paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0005274 | By REF MATH and MATH are singular vectors of MATH in the case MATH. Assume first that MATH. Let MATH and MATH be the MATH-submodules generated by MATH and MATH, respectively. We form the MATH-submodule MATH and consider MATH. The set MATH is a set of MATH-generators for MATH, since MATH has MATH-weight MATH. We have MA... |
math/0005274 | As a module over MATH we have MATH is a direct sum of MATH copies of MATH, generated by the highest weight vectors MATH, where MATH. Since the MATH-weights of the MATH's are all distinct for distinct MATH's it follows that these modules as MATH-modules are all non-isomorphic. Therefore if MATH is irreducible over MATH,... |
math/0005274 | Since as a MATH-module MATH is a direct sum of MATH copies of MATH we obtain a description of the vector space spanned by all proper MATH-singular vectors by virtue of REF. But as a MATH-module MATH is also a direct sum of MATH copies of MATH, from which we obtain similarly a description of the vector space spanned by ... |
math/0005274 | By REF MATH is a singular vector in MATH. Consider MATH, the MATH-submodule generated by MATH. Then we have MATH, where MATH is the irreducible MATH-submodule generated by MATH. Let us compute the space MATH, the space of MATH-invariants inside MATH. Since the MATH-weight of MATH is MATH, we know that MATH is a free MA... |
math/0005274 | By REF MATH is a singular vector in MATH. Let MATH be the MATH-submodule generated by MATH so that MATH, where MATH is the irreducible MATH-submodule generated by MATH. Consider MATH, the subspace in MATH of MATH-invariants. Now the MATH-weight of MATH is MATH and so MATH is a free MATH-module generated over MATH by MA... |
math/0005278 | Let MATH be MATH-narrow. If MATH has a zero, we have to show that, given MATH, there is some MATH such that both MATH and MATH. Now, if MATH, which is a proper subset of MATH, and MATH such that MATH, then MATH as well. Conversely, if a closed proper subset MATH is given, pick some MATH such that MATH on MATH, MATH off... |
math/0005278 | Let us denote MATH and MATH. If MATH is a left semi-Fredholm operator, then, since MATH is complemented by a finite-codimensional subspace MATH, MATH is bounded from below, because MATH acts as an isomorphism from MATH onto MATH. On the other hand, if MATH is bounded from below on some finite-codimensional subspace MAT... |
math/0005278 | REF is clear from the definition. For REF we note first that MATH . Indeed, by definition of MATH we have MATH whence MATH . On the other hand, an application of REF with MATH replaced by MATH gives MATH in REF . Now, by elementary arithmetic involving MATH and MATH we have, writing MATH for short, MATH . Because MATH ... |
math/0005278 | Let MATH be a maximal subsemigroup of MATH. Put MATH. We have proved above in REF that MATH is a subsemigroup, so MATH is a subsemigroup, too. By definition of MATH we have MATH. So the maximality of MATH implies that MATH. This proves the inclusion MATH. Let us now prove the inverse inclusion. Let MATH. Then there is ... |
math/0005278 | Let MATH. Then for every MATH we have MATH. This means that for every MATH and MATH there is an element MATH such that MATH. This in turn implies that MATH and MATH. So MATH. Now let MATH. Then for every MATH, every MATH and MATH there is an element MATH such that MATH. But by the definition of tubes, MATH. So MATH and... |
math/0005278 | We assume without loss of generality that MATH. Given MATH pick MATH such that MATH. If MATH and MATH is chosen according to REF , then MATH hence MATH which proves the lemma. |
math/0005278 | MATH if and only if for every pair MATH and MATH there is an element MATH such that MATH. This in turn is equivalent to the following condition: for every pair MATH and MATH there is an element MATH such that MATH, MATH and MATH belongs to the tube MATH (just put MATH). Evidently, the last equation coincides with the s... |
math/0005278 | The fact that every MATH-narrow operator is a strong NAME operator has been proved in a slightly different form in CITE. Consider the converse implication. Let MATH. Fix a closed subset MATH and MATH. According to the definition it is sufficient to prove that there is a function MATH for which the restriction to MATH i... |
math/0005278 | Since a narrow operator is a strong NAME operator and a strong NAME operator is MATH-narrow REF , it is left to prove that a MATH-narrow operator MATH on MATH is narrow if MATH is perfect. Let MATH be a functional, represented by a regular NAME measure MATH; we have to show that MATH is a strong NAME operator. Thus, le... |
math/0005278 | Fix some MATH and find an element MATH in norm-interior of MATH such that MATH. By REF , for every MATH there is an element MATH such that MATH. If MATH is small enough, then MATH. So, if in turn MATH is small enough, then MATH satisfies our requirements. |
math/0005278 | Select MATH. There are two symmetric cases: MATH or MATH. Consider for example the first of them. If we assume that our statement is not true, then we obtain MATH a contradiction. |
math/0005278 | CASE: First of all let us fix elements MATH such that MATH. Applying repeatedly REF with sufficiently small MATH to MATH, MATH and MATH we may select elements MATH with MATH, MATH, in such a way that for every MATH (to get the last inequality, we need to apply the previous lemma at each step). The element MATH will be ... |
math/0005278 | It only remains to show that the above condition is sufficient for MATH to be narrow. We first note that an operator satisfying that condition will also satisfy the conclusion of REF ; see the proof of that lemma. Now, if MATH, MATH and MATH, consider the relatively weakly open set MATH . By REF there exists some MATH ... |
math/0005278 | The convexity is evident. To prove the density we need to show, by the NAME theorem, that for every MATH and every MATH there is an element MATH such that MATH (in other words, MATH). Let us fix an element MATH with MATH and consider the operator MATH. Consider further MATH and a MATH-neighbourhood MATH of MATH in MATH... |
math/0005278 | Let us fix MATH, MATH and MATH such that MATH. According to the definition of MATH there exists a convex combination MATH of slices of the unit ball such that MATH and MATH. By REF there is an element MATH such that MATH and MATH. But the inclusion MATH means that MATH. So MATH . Because MATH is arbitrarily small, the ... |
math/0005278 | REF follows from the previous theorem, because every finite-rank operator is a strong NAME operator. For REF use REF and note that MATH . REF is a restatement of REF . |
math/0005278 | Consider an auxiliary space MATH and an auxiliary matrix MATH, MATH. Since MATH contains no copies of MATH either and since MATH is a MATH-limit point of MATH, there is, according to REF , a sequence of the form MATH which converges to MATH in MATH. This means in particular that MATH in MATH and MATH. So MATH and MATH ... |
math/0005278 | Using inductively REF we can select a doubly indexed sequence MATH in MATH with the following properties: CASE: for every MATH, MATH is a MATH-limit point of every column MATH; CASE: for every MATH, if MATH, then MATH. Applying REF and passing to a subsequence if necessary, we obtain strictly increasing sequences MATH,... |
math/0005278 | Assume there exist MATH and MATH such that for every finite-dimensional subspace MATH there is a slice MATH containing MATH with MATH for all MATH. Take a weak-MATH cluster point MATH of the net MATH and let MATH. We have MATH since MATH and therefore MATH. Now if MATH, then MATH and therefore MATH for some MATH that c... |
math/0005278 | Suppose MATH has the NAME property. Let MATH be a dense sequence in MATH. We select a sequence MATH of finite-dimensional subspaces of MATH by the following inductive procedure. Put MATH. Suppose MATH has already been constructed. Fix a MATH-net MATH, MATH, in MATH provided with the sum norm, select by REF finite-dimen... |
math/0005278 | First, let MATH be a perfect compact metric space. It follows from REF and CITE that the set of all narrow operators on MATH is stable under the operation MATH, that is, it is a semigroup. (In fact, CITE only deals with MATH, but the arguments work as well for a metric MATH.) We shall now reduce the general case to the... |
math/0005278 | We shall argue by induction on MATH. First of all consider MATH. Every MATH-neighbourhood of MATH can be represented as MATH, where MATH. Since MATH is a strong NAME operator by definition of the central part, there is an element MATH such that MATH and MATH. The last inequality means, in particular, that MATH and MATH... |
math/0005278 | REF implies that every operator which does not fix a copy of MATH can be factored through a space without MATH-subspaces. So every operator which does not fix a copy of MATH can be majorized by an operator which maps into a space without MATH-subspaces. Since the class of narrow operators is an order ideal, it is enoug... |
math/0005278 | In each MATH there is a separable subset whose closed convex hull contains MATH. So, passing to the linear span of these separable subsets we may assume that MATH is separable. Introduce a directed set MATH as follows: the elements of MATH are of the form MATH where MATH, MATH, MATH, MATH, MATH. Define MATH as follows:... |
math/0005278 | Consider elements MATH, MATH, a slice MATH and MATH. According to our assumption the quotient map MATH is a narrow operator. So there is an element MATH such that MATH and MATH. The last condition means that the distance from MATH to MATH is smaller than MATH, so there is an element MATH with MATH. The norm of MATH is ... |
math/0005278 | REF follows from REF from REF , and REF follows from REF . |
math/0005278 | Due to REF every finite-codimensional subspace of a space with the NAME property has the NAME property itself (see also CITE); this is the reason for the equivalence of REF . The implication MATH follows immediately from the definition of a wealthy subspace; REF are consequences of REF . |
math/0005278 | First of all let us fix a slice MATH from the definition of a MATH-fine pair and fix a MATH such that the set MATH, MATH still has diameter less than MATH. Now let us find a finite-codimensional subspace MATH such that CASE: MATH on MATH, CASE: if MATH and MATH, then MATH; the last condition can be satisfied by a varia... |
math/0005278 | Let MATH be the quotient map and let MATH; further let MATH and let MATH be the corresponding quotient map. Then MATH or MATH is MATH-codimensional in MATH. Now, in either case we have MATH for all MATH. Since MATH is a strong NAME operator by assumption, so is MATH, and MATH is narrow. |
math/0005278 | If MATH intersects all the elements of MATH, then the quotient map MATH is unbounded from below on every element of MATH. So the quotient map belongs to MATH which coincides with the class of strong NAME operators by REF . Now consider the converse statement. If MATH is almost rich, then for every MATH the map MATH is ... |
math/0005278 | According to REF we need to prove that for every positive MATH and every pair MATH the subspace MATH intersects MATH. To do this, according to REF , it is enough to show that for every MATH and every pair MATH there is a MATH-fine pair MATH which approximates MATH well; that is, MATH. Let us fix a positive MATH and sel... |
math/0005278 | It is clear that REF , see the remark following REF . Now suppose REF . Every MATH-codimensional subspace of MATH is wealthy by REF and is hence almost rich by REF . An appeal to REF completes the proof. |
math/0005278 | Let MATH, MATH, and MATH. Consider a slice in MATH of the form MATH . Applying REF to this slice, the elements MATH, MATH and MATH we get a function MATH such that MATH . Denote by MATH the set MATH. The condition MATH implies that MATH, so MATH . Next, introduce MATH. By the last inequality MATH and MATH to see this o... |
math/0005278 | Denote MATH . Since MATH and MATH coincide off MATH, we clearly have MATH . We also have that MATH . Indeed, we can write MATH as a countable union of disjoint (half-open) intervals; denote by MATH any one of these. Then MATH is constant on MATH, and MATH. Hence MATH . Summing up over all MATH gives the result. Next, w... |
math/0005278 | Let us fix MATH and MATH. Without loss of generality we may assume that MATH for a big enough MATH to be chosen later. Put MATH. Then MATH with MATH. So MATH and MATH. Since MATH we can pick MATH big enough to satisfy MATH. This shows that MATH is a strong NAME operator. To show that MATH fails the NAME property if MAT... |
math/0005280 | We need to show that MATH is stably congruent to a unidiagonal matrix. In fact MATH is congruent to MATH, which is congruent to MATH and which, in turn, since MATH is almost even, is congruent to MATH, where MATH is a diagonal matrix all of whose entries are MATH or MATH. But this is congruent to some unidiagonal matri... |
math/0005280 | MATH represents an element in MATH and so its bordism class is determined by the degree of MATH. It follows that MATH is bordant to the identity map of MATH. This gives us the manifold MATH and map MATH without the desired handlebody structure. We must eliminate the handles of index not equal to MATH. First of all we c... |
math/0005280 | This follows from CITE. The isomorphism MATH in REF is determined, since the maps MATH are isomorphisms. The commutativity of MATH corresponds to the diffeomorphism equivalence of the pairs MATH. |
math/0005280 | Suppose that MATH is bordant to the identity on MATH by MATH, where MATH consists of MATH - handles adjoined to MATH. Let MATH be an associated matrix. Suppose MATH is bordant to MATH by a MATH - homology bordism MATH. By pasting MATH together we create a bordism MATH from the identity map on MATH to itself. The inters... |
math/0005280 | Since each component MATH of MATH is null-homotopic in MATH, it is homotopic in MATH to a product of meridians of MATH. Thus we can connect sum several meridians of MATH to MATH to get a new knot MATH which is null-homotopic in MATH and is clearly isotopic to MATH in the complement of the other components of MATH. To s... |
math/0005280 | Suppose MATH is an algebraically split link determining MATH in MATH and MATH determines MATH in MATH. We can apply REF to MATH and the meridians MATH of MATH in MATH to allow us to assume that the components of MATH are null-homotopic in MATH. Now MATH is algebraically split in MATH and determines MATH in MATH. Since,... |
math/0005280 | Suppose that MATH is an algebraically split link which defines MATH. Now let MATH be any algebraically split link - NAME can assume that MATH is disjoint from the meridians of MATH and so lies in MATH. In fact, by REF applied to MATH and the meridians of MATH, we can assume that the components of MATH are null-homotopi... |
math/0005280 | Pick a maximal forest MATH for MATH, that is, a maximal tree for each connected component of MATH. If MATH, then we can assume that MATH, a base point of MATH, that is, that it factors though a map MATH. This map is determined by the induced one on the level of MATH. A different choice of a maximal forest or a differen... |
math/0005280 | It suffices to consider a vertex-oriented, edge-oriented connected graph MATH. To a map MATH, we will associate a map MATH and vice versa. Given a map MATH, (which in view of relation MATH we may assume that it is a decoration of the edges of MATH by elements of MATH) we define a map MATH as follows. For a closed path ... |
math/0005280 | Choose a base point MATH of MATH and a basing MATH of MATH, that is, a choice of disjoint paths MATH in MATH from MATH to points MATH, one for each component of MATH. Choose a framing of MATH and a lift MATH of MATH. Then, there is a unique lift MATH in MATH of MATH that contains MATH and a well-defined linking matrix ... |
math/0005280 | MATH . |
math/0005280 | Since MATH, we need only show the opposite inclusion. REF implies that for every MATH - AS link MATH in MATH, there exists a trivial MATH - link-MATH that ties a trivial unimodular link MATH such that MATH. Let MATH denote the image of MATH under surgery on MATH. MATH is a MATH - link (with nullhomotopic leaves) and MA... |
math/0005280 | This follows by elementary properties of NAME 's calculus applied to the unit-framed knots MATH shown in the pictures above. |
math/0005280 | First of all we prove the invariance of the MATH under surgery equivalence. If MATH is chosen so that MATH is MATH - AS, then we can, by tubing, arrange that the surfaces MATH used to define MATH are disjoint from the lifts of MATH and so pass unchanged into the MATH - covering of the surgered link. In particular the i... |
math/0005280 | Concordance, just as in the classical case, is generated by the following ribbon move MATH. Given a MATH - AS link MATH, consider also a finite number of disks MATH in MATH, disjoint from each other and MATH. For each MATH choose a band MATH connecting MATH to a component, which we denote MATH, of MATH. The band cannot... |
math/0005281 | The metric introduced in REF is equivalent to the metric described in CITE. The induced topologies are therefore the same. The result follows therefore from CITE. |
math/0005281 | Let MATH be a minimal basic encoder of MATH and let MATH be the first column of MATH. Note that MATH and that there is at least one entry of MATH which does not contain the factor MATH. Let MATH and consider the sequence of code words MATH. For each MATH one has that MATH. However MATH is in MATH. This shows that MATH ... |
math/0005281 | CITE. |
math/0005281 | The proof of the completeness part of the Theorem is analogous to the proof of REF . In order to show controllability, let MATH be an encoding matrix for a code MATH and consider two code words MATH and MATH. The codeword MATH required by REF can be constructed in the form MATH . |
math/0005281 | Let MATH. If MATH is not controllable, then MATH is not left prime and one has a factorization MATH, where MATH is left prime and describes the controllable sub-behavior MATH. Since MATH is an autonomous behavior it follows that MATH . It follows (compare with REF ) that the completion MATH. |
math/0005287 | For simplicity, assume MATH, the general case being quite similar. The diagonal MATH is obviously an invariant subset for the group MATH, where MATH runs over the set of all MATH-preserving transformations of the space MATH. Thus it suffices to show that if MATH is concentrated on the set MATH, then MATH, and if MATH i... |
math/0005287 | Let MATH be a MATH-preserving transformation of MATH. This transformation acts on the space MATH by substituting coordinates, that is, MATH, and it is clear that the law MATH of a homogeneous NAME process on MATH is invariant under MATH. Denote by MATH the conditional measure of MATH given the conic part equal to MATH.... |
math/0005287 | Given fixed MATH and a probability vector MATH, consider a partition MATH of the space MATH such that MATH, MATH. Let MATH be the sequence of i.i.d. variables with common distribution MATH and assume MATH, if MATH. Then the random variables MATH form a sequence of i.i.d. variables, and MATH. Consider a random process M... |
math/0005287 | Fix MATH and let MATH. Consider an arbitrary function MATH. Then MATH. Thus, in view of REF , the NAME transform MATH equals MATH . Using REF once more, we may consider the last factor as the NAME transform of MATH calculated on the function MATH. Denote MATH. Then we have MATH and REF follows. |
math/0005287 | Let MATH be a MATH-measurable functional on MATH which is invariant under all MATH that is, MATH a.e. with respect to MATH. Consider an arbitrary NAME function MATH. Then for each MATH where MATH denotes the expectation with respect to MATH. But in view of REF the last factor equals MATH hence we have MATH . Thus MATH ... |
math/0005287 | The representation is correctly defined and its unitarity follows from the invariance property of MATH. The irreducibility follows from the ergodicity of the action of the group MATH of multiplicators. |
math/0005287 | In case of a MATH-stable process, we have MATH, and REF follows immediately from REF . |
math/0005287 | It follows from REF that the NAME transform of the measure MATH equals MATH . But MATH as MATH, hence MATH and REF follows. |
math/0005287 | Easy calculation. |
math/0005287 | Using REF and the NAME theorem we obtain that the right-hand side of REF equals MATH and Theorem follows. |
math/0005287 | CASE: Denote the left-hand side of the desired identity by MATH and the right-hand side by MATH. Using the identity MATH we obtain MATH . By the NAME transform REF , the expectation equals precisely MATH, thus MATH and REF follows by changing variables. CASE: Follows from REF by letting MATH. |
math/0005292 | Let MATH be a smooth path of representations starting at the inclusion MATH corresponding to MATH. MATH where the sign equals MATH. Applying REF to the last expression gives MATH . Thus MATH has the same sign as MATH as claimed. To prove REF, apply REF and the chain rule to obtain: MATH . Since MATH REF follows from RE... |
math/0005292 | Here are two constructions for MATH, the first based on the Riemannian geometry of MATH with the NAME metric and the second based on NAME 's earthquake flows. Let MATH be the NAME geodesic satisfying REF. By REF, the geodesic length function MATH is strictly convex along MATH and the directional derivative MATH, for an... |
math/0005301 | First we will show homotopy equivalence and remark on MATH-homotopy equivalence later. We work with MATH, the barycentric subdivision. Notice that the associated poset of MATH contains the poset MATH of nontrivial abelian subgroups of MATH as a subposet, they are merely the commuting sets whose elements actually form a... |
math/0005301 | If MATH has a nontrivial center MATH, then MATH is conically contractible via MATH for any abelian subgroup MATH of MATH. Thus MATH is also contractible as it is homotopy equivalent to MATH. Let MATH be the natural inclusion of posets. Take MATH and let us look at MATH. However since MATH is nilpotent, it has a nontriv... |
math/0005301 | First we show that MATH is path-connected. Take any two vertices in MATH, call them MATH and MATH, then these are two noncentral elements of MATH, thus their centralizer groups MATH and MATH are proper subgroups of MATH. It is easy to check that no group is the union of two proper subgroups for suppose MATH where MATH ... |
math/0005301 | The remarks about NAME characteristics follow from the fact that if a finite group MATH acts freely on a space where the NAME characteristic is defined, then MATH must divide the NAME characteristic. So we will concentrate mainly on finding such actions. First for the action of MATH. MATH acts by taking a simplex MATH ... |
math/0005301 | Follows from the proof of REF , once we note that left multiplication by a central element of order MATH takes the subcomplex MATH of MATH to itself and that the map MATH maps MATH into itself, as the inverse of an element has the same order as the element. For odd primes MATH, MATH is fixed point free on MATH always a... |
math/0005301 | Since MATH does not commute with any nontrivial element, the center of MATH is trivial and MATH. Furthermore, it is clear that MATH is a cone with MATH as vertex and hence is contractible. |
math/0005301 | We first recall that the map MATH from REF has order REF as a map of MATH. However it might have fixed points, in fact from the proof of that proposition, we see that MATH fixes a simplex MATH of MATH if and only if each element MATH has order MATH and it fixes the simplex pointwise. Thus the fixed point set of MATH on... |
math/0005301 | First, from the condition that MATH for MATH we see that no nonidentity element of MATH commutes with anything outside of MATH. Thus conjugating the picture, no nonidentity element of any MATH commutes with anything outside MATH. Thus if MATH is a complete list of the conjugates of MATH in MATH, we see easily that MATH... |
math/0005301 | The order of MATH is MATH of course. It is easy to check that there are five NAME REF-subgroups MATH which are elementary abelian of rank REF and self-centralizing, that is, MATH, and are ``disjoint", that is, any two NAME subgroups intersect only at the identity element. Thus the picture for the vertices of MATH is as... |
math/0005301 | Let MATH be the non-commuting complex associated to the core of MATH. Let us define a facet to be a face of a simplicial complex which is not contained in any bigger faces. Thus the facets of MATH consist exactly of maximal non-commuting sets of noncentral elements in MATH. The first thing to notice is to every facet M... |
math/0005301 | This is because it is easily seen that MATH is a cone on the vertex MATH as everything outside MATH does not commute with MATH as MATH. |
math/0005301 | The cycle MATH in MATH has MATH by a simple calculation. Thus MATH and so the result follows from REF . |
math/0005301 | If MATH is an equivalence relation, it is easy to see that the centralizer classes are exactly the MATH equivalence classes. Thus one sees that the non-commuting complex for the core, MATH, is a simplex. Thus in REF , all terms drop out except that corresponding to the maximum face in MATH where the link is empty. This... |
math/0005301 | Follows from a direct interpretation of the inequalities in CITE, page REF . One warning about the notation in that paper is that MATH means the number of vertices in MATH and the empty face is considered a simplex in any complex. |
math/0005301 | By CITE, if MATH are the components of MATH, then under the MATH-action, MATH acts transitively on the components with isotropy group MATH under suitable choice of labelling. Thus the components are all simplicially equivalent and there are MATH many of them. However we have seen that MATH is MATH-homotopy equivalent t... |
quant-ph/0005013 | Write the four-qubit state MATH as MATH . The tensor MATH can be regarded as a MATH matrix in three different ways: MATH and the requirement of maximal entanglement is that MATH and MATH should all be unitary matrices. We show first that by local unitary transformations we can arrange that the coordinates of MATH satis... |
quant-ph/0005018 | The convergence of MATH to MATH is understood to use some kind of norm for the density matrices that is continuous in the matrix entries MATH. (The operator norm MATH, for example.) The entropy MATH is a continuous function of the finite set of eigenvalues of MATH. These eigenvalues are also continuous in the entries o... |
quant-ph/0005018 | This follows from the fact that the fidelity between two mixed states MATH and MATH equals the maximum `pure state fidelity' MATH, where MATH and MATH are `purifications' of MATH and MATH. (See CITE for more details on this.) |
quant-ph/0005018 | This follows directly from the definition MATH. |
quant-ph/0005018 | The proof follows from the existence of a universal quantum NAME machine, as proven by CITE. Let MATH be this UTM as mentioned above. The constant MATH represents the size of the finite description that MATH requires to calculate the transition amplitudes of the machine MATH. Let MATH be the state that witness that MAT... |
quant-ph/0005018 | Consider the quantum NAME machine MATH that moves its input to the output tape, yielding MATH. The proposition follows by invariance. |
quant-ph/0005018 | This is clear: the universal quantum computer can also simulate any classical NAME machine. |
quant-ph/0005018 | Take MATH and their minimal programs MATH (and hence MATH). Let MATH be the completely positive, trace preserving map corresponding to the universal QTM MATH with fidelity parameter MATH. With this, we define the following three uniform ensembles: CASE: the ensemble MATH of the original strings, CASE: MATH the ensemble... |
quant-ph/0005018 | Apply REF to the set of classical strings of MATH bits: MATH for all MATH. All MATH are pure states with zero NAME entropy, hence the lower bound on MATH reads MATH. The average state MATH is the total mixture MATH with entropy MATH, hence indeed MATH. |
quant-ph/0005018 | All the pure states have zero entropy MATH, hence by REF : MATH. Because all MATH-s are mutually orthogonal, this NAME entropy MATH of the average state MATH equals MATH. |
quant-ph/0005018 | First we sketch the proof, omitting the effect of the approximation. Consider any qubit string MATH whose minimal-length program is MATH. To produce MATH copies of MATH, it suffices to produce MATH copies of MATH and make MATH runs of MATH. Let MATH be the length of MATH; we call MATH the MATH-dimensional NAME space. C... |
quant-ph/0005018 | Fix MATH and MATH and let MATH be the MATH-dimensional NAME space. Consider the (continuous) ensemble of all MATH-fold tensor product states MATH: MATH, where MATH is the appropriate normalization factor. The corresponding average state is calculated by the integral MATH. This mixture is the totally mixed state in the ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.