paper stringlengths 9 16 | proof stringlengths 0 131k |
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quant-ph/0008047 | We have MATH . Since MATH, the result follows. |
quant-ph/0008047 | Suppose we had equality. A protocol MATH attaining this bound would certainly have to be p.p.t.; thus, if we apply this protocol to MATH, the output fidelity will take the form MATH for some constants MATH and MATH, or equivalently MATH for constants MATH, MATH. Evaluating this at MATH, MATH, we find: MATH . On the oth... |
quant-ph/0008055 | We proceed by assuming that another product state MATH orthogonal to all these exists, and showing that a contradiction results. Expanded in the basis above, MATH must have at least one non-zero term, which we will call MATH. That is MATH with MATH. Now, since simultaneous cyclic permutations of the MATH and MATH . NAM... |
quant-ph/0008055 | We show that this set is a UPB, for all MATH and MATH, by the same methods as before. Since GenTilesREF has less symmetry than GenTilesREF we will have to examine more cases, but the methods will be the same. We will number the following paragraphs to indicate the structure of the cases being considered. In all cases w... |
quant-ph/0008059 | (See CITE for the original proof.) First, attach to the superposition of MATH the (two qubit state) MATH . Then, in superposition, add modulo MATH the values MATH to this register (step MATH). Finally, apply a general phase change MATH (step MATH). It is straightforward to see that this yields the desired phase change ... |
quant-ph/0008059 | See CITE for the original proof by NAME, and CITE for the single query version of it. |
quant-ph/0008059 | See the original article by CITE, or better yet, the excellent analysis of it by NAMECITE . |
quant-ph/0008059 | See the original article CITE, or any other standard text on combinatorial objects CITE. |
quant-ph/0008059 | First, we use the amplitude amplification process of NAME 's search algorithm CITE described in REF to create exactly the state MATH with no more than MATH queries to MATH. (See the article by CITE for a derivation of this upper bound. Obviously, no queries are required if MATH.) After that, following REF , one additio... |
quant-ph/0008059 | First, prepare the state MATH with MATH queries to the function MATH REF . After that, measure the state in the basis spanned by the vectors MATH. Because MATH is a weighing matrix, this basis is orthogonal and hence the outcome of the measurement gives us the value MATH (via the outcome MATH) without error. |
quant-ph/0008059 | We will prove these bounds by considering the decision trees that describe the possible classical protocols. The procedure starts at the root of the tree and this node contains the first index MATH that the protocol queries to the function MATH. Depending on the outcome MATH, the protocol follows one of the three outgo... |
quant-ph/0008059 | See the proof of REF in the appendix. |
quant-ph/0008059 | Consider, like in the proof of REF , a decision tree with nodes MATH and corresponding query indices MATH. For every index MATH there is exactly one function with MATH. For the tree this implies that every node MATH can only have two proper subtrees (corresponding to the answers MATH and MATH) and one deciding leaf (th... |
quant-ph/0008059 | Let MATH be the set of possible values of MATH at a given moment during the execution of the protocol. (Thus, initially we have MATH, and we want to end with the unique answer determined by MATH.) Below we will show that if MATH has at least MATH elements, then there always exists an index MATH such that all three poss... |
quant-ph/0008059 | We exhibit the quantum algorithm in detail. We start with the superposition MATH . (The reason for the ``dummy" part of state that we use will be clear later in the analysis.) The first oracle call is used to calculate the different MATH values for the non-dummy states, giving a superposition of states MATH. At this po... |
quant-ph/0008059 | (For the quadratic character, these are simple instances of so-called NAME sums CITE; see for example REF.) Rewrite MATH with MATH. If MATH this sum equals MATH. Otherwise, we can use the fact that MATH (for MATH) to reach MATH . Earlier we noticed that MATH, and therefore in the above summation (where the value MATH i... |
quant-ph/0008095 | Let MATH, then the NAME representation of MATH is MATH where MATH. By simple calculation, MATH . Since MATH approximates MATH, MATH. |
quant-ph/0008095 | By induction. It is true for MATH. Assume the statement is true for all natural numbers smaller than MATH, and let's examine the case MATH. Pick a coordinate MATH such that both the subcubes of MATH and MATH have nonempty subsets MATH and MATH of MATH. Then MATH. We can assume without loss of generality that MATH. Then... |
quant-ph/0008095 | Let MATH be the number of true assignments. By REF , in the Boolean cube, the number of edges that connect two true assignments is less than MATH, and the number of edges that connect two false assignments is less than MATH. Therefore, MATH . |
quant-ph/0008095 | For each MATH, let MATH be the characteristic function of MATH, that is, MATH . Let MATH be the probability that MATH is observed as the output when the input is MATH. Then by REF , MATH is a nonnegative polynomial of degree no more than MATH, and MATH approximates MATH. Furthermore, for any MATH, MATH. For simplicity ... |
cs/0009006 | We form a bipartite graph, in which the vertices correspond to the variables and components of the instance. We connect a variable to a component by an edge if there is a (variable,color) pair using that variable and belonging to that component. The instance is solvable iff this graph has a matching covering all variab... |
cs/0009006 | We employ a backtracking (depth first) search in a state space consisting of MATH-CSP instances. At each step we examine the current state, match it to one of the cases above, and recursively search each smaller instance. If we reach an instance in which REF applies, we perform a matching algorithm and either stop with... |
cs/0009006 | Randomly choose a subset of four values for each variable and apply our algorithm to the resulting MATH-CSP problem. Repeat with a new random choice until finding a solvable MATH-CSP instance. The random restriction of a variable has probability MATH of preserving solvability so the expected number of trials is MATH. E... |
cs/0009006 | As described, we find a maximal bushy forest MATH, then cover the remaining vertices by height-two trees. We choose colors for each internal vertex in MATH, and for certain vertices in the height-two trees. Vertices adjacent to these colored vertices are restricted to two colors, while the remaining vertices form a MAT... |
cs/0009006 | We apply REF MATH times, resulting in a set of MATH constrained MATH-edge-coloring problems each having only MATH edges. We then treat these remaining problems as MATH-vertex-coloring problems on the corresponding line graphs, augmented by additional edges representing the constraints added by REF . The time for this a... |
cs/0009013 | There are MATH edges. For each edge MATH, the complexity of the associated region can be at most MATH. Since any pair of constant degree polynomials intersect in a constant number of points, the overall complexity of MATH is given by MATH. |
cs/0009023 | According to CITE there are exactly two different rectilinear drawings of MATH, of which the convex hull is either a triangle or a quadrilateral. The former has no crossings and corresponds to the concave kite. The latter has one crossing and corresponds to the convex kite. Since the drawing is comprised of nested tria... |
cs/0009023 | Let MATH be a concave kite in a nested triangle drawing with the standard vertex labels MATH, MATH, MATH, and MATH. Since MATH is concave, the middle vertex MATH is within the triangle MATH. The vertices MATH and MATH determine a line that defines a half-plane MATH that does not contain MATH. Since the vertices MATH, M... |
cs/0009023 | If both edges MATH and MATH each cross at least one kite edge, then we are done. Without loss of generality, assume that MATH does not cross any kite edges. Let MATH and MATH be the other two inner vertices, and consider the path MATH. Since edge MATH does not intersect the path, MATH creates a barrier on the other sid... |
cs/0009023 | Since the two concave kites share the same middle vertex, there are two possible cases. Either the labels of the internal vertices match, in which case we are done. Otherwise, the left and right labels are interchanged. By way of contradiction, assume that they are interchanged; this implies that the kites are disjoint... |
cs/0009023 | Either vertex MATH is contained in kite MATH or not. If MATH is inside MATH, then, because kite MATH is concave, a barrier path MATH is created between vertex MATH and vertex MATH. Hence, edge MATH must cross the path MATH, intersecting one of the path's two edges. If vertex MATH is not contained in kite MATH, then ass... |
cs/0009023 | Using the kite edges we construct two polygons MATH and MATH. Since both polygons contain region MATH and since the only shared edge, is a middle edge, edge MATH must cross into both polygons, contributing at least one kite edge crossing from each. |
cs/0009023 | Select a pair of green vertices and remove all other green vertices from the drawing. This forms a MATH with exactly one MATH-rgedge crossing that is uniquely identified by the two green vertices. Since there are MATH pairs of green vertices, there must be MATH-rg edge crossings. |
cs/0009023 | That the convex hull of an optimal rectilinear drawing of MATH is a triangle has been shown in CITE and CITE. Using a counting technique similar to CITE, consider a drawing composed of a red triangle that contains a green convex quadrilateral that contains two blue vertices. By the MATH principle there are MATH-rg cros... |
cs/0009023 | Select two of the three red, green, and blue triangles. These two triangles form a nested triangle drawing of MATH with three REF-colour crossings. Hence, there are three REF-colour edges of each type. |
cs/0009023 | Let the outer triangle be red and the inner green. By the MATH . Principle REF there are three MATH-rg edge crossings. If the two triangles are non-concentric then there is at least one MATH-gg crossing. |
cs/0009023 | The red triangle contains the green triangle and the green triangle contains the blue triangle. Therefore, every red-blue edge must cross into the green triangle. Since there are nine red-blue edges, there are nine MATH-gg crossings. |
cs/0009023 | The are three green and three blue vertices, thus there are nine unique green-blue pairs of vertices. By the MATH principle, each pair contributes exactly one MATH-rb crossing. Hence, a nested triangle drawing of MATH has exactly nine MATH-rb crossings. |
cs/0009023 | We make use of the fact that the green and blue triangles form a MATH and that any rectilinear drawing of MATH falls into one of the five configurations: CCC, VVV, CVV, binary CCV, and unary CCV. The proof is by case analysis on the green-blue MATH sub-drawing. The green-blue MATH is drawn in one of the five configurat... |
cs/0009023 | The first part of the statement is proven in CITE and the counting argument in REF . Putting REF , and REF together accounts for REF of REF crossings in an optimal drawing. REF states that there are at least nine internal crossings. Since MATH, the number of MATH-gg and MATH-bb crossings must be zero; this implies conc... |
cs/0009023 | By REF , an optimal drawing of a MATH has REF crossings. Referring to REF , an optimal drawing has at least REF non-internal edge crossings (REF , and REF). By REF , there are at least nine internal edge crossings and hence, an optimal drawing has exactly nine internal edge crossings. NAME MATH-bb crossing occur if the... |
cs/0009023 | By way of contradiction, assume that there exists a rectilinear drawing of MATH with REF crossings. Since each edge crossing comprises of four vertices, the sum of responsibilities of each vertex totals MATH. Therefore, the average responsibility of each vertex is MATH. Furthermore, each vertex in the drawing is respon... |
cs/0009023 | At least two of the red-white edges must cross into the green triangle on distinct green-green edges as a consequence of the nested triangle requirement and the containment argument. Select two of the three red-white edges such that they cross into the green triangle on distinct green-green edges and such that the tota... |
cs/0009023 | By way of contradiction, assume that there exits an optimal rectilinear drawing of MATH whose convex hull is not a triangle and REF edge crossings. By the same averaging argument used in REF , at least four of the vertices are responsible for REF edge crossings; removing any of them yields an optimal drawing of MATH wi... |
cs/0009023 | By REF the outer two hulls of the optimal drawing MATH must be triangles. We must still account for the four internal vertices. If the four vertices form a convex quadrilateral then we are done; otherwise, assume the tenth vertex is inside the third nested triangle. Colour the tenth vertex white. Now count the number o... |
cs/0009023 | By way of contradiction assume that MATH. By REF the inner hull must be a convex quadrilateral. Repeat the argument from REF disregarding the MATH-bb and MATH-bb edge crossings (because there is no blue triangle). This gives us an initial count of MATH edge crossings. Let the entire inner convex quadrilateral be colour... |
cs/0009023 | NAME 's rectilinear drawing of MATH with REF edge crossings CITE is exhibited in CITE, and hence MATH. By REF MATH. The result follows. |
cs/0009023 | This drawing is coloured by only two colours: red and green. By the MATH principle there are MATH-rg crossings. Since the six green vertices comprise the second hull, there are MATH-gg crossings. REF crossings counted so far include all except the MATH-gg crossings. We now consider the MATH-gg crossings. Select four of... |
cs/0009023 | As before, the single vertex inside the second hull is coloured blue. By the MATH principle there are MATH-rg crossings and MATH-rb crossings. By the same argument used in the previous lemma there are MATH-gg crossings. There are at least five MATH-gg crossings. Thus, we reach a count of REF crossings without having co... |
cs/0009024 | We apply a randomized incremental arrangement construction algorithm. Each cell in the recursive decomposition is an arrangement cell at some stage of the construction. The bound on the representation of a halfspace comes from applying the methods of CITE to the zone of the boundary hyperplane. |
gr-qc/0009004 | Let MATH be REF-dimensional hyperbolic pair of pants described in the previous section. Turn the pair of pants up-side-down and think of MATH as being in the shape of a ``Y". Each endpoint of ``Y" represents a component of MATH isometric to the hyperbolic REF-manifold MATH described in the previous section. We call the... |
gr-qc/0009004 | Let MATH be as in REF and let MATH be as in the proof of REF . Let MATH be the compact hyperbolic REF-manifold obtained by doubling MATH along the union of the right and left components of MATH. Then MATH has a totally geodesic boundary consisting of two copies of MATH corresponding to two copies of the bottom componen... |
gr-qc/0009105 | In terms of the function REF , the lens map can be written in the form MATH . As MATH is an immersion transverse to MATH at MATH and MATH is a submersion, the differential of MATH at MATH has rank REF if and only if the differential of MATH at MATH has rank REF. This proves the first claim. For proving the second claim... |
gr-qc/0009105 | By contradiction, let us assume that there is a sequence MATH with pairwise different elements in MATH. By compactness of MATH, we can choose an infinite subsequence that converges towards some point MATH. By continuity of MATH, MATH, so the hypotheses of the proposition imply that MATH. As a consequence, MATH is a reg... |
gr-qc/0009105 | The result MATH can be read directly from REF , choosing the regular value MATH which has exactly one pre-image point under MATH. This implies that MATH must be surjective since a non-surjective map has degree zero. So MATH being the continuous image of the compact set MATH under the continuous map MATH must be compact... |
gr-qc/0009105 | We fix a trivialization for the bundle MATH and identify MATH with MATH. Then we consider the bundle MATH over MATH, where MATH is, by definition, the subspace of all lightlike directions that are tangent to past-oriented lightlike geodesics that leave MATH transversely at MATH. Now we choose for each MATH a vector MAT... |
gr-qc/0009105 | In the proof of REF we shall adapt techniques used by CITE in their study of asymptotically simple and empty spacetimes. To that end it will be necessary to assume that the reader is familiar with homology theory. With the sphere bundle MATH, introduced in REF , we may associate the NAME homology sequence MATH where MA... |
gr-qc/0009105 | As usual, let MATH denote the chronological past of MATH in MATH, that is, the set of all points that can be reached from MATH along a past-pointing timelike curve in MATH. To prove REF , fix a point MATH. Choose a sequence MATH of points in MATH that converge towards MATH in such a way that MATH for all MATH. This imp... |
gr-qc/0009105 | In the first step we construct a MATH vector field MATH on MATH that is timelike on MATH, has MATH as an integral curve, and coincides with MATH on MATH. To that end we first choose any future-pointing timelike MATH vector field MATH on MATH. (Existence is guaranteed by our assumption of time-orientability.) Then we ex... |
gr-qc/0009105 | REF is obvious from REF was just established. The proof of the remaining two conditions is based on the fact that on MATH the MATH-lightlike geodesics coincide with the MATH-lightlike geodesics (up to affine parametrization). REF is satisfied since every lightlike geodesic in MATH has past end-point on MATH and future ... |
hep-th/0009124 | We show how to derive the first relation. Inserting the definition we find: MATH where we used NAME rule and rearranging of the summation of the second term in the second line and the the relation MATH on the last line. The second equation follows from a similar calculation. |
hep-th/0009145 | We can get rid of the term MATH in REF by an appropriate reparametrization MATH of the dilaton. With REF we see that MATH has to obey MATH . After one integration this may be verified to become MATH . We choose MATH to be orientation preserving, that is, MATH. Our assumption MATH and the regularity of MATH within MATH ... |
hep-th/0009145 | The equations of motion of the class of models under consideration are given by MATH . By contracting this equation with the generator MATH of the horizon, we obtain MATH where this equation is assumed to be evaluated on the horizon. By integrating REF is now established as in the previous section provided the late-tim... |
math-ph/0009002 | The fact that MATH and MATH are annihilated by MATH follows trivially from the fact that MATH and MATH are annihilated by each pairwise interaction MATH. So, in fact these states are frustration-free ground states. Next, MATH by REF . We observe that each MATH is an orthogonal projection. Moreover MATH commutes with MA... |
math-ph/0009002 | First, MATH because MATH and MATH. It is also clear that MATH, and MATH, in agreement with REF . Because MATH and MATH are eigenvectors of the self-adjoint operator MATH, all that remains is to check that REF holds on MATH. But this is true by REF , since MATH and MATH on MATH. |
math-ph/0009002 | Define MATH. Define MATH an infinite matrix such that MATH. Let MATH be an orthonormal family in any NAME space, and let MATH. Then MATH and MATH. For simplicity let MATH, and let MATH. We consider MATH. Then we calculate MATH . Since MATH, this shows that MATH is bounded and MATH is invertible. Under the invertibility... |
math-ph/0009002 | Partition MATH into MATH intervals MATH each of length MATH. If MATH then MATH . By REF , MATH . So MATH . In other words, MATH for some MATH. Since MATH, MATH. Let MATH, then REF holds. Note that for any orthogonal projection MATH and any operator MATH we have the decomposition MATH . If MATH is nonnegative, then MATH... |
math-ph/0009002 | Since MATH, it is clear that MATH . So REF implies REF . To prove REF notice that for any operator MATH, any orthogonal projection MATH, and any nonnegative operator MATH, MATH . So, for any nonzero MATH, MATH . Setting MATH, MATH and MATH we have MATH . Since MATH the corollary is proved. |
math-ph/0009002 | We first prove that MATH . It is easy to see that, just as for the droplets on an interval, MATH where MATH. In fact, using the same tools as in REF, we can calculate exactly, for MATH, MATH . It is verifiable that this satisfies the bounds above. The other expectations MATH and MATH are similar. Applying REF , proves ... |
math-ph/0009002 | The proof that MATH is essentially the same as in REF. One fact we should check is that for each MATH, MATH. We observe that MATH . But as before, MATH . An obvious fact is MATH . Taking MATH, yields the desired result. We have the usual orthogonality estimates MATH . In fact, the estimate of MATH follows by REF , taki... |
math-ph/0009009 | NAME: Getting an upper bound for MATH in terms of MATH is relatively easy, as before. The problem is the lower bound. One might suppose that one decomposes MATH into small boxes as before, with NAME b.c. on each box, and in each box one approximates MATH by a constant. This will NOT work, even if MATH, because all the ... |
math-ph/0009010 | The transform for MATH comes directly by taking the logarithmic derivative of MATH with respect to MATH. Then the result for MATH follows as it is adjoint to MATH. For MATH, convert MATH to MATH to find the stated result. |
math-ph/0009010 | We take several steps to pull the MATH-factors across all terms. CASE: Apply the NAME formula for sl REF : MATH with MATH. CASE: Recall the HW formula, REF MATH . Now the adjoint action gives MATH and hence from the above HW formula, MATH CASE: Next, since MATH acts a dilation on MATH, MATH CASE: Now for the MATH-facto... |
math-ph/0009010 | Apply the NAME formula in MATH and use the fact that appropriate elements of MATH and MATH are mutually adjoint, specifically, MATH and MATH. |
math-ph/0009010 | These follow readily from the commutation relations for the MATH and MATH operators. |
math-ph/0009010 | Use the hat-representation from REF in the dual form acting on the NAME function. Setting MATH, MATH, and MATH as in the statement of the Theorem, MATH and compute accordingly. |
math-ph/0009010 | Use the hat-representation found in REF . With MATH and MATH, it is readily checked that each of the sl REF operators commutes with MATH and MATH. |
math-ph/0009020 | Due to NAME 's formula (there is a typo at the NAME 's approximation (see REF ) in the NAME 's book) MATH holds MATH . Taking into account that MATH, and the first two assumptions of the NAME 's limiting process give MATH where the Right-hand side is MATH, which leads thanks to the third assumption of the limiting proc... |
math-ph/0009020 | CASE: MATH . For the purpose of maximization MATH can be MATH-transformed, into MATH where MATH, MATH is gamma-function, and MATH. Necessary condition for maximum of MATH than is MATH since, according to the assumed adding-up constraint, MATH. First, it will be proved that MATH for any MATH, and any MATH. The first der... |
math-ph/0009020 | MATH. |
math-ph/0009026 | Before we proceed to prove this theorem, let us consider in more details a matrix (spinor) representations of the antiautomorphisms MATH and MATH. According to NAME - NAME theorem the antiautomorphism MATH corresponds to an antiautomorphism of the full matrix algebra MATH: MATH, in virtue of the well - known relation M... |
math-ph/0009026 | Let us consider first the types with the ring MATH. As follows from REF , the type MATH admits the Abelian automorphism groups MATH if MATH is the product of MATH skewsymmetric matrices REF and MATH is the product of MATH symmetric matrices (MATH). Therefore, MATH . Further, the type MATH admits the non - Abelian autom... |
math-ph/0009031 | The statement is clearly true for MATH. Assume it to hold up to MATH. One calculates, using the lemma with MATH, MATH . Hence the proof follows by induction. |
math-ph/0009031 | Each normalized element MATH of MATH defines a covariant state MATH by REF . In particular, one has MATH with MATH . Using this result covariance follows from MATH . |
math-ph/0009031 | Let MATH denote the linear space of continuous functions with compact support in MATH and with values in MATH. A sesquilinear form is defined by MATH for all MATH (MATH is the modular function of MATH). From the positivity of MATH follows that MATH is a positive form. Let us assume for simplicity of notations that MATH... |
math-ph/0009031 | Positivity is verified as follows. MATH . For each neighborhood MATH of the neutral element of MATH let MATH be a positive normalized function with support in MATH. Then MATH is an approximate unit of MATH. One has MATH . The latter tends to REF because of continuity of MATH and MATH in the vicinity of MATH. One conclu... |
math-ph/0009033 | We assume MATH is real-valued; the proof for complex-valued MATH is similar. Let MATH with MATH. Further restrictions will be imposed later. Pick MATH, non-decreasing, with MATH, MATH if MATH, and MATH if MATH. The function MATH is a smooth version of MATH. By construction of MATH, MATH . Henceforth MATH and MATH are a... |
math-ph/0009033 | Since MATH is bounded uniformly in MATH it suffices to prove existence of REF for vectors MATH from the dense subspace MATH. We only consider creation operators, the proof for annihilation operators is similar. For given MATH let MATH where MATH. By NAME 's argument the existence of the limit REF follows if MATH . In t... |
math-ph/0009033 | CASE: The existence of the limit for MATH follows from REF . The operator bound REF follows from MATH and from the boundedness of MATH. Finally if MATH the existence of the limit follows by an approximation argument, because MATH is dense in MATH. CASE: Follows from REF and the CCR for MATH and MATH. CASE: For MATH and... |
math-ph/0009033 | By REF and because of the assumption MATH, it follows that, for each MATH the vector MATH is in the range of a spectral projector MATH, for some MATH. Thus MATH is well defined and MATH . Now we want to prove the equality REF . We consider only the case with MATH creation operators, the other cases being similar. Assum... |
math-ph/0009033 | We proceed by induction over MATH and assume the statement holds for MATH replaced by MATH. Let MATH where MATH MATH as MATH. Hence by REF the limit MATH exists and by the induction hypothesis MATH . Since MATH is closed this proves MATH and the first equation from MATH . The second equation follows from REF and the la... |
math-ph/0009033 | We use MATH as an abbreviation for the operator MATH. For each MATH we have MATH which shows that MATH . The corollary now follows, because, by REF , MATH for MATH sufficiently large. |
math-ph/0009033 | This proof follows a similar pattern as the proof of REF . Some of the explanations given in that proof are not repeated here. We may assume MATH is real-valued. Pick MATH as in the proof of REF with the choice MATH. That is, MATH is non - decreasing, MATH, MATH if MATH and MATH for MATH. Here MATH and small. Again MAT... |
math-ph/0009033 | Using the representation MATH we get MATH . The statement of the lemma follows from the last equation since the factor in the braces, on the right-hand side of the last equation, is equal to MATH. |
math-ph/0009033 | Since MATH is bounded uniformly in MATH it suffices to prove existence of REF for MATH. We only consider creation operators, the proof for annihilation operators is similar. For given MATH let MATH where MATH. By NAME 's argument the existence of the limit REF follows if MATH . In the following we will use the notation... |
math-ph/0009033 | CASE: Follows from MATH dense, from MATH and because MATH is a form-core for MATH. CASE: Follows from REF and the CCR for MATH and MATH. CASE: For MATH, by REF, MATH . Now, by REF , MATH . Since MATH for MATH, by REF , it follows that MATH as MATH. This proves the first pull through formula, the proof of the second is ... |
math-ph/0009033 | CASE: Suppose first MATH and pick MATH. Then existence of MATH is proved as in REF with only small modifications: rather than MATH as in REF one now has MATH for which we use the formula MATH . New is the factor MATH in front, which does not affect the subsequent estimates since MATH, and the additional term MATH which... |
math-ph/0009033 | REF follows from MATH and the spectral theorem. REF follows from MATH . |
math-ph/0009033 | Let MATH, MATH and MATH . Since MATH we have MATH and hence it suffices to show that the lemma holds with MATH replaced by MATH. For the case MATH see for example, CITE. If MATH let MATH and note that MATH . Next, in each factor MATH commute as many factors of MATH to the right as possible. For MATH even one gets MATH ... |
math-ph/0009033 | It is enough to show that MATH is form-bounded with respect to MATH. Since MATH by REF , and from MATH it follows that MATH for some MATH. Combined with the assumption on MATH this shows that MATH is form-bounded with respect to MATH and hence proves the lemma. |
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