1 π = j 1 j = 1 l j=1 j dk k 2 + j 2 (1.179) Therefore, we conclude that the determinant (1.163) for the cubic oscillator and (1.155) for l = 3, m = 2 should be related by det M = 4 det M 3,2, (1.180) detm 0 detm 0,2 which is indeed the case. 1.7 Instantons in the double well The double-well illustrates one of the most important applications of instantons: their ability to lift perturbation theory degeneracies. Indeed, the double-well potential has, in perturbation theory, two different ground states located at the two degenerate minima. This implies, in particular, that parity symmetry is spontaneously broken in perturbation theory. However, in the full theory, this cannot be the case: we know from elementary quantum mechanics that the spectrum of the Schrödinger operator in this bound-state problem must be discrete, and that the true vacuum is described by a symmetric wavefunction. This wavefunction corresponds, in the limit of vanishing coupling, to the symmetric combination of the two perturbative vacua. The energy split between the symmetric and antisymmetric combination is however invisible in perturbation theory and goes like exp( 1/g) a typical instanton effect. This energy split can be computed with the WKB method. In this section we will derive it with instanton techniques in the path-integral approach. Consider the double well potential with a Hamiltonian of the form (1.4) and W(q) = g 2 ( q 2 1 ) 2, g > 0. (1.181) 4g In perturbation theory one finds two degenerate ground states, located around the minima q = ± 1 2 g. (1.182) The frequency of oscillations around this minima has been normalized to be ω = 1, and the ground state energy obtained in stationary perturbation theory is a formal power series of the form E 0 (g) = 1 2 g 9 2 g g3 (1.183) The Hamiltonian is invariant under the parity symmetry q q, (1.184) and thus it commutes with the corresponding parity operator P, whose action on wave functions is P ψ(q) = ψ( q). (1.185) 28

30 The eigenfunctions of H satisfy H ψ ǫ,n (q) = E ǫ,n (g)ψ ǫ,n (q), P ψ ǫ,n (q) = ǫψ ǫ,n (q), (1.186) where ǫ = ±1 is the parity. The quantum number N can be uniquely assigned to a given state by the requirement that, as g 0, E ǫ,n (g) = N + 1/2 + O(g), (1.187) i.e. it corresponds to the N-th energy level of the unperturbed harmonic oscillator. Let us now focus on the grounds state energy. As we know, the energy levels E ǫ,0 (g) are degenerate in perturbation theory, but they are split by non-perturbative effects. Therefore, they will have a perturbative contribution, given by (1.183), and they will differ in the nonperturbative corrections. In order to study non-perturbative effects, it is always convenient to focus on a quantity which is purely non-perturbative, i.e. which vanish in perturbation theory. In our case, this quantity is clearly E +,0 E,0. (1.188) This is, morally speaking, the analogue of ImE 0 in the case of unstable potentials, which is also purely non-perturbative. We would like to find now a quantity which can be computed in the path-integral formalism and which is sensitive to the difference of energies (1.188). Clearly, the thermal partition function is not the most appropriate quantity. However, one can consider the twisted partition function ( Z a (β) = Tr P e βh) (1.189) where P is the parity operator (1.185). For large β and small coupling constant one finds that, Z a (β) e βe +,0 e βe,0 βe βω/2 (E +,0 E 0, ), (1.190) so in principle we can use this twisted partition function to extract (1.188). In addition, Z a (β) can be written in terms of a path integral with twisted boundary conditions, Z a (β) = Dq(t) exp [ S ( q(t) )], (1.191) q(β/2)=p(q( β/2)) In the case of the double well potential we are studying, the boundary condition reads q( β/2) = q(β/2). (1.192) In the infinite β limit, the leading contributions to the path integral come from paths which are solutions of the Euclidean equations of motion and have zero energy. In the case of Z a (β), constant solutions of the equation of motion do not satisfy the boundary conditions. Therefore we have to sum over paths which connect the two minima of the potential (1.182), like in Fig. 12. These correspond to nontrivial instanton configurations. In the example of the double well potential (1.181), such solutions are q t 0 ± (t) = ± 1 2 g tanh ( ) t t0 2 (1.193) The solutions q t 0 ± are called (anti)instantons of center t 0. They are represented in Fig. 12 for g = 1/4. Since both solutions depend on an integration constant t 0, there are two one-parameter families of degenerate saddle points. 29