paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math-ph/0007007 | As for MATH in the proof of REF , with MATH in REF changed to MATH. |
math-ph/0007007 | We apply the variational principle, using MATH-fold products of MATH-particle states MATH as MATH-particle states MATH for all MATH with MATH. Setting MATH we see that the inequality holds, for each MATH and MATH. |
math-ph/0007007 | This results from combining REF with REF . |
math-ph/0007007 | We use the NAME inequality CITE MATH (MATH is normalized as MATH, the constant can be chosen to be MATH), which, together with NAME 's inequality MATH, implies that MATH where MATH is the one-particle reduced density matrix of MATH, and MATH is its density. Now if MATH, which we can of course assume, then MATH with MATH defined in REF . Together with REF this implies that MATH so finally MATH . |
math-ph/0007007 | For MATH consider the wave functions MATH. By REF there holds for each MATH the inequality MATH . Both sides of this inequality are NAME measurable functions of MATH. Observe MATH, MATH, and integrate both sides of REF with MATH. Divide by MATH, and use MATH in front of the kinetic energy term. |
math-ph/0007007 | From MATH we borrow a part of MATH to use REF , MATH sum over all pairs MATH , and multiply with MATH. This gives MATH . Inserting REF into REF completes the proof. |
math-ph/0007007 | This follows from the convergence of the energies, by analogous arguments as in CITE or CITE. |
math-ph/0007007 | Let MATH be the density of MATH. We have MATH where we used again the NAME inequality for the last step. By an analogous argument as in REF we see that the right side of REF converges to MATH as MATH. |
math-ph/0007007 | See REF . |
math-ph/0007007 | Since MATH is unique, MATH converges weakly to MATH by REF and the proof of REF . Apply REF to MATH and MATH, and note that MATH and MATH. |
math-ph/0007007 | We use the notations of the proof of REF . The first assertion is easily proved, using that MATH and MATH. To prove the second we denote MATH and MATH, and compute MATH where we have used MATH in the last step. Therefore MATH as MATH, which gives immediately the desired result. |
math-ph/0007009 | The symmetry and positive definiteness of MATH implies that MATH is also symmetric and positive definite. Hence there exists an orthogonal matrix MATH, MATH, such that MATH where MATH is a diagonal matrix with all positive eigenvalues MATH. Thus MATH which has the same MATH's as eigenvalues. Hence it is nonsingular. Substituting REF into REF , we have MATH where MATH, MATH, and MATH. REF is completely diagonalized. Its solution are MATH . NAME processes each with transition probability MATH and stationary distribution MATH where MATH . Furthermore, the two-time covariance matrix for the stationary process is MATH . Therefore, by substituting MATH back into the above equations, we have MATH in which matrix MATH . The stationary covariance MATH gives a stationary Gaussian distribution. Combining with the transition probability in REF and NAME properties, a Gaussian stationary process with covariance MATH is obtained for REF . Finally, substituting MATH back into REF , the stationary two-time covariance matrix MATH . We now show that MATH is symmetric. MATH immediately leads to MATH, which gives: MATH . That is MATH for any interger MATH. Therefore matrix MATH is symmetric. REF . A Gaussian process MATH is completely determined by its expectation MATH and covariance MATH. A stationary NAME Gaussian process is completely determined by the joint distribution MATH. The Gaussian MATH is determined by all its first and second moments, among which are the two-time covariance matrix MATH. Therefore, the symmetric matrix MATH leads to MATH; and with the NAME property the stationary Gaussian process is then time reversible. CASE: For a stationary NAME process MATH, the joint probability distribution MATH is uniquely determined by the stationary distribution MATH and the transition probability MATH. By the standard method in probability, the time reversibility implies CITE MATH being valid for arbitrary MATH where MATH is the space of smooth functions with compact supports. Note that MATH is the fundamental solution to REF , MATH where the differential operator MATH . NAME both sides of REF with respect to MATH at MATH, we have MATH let MATH and MATH, then MATH and MATH are two arbitrary functions in MATH. Since MATH, MATH . Therefore, the operator MATH is symmetric with respect to the reciprocal of its stationary distribution MATH: MATH. This result is known to physicists. CASE: The differential operator MATH in REF can also be rewritten as MATH where MATH is the trace of the matrix MATH. The statement REF MATH in which the positive MATH, MATH and MATH are arbitrary functions, leads to MATH . Through integration by parts, the first term on the left-hand - side (and similarly for the right-hand-side) MATH and we have MATH . By a simple rearrangement, we have MATH . Since MATH and MATH are arbitrary, we have MATH in which MATH. Therefore MATH that is MATH. CASE: In REF MATH is positive definite. Hence the integrand is always positive, therefore MATH implies MATH. Therefore MATH is a symmetric matrix. Furthermore, MATH is a normal density and hence, MATH is positive definite. |
math-ph/0007009 | The proof relies on a direct verification of the Gaussian function with REF as a solution to REF . This has been done many times by physicists CITE. Hence we will not repeat the lengthy computation. With the verification of the solution, and uniqueness of the fundamental solution to REF , the lemma is proven. |
math-ph/0007009 | Necessity: If the solution to REF is nonsweeping, then by REF and the sweeping theorem from REF, its fundamental solution has a stationary limit MATH. Since MATH has Gaussian distribution for MATH, the limit distribution is also NAME with finite variance. Therefore, MATH . By the general formula for MATH CITE and using MATH, the convergence of REF implies that all the eigenvalues of MATH in REF must have positive real parts. Sufficiency: If all the eigenvalues have positive real parts, the nonstationary Gaussian solution in REF has a unique Gaussian density as its limit when MATH. With the help of the fundamental solution for REF and the NAME property, a stationary Gaussian process related to the quantities in REF can be constructed. Hence the solution of REF are nonsweeping. |
math-ph/0007009 | By REF with reversibility has symmetric MATH; and by REF its stationary process has covariance MATH. Therefore, MATH is symmetric, and MATH . Hence, the standard fluctuation-dissiplatin relation follows, and furthermore REF can be simplified into MATH. On the other hand, the standard fluctuation-dissipation relation is satisfied by any stationary Gaussian processes with or without the symmetric MATH: From REF we have MATH . Let MATH, the upper limit vanishes and we have MATH . Hence by REF it is not a sufficient condition for reversibility. |
math-ph/0007009 | The proof of this result is contained in the proof of REF , MATH. |
math-ph/0007013 | With REF in the hand, the proof goes along very much the same lines as in CITE. Indeed, let MATH be as in REF and set MATH in REF to be MATH, where MATH will be chosen later and MATH will tend to MATH. Let MATH be in the MATH-class and recall that MATH with MATH. Abbreviate MATH and MATH, and note that MATH. Then, using also that MATH, we have from REF that MATH-almost surely. Abbreviating MATH, we have to prove that, for any MATH, MATH for some appropriate MATH and sufficiently large MATH. Note that the eigenvalues MATH have identical distribution. Furthermore, their exponential moments can be estimated by MATH where MATH is a constant related to MATH, see CITE. Since MATH is increasing and MATH slowly varying, it suffices to prove REF for MATH taking only a discrete set of values; the main difference compared to CITE is that now we take MATH . Let MATH and note that REF implies that MATH. The proof now proceeds exactly as in CITE: We let MATH and use the NAME inequality and REF to derive that MATH is summable on MATH for all MATH, provided MATH is chosen appropriately and MATH is sufficiently large. The claim is finished using the NAME lemma. |
math-ph/0007013 | Let MATH be the embedded discrete-time simple random walk on MATH and let MATH be its local times defined by MATH. Let MATH denote the expectation with respect to the discrete-time walk, starting at MATH. Abbreviate MATH and MATH. Then, by REF, MATH where MATH, and MATH is a shorthand for MATH. Fix MATH and define MATH . Let MATH be the set of all the times at which the walk visits a point MATH with MATH. By relaxing the constraint MATH in MATH to MATH for every MATH, neglecting the terms with MATH in the exponential, and integrating out MATH, we get MATH . Neglecting the first indicator and the restriction to MATH, we can carry out the sum over MATH in REF and find that MATH . On MATH, the walk visits either all sites in MATH or all sites in MATH. Hence, we can estimate MATH . REF then follows from the assertion MATH where we used that MATH whenever MATH. (The term with MATH in REF can be added or removed at the cost of changing MATH by a finite amount.) Let us prove that REF holds. First we note that MATH contains in every sufficiently large interval in MATH a positive fraction of sites. Indeed, put MATH and note that by NAME 's theorem we have MATH for every bounded interval MATH and some MATH independent of MATH. A routine application of the NAME lemma implies that MATH for MATH large enough, MATH-almost surely. Now we prove that with high probability, there are sufficiently large intervals which are traversed from one end to the other at least twice by the random walk MATH. Fix MATH and abbreviate MATH. We divide the walk into MATH pieces MATH (neglecting a small overshoot) with MATH for MATH. Note that these MATH walks are independent copies of each other. Let us introduce the events MATH where MATH. It is elementary that MATH as MATH. Furthermore, with the help of a concatenation argument and convergence of simple random walk to Brownian motion we derive that MATH, whenever MATH is large enough. Now we estimate MATH . Note that, on MATH, there is an interval MATH with MATH such that every point of MATH is visited by at least two of the subwalks, that is, we have MATH for any MATH. If MATH is sufficiently large, we deduce from REF that there are at least MATH points MATH with MATH. By using this in REF, we have MATH . The right hand side is clearly summable on MATH since MATH. This finishes the proof. |
math-ph/0007013 | Abbreviate MATH and let MATH. Then MATH . Using this estimate and the NAME inequality, we have for any MATH that MATH for any MATH. Set MATH and note that we have MATH as MATH, due to MATH. Consequently, the term with MATH is negligible and the right-hand side of REF is bounded by MATH. The claim is finished by the NAME lemma. |
math-ph/0007013 | Pick any MATH. Let MATH be so large such that the sum in REF with MATH for all MATH exceeds MATH. Note that MATH. Combining the results of REF for MATH and REF , we derive, for sufficiently large MATH respectively,MATH, the bound MATH where MATH is the constant from REF. But MATH by our choice of MATH, which means that MATH, where MATH. The rest of the argument does not involve the particular form of MATH and can directly be taken over from CITE. |
math-ph/0007013 | The argument is based on the asymptotic sublinearity of MATH at infinity. However, in order to have sublinearity on the whole interval MATH, we first construct an auxiliary modification of MATH. Let MATH be such that MATH is positive, increasing, and concave on MATH. Let MATH to be the right derivative of MATH at MATH. Define MATH by the formula MATH . Note that MATH is positive, increasing, concave and hence sublinear on MATH. Moreover, REF holds true for MATH replaced by MATH. For MATH, abbreviate MATH. Choose MATH and estimate, for MATH, MATH where we used the fact that MATH almost surely, and sublinearity of MATH. Since we have that MATH, the NAME of Large Numbers tells us that the right-hand side of REF is almost surely no more than MATH, where for MATH we can take, for instance, MATH . Hence, we derive MATH which directly yields the desired claim. |
math-ph/0007013 | We begin by formalizing the event in REF; in order to later approximate continuous MATH by a discrete variable, we write MATH instead of MATH: MATH . Note that the probability of MATH does not depend on MATH and note that different MATH's are independent if the MATH's have distance larger than MATH from each other. The proof of REF shows that MATH as MATH (the only modification required is to replace every occurrence of MATH in the meaning MATH by MATH). In order to prove our claim, it is sufficient to show the summability of MATH over all MATH such that MATH. (The sufficiency follows from the facts that MATH as MATH and that MATH and MATH are increasing. The error terms are absorbed into an extra MATH in REF compared to REF, see CITE.) Using the independence of MATH for MATH and the bound MATH, we easily derive MATH where we used that MATH and then applied the definition of MATH. Use concavity of MATH to estimate MATH and use REF to bound MATH by MATH. Furthermore, since MATH is bounded from above by a positive power of MATH, we see from REF that MATH. Applying all this reasoning on the right-hand side of REF, we see that MATH as MATH, which is summable on the sequence of MATH such that MATH. This finishes the proof. |
math-ph/0007013 | Let MATH and fix MATH and MATH such that MATH. Let MATH, define MATH as in REF and let MATH be as in REF ; suppose MATH without loss of generality. Let MATH. As in CITE, the lower bound will be obtained by restricting the walk in REF to perform the following: The walk keeps jumping toward MATH, spending at most time MATH at each site MATH such that it reaches MATH before time MATH. Then it stays at MATH until time MATH and then within MATH for the remaining time MATH. Inserting this event into REF and invoking NAME property at time MATH we get MATH where the same argument as in CITE shows that MATH for large MATH, while for MATH we have MATH where we recalled the notation of REF. Now MATH, so using REF we have that MATH where we used the definition of MATH. Since MATH is asymptotically concave, MATH and the exponent is MATH. Consequently, MATH where MATH still depends on MATH. The proof is finished by letting MATH (which eliminates the dependence on MATH), optimizing over MATH and MATH with MATH and letting MATH. |
math-ph/0007014 | It suffices to prove that MATH for all MATH. There is a normalized vector MATH such that MATH. (MATH is antisymmetric according to the NAME principle.) We use the notation MATH to denote the inner product in NAME space and spin space. Then we can write MATH with MATH . As a (unnormalized) variational trial vector we take the vector MATH. Recall that MATH since MATH is the ground state of MATH. We also recall the NAME inequality MATH . Using REF , integration by parts, and the fact that MATH satisfies the NAME equation MATH, we easily find that MATH . When MATH the last term in REF is not present so no assumption about the potential MATH is needed. When MATH we can omit the last term because it is negative by assumption. Now, by the MATH-translation invariance of MATH, for every MATH there is a `translated' vector MATH so that MATH and MATH. (This is accomplished by the unitary operator on MATH that takes MATH for every MATH and MATH.) Thus, if we denote the quantity in MATH in REF by MATH, and if we define MATH by replacing MATH by MATH in the definition of MATH, we have MATH . Note that MATH. But MATH and this is strictly negative by assumption. Hence, for some MATH we have that MATH and thus MATH, which proves the theorem. |
math-ph/0007014 | Let us first show that it suffices to prove that for any normalized sequence MATH, MATH, (not necessarily minimizing) tending weakly to zero MATH . To prove this let MATH be some minimizing sequence, that is, assume that MATH and that MATH . By the NAME Theorem we can assume that this sequence, as well as the sequence MATH converge weakly in the sense that for any MATH with MATH we have that MATH where MATH is the weak limit of MATH. Our goal is to show that MATH and that MATH. Write MATH. Obviously MATH as well as MATH go weakly to zero. Thus MATH where we used that the cross terms vanish. Since MATH this shows that MATH minimizes the energy, and, furthermore, that MATH for some positive constant MATH. The second inequality is trivial if MATH and otherwise follows from our REF . This proves that MATH converges strongly to zero along a subsequence, which implies that MATH. Hence MATH is a normalized ground state. Thus, it suffices to prove REF . The steps that lead to a proof of REF are quite standard. The only difficulty is that one has to localize in NAME space, which we describe first. We follow CITE with some necessary modifications and some simplifications. Recall that, when the MATH operators are viewed in MATH-space MATH they obey the commutation relations MATH . Consider now two smooth localization functions MATH and MATH that satisfy MATH and MATH is supported in a ball of radius MATH. The first derivatives of MATH and MATH are of order MATH. The operators MATH act both on the space MATH. Note that MATH . Thus, these new creation and anihilation operators create another NAME space MATH that is a subspace of MATH and is isomorphic to the old NAME space MATH. Hence, there exists a map MATH that is an invertible isometry between NAME spaces. It is uniquely specified by the properties MATH and the vacuum in MATH is mapped to the vacuum in MATH. The map MATH is defined on MATH only, but we can extend it to all of MATH by setting MATH whenever MATH is perpendicular to MATH. In other words MATH is a partial isometry between NAME where MATH on MATH, and where MATH is the orthogonal projection onto MATH. We continue to denote the extended map by MATH. Let MATH and MATH be smooth nonnegative functions, with MATH, MATH identically one on the unit ball, and vanishing outside the ball of radius MATH. Set MATH. It is a standard calculation to show that for any MATH with finite energy MATH . The last two terms in REF are bounded by MATH. One goal will be to show that for any MATH with finite energy MATH . The error term MATH vanishes as both MATH and MATH go to infinity and depends otherwise only on the energy of MATH. Notice that the invertible map MATH depends on the cutoff parameter MATH as well. REF will be proved in REF in the Appendix. The intuition behind the estimate REF is that localized electrons interact only weakly with far away photons. Those photons are described solely by their own field energy. An immediate consequence of REF is the estimate MATH which is obtained by noting that the field energy in the second factor can be estimated from below by MATH where MATH is the projection onto the vacuum of the second factor of MATH. In a further step we prove in REF that the sequence MATH as MATH. Returning to REF , using REF we have that MATH with MATH going to zero as MATH. Roughly speaking MATH forces some of the particles to be far away from the origin. Any such particle configuration can be described by two clusters with no interaction between them. In particular the interaction between these clusters via the radiation field is turned off. This means that each cluster carries its own field energy. To prove this the localization in NAME space is used. Moreover the cluster that is far away from the origin does not interact with the external potential although the repulsion among its particles is still present. To summarize, by combining REF we have proved that MATH where MATH and MATH tends to zero as MATH and MATH. |
math-ph/0007014 | First, note that MATH if MATH, because MATH has this same monotonicity property. Therefore, for any sequence of MATH, MATH is monotonically decreasing and has a sequence-independent, finite limit, which we call MATH, and we note that MATH. To prove the opposite, namely MATH, we shall prove that MATH for every MATH. Let MATH be normalized and such that MATH. We note that MATH, where MATH is the number operator MATH . Thus, if we use MATH as a variational function for MATH we have MATH, and our goal is accomplished provided that MATH can be chosen so that MATH, in addition to MATH. If a way can be found to modify MATH to another vector MATH so that MATH, in addition to MATH the proof will be complete. A suitable choice is MATH, with MATH being the projector onto the subspace of MATH, with MATH or fewer photons, that is, MATH, for an appropriately large MATH. It is easy to see, with the help of REF that MATH, that MATH, and MATH as MATH. The following REF , shows that the other terms converge as well. The same proof works for MATH. |
math-ph/0007014 | Write MATH where MATH contains the annihilation operators and MATH contains the creation operators. We omit MATH and we omit the vector index of MATH for simplicity. Since MATH we learn from REF that MATH and MATH are in MATH. Since MATH strongly, MATH . The same holds if MATH is replaced by MATH and MATH by MATH. This proves the statement for MATH. The statement for MATH is proved in the same way. |
math-ph/0007014 | MATH . |
math-ph/0007014 | Each of the various energies converges as MATH by REF . |
math-ph/0007014 | Let MATH any smooth, bounded function on MATH with bounded first derivative. We easily compute MATH . We use the NAME REF to compute MATH . Now we choose MATH to be MATH and where MATH is a smooth cutoff function that is identically equal to REF outside the ball of radius REF, and identically zero inside the ball of radius REF. We let MATH at the end. Next, we calculate MATH and note that the first and second terms are compactly supported in MATH and each is bounded by a constant MATH that depends on MATH and MATH. Returning to REF , we obtain, after rearranging terms, MATH . Since MATH we know by REF that MATH . In conjunction with REF this shows that MATH for MATH large enough. After letting MATH by monotone convergence a similar bound with MATH replaced by MATH is obtained. |
math-ph/0007014 | This proof is a slight modification of the one in CITE. The basic idea is to show that there is effectively no interaction between localized particles and low momentum photons. To make this idea explicit we write our Hamiltonian in a gauge different from the usual NAME gauge. To be precise, define MATH which is well defined owing to the ultraviolet cutoff. The unitary operator that accomplishes this is MATH. This is an `operator-valued gauge transformation'. It commutes with MATH for all MATH, but not with MATH or with MATH. Define MATH with MATH. The transformed Hamiltonian MATH is MATH . To estimate MATH, write MATH where MATH, and note that MATH . It remains to estimate MATH. By the NAME equation for MATH . While this equation is correct, the derivation is somewhat formal. This is rigorously justified in REF. Now add MATH on both sides. Since MATH is the ground state energy and MATH the operator MATH has a bounded inverse MATH and hence MATH . For consistency, note that for MATH and hence MATH, that is, for these modes MATH is the vacuum as it should be for a minimizer. Since MATH the norm of the last term is bounded by MATH . To bound the norm of the first term we need to estimate MATH . Next estimate the square of the first factor to get MATH where MATH and MATH are independent of MATH. Since MATH the constant MATH is also independent of MATH. Finally the second factor in REF is bounded by MATH. The term containing the NAME matrices in REF is estimated similarly. In conjuction with REF this shows that MATH . This, together with REF , proves the theorem. |
math-ph/0007014 | We differentiate REF with respect to MATH and obtain MATH . The norms of the first and third terms can estimated precisely the same way as in REF , and yields a bound of the form MATH . For the second term, a straightforward calculation shows that MATH . The last term is dealt with in a similar fashion as the previous ones. Using the steps in REF , this leads to the bound MATH . The fourth term can be estimated in the same fashion to yield a similar result. Differentiating REF leads to the same estimate with MATH replaced by MATH. This, together with REF , proves the theorem. |
math-ph/0007014 | The Hamiltonian MATH has a normalized ground state MATH, by REF . Pick a sequence MATH tending to zero and denote the corresponding eigenvectors by MATH. This sequence is a minimizing sequence for MATH by REF . Since MATH is bounded there is a subsequence (call it again MATH) which has a weak limit MATH. Since MATH and by the lower semi-continuity of non-negative quadratic forms (in our case, MATH) MATH . Hence MATH will be a (normalized) ground state if we show that MATH (that is, MATH strongly). It is important to note, however that if we write MATH, where MATH is the MATH-photon component of MATH then it suffices to prove the MATH norm-convergence of each MATH. The reason is the uniform bound on the total average photon number; see the remark after REF which implies MATH . Likewise, it suffices to prove the strong MATH convergence in the bounded domain in which MATH for each finite MATH. The reason for this is the exponential decay given in REF , which is uniform by REF . Finally, by REF MATH vanishes if MATH for some MATH. So it suffices to show MATH convergence for MATH restricted to MATH for each MATH. CASE: For each MATH and MATH we show that MATH restricted to MATH is a bounded sequence in MATH. The key to this bound is REF and MATH where the arguments MATH and the spin indices have been supressed. By the symmetry of MATH, REF , NAME 's inequality and REF MATH independent of MATH. The constant MATH depends on all the parameters, but is finite because MATH in the integration. Similarly, by NAME 's inequality MATH which is uniformly bounded by REF . Since the classical derivative of MATH is not defined in all of MATH one has to check that the weak derivative coincides with the classical derivative a.e. Because of our REF , the classical derivative is not defined along the MATH-axis. One has to show that MATH for any test function MATH. Here MATH is MATH with a MATH cylinder around the MATH-axis removed in each MATH-ball. The first equality is trivial; it is the second equality that has to be checked. This amounts to showing that the boundary term, coming from the integration by parts vanishes in the limit as MATH tends to zero. But this follows immediately from REF . This shows that MATH as a function of all its MATH variables, is in the NAME space MATH and that MATH. Since MATH converges weakly in MATH it converges weakly in MATH and since the sequence is bounded in MATH, MATH converges weakly to MATH. The NAME theorem (see REF ) states that such a sequence converges strongly in MATH if MATH. The boundedness of MATH is crucial here. For our purposes we need MATH, and hence we have to pick MATH such that MATH which is possible for each MATH and MATH. We conclude that MATH strongly in MATH as MATH, for each MATH and MATH. This proves the theorem. |
math-ph/0007014 | Our immediate aim is to compare the field energy MATH with the localized field energy MATH. For simplicity the various indices are supressed and MATH is replaced by MATH. The variable MATH plays no role here. Pick an orthonormal basis MATH of MATH in MATH. States of the form MATH where MATH is finite, form an orthonormal basis in the NAME space. The field energy acts on such states as MATH . Thus, we have that MATH and MATH where the error MATH is given by MATH . Thus MATH is given by the operator (the MATH's are omitted) MATH . The expression for the operator MATH does not look hermitian but it is, remembering that MATH is a partial isometry. Standard estimates lead to MATH where MATH is the number operator. Here MATH denotes the operator norm associated with the kernel MATH. This norm can be estimated using the formula MATH . Recalling the definition of MATH, the operator norm of the first term is easily seen to be bounded by a MATH. Likewise, the second term, using the formula MATH can be estimated by MATH. The term MATH is estimated in a similar fashion. The estimate REF immediately shows that for a general state MATH we have that MATH . Since the photons have a mass we can estimate the number operator in terms of the field energy. The field energy is relatively bounded with respect to the Hamiltonian, that is, MATH for some positive constants MATH and MATH, and thus we obtain MATH . Note that this estimate had nothing to do with the electron, in particular the MATH - space cutoff is not present in the calculation. Next we have to compare MATH with MATH . This time the MATH - space cutoff is important. We would like to estimate the difference MATH . It suffices to treat the first term, the other is similar. It can be easily expressed as MATH where MATH . Using the form boundedness of the kinetic energy with respect to the full Hamiltonian, we have MATH for positive constants MATH and MATH. Thus, using NAME 's inequality it suffices to show that MATH as MATH (the localization radius for the electrons) tends to infinity. Denote by MATH . Explicitly, MATH is given by MATH and it suffices to estimate each of these terms separately. Each of the last two terms can be brought into the form MATH where MATH is one of the functions MATH . The terms REF are estimated by MATH respectively MATH . In both formulas the index MATH is in MATH. The function MATH lives in the region where MATH for MATH and MATH. The function MATH (and likewise MATH) is not zero only if MATH for some MATH. Thus, MATH and MATH are nonzero only if MATH. As MATH gets large only the tail of the function MATH contributes to the integral which can be made as small as we please. The number operator is bounded by the field energy times MATH which in turn is bounded by the full energy. To estimate the first term in REF we calculate MATH . Note that the tensor product in the first line is different from the second. In the first the identity acts on MATH while in the second MATH indicates the tensor product of the NAME spaces only. The functions MATH indicates a basis of MATH. The operators MATH and MATH have unit norm. Thus MATH where MATH indicates that the operator norm has been taken. The norms of the commutators are of the order MATH and hence vanish as MATH. Since the photons have a mass we can estimate the number operator in terms of the field energy. Similar consideration apply to the MATH term in REF . The only difference is that instead of REF we have MATH respectively MATH with MATH. Again this terms tend to zero as MATH. The proof for the case where MATH is similar but simpler since the operator MATH does not depend on MATH. Finally, we have to compare the MATH term with its localized counterparts. The estimates are similar to, but much easier than the estimates for MATH and are omitted for the convenience of the reader and authors who, by now, are exhausted. |
math-ph/0007014 | By the IMS localization formula we have that MATH where the second term goes to REF as MATH. With our assumption on MATH only the sets MATH with MATH contribute. From REF we get that MATH as first MATH then MATH. Certainly MATH and MATH from which the statement immediately follows. |
math-ph/0007014 | Since the energy of MATH is uniformly bounded we also know that MATH is uniformly bounded Let us describe the operator MATH in more detail. Recall that MATH where MATH denotes the vacuum vector in NAME space and MATH. Hence, using the definition of MATH, we find that MATH . The projection MATH annihilates the photons in the second factor. In other words, the operator MATH when acting on a state MATH produces the localized state MATH where MATH . It follows that MATH . Next, we show that REF implies that MATH . To achieve that we note first that on account of the positive mass we have that MATH is uniformly bounded. Since MATH is of the form MATH we know that MATH. It is therefore sufficient to prove REF for each function MATH . From the lemma below we learn that MATH is uniformly bounded. Thus, we can write REF as MATH which vanishes as MATH since MATH is uniformly bounded and since MATH is compact on every finite particle subspace. Compactness follows from the fact that for continuous functions MATH and MATH vanishing at infinity the operator MATH is compact. |
math-ph/0007014 | We write MATH with MATH. There are thus four terms in MATH . Using the NAME inequality, the MATH term can be bounded above by MATH. On the other hand, MATH, where MATH is the commutator MATH; the factor REF comes from the two polarizations MATH. Altogether, we obtain MATH . Finally, we use the NAME inequality again to obtain MATH . In our case, MATH, so MATH. For MATH, replace MATH by MATH and replace MATH by MATH. |
math-ph/0007014 | In addition to REF , use the facts that for any MATH, MATH and MATH. |
math-ph/0007014 | We shall use the abreviation MATH . For our special choice of gauge MATH where we set MATH . Next, pick any MATH in MATH and calculate (recalling the definition of MATH in REF ) MATH with MATH. This extends, using an approximation argument, to all MATH and, in particular, to MATH. Here we note that, on account of REF and the assumption on the potential, MATH implies that MATH. Hence MATH which yields the inequality MATH for all MATH in MATH. Pick MATH of the form MATH where MATH is an orthonormal basis of MATH and MATH a bounded function. Summing over this basis, we get on the left side of REF MATH and on the right side MATH . Hence MATH . The right side can be written as MATH and, applying NAME 's inequality, this is bounded above by MATH . Hence we obtain the bound MATH . Since, MATH we have that MATH. Moreover, MATH is relatively form bounded with respect to MATH. But, as in the proof of exponential decay REF , we have for each MATH and we arrive at the bound MATH where MATH is some constant independent of MATH. Since MATH is arbitrary we obtain for almost every MATH and each MATH that MATH . The right side is bounded by MATH which is finite on account of the exponential decay of MATH. |
math-ph/0007014 | First some notation: For any function MATH define MATH and MATH . Returning to REF with MATH replaced by MATH we have MATH which can be rewritten as MATH . Notice that without the first term on the right of the inequality sign, the structure of this inequality is the same as REF , except that, of course, MATH plays the role of MATH, MATH plays the role of MATH and MATH plays the role of MATH . Thus, without this term we would obtain immediately the estimate analogous to REF , MATH . The remaining term in REF , after summing over the functions MATH, turns into MATH which, by NAME 's inequality, is bounded above by MATH . This, together with REF , yields MATH . Again, since MATH is arbitrary we obtain for every fixed MATH . Combining this with REF we get MATH . The polarization vectors defined in REF , are differentiable away from REF-axis. The same straightforward estimates as in REF lead to MATH which hold for all MATH and small MATH with a constant MATH that is independent of MATH. Next, we observe that for MATH fixed, there exist a sequence of MATH values, say MATH, tending to zero so that MATH converges weakly to some element MATH which satisfies the estimate MATH . Here MATH is the MATH-th canonical basis vector. Next we identify MATH as the weak derivative of MATH. This weak derivative, by definition, can be computed via the expression MATH where MATH is any state in MATH and MATH is any test function in MATH. Clearly the above expression equals MATH . But along the sequence MATH which identifies MATH as the (negative) weak derivative of MATH. |
math-ph/0007020 | We first consider the case when MATH is self-adjoint, that is, MATH. Write MATH with MATH and MATH. Here MATH are the positive and negative parts of MATH, obtainable from the spectral representation of MATH or more explicitly as MATH. Obviously MATH. Since MATH, MATH is self-adjoint. For arbitrary MATH write MATH with MATH, such that MATH are self-adjoint with MATH whenever MATH. More precisely we have the a priori bound MATH. This gives the decomposition MATH and hence by the linearity of MATH with MATH. In particular REF shows that indeed REF holds. |
math-ph/0007020 | We adapt a standard proof (see for example, CITE, p. REF) used in the context of positive maps on non-unital MATH-algebras. We first claim that it suffices to prove boundedness on MATH. Indeed, this follows from the decomposition REF and the related bounds. Assume that MATH is not bounded on MATH. Then there are MATH with MATH and MATH. Let MATH, such that MATH holds for all MATH. We need the fact that MATH holds for any MATH. The case MATH is trivial. When MATH take MATH to be a complete orthonormal basis in MATH diagonalizing MATH. Then we have MATH proving REF . Here we have used the estimate MATH valid for any MATH, any normalized MATH and any MATH (see for example, CITE, p. REF). Since MATH is positive REF gives MATH, which is a contradiction. |
math-ph/0007020 | Assume that there is MATH with MATH, such that MATH for all MATH. If MATH is not positive definite, then we have a contradiction, so let MATH. We claim there is MATH which is not positive definite with MATH. In fact, we may take MATH to be a one-dimensional orthogonal projection onto an eigenvector of MATH times the corresponding eigenvalue, which we choose not to be zero. This gives MATH and we are back to the first situation. |
math-ph/0007020 | Assume that MATH holds such that there is MATH with MATH. Then MATH holds for all selfadjoint MATH due to the bound MATH and therefore MATH for all MATH and all MATH. This in turn implies MATH for all MATH and all MATH. Since MATH by definition, there is MATH with MATH (see the decomposition REF ), so MATH is not ergodic. |
math-ph/0007020 | It suffices to consider MATH, where MATH is any one-dimensional orthogonal projection (compare the proof of REF ). Let MATH be a unit vector, so we have to consider MATH and the claim follows. |
math-ph/0007020 | We use the same notation as in the proof of the previous lemma. In addition write MATH for MATH and set MATH. Then MATH and again the claim follows. |
math-ph/0007020 | The estimate REF for the case MATH is trivial, so let MATH. Since MATH is self-adjoint, there is a complete set orthonormal eigenvectors MATH with real eigenvalues MATH of MATH. With MATH we obtain MATH . Observe that MATH holds, since MATH and MATH. In the last estimate in REF we used the estimate REF . Furthermore MATH holds for all MATH. Also MATH for at least one MATH. Otherwise we would also have MATH for all MATH by NAME inequality and hence MATH, contradicting the assumption. Thus the first inequality in REF is actually strict for at least one MATH. Inserting REF into REF and using this last observation proves REF . |
math-ph/0007020 | Here and in the proof of the next lemma we will mimic and extend REF, page REF in the present context. Set MATH such that MATH are all in MATH. Consider MATH in MATH, which defines an element in MATH. By REF Consider the following linear transformation MATH on MATH (equipped with the natural scalar product) given as the MATH matrix MATH such that for all MATH with the obvious notation for the scalar product in MATH. A linear transformation in MATH is MATH if and only if its trace and its determinant are both MATH. Hence the bound REF follows. This discussion also easily gives the last claim in REF . Indeed, if MATH, its determinant is equal to zero if and only if there is a non-zero eigenvector of MATH with eigenvalue zero and this is the case if and only if equality in REF holds and then such an eigenvector of MATH is given either as MATH or as MATH . More precisely, if all matrix elements of MATH are non-vanishing then these two eigenvectors have non-zero entries and are proportional. If MATH with MATH and if one diagonal element is vanishing, such that also the off diagonal elements are vanishing, then exactly one of these vectors is the null vector. In view of REF this concludes the proof of REF . |
math-ph/0007020 | Choosing MATH in REF we have the estimate MATH for all MATH. Thus REF follows trivially from REF in case MATH. So let MATH. Take MATH to be an orthonormal basis in MATH diagonalizing MATH with eigenvalues MATH. Then we obtain MATH . Here we have used REF for the first inequality. The second inequality is a consequence of NAME inequality and REF . In what follows this will become important, since we will encounter the situation when all inequalities are actually equalities. Also we have used the fact that with MATH being an orthonormal basis also MATH is a possibly incomplete set of orthonormal vectors and which is responsible for the last inequality. |
math-ph/0007020 | We want to prove that MATH is also an eigenvector with eigenvalue MATH, that is, MATH. Since MATH (such that also MATH and hence MATH by the uniqueness of MATH) it suffices to show that MATH holds for all MATH, where the MATH are as in the previous lemma. Also by the previous lemma we have MATH . Here we have used the fact that MATH. By the assumption MATH we must have equality. In particular this implies that MATH . Inspection of the proof of REF shows that we must have equality in REF when MATH for all MATH. The relations REF then imply that we must have MATH and MATH for all MATH with MATH. Assume MATH to be such that MATH. Since MATH we have MATH and MATH for all such MATH. Similarly MATH. Hence we may rewrite REF as MATH and MATH whenever MATH. Furthermore we recall that we made use of NAME inequality in the estimate REF . For this to be an equality we claim that we actually must have MATH for all MATH. Indeed, the NAME inequality in this context is an equality if and only if both sides of REF are proportional with a common proportionality factor for all MATH. By the equality REF this proportionality factor has to be equal to REF, thus proving REF . If MATH then MATH and hence by REF MATH. Thus we have established REF in the case when MATH. If MATH then the left hand sides of REF are both non-vanishing since then MATH. Therefore MATH is an eigenvector of MATH with eigenvalue MATH . This gives REF in the case when MATH, thus concluding the proof of the theorem, since for compact operators the spectrum is pure point. |
math-ph/0007020 | Assume there are two eigenvectors MATH and MATH in MATH with eigenvalue MATH, normalized to MATH. By REF , its proof and by REF we have that MATH is an eigenvector with eigenvalue MATH. In particular both MATH and MATH are unitaries. Since MATH is also an eigenvector for the same eigenvalue for all MATH by the same argument we must have MATH . The function MATH is easily seen to be continuous. Hence there is MATH such that MATH . Take the square of REF and use REF to obtain MATH . Due to the relation MATH this gives MATH which written out in turn gives MATH for all MATH and all MATH. We distinguish two possible cases. First if MATH and hence MATH there is nothing to prove. Otherwise MATH is a unitary operator MATH and hence by the spectral theorem for unitary operators there is MATH such that MATH. Choose MATH such that MATH. This contradicts REF , thus concluding the proof of the first part of the theorem. As for the last part MATH is unitary, as already noted. Relation REF shows that MATH is also an eigenvector of MATH with eigenvalue MATH. Hence MATH by uniqueness and normalization. So MATH commutes with MATH and therefore MATH is normal with MATH and MATH, where we recall that MATH is an eigenvector with eigenvalue MATH. |
math-ph/0007020 | We use familiar arguments. Let the MATH form an orthonormal basis of eigenvectors for MATH arranged such that MATH and MATH. Here the MATH's are the eigenvalues, that is, MATH. Since MATH is a simple eigenvalue, MATH. By NAME 's theorem we have MATH with MATH and MATH. Obviously MATH holds such that MATH and the claim follows. |
math-ph/0007020 | We recall that MATH is the norm limit of MATH as MATH. Since the space of compact operators is closed with respect to the norm topology it suffices to prove that all MATH are compact. By assumption to each MATH there is a sequence MATH of finite rank operators in MATH such that MATH . Since MATH is of finite rank MATH with MATH is also of finite rank and in particular compact. Using the a priori estimate MATH and a standard MATH argument gives MATH . |
math-ph/0007020 | Take the MATH's to form an orthonormal basis in which MATH is diagonal, that is, such that MATH takes the form REF and choose MATH with MATH as above with MATH. Then we have MATH for all MATH, such that MATH is idempotent. Also the kernel of MATH consists of all elements which may be written in the form MATH. Moreover let MATH. Then for given MATH the equation MATH has a solution in MATH of the form MATH . |
math-ph/0007022 | Given that the argument is not that different from the derivation of REF , we permit ourselves to present here just a brief summary. We start by introducing a convenient extension of the space MATH over which the random process MATH is defined. Let MATH and let MATH be a probability measure on MATH having MATH-marginal MATH . Writing the points in MATH as MATH, we extend the translations MATH, originally defined on MATH, into a flow on MATH by: MATH . Notice that on the enlarged space the covariant functional defined by MATH with MATH the second coordinate of MATH, provides an antiderivative of MATH, since MATH . Moreover, if MATH is stationary under the flow MATH, then the random field MATH, on the probability space MATH, is stationary under MATH. This field, however, need not yet provide us with the covariant functional MATH of REF , since it may depend not only of MATH but also on the entire MATH. The proof of REF is in two steps. First, it is established that there indeed exists a probability measure MATH on MATH such that: i. the MATH marginal of MATH is MATH, and ii. MATH is stationary under MATH. Then, using the stationarity - which permits the application of the ergodic theorem - we construct a modified field on MATH which depends only on the first coordinate, MATH. It is this field which yields the desired MATH. To prove the first claim, we let MATH be the probability measure for which MATH has the distribution MATH and the conditional distribution of MATH, given MATH, is that of the random variable MATH with MATH sampled over MATH with the uniform probability distribution ( MATH ). It is easy to see that MATH, where MATH, is close to being invariant under MATH, with the variational distance between MATH and MATH bounded by MATH. It is also easy to see that the distribution of MATH under the measures MATH is tight: MATH . Under the tightness assumption of REF , the right hand side vanishes as MATH. The above considerations suggest that compactness and continuity arguments can be invoked to prove that the sequence of probability measures MATH has a convergent subsequence, as MATH, and that the limiting measure is strictly stationary under the flow MATH. The existence of a suitable limiting measure can indeed be proven using a topology which is natural for MATH in the present set-up (it is at this point that we leave the argument at the level of a sketch), for which REF implies that the sequence of measures MATH is tight, and thus has a convergent subsequence, with limit MATH. Clearly the MATH-marginal of MATH is MATH. That MATH is also stationary would be immediate were MATH continuous on MATH. However, there is here a slight complication in that the second component on the right-hand-side of REF is not continuous in MATH. Nevertheless, the set of configurations MATH for which MATH is discontinuous - those configurations having atoms at MATH or at MATH - has MATH-measure MATH, (an easy consequence of the stationarity of MATH). Using this fact, the conclusion of stationarity follows. The above implies that in the larger space MATH, the process MATH has a stationary antiderivative MATH. While the antiderivative thus obtained need not be determined by MATH alone, its stationarity implies that it has the necessary asymptotic statistical regularity to yield an antiderivative of the field MATH which is a covariant function of MATH alone. This can be done by selecting from among the one-parameter family of possible antiderivatives (differing only by an overall shift) the one whose median value, averaged along the positive MATH-axis, is set at MATH. The median level for MATH is defined by MATH provided the limit exists. (The symbol MATH represents the indicator function.) For MATH, which is sampled with a stationary distribution, the existence of the limit (simultaneously for a countable collection of MATH) follows by the ergodic theorem. Using the median, we define MATH . Because MATH . , MATH depends only on MATH (and not on the other coordinate of MATH), and thus it defines a suitable covariant antiderivative of MATH. |
math-ph/0007022 | It suffices to consider the point-charge processes, since any point process may be regarded as a special case with unit charges. Under the assumption of tight fluctuations, REF says that there is a measurable function MATH of the charge configuration such that MATH . Since the charge MATH assumes only values which are integer multiples of MATH, we may conclude that MATH defines a cyclic factor as described in REF . |
math-ph/0007022 | The argument in the discrete case is similar to the above, with the only notable difference being that the charge MATH now includes MATH in addition to an integer multiple of MATH. Thus in this case the ``antiderivative functional" MATH provided by REF satisfies MATH and hence MATH defines a cyclic factor as described in REF . |
math-ph/0007026 | If MATH is linear and MATH in REF is not zero, then MATH which is inconsistent. |
math-ph/0007029 | Consider the NAME operator defined on MATH by MATH . Since MATH is compact, the spectrum of MATH is discrete and its first (simple) eigenvalue and the corresponding (normalized) eigenfunction are analytic functions of the (real) parameter MATH CITE. We thus expand MATH and the corresponding eigenfunction MATH as a power series of MATH around zero: MATH . On the other hand, we also have that MATH, where MATH . Substituting these expressions in the equation giving the eigenvalues for MATH we obtain, equating like powers in MATH, MATH . From the first equation it follows that MATH and that MATH is constant, which we take to be one. Substituting this in the equation for MATH and integrating over MATH gives that MATH vanishes and MATH satisfies MATH . Substituting now this in the last equation gives that MATH satisfies MATH . Again integrating over MATH gives MATH . Taking squares on both sides of REF we get MATH. On the other hand, multiplying the same equation by MATH and integrating over MATH gives that MATH . Substituting these two expressions into REF we finally obtain MATH and it follows from REF below that MATH is always positive for MATH. To give an example of a function MATH for which MATH becomes negative when MATH it is sufficient to take MATH. We obtain from REF that in this case MATH where MATH is an arbitrary constant. Substituting this into the expression for MATH yields MATH which is negative for MATH. |
math-ph/0007029 | The spectral problem corresponding to MATH is MATH which has discrete spectrum MATH. We will prove that if MATH then MATH for all MATH. To this end rewrite REF as MATH . It is not difficult to see that MATH is an eigenfunction if and only if MATH for some real number MATH different from zero. For MATH the operator MATH is invertible and thus this last equation has one and only one solution given by MATH. Substituting this into REF gives MATH from which the result follows. |
math-ph/0007029 | It only remains to show the result for MATH larger than MATH. Clearly in this case MATH. Consider now the family of potentials given by MATH . Note that although MATH is not continuous on the circle, it can be approximated by continuous functions without affecting our results. For this family of potentials we obtain the functional MATH where MATH is normalized. We now take MATH to obtain MATH and since MATH it follows that MATH can be made to be arbitrarily close to MATH. |
math-ph/0007029 | The first inequality follows directly by writing the eigenvalue problem as MATH and applying the previous corollary with MATH. The second part is a consequence of the fact that for MATH larger than MATH the principal eigenvalue must be larger than MATH. |
math-ph/0007029 | Fix a point MATH in MATH and denote by MATH the geodesic ball centred at MATH with radius MATH. Let now MATH and define the potential MATH (As before, this is discontinuous but can be approximated by continuous functions without changing the results.) By subtracting MATH on both sides of the equation for the eigenvalues, we can, without loss of generality, take MATH to be zero. We are thus lead to the functional MATH . Consider now the auxiliary eigenvalue problem defined by MATH and denote its eigenvalues by MATH, with corresponding normalized eigenfunctions MATH. From the results in CITE we have that MATH . We now build test functions MATH, MATH defined by MATH for which MATH and, by the result from CITE mentioned above, this converges to MATH, MATH as MATH goes to zero. Finally, note that for each MATH the set MATH satisfies the necessary orthogonality conditions, since this is the case for MATH . |
math-ph/0007030 | The linearity and antisymmetric properties are obvious. Two other properties are secured because CASE: MATH commutes with convolutions REF and sends zero function to itself REF ; CASE: The commutator MATH satisfies both to NAME and NAME identity. For example the NAME identity could be verified as follows: MATH where REF follows from the linearity of MATH and REF is a consequence of REF . |
math-ph/0007030 | The proof is a straightforward calculation using REF. We will carry them separately for cases of MATH and MATH. Let MATH, MATH. Then: MATH where the line REF follows from the first case in REF is exactly the first case in REF. The second case MATH (symbolically corresponding toMATH) is also not difficult but somehow longer: MATH . We use the second case of REF to obtain REF. Now let us change variables and continue the above calculations: MATH . Interchanging primed and double primed variables in REF - REF we conclude that the integral is equal to itself with the opposite sign and thus vanish. In contrast such an interchange in the integral REF - REF lead to a continuation of REF - REF: MATH . This finishes the proof. |
math-ph/0007030 | The properties follows from the corresponding properties of MATH-mechanical brackets REF and conservation of algebraic identities by representations REF . |
math-ph/0007030 | It is known (see CITE) that the above four properties are a direct consequence of those from REF . Again the properties are very well known for the quantum commutator and the NAME brackets. |
math-ph/0007030 | To establish first statement one verifies images of MATH under representations MATH REF and MATH REF by a direct calculation. We proceed with a derivation of REF . Let CITE MATH be the left and the right invariant vector fields on MATH correspondingly. They generate the right MATH and the left MATH shifts on MATH correspondingly (left invariant vector fields generate right shifts and vise verse): MATH . Then we could express convolutions REF with MATH as second order differential operators: MATH . Therefore the commutator MATH is MATH . We substitute values from REF in order to obtain REF. Finally the MATH-brackets REF are MATH . Substitution of the last REF into MATH-mechanical REF proves REF. |
math-ph/0007035 | Proof can be found in CITE. |
math-ph/0007035 | It is obvious that for MATH the operator MATH is a contraction. The second inequality will be proven in a more general context in REF. Since MATH are mutually orthogonal invariant spaces of MATH from REF follows that MATH . Proof of the fact that MATH and MATH are adjoint can be found in CITE. Proofs of the other two equations are straightforward. |
math-ph/0007035 | For any polynomial MATH we have MATH therefore approximating the square root by polynomials we obtain MATH. Note that for MATH we have MATH therefore MATH . If in the preceding calculations we replace MATH by MATH (and vice versa) and replace MATH by MATH we obtain the opposite inequality. |
math-ph/0007035 | Proof is straightforward, we shall omit it. |
math-ph/0007035 | Note that an expression is in the mentioned form if and only if it does not contain any subexpression being the left hand side of one of REF - REF . If it does not hold by replacing the left hand side of appropriate equation by the right hand side we obtain an expression (or a sum of expressions) which is either shorter or has the same length but smaller number of disorderings. We can easily see that this procedure has to stop after finite number of iterations. |
math-ph/0007035 | Let MATH be an isometry such that MATH limited to MATH is equal to identity. Such isometry exists because MATH and MATH have the same dimension. We define now an second quantization of MATH, that is, an isometry MATH by the formula MATH. The wanted MATH-isomorphism is MATH . The uniqueness of such isomorphism follows from the fact that it is uniquely defined on a dense subspace MATH . |
math-ph/0007035 | Proof of the first two inequalities can be found in CITE. Now we show the third one. MATH . The last inequality can be proven similarly. |
math-ph/0007035 | The last inequality holds since for each MATH we have MATH . |
math-ph/0007035 | Let MATH where MATH are disjoint intervals. Since MATH is left - adapted for any MATH we have MATH therefore MATH what proves claimed theorem. |
math-ph/0007035 | Proof of an analogue fact can be found in the paper of CITE. |
math-ph/0007035 | Let us consider a NAME space MATH such that there exists an operator MATH which restricted to MATH is a unitary operator MATH and restricted to MATH is equal to MATH. We introduce a process MATH, where MATH denotes the orthogonal projection on the subspace MATH for a set MATH. For any operator MATH we define MATH. Of course this operator is not bounded and therefore any manipulations with it have to be done carefully. REF ensures that MATH on MATH is bounded and its norm equals to MATH. Let MATH, MATH, where MATH. For any measurable set MATH we have MATH . Let MATH where MATH are disjoint intervals. For different values of MATH operators MATH have mutually orthogonal images and cokernels. MATH where in the last equality we used REF and in the last but one equality we used that second quantization MATH of unitary operator MATH defined as MATH is again unitary and MATH . Hence MATH . |
math-ph/0007035 | It is enough to prove that for each MATH and all vectors MATH . Let MATH. We have MATH and the operator norm of the second summand does not exceed MATH. We shall use the notation introduced in proof of REF . It follows from the fact that MATH has measure MATH that MATH for any fixed vector MATH. If we rewrite REF replacing MATH by MATH and MATH by MATH we see by majorized convergence theorem that the first summand in REF tends strongly to MATH. |
math-ph/0007035 | MATH because MATH and MATH. |
math-ph/0007035 | It is enough to notice that MATH and recall the definition of the norm MATH and REF . |
math-ph/0007035 | For a unital vector MATH orthogonal to MATH we have (see REF ) MATH and therefore for some constants MATH which depend only on MATH and MATH we have MATH . The second part of the lemma follows from the majorized convergence theorem. |
math-ph/0007035 | For MATH functions MATH uniformly tend to MATH. Therefore by preceding lemmas appropriate norms of biprocesses MATH tend to MATH what proves that the limit in the first equation holds in the operator norm. For MATH and MATH for each MATH functions MATH by REF uniformly tend to MATH and REF shows that the limit in the first equation holds in the strong operator topology. For MATH and MATH REF assure that the limit in the second equation holds in the strong operator topology. For MATH, MATH we introduce a NAME space MATH such that there exist operators MATH, MATH which restricted to MATH are equal to MATH and which map isometrically MATH respectively onto MATH and MATH. In the following for MATH and MATH will denote the orthogonal projection respectively onto MATH and MATH, furthermore MATH and MATH. We have MATH . Note that even though MATH, MATH are unbounded operators, the right - hand side of this equation is well defined on domain MATH (see proof of REF ). For each MATH we consider a sequence MATH and a bounded operator MATH defined as MATH where MATH. It is easy to see that the sequence MATH strongly tends to MATH and since MATH then REF strongly tends to MATH, what proves the first equation. All the other cases we obtain by taking adjoint of considered cases. |
math-ph/0007035 | For MATH let MATH be simple adapted biprocesses. We assume that intervals MATH form a partition, that is, that they are disjoint. Note that we can replace partition MATH by a refined partition MATH so that MATH. MATH . The second and the third summands tend by REF to the second and the third summands of the right hand side of REF . We shall find the weak limit of the first summand when the grid of the partition tends to MATH. If MATH or MATH then it is easy to see that the term REF tends strongly to MATH. If MATH REF gives MATH what tends to MATH as the grid of the partition MATH tends to MATH. By taking the adjoint we see that if MATH then the term REF weakly tends to MATH as the grid of the partition MATH tends to MATH. The cases we have already considered show that REF holds. If MATH and MATH then we can split the normally ordered form of the expression MATH into two parts: the first which does not contain operators MATH, MATH and is equal to MATH and the second, which contains these operators in this order. The sum over MATH of the second part tends in operator norm to MATH as the grid of the partition tend to MATH because it is in form REF for MATH and MATH, what proves REF . If MATH and MATH then we can split the normally ordered form of the expression MATH into two parts: one equal to MATH and the second part which contains operators MATH, MATH in this order. The sum over MATH of the second part is in the form REF with MATH and MATH therefore tends strongly to MATH as the grid of partition tends to MATH, what proves REF . By taking the adjoint of REF we obtain REF , that is, the case MATH, MATH. If MATH, MATH we introduce a NAME space MATH such that there exist operators MATH, MATH which restricted to MATH are equal to MATH and which map isometrically MATH respectively onto MATH and MATH. MATH . It is easy to see that as the grid of the partition MATH tends to MATH that the operators MATH strongly tend to MATH, therefore the first summand strongly tends to MATH, what proves REF . Now it is enough to notice that any biprocesses can be approximated by simple biprocesses. |
math-ph/0007042 | It follows easily from the relations REF , taking into account REF . Indeed the matrix NAME REF is linearizable via the position MATH which yields for MATH the linear ODE MATH . Hence, if REF matrices MATH are known, the matrix MATH can be evaluated, via REF , by solving the linear (generally nonautonomous) matrix ODE REF . Then the MATH matrices MATH can be obtained from the, assumedly known, MATH matrices MATH via REF . The first part of the Lemma is thereby proven. The second part of the Lemma is implied by the formulas MATH where MATH stands for MATH or MATH , and correspondingly MATH stands for MATH or MATH, and MATH which, via REF , clearly entail the equivalence of REF . As for the validity of these formulas, REF , they are an immediate consequence of REF and of REF , as can be easily verified by time-differentiating them and using REF . The Lemma is thereby proven. |
math-ph/0007042 | The second Proposition of the Lemma entails that the solutions of REF can be obtained by first solving the ODEs REF to determine the MATH matrices MATH and by then obtaining from these, according to the first Proposition of the Lemma, the MATH matrices MATH. |
math/0007005 | First, recall from the NAME theorem that the decomposition MATH is given by MATH . Now fix MATH and endow a vector space MATH with the flag module structure defined by MATH . If we replace MATH by MATH for some MATH, the expression for MATH is just multiplied by MATH, so up to isomorphism (of graded modules), the flag module thus obtained only depends on the class MATH. Conversely, assume that MATH is a flag module of MATH, and choose a graded basis MATH of MATH. For each MATH, let MATH be defined by MATH . Since the algebra MATH is commutative, we have MATH for every MATH, MATH, hence MATH . It follows that MATH is a multiple of MATH, say MATH. Inserting back into REF yields MATH, for all MATH. Since MATH, we get MATH for all MATH. Therefore, MATH . The collection MATH defines a linear form MATH on MATH. Furthermore, we have MATH for all MATH, MATH, so MATH, which shows that the linear form MATH is a character on MATH, corresponding to a point MATH of the affine variety MATH. Moreover, since MATH is cyclic, each MATH must be nonzero, so MATH actually lies in MATH, say MATH. This yields an element MATH. It is clear that these two constructions are inverse to each other. |
math/0007005 | It remains to show that the above construction can be reversed, so assume that MATH is a collection of points in MATH satisfying REF . Choose a (nonzero) representative MATH for each MATH, and endow a vector space MATH with the flag module structure defined by REF . By REF , this rule is compatible with relations of type REF in MATH. By REF , it is also compatible with relations of type REF , provided that, for each MATH and each MATH, we suitably rescale one of MATH, MATH, MATH, MATH. Proceeding by induction over the height MATH, we may perform this rescaling in a consistent way. It is clear that the two constructions are inverse to each other. |
math/0007005 | First, let MATH be monogressive and MATH-effective. If a point MATH has homogeneous coordinates MATH (see REF ), then by MATH-effectiveness, the coordinates of MATH are MATH. Using REF, we therefore see that MATH sends MATH to the following point in MATH: MATH . Using the change of ``variables" MATH (symmetric difference), MATH, and MATH, this expression may be rewritten as MATH where MATH is shorthand for the set MATH. By induction over MATH, we may assume that MATH for all monogressive MATH-effective orthocells MATH of rank smaller than MATH (the case MATH being trivial). Since the above sum is in MATH by definition, the only remaining term, namely MATH, is in MATH as well. We still need to show that the image of a point MATH is in the span of the MATH (for MATH monogressive and MATH-effective) even if MATH is not MATH-effective (but still monogressive). Reordering the MATH if necessary, we may assume that, for some MATH, they satisfy MATH . Let again MATH be homogeneous coordinates for a point MATH. This time, the coordinates for MATH are obtained by multiplying MATH by MATH only for MATH. Furthermore, we have MATH and a similar expression for MATH, with MATH replaced by MATH. A computation similar to the one above shows that MATH now sends MATH to MATH (where we have used the fact that MATH and MATH). |
math/0007005 | Consider a monogressive MATH-effective cell MATH. For each MATH, write again MATH, MATH (hence MATH by monogressivity). Reorder the MATH in such a way that MATH. By MATH-effectiveness, each MATH must appear in the subarray MATH, and each MATH in the subarray MATH. Now let MATH, and note that for each MATH, replacing MATH by MATH in MATH leaves MATH invariant up to a sign. Choosing MATH appropriately, we may assume that MATH takes the following form: MATH with, say, the following orderings: MATH . This rearrangement does not affect the monogressivity of MATH (nor, for that matter, its MATH-effectiveness). Indeed, the only nonobvious point here is the relative ordering of the MATH and the MATH: by monogressivity, we have MATH and MATH, so if MATH, then, whatever the order in which MATH appear in the original array MATH, we must have either MATH, or MATH. In both cases, MATH and MATH are outside of the (numerical) interval MATH, so they may indeed appear between MATH and MATH in the new array without affecting monogressivity. An orthocell thus modified will be called MATH-normal. The proof will be finished if we show that there are MATH-normal orthocells in MATH. Since we have the recursion rule MATH it is enough to show that the number of MATH-normal orthocells in MATH satisfies the same recursion rule. If MATH is such an orthocell, there are two possibilities. CASE: Either each MATH fixes MATH. Removing MATH from the array MATH, we then obtain an orthocell in MATH, which is MATH-normal, MATH-normal, or MATH-normal, according to the position of MATH in the array, relative to MATH and MATH. CASE: Or some MATH involves MATH: necessarily, MATH and MATH. Removing again MATH from the array, and discarding MATH from MATH, we then obtain a MATH-normal orthocell in MATH. Clearly, this procedure may be reversed, starting from a normal orthocell in MATH and inserting MATH at all possible places in the corresponding array. Hence the desired recursion rule. |
math/0007005 | The vector MATH is a highest weight vector, of weight MATH. Now apply REF , noting that the dimension of the simple module of highest weight MATH is precisely MATH. |
math/0007005 | This is immediate from the action of the generators of MATH on the basis elements of MATH and MATH, as described in REF: the formulas obtained there are symmetric in MATH and MATH. |
math/0007005 | The first statement amounts to the commutativity of REF , which immediately follows from REF. For the second statement, consider the composite map MATH corresponding to the line bundle MATH, and denote by MATH the linear span of the image of this map. Claim A. The subspace MATH is contained in the unique simple MATH-submodule MATH of MATH of highest weight MATH. Indeed, let MATH be the kernel of a projection MATH, and define similarly MATH, MATH. Since MATH is quadratic (compare CITE), we have MATH, so dually, MATH, and MATH is clearly contained in the right hand side. This shows Claim A. The proof will be finished if we show the following Claim (from which REF follows): Claim B. Consider the maps MATH and MATH from MATH to MATH. Their restrictions to MATH agree. By Claim A, it will be enough to show that the restrictions to MATH agree. Since both maps are morphisms between the simple MATH-modules MATH and MATH, they must be equal up to a constant. But they both send the (highest weight) vector MATH to MATH, so this constant is equal to MATH. This shows Claim B. |
math/0007006 | For all MATH, let MATH and MATH. We need to show that MATH. By induction over MATH, we may assume that the left hand side is equal to MATH, so we must show that MATH. We decompose MATH, where MATH and MATH. Orthogonality of MATH already implies that for each MATH, MATH commutes with all elements of MATH and of MATH. Since MATH for each MATH, it follows that MATH. To study the remaining factor MATH, we consider the subgroup MATH (where MATH denotes the trivial group whenever MATH is not a root). Fix MATH. Let MATH for some MATH, MATH. For all MATH and all MATH, a well known commutation rule (see CITE) implies that MATH . Similarly, MATH for all MATH. Using again that MATH, it follows that MATH. In particular, MATH. To complete the proof, it remains to show that MATH, or, in other words, that MATH cannot be a negative root if MATH. Write MATH for all MATH in the root lattice. By orthogonality, we have MATH so there are two cases: CASE: if MATH is long (in its component), then MATH cannot be a root (except for MATH itself); CASE: if MATH is short, then MATH can only be a root if it is of the form MATH for some MATH. But by assumption, MATH, so MATH and MATH are either positive roots (when MATH) or nonroots (when MATH are incomparable). |
math/0007006 | Use REF and the fact that the MATH are complete. |
math/0007006 | Let MATH be another nonincreasing numbering of the elements of MATH. For any incomparable roots MATH, the commutator MATH is trivial, so it is enough to show that the sequence MATH can be rearranged into MATH by successively swapping adjacent pairs of incomparable roots. We have MATH for some MATH, and by assumption, this element must be maximal in MATH. Therefore, MATH is incomparable with each of MATH, so we may move it past these roots to the beginning of the sequence. Now we have got two sequences with a common first term; discarding it, we may reapply the same procedure inductively. |
math/0007006 | First, note that if MATH is MATH-stable, then for every MATH, MATH is also MATH-stable; we may therefore restrict the proof to the case where MATH. Since for each root MATH, the quotient MATH is a projective line, the last statement follows from the first one and from REF . Now choose a representative MATH for the reflection MATH and recall that the NAME decomposition MATH induces an open covering MATH of MATH by two affine lines. Taking products, we get an open covering of MATH by MATH affine sets. More precisely, for each subset MATH, we define MATH where MATH . The set MATH is dense in MATH, so its image MATH is also dense in MATH. Since MATH normalizes each root group MATH, it follows that MATH is MATH-stable. It remains to show that MATH is an embedding. Since the variety MATH is complete, it is enough to show that CASE: the restriction of MATH to each open affine set MATH is an embedding, and CASE: MATH is injective. REF holds for MATH: indeed, the multiplication map MATH is well known to be an embedding (see for example, CITE), and so is the canonical map MATH (by the NAME decomposition of MATH). To show REF for an arbitrary subset MATH, we factor the restriction of MATH to MATH through an isomorphism MATH, as follows. Consider an element MATH. Recall that for any two orthogonal roots MATH, MATH normalizes MATH (see for example, CITE), so we may rewrite MATH for some MATH. It is clear that the map MATH is an isomorphism, and by construction, the restriction of MATH to MATH is equal to the composition MATH, where the last arrow is (left) multiplication by MATH. This shows REF . Finally, we must check REF , so let MATH and MATH be two points in MATH. Let MATH be maximal (with respect to set inclusion) such that MATH: then for each MATH, MATH with MATH, and for each MATH, we must have MATH. It follows that MATH for some MATH, MATH, MATH. Therefore, MATH lies in the NAME cell MATH. Similarly, if MATH is maximal such that MATH, then MATH. Now assume MATH; there are two cases: CASE: if MATH, then MATH because they lie in different NAME cells of MATH; CASE: if MATH, then MATH by REF . This shows REF . |
math/0007006 | The last part follows from MATH-stability (see REF ). For the first part, recall again CITE that if MATH are orthogonal roots, then MATH; since MATH, we then also have MATH. Now use this fact in the definition of MATH. |
math/0007006 | For each MATH, choose an isomorphism MATH such that MATH for all MATH and all MATH CITE. Consider first the case where MATH (that is, MATH). The set of all MATH, MATH, is an open dense subset of MATH (compare the proof of REF ), and we have MATH . By orthogonality, the MATH are linearly independent, hence the morphism MATH is surjective. Therefore, by continuity, this morphism turns the MATH-action on MATH into the natural componentwise action of MATH on MATH. The orbit structure is now clear. The general case MATH is obtained similarly, by multiplying MATH by MATH. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.