0-brane Quantum Chemistry VA´CLAV KARESˇ1 Institute for Theoretical Physics, Masaryk University, Kotla´ˇrska´ 2, 611 37, Brno, Czech republic 4 0 0 2 Abstract n a We apply two different numerical methods to solve for the boundstate J oftwo0-branesinthreedimensions. Onemethodisdevelopedbyusinthis 4 2 work and we compare it to a method existing in the literature. In spite of considering only three dimensional Minkowski space we obtain interesting 1 results which should give some basic understanding of the behaviour of v 9 0-branes. 7 1 1 0 4 0 / h 1 Introduction t - p e D-branes are very important non-perturbative objects in string theory. Their h existence is essential since they are needed, among other things, for various string : v theory dualities to work. The most primitive definition of a D-brane is as a hyper i X surface on which open strings end. r In this work we are interested only in D0-branes. They play a particularly a important role since they are the basic constituents of Matrix theory [1], which, being a suggestion for a non-perturbative definition of string theory (or M-theory [2]), deserves particular attention. It is also possible to think about them as bound states of higher dimensional unstable D-branes [3, 4]. Furthermore, as was shown by Myers [5], they can also form bound states with all the properties of higher dimensional branes. It is therefore clear that D0-branes have many interesting properties that make them worth studying in detail. Concretely this means studying supersymmetric quantum mechanics which is aninteresting topic in itself. Although this topic has been studied before [6]-[14] we feel that there are still unresolved issues. In particular one could use a computer to compute properties of D0-branes which are not possible to address analytically because of the complexity of the theory. This is the goal of this work, to develop a “0- brane Quantum Chemistry”. It should be mentioned that similar issues have 1E-mail: [email protected] 1 been addressed in [6]. However, we use a different method which we compare to the method developed in [6] (to which we also suggest certain improvements). More concretely, in this work we try to find the bound state of two D0-branes in three dimensional Minkowski space (which is really a toy-model for the real situation, D0-branes in ten dimensional Minkowski space). It should also be noticed that the really interesting cases where the D0-brane theory is thought to describe macroscopic supergravity states is achieved by taking the number of zero branes to infinity. 2 The model The low energy physics of N parallel Dp-branes is governed by the dimensional reduction of 9 + 1 dimensional = 1 supersymmetric Yang-Mills theory with N U(N) gauge group to p+1 dimension [15]. The center of mass motion is governed by the overall U(1) factor so if we are interested in relative motion only, we can choosethegaugegrouptobeSU(N). Inthecasewe areinterested in, therelative motion of two 0-branes, we thus choose the gauge group to be SU(2). The action is [7] 1 i 1 2 i S = dt X˙aX˙a + ψaψ˙a ǫabcXbXc + ǫabcXaψb(γi) ψc 2g i i 2 A A − 4g i j 2 i A AB B Z (cid:20) s s (cid:16) (cid:17) 1 1 2 i + ǫabcX˙aAbXc + ǫabcAbXc ǫabcAaψbψc (1) g i 0 i 2g 0 i − 2 0 A A s s (cid:16) (cid:17) (cid:21) and the Hamiltonian derived from this is [7] 7 g 1 2 H = s (πa)2 + ǫabcXbXc ǫabcXaχ¯b (γ˜i) χc 2 i 4g i j − i A AB B s (cid:16) (cid:17) Xi=1 1 i ǫabcXa χb χc χ¯b χ¯c ǫabcXa χb χc +χ¯b χ¯c (2) −2 8 A A − A A − 2 9 A A A A (cid:16) (cid:17) (cid:16) (cid:17) together with the constraint one gets from varying (1) with respect to A 0 Ga ǫabc(Xbπc iχ¯b χc ) = 0 . (3) ≡ i i − A A On quantum level, we restcrict our Hilbert space to vectors which satisfy Ga Ψ = 0 (4) | i that is, our physical space is gauge invariant because Ga are gauge generators. We will only study motion of two 0-branes in three dimensional Minkowski space. We hope that this gives us the basic behavior of 0-branes and also an understanding of the full problem. This problem of two branes is also described in [7] but we study it in a different way. In the three dimensional case the action 2 is given by dimensional reduction of = 1 supersymmetric Yang-Mills theory N with SU(2) gauge group in 2+1 dimension to 0+1 dimension. The Hamiltonian takes the slightly simpler form g 1 2 H = s (πa)2 + ǫabcXbXc 2 i 4g 1 2 s (cid:16) (cid:17) 1 i ǫabcXa χbχc χ¯bχ¯c ǫabcXa χbχc +χ¯bχ¯c (5) 1 2 −2 − − 2 (cid:16) (cid:17) (cid:16) (cid:17) whereX arefieldsintheSU(2)adjointrepresentationandχisacomplexfermion i also in the adjoint representation. Of course we still have to impose gauge in- variance (4). In fact, the gauge invariance complicates things somewhat since we would like to separate out gauge invariant degrees of freedom from pure gauge degrees of freedom in our basic quantum mechanical operators Xa and πa. Let i i us focus on physical content of the Xa. It contains six components (the gauge i index runs over three values and the space index i runs from 1 to 2). We know that we can remove three of these variables using gauge transformations so only three variables are observable. These three variables should describe the relative positionoftwo pointlike objectsintwospace dimensions. Wedrawtheconclusion that one of the physical variables do not have the interpretation of a coordinate but rather as some internal auxilliary degree of freedom. To get some further insight into this problem it is neccesery to investigate the bosonic vacuum of the theory. It is possible to explicitly separate the gauge degrees of freedom from X a by decomposition it in matrix form [12] i (X) = (ψ) (Λ) (η) . (6) ai ar rs si Here the matrix ψ is an group element in the adjoint representation of SU(2). Thus when the gauge group acts on Xa, ψ just changes by ordinary gauge group i multiplication (from the left). We will parametrise the group element ψ by the ”angles” α,β,γ (A.5). This decomposition has the advantage that all the gauge dependence sits in ψ and all the other matrices are gauge invariant. In an analogous way we have separated out the dependence on rotations in space. Namely, performing an SO(2) rotation in space we have an element of SO(2) acting from the right on the matrix X . Thus we can separate out the ai dependence on the angle in space (we will call it φ) by saying that η is a group element of SO(2). We are left with the matrix Λ (A.2) which by construction is both gauge and space rotation invariant λ 0 1 Λ = 0 λ . (7)  2 0 0     The bosonic potential in (5) is gauge and rotation invariant and in the new decomposition coordinates depends only on two λ which have length dimension i (Fig. 1). 3 Figure 1: Bosonic potential The parametrisation λ = rcosθ; λ = rsinθ (8) 1 2 is the only way how to obtain exactly one variable, r, with the dimension length, which could represent relative distance of two branes. The dimensionless θ is the auxilliary coordinate. The potential in this coordinate reads 1 4 2 r sin 2θ . (9) 8g s Looking at the picture we can draw some interesting conclusions. If we fix a point on the bosonic vacuum (a classical static configuration with minimum energy), that is on the axes, we can study the behavior of the potential for small fluctuations of the auxilliary variable θ. We see that for large r the θ fluctuation are very much suppressed but at small r, θ will be allowed to fluctuate. This can be interpreted to mean that when the branes get close to each other, they can start to move also in the θ direction. Thus, θ is an auxilliary coordinate which is visible only when the branes come close together. The above discussion included only the bosonic degrees of freedom, we should keep in mind that the fermionic degrees of freedom can (and will) change this behavior somewhat. In essence, the Pauli repulsion will try to spread out the wavefunction as much as possible. When we use the standard operatorrepresentation ofπa andXa the first term i i in the Hamiltonian (5) is proportional to the Laplacian which we have to rewrite in the decomposition coordinates above to separate out gauge using formula 1 ∂ (√ggij∂ ) (10) i j √g where the g is the metric which is trivial in Xa coordinates. It is also good idea i 4 to rewrite the Laplacian in terms of gauge angular momenta La ǫabcXbπc: ≡ i i ∂ ∂ ∂ 1 L = icotγsinα +icscγsinα +icosα − ∂α ∂β ∂γ ∂ ∂ ∂ 2 L = icotγcosα +icscγcosα isinα − ∂α ∂β − ∂γ ∂ 3 L = i , (11) ∂α and physical angular momentum ∂ L0 ǫijXaπa = i (12) ≡ i j − ∂φ since their action on states with given gauge and rotational properties is simple. In particular, the (bosonic) ground state should have total spin equal to zero and be gauge invariant. All angular momenta are Killing vectors of the metric. In Appendix B we derive the lagrangian expressed in terms of gauge invariant variables and angular momenta 1 ∂ ∂ 1 ∂ ∂ L = r5 + sin4θ (ΨΠ−1Ψ−1) LµLν . (13) r5∂r ∂r r2sin4θ∂θ ∂θ − µν Here Lµ µ = 0,1,2,3 is a compact notation for the (physical and gauge) angular momenta defined above. Furthermore, we have defined 1 Ψ = − (14) ψ (cid:18) (cid:19) and Π r2 r2sin2θ r2sin2θ   . (15) r2cos2θ r2sin2θ r2      To find the (bosonic) ground state we will find all gauge invariant states with spin zero. The first state is the vacuum state 0 . Then we may act with the | i fermionic creation operators χa on the vacuum to find new states. The following states has total spin zero and they are gauge invariant 1 1 r = ψ a = eiφψ ǫabcχbχc 0 . (16) ar ar | i 2 | i 2 | i That is, they satisfy Ga r = 0 . (17) | i 5 The most general gauge invariant wavefunction with total spin zero can then be written g(r,θ) 0 +f (r,θ) r . (18) r | i | i It would also possible to construct the superpartner ground state with the help of r′ = eiφψ χa 0 ar | i | i 1 0′ = e2iφǫabcχaχbχc 0 . (19) | i 6 | i Now we study what happens when we act with the Laplacian (13) on this wavefunction and using the result we write the Hamiltonian matrix in this base. Let us start with second part of the Laplacian which is the relevant bosonic piece (ΨΠ−1Ψ−1) LaLb r = s Π−1 r TrΠ−1 − ab | i | i sr −| i 2(ΨΠ−1Ψ−1) LLa r = 2i u Π−1ǫ − 0a | i | i 0s usr (ΨΠ−1Ψ−1) LL r = r Π−1 (20) 00 00 − | i −| i where Tr is only on the gauge indexes. Finally the fermionic interaction in the Hamiltonian gives us H r = 0 (Λη) +i 0 (Λη) F r1 r2 | i −| i | i H 0 = r (Λη) i r (Λη) . (21) F r1 r2 | i −| i − | i So the full Hamiltonian matrix elements are g 1 H = s h+Π−1 δ TrΠ−1 +2iΠ−1ǫrus δ Π−1 + r4sin22θ rs − 2 rs − rs 0u − rs 00 8g s (cid:16) (cid:17) H = (Λη) +i(Λη) 0r r1 r2 − g 1 s 4 2 H = h+ r sin 2θ (22) 00 − 2 8g s where 1 ∂ ∂ 1 ∂ ∂ 5 h = r + sin4θ . (23) r5∂r ∂r r2sin4θ∂θ ∂θ Notice that the Hamiltonian we have obtained is the same as in [14] but differs from the one used in [7]. 3 Numerical calculation I This section gives an overview of solving the Hamiltonian eigenvalue problem (5) using the numerical renormalized Numerov method [16]. With it one can solve for the discrete spectra of a one dimensional operator of the form 1 ∂2 1+V(x) (24) − 2∂x2 6 which acts on L[a,b] Cn with Dirichlet boundary conditions. However, our ⊗ Hamiltonian depends on the two coordinates r and θ (the other coordinates, the angles, are fixed by the requirement that we are studying only gauge invariant states with spin zero). The “kinetic” term in our Hamiltonian (i.e. the term which contains the derivatives) is 1 ∂ ∂ 1 ∂ ∂ 5 r + sin4θ (25) r5∂r ∂r r2sin4θ∂θ ∂θ and we see that it is naively not of the form required above. Let us sketch briefly how to modify our problem to be able to apply the method. An arbitrary wavefunction can be written in the form Ψ = Ψ (r)Y (θ)e (26) ij ij i | i i,j X where Y (θ),j = 1,... is a complete basis of functions (with apropriate bound- ij { } ary conditions) in θ (the explicit choice of basis does not have to be the same for different values of the index i but can be chosen to optimize the numerics). The functions Ψ ,j = 1,... (which we will compute by the Numerov method) ij { } one can think of as being “combination coefficients” depending continuously on r. e is the standard base spanning Cn, in our case it is the four dimensional i { } complex vector space on which the Hamiltonian matrix acts. Choosing a con- crete basis depends only on the Hamiltonian domain which depends on one of the selfconjugated extensions of the Hilbert space (an extension of the Hilbert space such that the Hamiltonian operator is hermitian) but we will rather apply a physical principle which will be described later. The expectation values of the Hamiltonian in the base Y e (no sum) (27) ij i gives the same number of coupled equations for the radial part as the number of Ψ in (26). Thus the problem is now correctly defined and the matrix repre- ij sentation of the expectation value of the Hamiltonian in the basis above forms the potential which is used in the one dimensional Hamiltonian (24). To get 1 in the kinetic term we have to orthonormalize the base. Furthermore, to be able to use the Numerov method on a computer we need a finite basis which means that we need to “cut off” or restrict the base to be finite. The rescaling of the wavefunction Ψ by the factor r5/2 transforms the radial part of (25) to | i the operator ∂2 15 2 r (28) ∂r2 − 4 and changes boundary condition to Dirichlet. 7 Let us now describe the physical principle we use to choose the basis functions Y (θ) for our Hamiltonian (5). ij We define the wavefunction on the interval θ [0,π/4] (A.27). It is not ∈ necessary to choose this particular interval, one could, for instance, select the ˜ interval[ π/4,0]insteadoftheabovementioned. Usingtheidentificationθ = θ − ˜ − the Hamiltonian defined on the interval θ [ π/4,0] acting on states with total ∈ − spin zero is connected to our original Hamiltonian on the interval θ [0,π/4] by ∈ the unitary transformation H(θ˜) = U†H(θ)U (29) where 1 1 U =  . (30) 1  −   1     ˜ Itisalsopossibletoconsider otherintervalθ [π/4,π/2]. Usingtheidentification ˜ ∈ θ = π/2 θ the corresponding Hamiltonian can be also obtained by a unitary − transformation with the matrix 1 i U =  − . (31) i    1     The wavefunctions of course also transform under the unitary transformation ˜ Ψ(θ) = U Ψ(θ) . (32) | i | i If we require that the wavefunctions be everywhere smooth, the above condition severely restricts the possible wavefunctions and in particular the basis wavefunc- tions that we can use. We note that this principle was not used in [7] and hence the θ derivative of their groundstate wavefunction at θ = π/4 is not well defined. For the bosonic part of the wavefunction we will use the basis states cos(4jθ)e j = 0,... (33) 1 and for the fermionic part we will use icosθcos(2jθ)e +( 1)jsinθcos(2jθ)e j = 0,... (34) 2 3 − − Theynicelycancelthedivergences intheHamiltoniancomingfromtheLaplacian. Of course these basis functions satisfy the boundary condition above and they also form a complete basis for the functions with the given boundary conditions. 8 However, the basis isnot orthonormaldue to the nontrivialθ part inthe measure (A.26). So we need to orthonormalize these basis functions to get 1 in (24). Doing so, using the Gramm-Schmidt procedure, we diagonalize the θ part of the Laplacian. Notice that these two sets of basis functions also correspond to the bases which are implicitly used in the Wosiek method [6] when one are doing calculations with states with total spin zero (the basis functions above depend on the total spin). This will be shown in section 5. Thus we will be able to compare our results and our method very directly with the results obtained by the Wosiek method [6]. It is not obvious how much our results for a fixed number of basis functions fit the exact solutions which one would get using the complete basis. To get some intuition for how the general solution would look like we will repeat the calculations increasing the number of basis functions each time and hopefully one can extrapolate the result to the exact case. At least we should be able to make an intelligent guess at the properties of the exact solution. To study the groundstate of our Hamiltonian (which, because of supersym- metry, should have energy zero [11]) is a good test for the method described above. There are some results for coupling constant g = 0.1 in the table 1 (p. 17) s where N is the number of the test functions (33,34) and E is their corresponding groundstate energy. The dependence of the energy on the number of functions we include in the basis given in the last two columns of the table is as follows 1.44 E = . (35) N As was claimed in [6], one can therefore predict the groundstate energy as a function of the number of basis functions with very high accuracy. We therefore see that in any concrete numerical calculation (using a finite basis) we do not expect to get zero energy. Only in the (numerically unobtainable) case of infinite basis do we get zero energy. The picture (Fig. 2) is the probability density for the case with the high- est number of basis functions in the table. The following two pictures show the bosonic contribution coming from (33) and the fermionic one coming from (34). The domain of these plots is (r,θ) [0,2.2] [ π/4,π/4]. The hill of the prob- ∈ × − ability density is located at the boson potential valley (9) and isolines represents sections for fixed r, θ and one for fixed density on each picture. The maximum of the probability density of any constant r section is in the potential valley (θ = 0). Notice that the global maximum is not at r = 0. This is probably an effect of the fermion Pauli repulsion which can be seen from the purely fermionic contribution to the probability density which is zero at r = 0. Increasing the number of basis functions, the only thing that happens is that the global maximum moves slowly to larger and larger r at the same time as the whole wavefunction becomes more spread out in r but more peaked in θ. One 9 Figure 2: Probability density Figure 3: Bosonic part of the probability density 10