Chapter 14. Hamiltonian Formulation of Local Symmetry

Chapter 14 Hamiltonian Formulation of Local Symmetry For several purposes, it is sometimes preferable to work on a lattice with different lattice spacings in different directions, as we have done in analyzing the plaquette We can write the action as a sum over sites with the appropriate space-time volume, d 4 x a x a y a z a t n U n,n+µ = e igaµ A µ on each link Fµν 1 g a µ a Tr U ν pµν + U p µν 1I µν for each placquette and so 1 S = a x a y a z a t n +i µ { 1 1 4g a µ a Tr U ν pµν + U p µν 1I 1 δ µ+δ ν µν 1 a µ ψ n+µ γ µ U n,n+µ ψ n m ψψ } We now turn to a discussion of the Hamiltonian formulation, as well as the Lagrangian This will give us a quantum mechanics in the usual language, 1 I am leaving out a discussion of fermion doubling In general we shall concentrate on the gauge field U 151

15 Last Latexed: April 5, 17 at 9:45 Joel A Shapiro but of a rather unusual form, because the kinetic energy term will not be the usual p i/m, because the degrees of freedom, the group elements on each link, do not live in a Euclidean space, but instead live on the group manifold The way the derivative operators enter the Hamiltonian will lead us to a metric and a measure on the group manifold We will derive the Hamiltonian formulation of the lattice gauge theory by considering the a t limit First we must use gauge invariance to remove some of the degrees of freedom Quantum mechanics is an integral of the action over all possible configurations of the physical degrees of freedom Gauge degrees of freedom have no effect on the action, and are therefore not physical degrees of freedom Given any configuration U, ψ on all the links and sites, it is possible to find a new configuration, gauge transformed, satisfying some imposed gauge requirements but representing the same physical configuration In our case, we pick a particular time, say t = We now perform gauge transformations at each site with t in order to make all U n,n+ˆt = 1I This can be done iteratively For example, at t = a t, at site n = r, a t we examine U r,,ˆt = e iω r, and then make a gauge transformation at the site r, a t with ω r, with no change at r,, so the new U on the timelike link is U r,,ˆt = e iω r e iω r e = 1I This set of transformations changes all the spacelike U s at t = a t to new values, but it does not affect the integral over all such U s, provided the weight of integration is group invariant We shall say more about this later Reiterating this step forwards in time to t =, and also going backwards from t =, we wind up with a configuration of U s and ψ s equivalent to our original one, but subject to the gauge constraint U n,ˆt = 1I In the continuum limit, this means A = everywhere, which is called the temporal gauge Now let us consider the limit a t with a x = a y = a z = a held fixed The sum over times, times a t, is just dt, so S = dtl as usual, with L = { 1 g a a t TrUPj 1I + hc n x,n y,n z a 3 + 1 Tr 4g a 4 ij j U pij + U p ij 1I + fermion terms The first term is the sum over placquettes in a time-space plane, while the

618: Last Latexed: April 5, 17 at 9:45 153 second is in a space-space plane at a fixed time The first of these involves Expand so U pj = U n ν, U n }{{} ν +a t,j U n ν +a j, U n,j }{{} 1I 1I U n,t+a t,j = U n,t,j + a tu n,t,j + 1 t at U t, U 1 U n,t+a t,j U n,t,j 1I = a t n ν,j U n t ν,j + at U 1 n ν,j t U n ν,j Adding the hermitean conjugate gives U 1 a t n ν,j U n t ν,j + U 1 U n ν,j n ν,j t + at U 1 t U + U 1 U t The first term is a t U 1 U = a t 1I =, while from the second we may t t subtract a t t = at which leaves U 1 U + U 1U = at U 1 1 U U + t t t t t t + U 1 U t Tr U Pj 1 + hc = Tr U U t+a U t t 1I + hc = a t 1 Tr t U t U 1 1 U We can improve this by using our expression for zero again, U = U, t t U 1 1 U so = U U 1 and t t Tr U t+a U t t 1I + hc = a t Tr U 1 UU 1 U, and L = a g Tr U 1 1 UU U V {U}, n where V contains the spacelike placquette terms, without any time derivatives To get the Hamiltonian we need first to find the canonical momenta Π = In our case the coordinates q are the matrix elements of U on each δl δ q

154 Last Latexed: April 5, 17 at 9:45 Joel A Shapiro spacelike link U l ab, where l specifies the link both spacelike direction and spatial position and ab are the matrix indices For convenience we will define the matrix P as the transpose, so δl P l cb := δ U l bc = a U 1 g l U l U 1 l, cb H = l,c,b U l bc P l cb L = l g a Tr U lp l U l P l + V {U} Quantum mechanically, we write the momentum conjugate to q as i, q so P lcb = iu lbc But P only enters H together with U So let us define, for each link, E l ab = i U l P l ab = U l ac U l bc In terms of of the E s, the kinetic energy term in L becomes a l Tr E g l The E s are a scaled electric field ag times the usual field As a differential operator it is very interesting But before we pursue the group theoretic arguments over the interpretation of E, let us meet an objection to the above procedure, that the U ab are not independent real coordinates Instead, we should write U = e iωi L i and treat the real, independent ω i as the coordinates Then we get Π i = δl δ ω = δ U ab δl i δ ω i δ U = δu ab δl ab δω i δ U ab where I have made use of U ab = δu ab ω i δω δl i But δ U = a U 1 UU 1, so ab g ba [ Π i = a g Tr δu 1 δωiu j ] ω j δu U 1 δω j and ω i Π i = a g Tr ω i δu δω iu 1 ω j δu δω U 1 = ag Tr UU 1 UU 1 j Of course we have a U and an ω for each link, but we will drop the index l, or n, l that specifies which link, in what follows

618: Last Latexed: April 5, 17 at 9:45 155 in agreement with what we found by careless manipulation One sees, however, that there are only N 1 coordinates and momenta for SUN, not N as on might expect from P l cb The requirement that U be unitary and of determinant 1 means there are constraints on the P l cb s we can vary U only preserving the unitarity and determinant Interpreting the δ P l cb quantum mechanically as iδu l bc would be varying into a dimension for which the action is undefined, where det U 1 Our E l cb do not have that problem Consider any function f on n n unitary matrices u Each such matrix is u = e iφ U with U SUN, so we may write fu = fφ, U and f = f u cd iφ f u cd = e, φ φ u ab U ab U ab so if E ab = iu ac, E ab f = ie iφ f U ac = iu ac u bc u bc f U bc and has no depen- φ dence on how f varies off the det u = 1 subset So we see that the E ab does not involve a derivative in the direction away from detu = 1, but does form a complete set of N 1 first order differential operators on functions of SUN matrices The N 1 momenta conjugate to ω i are quantum mechanically Π i = i δ = i δu ab δ δω i δω i δu ab Collectively these clearly span the same space as the E ab To understand the connection, consider different ways of asking how a function on G varies as we move the element g G We see that Π i is not the same as E To understand the connections, consider a function f : G C defined on the group Then derivatives of f mean finding the change in fu under small changes in U If we write U = e iωi L i, then ω ifu = iπ if On the other hand, we could consider the change in f as U U ω = e iωi L i U, for infinitesimal ω i If we define the differential operators E j fu = i ω f e iωj L j U = i fu U ab = fu il j j ac U cb ω j U ab ω= = L j ac U cb U ab f = TrL j Ef U ab ωj =

156 Last Latexed: April 5, 17 at 9:45 Joel A Shapiro So we see that the electric fields E ab = U l ac P l cb which we defined earlier are just linear combinations of the left-handed differential operators E j, with E = j E jl j For completeness, let us also define a right derivative on the space of functions from G C by R j fu = i ω if Ue iω jl j = f ω= = f U cb U ca L j ab U cb i ω i Ue iω j L j cb ω= This is not exactly the same as E j If we make a matrix from R j in the same way as we did for E j, R ab = j R j τ j ab = U cb δ δu ca δ and recall E ab = U ac, δu bc so Rf = δf δu U = U 1 U δf T δu U = U 1 EfU T Note that the E j, R j and Π j are each sets of N 1 first order differential operators on the space of functions from G C, an N 1 dimensional space Thus they are linear combinations of each other The connection between them is associated with the adjoint representation of the group We will write [SL l ] jk for the adjoint representation of the generators, so [SL l ] jk = ic k jl Let us call the adjoint representation of the group S: [ ] S jk e iωl L l = e iωl SL l [Note: S and S were both previously called Γ adj Also it is hard to distinguish S from S ] These will come up in the connections we are seeking jk 141 Relation of R j to E j Now U 1 L l U is a linear combination of the L s, so U 1 L l U = M lm UL m We can show M lm = S lm

618: Last Latexed: April 5, 17 at 9:45 157 Proof: Let Uλ = e iλωj L j, Mλ := MUλ Then e iλωj L j L l e iλωk L k = M lm λl m Differentiate with respect to λ: e iλωj L j [L l, iω k iλω L k ] e j L j = dm lmλ L }{{} m = ω k clk n M nm λl m dλ ω k clk nln The L m are linearly independent, so dmλ = ω k clk n dλ M nmλ = iω k [SL k M] lm This is a differential equation with solution lm Mλ = e iλωk SL k K with initial condition M lm = δ lm = 1I = K, so Mλ = e iλωk SL k = SUλ Thus e iωk L k L l e iωk L k = S lm e iωb L b L m = e iωb SL b So U 1 L l U = S lm UL m and by exponentiation, U 1 e iωk L ku = e iω k S km UL m Now reconsider R l fu = i = i = i where ω m = ω k S km U 1 So Ue ω lf iωk L k ω= ω lf e iωk S km U 1 L m U ω n } ω {{ l } ω nf e iω m L m U S ln U 1 lm ω= ω= R l fu = S lm U 1 E m fu, R l = S lm U 1 E m, and E m = S ml UR l, as S 1 U 1 = SU L m

158 Last Latexed: April 5, 17 at 9:45 Joel A Shapiro Why is E l better than R l? In the Hamiltonian we have El, why not R l? E m = S ml UR l S mk UR k = S ml US mk UR l R k is ml [ ω ls mk Ue iωj L j ω= ] R k Note that S is hermitean and imaginary so S = e iωj SL j is unitary and real, hence orthogonal, and S ml US mk U = δ lk Also S mk Ue iωj L j = S mn US nk e iωj L j, so i S ω l mk Ue iωl ω= = S mj US jk L l, so the second term is is ml S mj S jk L l R k = δ lj ic k jl R k Thus E m = j R l R k δ lk + iδ lm cml R j = Rj, by the antisymmetry of c So in fact the Hamiltonian does not distinguish between El and Rl, as they are the same thing We can also ask what the connection is between E l and To find the ω l connection we need to develop a formula for x eax for an operator A which need not commute with its derivative The formula is x eax = αax Ax dα e x e1 αax 141 which you proved for homework Then around a fixed element U = e iωl L l, we may write e iωl L l = e iρl L l e iωl L l with ρ as ω ω Thus ω lf e iωl L l = ρm ω l ρ mf e iρm L m ρ m U = ω ie mf l The partial derivative matrix can be evaluated with this formula as applied to fu = U, or ω leiωk L k = i ρm ω l L me iωk L k = ρ m ω l L m = dα e iαωk L k L l e iαωk L k = dα e iαωk L k il l e i1 αωk L k dα S lm e iαωk L k L m,

618: Last Latexed: April 5, 17 at 9:45 159 or so ρ m ω l = [ ] [ ] dα e iαωk SL k 1 e iωk SL k = lm iω k SL k lm [ ] ω = 1 e iωk SL k E l ω k m SL k lm gives the connection of the left derivative with the ordinary partial derivative [Note: the fraction in square brackets is well defined even though ω k SL k is likely not an invertible matrix, so dividing by it is dubious But the expression in the bracket has a well defined power series expansion and thus a well defined meaning] We have given a great deal of discussion to our Hamiltonian operator We now turn to the states on which this operator acts A state can be specified by giving a at each site, the occupation number for the fermions at that site b on each link, a group element We focus on this We are not accustomed to having dynamics with canonical coordinates defined on a compact space like U G, except in advanced courses in classical mechanics eg the orientation of a rigid body In quantum mechanics we know that the momentum is related to translations, so the ordinary kinetic energy term i p i/m is connected with a space of q s which is translation invariant in each p i Our kinetic energy is not of that form, as the E l s don t commute and the /ω l s do not enter H in this simple fashion This is consistent with the fact that our space of U s is just not a Euclidean N 1 dimensional space Now quantum mechanics can be viewed as a functional integral over all possible configurations, which requires a weight for the configurations In a deep sense this weight is connected with the kinetic energy term in the Lagrangian density We can sidestep this issue temporarily by observing that the only sensible way to integrate over the U s is in a group invariant way In terms of eigenstates U ab of U on a given link, we need to be able to write 1I = dµu U ab U ab, or in terms of the parameters ω l, dµu = dµω l, the measure must be group invariant, dµωfuω = dµωfguω for any function f : G C

16 Last Latexed: April 5, 17 at 9:45 Joel A Shapiro and any fixed element G in the group This is also what we needed long ago to define the orthogonality of group representations So now we leave physical applications and turn to the problem of finding the Haar or Hurwitz measure dµω l satisfying the invariance Let us write dµω = hωdω 1 dω n for an n dimensional Lie group For a fixed G and ω, the group element U = GUω = Uω = e iω j L j, where the ω j are functions of the ω s and also of G, but we are holding G fixed If we change variables on the right hand side of our invariance equation, we get hωf Uω d n ω = hωf Uω ω j dn ω ω k where / is the Jacobian determinant Invariance for arbitrary functions f clearly requires hω = hω ω j ω k Let h = c Then h at any other point ω is determined by taking G = e iωj L j, hω = c ν k ν j where e iν k L k = e iω k L ke iν k L k This is a necessary condition for the invariance To show it is also sufficient, we need to show that the relation holds for all ω Consider the measure at two points, ω and ω We need to show that if Uω = GUω for a fixed G, hω = hω ν ν =ω ν 1 ν=, ν ω ν Let G = e iω k L k and G = e iω k L k and Uν = G Uv and Uν = G Uν Then Uν = G G 1 Uν = GUν, so G = G G 1 / / ν ν = ν ν c c ν =ω ν ν = = hω hω hω hω which verifies the invariance in general 1 ν ω

618: Last Latexed: April 5, 17 at 9:45 161 so Thus Now we need to evaluate the Jacobian at ν = As e iν k L k = e iω k L ke iν k L k, k L k ν = νj L k L k keiν ν= ν keiωk ν j eiνk = νj L k il ν= ν j keiωk ν j ν kl j = = dα e i1 αν k L k il j e iαν k L k ν =ω [ dαe iαωj L j L k e iαωj L j 1 e iω m SL m ] = iω m SL m hω = c det ν e iω m SL m 1 = c det ν iω m SL m For any fixed real ω, ω m SL m is a hermitean matrix and can be diagonalized If λ j are the real eigenvalues, we have hω = c j e iλ j 1 iλ j = j sinλ j / λ j / e iλj/ But ω m SL m is also imaginary, while still hermitean, so it is antisymmetric and has zero trace This is preserved by similarity, so j λ j =, and j kj L j hω = c j sinλ j / λ j / Let us consider an explicit example, SU Any U = e i ω τ/ is equivalent by similarity to one along the z axis, with the same ω Then ω j SL j = ω SL 3 S is the adjoint representation, so L 3 has the eigenvalues ±1, The zero eigenvalue gives 1 in the product recall the note on the definition of the [] expression and the others each give sin ω // ω, so all together hω = 4c sin ω / ω Choosing c = 1, we may take the invariant volume to be dω 1 dω dω 34 sin ω / ω ω π

16 Last Latexed: April 5, 17 at 9:45 Joel A Shapiro If we used spherical coordinates for ω, this gives π dω d Ω 4 sin ω /, where dω = sin θ dθ dφ as usual We note that the volume with ω π ǫ is 4π 3 ǫ3, the same as the volume with ω ǫ, and not ǫ times the ordinary area of the sphere, that is, 4ππ ǫ The metric says points with ω π, even in very different directions, are close together, because for ω = π, all 1 these ω s correspond to the same U = Note hω is exactly the 1 metric 3 on S 3 induced by embedding in E 4, with the angle with respect to the 4 th axis ω/ Some comments: a For SU3, not all directions are equivalent So there is no equivalent of ω b For many applications, we needed the existence of the invariant measure but not the actual expression for it For example, in our orthogonality proofs c Most discussions of SU parameterize the group in terms of Euler angles rather than in the way we did, the angle of the rotation and the direction of the axis d Note that the factor in h is just the determinant of the matrix which connects the left derivatives to the ordinary derivatives This is not an accident, but rather has to do with this relation defining a metric not just a measure on the group space, and the measure given by the determinant of the metric 3 See hypersph in Supplementary Notes