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PQM Supplementary Notes: Spin, topology, SU(2) SO(3) etc
- Jody Warren
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1 PQM Supplementary Notes: Spin, topology, SU(2) SO(3) etc (away from the syllabus, but interesting material, based on notes by Dr J.M. Evans) 1 Rotations and Non-contractible Loops Rotations in R 3 can be thought of as the group of real special orthogonal matrices Copyright 24 University of Cambridge. Not to be quoted or reproduced without permission. SO(3) = {R 3 3 real matrix : R T = R 1, detr = 1}. Each rotation can also be specified geometrically by an axis n (a unit vector) and an angle θ which can be combined ξ = θn. If we were to choose 0 θ π, we would have a parametrization which is almost one-to-one it is just slightly redundant in that rotations with axes ±n and θ = π are the same. Thus, we can picture the space of all rotations as a solid, three-dimensional ball ξ π, but with opposite or antipodal points on the boundary identified. As θ becomes larger than π, it appears on the other side of the ball. Thus the radius of position in the ball gives the rotation angle, minus some integer factor of π. This description of the rotation group reveals an important topological property: there are loops, or smooth closed paths, which cannot be smoothly shrunk to a point. Consider paths which start and end at the identity rotation, i.e. the centre of the sphere. Figure 1a shows a loop which can obviously be shrunk to a point it is contractible. Figure 1b shows a loop for which the angle of rotation starts at zero, smoothly increases to π (at the point A where it reaches the boundary and reappears at the antipodal point A ), and then continues to increase to 2π, back at the identity. It is clear intuitively that the loop in figure 1b cannot be shrunk to a point while keeping its ends fixed it is non-contractible. (a) (b) A A Figure 1: Examples of (a) a contractible loop, (b) a non-contractible loop. The ends of the loop at the origin are fixed. Now consider the loop in figure 2a, along which the angle of rotation increases from 0 to 4π, requiring the boundary to be crossed twice, once at A, reappearing at A, then again Mathematical Tripos II: PQM Supplementary Notes 1 Michaelmas 24
2 at B, reappearing at B. By moving the second antipodal pair B and B, as shown in figures 2b,c, the section of the path between A and B can be pulled across the boundary, smoothly deforming it to the situation in figure 1a. Thus after two complete rotations the situation becomes simple once more. In general, the angle along any closed path must increase by an integer multiple of 2π, and the loop will be contractible if this integer is even but non-contractible if it is odd. Copyright 24 University of Cambridge. Not to be quoted or reproduced without permission. (a) (b) (c) B A A B A B Figure 2: A contractible loop, as shown by the continuous deformations of (a) to (c). The ends of the loop at the origin are fixed. Why should any of this be important for Quantum Mechanics? We are used to thinking that a rotation by 2π is precisely the same as no rotation at all, but we have seen above that this is not quite right if the rotations arise as part of a smooth family. The structure of Quantum Mechanics is sensitive to this and is capable of keeping track of whether the number of complete rotations we perform is even or odd. To understand how, we must first consider symmetries in a little more generality. B A Symmetries and Projective Representations Symmetries in QM are represented by unitary operators: given a group G there is a unitary U(g) for each g G. There is a freedom or ambiguity in how these operators are defined, however, because states in QM are physically equivalent if they are multiples of one another, so a change U(g) α(g)u(g) (1) where α(g) is any complex number of unit modulus, does not affect the physical interpretation of the operator. This freedom is often sufficient to allow us to choose operators which obey U(g 1 )U(g 2 ) = U(g 1 g 2 ) for all g 1, g 2 G which is a representation of G on the states. More generally, though, U(g 1 )U(g 2 ) = ω(g 1,g 2 )U(g 1 g 2 ) for all g 1, g 2 G where the complex number ω(g 1,g 2 ) is called a cocycle; this is a projective representation on states. Mathematical Tripos II: PQM Supplementary Notes 2 Michaelmas 24
3 For G a continuous group actually smooth in an appropriate sense, a Lie group we would also like U(g) to depend smoothly on g. The topological properties of paths in G then become important when we consider the behaviour of U(g) as the group element g varies around a loop l beginning and ending at some fixed g 0. It is quite possible for U(g) to change smoothly with g along l in such a way that the operators at each end of the loop do not coincide: consistent with the ambiguity (1), they may differ by a change U(g 0 ) α l U(g 0 ) (2) Copyright 24 University of Cambridge. Not to be quoted or reproduced without permission. which depends l. It would obviously be simplest if we could arrange to have α l = 1 for all loops, and to investigate this we specialise from here on to the case in which the operators act on a finite-dimensional space of states. Let N be the dimension of the space of states, so that each U(g) can be regarded as an N N matrix. We can use up some of the freedom in (1) to impose the condition that each U(g) has unit determinant. This does not fix things completely, but the residual freedom in (1) or (2) is discrete: detu(g) = 1 α N = 1 (3) Notice too that once the determinant condition is imposed, any cocycle must also obey ω N = 1. We will assume (3) holds henceforth. If G is simply-connected, meaning that any closed loop is contractible, then the determinant condition implies that α l = 1 for all l. This follows straightforwardly using continuity: α l is restricted to a discrete set of values by (3) and it must therefore remain unchanged under any smooth variation in l. If, in addition, the loop l can be shrunk to the single point g 0, then we must have the same matrix at the beginning and end of the loop, so α l = 1. If G is not simply connected, the same argument still shows that α l = 1 for any contractible l, but there may also be non-contractible loops with α l 1. In this case, we can at least deduce, in a similar fashion, that α l = α l whenever the loops l and l are in the same topological class, meaning that they can be smoothly deformed into one another. Furthermore, we observe that if loops l and l are traversed successively, then the change in U(g 0 ) is α l α l, so there are self-consistency constraints. For a simply-connected group G, we can now give a unique definition of U(g) for all g, by using (3). Setting U(1 G ) = 1, define U(g 0 ) by choosing any path from the identity 1 G to g 0 and demanding that U(g) changes smoothly along it. The values along the path are unique (by the determinant condition and continuity) but the end result U(g 0 ) is also independent of the path, because any two paths can be combined to form a closed loop at g 0 (traversing one path backwards and then the other forwards) which is contractible. With this definition, continuity also ensures that all cocycles are equal to one, so that we have a genuine representation of G. Carrying out the same construction when G is not simply connected, we will encounter paths from 1 to g 0 which cannot be smoothly deformed into one another and, starting from U(1 G ) = 1, we will in general obtain different values for U(g 0 ) depending on the path we take. In this way we are forced to consider multi-valued functions on G, just as when defining a continuous square-root in the complex plane. The ambiguity or multi-valuedness in U(g 0 ) can be resolved only by keeping track of the path used to reach g 0, just as for the complex square root we must keep track of how many times we have encircled the origin to be sure which branch we are on. Such a multi-valued definition inevitably means that non-trivial cocycles appear. Mathematical Tripos II: PQM Supplementary Notes 3 Michaelmas 24
4 We are at last in a position to understand rotations properly. The group G = SO(3) is not simply connected, but there are just two topological classes of loops, depending on whether the net angle of rotation is an even or odd multiple of 2π. Any loop l in the first class is contractible, and so has α l = 1. Any loop in the second class is non-contractible, but if we traverse l twice the result becomes contractible again, and so α 2 l = 1, or α l = ±1. The finite-dimensional spaces on which the rotation operators act are nothing but the multiplets of angular momentum j with basis states { j m : m = j,j 1,..., j }, dimension N = 2j+1, for j = 0, 1 2, 1, 3 2,.... Copyright 24 University of Cambridge. Not to be quoted or reproduced without permission. From (3) we know that αl N = 1 but from above α l = ±1. If j is an integer then N is odd and it follows that α l = 1 for all loops, whether they are contractible or not. In this case we have a genuine representation of SO(3) on the states. If j is a half-integer then N is even and α l = 1 is possible for non-contractible loops. This is exactly the behaviour found by the explicit calculation in lectures for j = 1 states: under a continuous family of 2 rotations through a net angle which is an odd multiple of 2π the states change sign. The discussion above shows that the origin of this unexpected sign is the non-trivial topology of the rotation group in our new terminology, we have a projective representation of SO(3). 3 The Double Cover SU(2) SO(3) It is not too difficult to live with the multi-valuedness of U(g) for half integral spin representations of SO(3), since the ambiguity is just a sign ±1 in either the operators or the cocycles. Nevertheless it is natural to ask if we can do better. Returning to symmetries in general, one way of getting around the multi-valuedness of projective representations is to view this as a problem not with the representation but rather with the underlying group. From this standpoint, the true symmetry at work is not exactly G but something closely related to it, the covering group G. Technically, we require a homomorphism G G which is onto and many-to-one, but in just such a way that the multi-valued or projective representations of G descend from genuine, single-valued representations of G. Pursuing the analogy with the complex square root, this is like saying that the problem is not with the form of the function, as such, but rather with the domain of its definition. The square root becomes single-valued on a new domain obtained by gluing together two copies of the cut plane, and there is a two-to-one map from the new set to old. For the rotation group G = SO(3), the simply-connected covering group is G = SU(2), a special unitary group defined by SU(2) = { U 2 2 complex matrix : U = U 1, detu = 1}. There is a double cover, or two-to-one, surjective homomorphism Φ : SU(2) SO(3), ker Φ = Z 2 = {±1} which we will now construct with the help of the Pauli matrices. Let H denote the set of traceless, hermitian, 2 2 complex matrices, which is a real vector space with the standard definitions of addition and scalar multiplication of matrices. Given any X H and any U SU(2) we note that UXU is also hermitian and tr(uxu ) = Mathematical Tripos II: PQM Supplementary Notes 4 Michaelmas 24
5 tr(x) = 0, by cyclicity of the trace, so each U defines a map H H. This map is clearly linear, and has the property X UXU, Y UY U, tr(xy ) tr(uxu UY U ) = tr(xy ). (4) Now recall that the Pauli matrices σ = (σ 1,σ 2,σ 3 ) each belong to H and that they obey σ i σ j = δ ij + iǫ ijk σ k, tr(σ i σ j ) = 2δ ij. (5) The Pauli matrices provide a convenient basis which we can use to identify H R 3 by associating to any vector x R 3 a matrix X = x σ H, or vice versa. This one-to-one correspondence is an isomorphism of vector spaces, and from (5), Copyright 24 University of Cambridge. Not to be quoted or reproduced without permission. x X = x σ, y Y = y σ x y = 1tr(x σ y σ) = 1 tr(xy ), (6) 2 2 so there is also a natural way to carry over the standard inner-product from R 3 to H. With the identifications we have just made, a linear map H H defined by some U SU(2) corresponds to a linear map R 3 R 3 given by some real 3 3 matrix R: x Rx where x σ Ux σu = (Rx) σ for all x. (7) But by comparing (4) and (6) we see that (Rx) (Ry) = x y, so R is orthogonal. Furthermore, Tr{ (Re 1 ) σ (Re 2 ) σ (Re 3 ) σ } = Tr(Uσ 1 U Uσ 2 U Uσ 3 U ) = Tr(σ 1 σ 2 σ 3 ) implies, using (5), that ǫ ijk R i1 R j2 R k3 = detr = 1, so R SO(3). The double cover is the map Φ : SU(2) SO(3), Φ(U) = R (8) with U and R related by (7). It is easily checked to be a group homomorphism with the kernel {±1}, as claimed above. Finally, we return to spin half in Quantum Mechanics. The motivation for considering the covering group was to eliminate the need for multi-valued or projective representations. We know (from lectures and example sheets) that the unitary rotation operator for spin half states is easily calculated: U(θn) = exp( i 1 2 θn σ ) = cos 1 2 θ in σ sin 1 2 θ. (9) For a rotation R with axis n and angle θ, this formula gives, explicitly, an element U of SU(2) with Φ(U) = R. If we change θ by 2π then R is unchanged and U changes sign. But this is of course consistent with Φ( U) = R, the map Φ being two-to-one. Thus, the example which first led us to observe the multi-valued nature of the rotation operators on half-integral spin states is nothing but the simplest (or defining) representation for SU(2), the covering group of SO(3). Scattering and Cross-Sections from the Golden Rule (an alternative derivation of some results in Part II AQM) A scattering process involves a particle, or beam of particles, being sent towards a target, interacting with it, and then emerging along some new direction. The aim is to calculate the probability for scattering in a given range of final directions, and to understand how Mathematical Tripos II: PQM Supplementary Notes 5 Michaelmas 24
6 this depends on the energy of the incident particle or beam. Applications are wide-ranging: from α-particles incident on atoms, as in the original Rutherford scattering experiment, to X-rays or neutrons diffracting from crystals, scattering processes have provided crucial evidence concerning the nature of the targets, e.g. the existence of the atomic nucleus, or the lattice structure of crystalline solids. We can, in principle, consider an initial wave packet, localised in space, which is fired in towards a target and which scatters off it by evolving according to the time-dependent Schrödinger equation. In general, this approach is far too complicated to be tractable, however, and we deal instead with an idealisation where the initial and final states, a = ψ k and b = ψ k, are eigenstates of the free particle Hamiltonian H 0, with definite momenta hk and hk respectively. The full Hamiltonian Copyright 24 University of Cambridge. Not to be quoted or reproduced without permission. H 0 + V with H 0 = 1 2m p2 incorporates the interaction with the target through the potential V which we assume is independent of time and falls off rapidly V (x) 0 as x. Treating the potential by first-order time-dependent perturbation theory, the transition rate from a to b is given by Fermi s Golden Rule: 2π ρ(e) b V a 2 h where ρ(e) is the density of final states (with respect to energy) with certain specified parameters, and E = E a = E b, or k = k in our case. Let us choose coordinates with the z-axis along the direction of the incident particle or beam, so k = kẑ. Outgoing particle states have k = kn for some unit vector n, and we will consider scattering into a small solid angle δω, corresponding to a small variation in the direction of n. The number of such final states with energy in the range E to E +δe is, by definition, ρ(e,δω)δe. We expect this to be proportional to δω and this will be confirmed shortly. In addition, we expect the transition rate to our specified set of final states to be proportional to the incident particle flux, as calculated from the initial state. In the Golden Rule, the initial and final states must be normalised, and we also need to be able to count states to determine ρ(e,δω). It is very convenient, therefore, to put the entire system in a large cubical box with sides of length L, parallel to the coordinate axes, so that the momenta and eigenvalues of H 0 become discrete (we could think of this box as the laboratory). Rather than demanding that the wavefunctions vanish on the faces of the box, however, it is simpler to impose periodic boundary conditions, which imply that the allowed eigenstates of H 0 and their eigenvalues are ψ κ (x) = 1 3/2eiκ x with E κ = h2 κ 2 L 2m It is easy to check box ψ κ (x) ψ κ (x)d 3 x = where κ = 2π L (N 1,N 2,N 3 ) for N i Z. { 1 if κ = κ 0 else Our final results must be independent of the choice of boundary conditions and of the size of the box, which we will eventually remove by taking L. With the values for κ permitted above, the number of states lying within a certain volume in κ-space is approximately (L/2π) 3 times the volume, and this approximation becomes Mathematical Tripos II: PQM Supplementary Notes 6 Michaelmas 24
7 Copyright 24 University of Cambridge. Not to be quoted or reproduced without permission. better and better the larger L becomes. We are interested in final states with κ = k = kn, the direction n constrained within δω, and the size of the wave vector in the range k to k+δk, corresponding to a range of energies E to E+δE. The volume in κ-space is therefore k 2 δkδω and so the number of states is which determines the density ρ(e,δω). ( L ) 3k 2 δk δω = ( L ) 3mk 2 δe δω = ρ(e,δω)δe, 2π 2π h The Golden Rule also requires us to calculate ψ k V ψ k = ψ k (x) V (x)ψ k (x)d 3 x = 1 L 3 V (x)e iq x d 3 x = 1 L 3 Ṽ (q), where hq = hk hk is the momentum transfer. Plugging this matrix element and the density into the formula, we get the rate for scattering into solid angle δω at energy E: 2π h ρ(e,δω) ψ k V ψ k 2 = 1 mk (2π) 2 h 3 L Ṽ 3 (q) 2 δω. The last step is to divide this rate by the particle flux j = hk/ml 3 corresponding to the initial wavefunction ψ k (x). The result is the differential cross-section δσ for scattering into the solid angle δω at energy E and is usually written, in the limit δω 0, dσ dω = ( m ) 2 2π h 2 Ṽ (q) 2, q = k k. ( ) With the rate correctly normalised relative to the incident flux, the resulting cross-section is independent of L and we can remove the box by taking L. The formula ( ), giving the cross-section to leading order in the potential V, is known as the Born approximation for scattering. In applying the Golden Rule we have, of course, used first-order time-dependent perturbation theory. There are other ways of getting the same result for instance, by modelling the scattering process as a stationary state of the entire Hamiltonian with an appropriate asymptotic form for the wavefunction as x (see Part II Applications of Quantum Mechanics). One nice feature of the derivation given here is that it is fairly easily adapted to cases in which the scatterer, i.e. the source of the potential V, also undergoes a change of state, e.g. a particle scattering off an atom and causing a transition. An important application of ( ) is to scattering in the screened Coulomb potential V (x) = λ 4π 1 x e µ x Ṽ (q) = λ q 2 + µ 2. The exponential factor is convenient mathematically since it allows a sound calculation of the Fourier transform (it ensures that V tends to zero sufficiently rapidly at large distances). It can also be motivated physically, e.g. if we are considering scattering off an atomic nucleus, it gives a way of representing the screening effects of the surrounding electrons (which should indeed become negligible for x very small, where the nucleus dominates). If θ and φ are the usual polar angles (with our choice of z-axis along the direction of the incident beam) then a simple calculation reveals ( hq) 2 = 8mE sin 2 1 θ, and the scattering is independent of 2 Mathematical Tripos II: PQM Supplementary Notes 7 Michaelmas 24
8 Copyright 24 University of Cambridge. Not to be quoted or reproduced without permission. φ. This is to be expected for a spherically symmetric potential, since the scattering problem is then axis symmetric about the direction of the incident beam. To study scattering in a pure Coulomb potential we take µ = 0 to obtain the Rutherford cross-section dσ dω = λ 2 (16πE sin θ)2 This is the theoretical basis for understanding the scattering experiments which first proved the existence of a compact, massive nucleus within the atom (α-particles, or Helium nuclei, incident on a thin metal film). There is an historical twist here, however, because the result above is independent of h, reflecting the fact that this cross-section can be obtained by a purely classical scattering calculation for the Coulomb potential (this is essentially because when µ = 0 the answer is severely restricted by dimensional analysis). So Rutherford was able to get the right answer using Classical Mechanics but then the concept of the nuclear atom which he introduced led to the development of Quantum Mechanics by Bohr, Heisenberg, Schrödinger, Dirac and others. Mathematical Tripos II: PQM Supplementary Notes 8 Michaelmas 24
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Vogue 7731; 1960s; Misses' One-Piece Jumper or Dress. Fitted jumper or dress has cut-away armholes and jewel neckline with front slit. Skirt has front and back darts and gathers. Pockets in side seams. Purchased belts and blouse.
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NBA, Vimeo, BBC and many OTT platforms join our adventure into the digital revolution
Organizations seeking to sell digital content to online viewers have a number of challenges. It must be easy for users to find and buy content. The technical solution for selling the content should be quick and inexpensive to deploy, and have the agility and flexibility to support many offer types as well as rapid launch of new offers. And the business proposition should not incur risk of excessive costs up front. Examples of such organizations could be any OTT, entertainment, media or content company like Vimeo, NBA, NFL, NASCAR, BBC and many others who have chosen an OTT SaaS-based solution […] More >>
Challenges of the Hyper-Connected World
Today, Amdocs announced the release of CES 9.2, our latest suite of solutions designed to help service providers "succeed in the hyper-connected world". But what exactly is the hyper-connected world, and what are the new challenges it brings? It's not new that we're living a connected world. But it's getting a lot more connected very quickly. Only 394 million people worldwide had Internet access in the year 2000. Today, that number stands at nearly 3 billion. And it's not only people who are connecting more. 16 billion devices are currently communicating through the network, and that's expected to reach 50 billion by 2020. Together, man and machine are ushering in a hyper-connected world. More >>
Innovation: A Lesson in Survival
Industries that innovate are the ones that stand the test of time. I was reminded of this again at a recent conference on innovation by a presentation which summarized 500 years of the Royal Mail Group, from its formation by King Henry VIII (a challenging CEO, apparently) in 1517 to the privatization by the UK Government exactly a year ago. The presentation was delivered by Nick Pendleton, an ex-colleague from our joint days at BT, who now leads the Royal Mail's strategy and innovation. What struck me was how the Royal Mail had, across the centuries, absorbed a succession of innovations into their business model. More >>
Breathing new life into Pay TV
Is Pay TV an endangered species, or does it have the ability to offer customers the content they want using its infrastructure and loyal audience? More >>
What do you want from your Pay TV provider?
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Partnering with OTT from a position of strength
Should we pity today's service provider? Customers are demanding more, faster, content thanks to our enthusiasm for smart devices combined with over-the-top (OTT) services from players such as Skype, YouTube, Google, and most recently, Facebook. Service providers are expected to deliver it to consumers in a high-quality format, across multiple channels, and without receiving any corresponding increase in revenues. Although many fear this will cast service providers in a "dumb pipe" role where all they do is provide connectivity, all is not doom and gloom. In fact, according to a recent Coleman Parkes survey of service providers, over-the-top players and device […] More >>
Is a faceoff with Facebook imminent?
For the past couple of years we have been regarding Google as the biggest potential OTT (or come from behind) threat to the traditional service providers. And deservedly so. They had in essence become the internet, and were offering popular voice, chat and video content. They were augmenting that with fiber networks and talk of becoming an ISP. They beat the telecom industry to the mobile payment punch with their m-wallet. They had delivered the worlds most popular smartphone/tablet operating system and they even went out and bought a handset manufacturer. Oh, and all those fancy value-added services telecom operators are talking about – enterprise solutions, cloud? Check and Check. Oh, […] More >>
Losing the hold on the customer
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Everybody loves connected devices…well almost everybody
Having recently returned from CES, I was amongst the tens of thousands of people drooling over the thousands of consumer electronic devices on display. What is immediately noticeable at such conventions is the open, symbiotic relationship between service providers and device manufacturers. Almost every major wireless carrier announced a slew of new connected devices including smartphones, tablets, laptops and even automobiles. The relationship is clearly advantageous to both sides with carriers looking to drive network and service consumption while device manufacturers get to sell more devices. It seems that everybody loves connected devices…well almost. There was one group of companies […] More >>


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By Sandra Block, USA TODAY
The custom of making New Year's resolutions dates back more than 4,000 years, when the Babylonians started the year by returning borrowed farm equipment.
You could continue that tradition by resolving to return your neighbor's leaf blower. Or here's a better idea: Resolve to take some simple steps that will save you money. Some suggestions for 2009:
•Order your free credit reports. You're entitled to a free credit report every 12 months from each of the three major credit bureaus. You can order credit reports, either online or through the mail, at www.annualcreditreport.com.
JOIN THE DISCUSSION: Share your money-saving tips
Don't be misled by other websites purporting to offer free credit reports, says Evan Hendricks, author of Credit Scores and Credit Reports: How the System Really Works. These companies often require consumers to sign up for credit monitoring or other services — for a fee — to receive a credit report.
Checking your credit reports is critical this year, says John Ulzheimer, president of educational services for Credit.com. Interest rates on home mortgages and car loans have dropped to record lows, but you need spotless credit to qualify.
Once you receive your credit reports, review them carefully for omissions that could hurt your credit score. For example, if you've been making regular payments on a loan, and the loan doesn't appear on your reports, contact the lender, Ulzheimer says. Otherwise, "You're not getting credit for good credit management," he says. "These days, you need every good account on your credit report you can possibly muster."
If you find an error, such as an inaccurate account balance or an account that doesn't belong to you, contact the credit bureau. All three credit bureaus provide a way to dispute information on your report. Ulzheimer also recommends contacting the lender that supplied the inaccurate information.
And what about negative information that's accurate? "There's not a whole lot you can do except compensate by doing things right," Ulzheimer says. That means reducing your credit card balances and scrupulously paying your bills on time.
•Submit claims for your health care flexible-spending account. Flex accounts allow workers to contribute pretax dollars to pay for unreimbursed medical and dental expenses. At most companies, you're required to use the money by Dec. 31 or forfeit the balance. However, most large companies also give their employees until March 31 to submit claims for expenses incurred before the end of the previous year.
But there's no reason to wait until March to get your money — especially if you have holiday bills to pay. You can also speed the process by submitting the right documents, says Scott Keyes, a health benefits consultant with Watson Wyatt.
If you made a co-payment, send in the receipt, Keyes says. But if you made a payment that was put toward your deductible, or you have co-insurance, wait until you receive an Explanation of Benefits from your insurance company, Keyes says. That's the document your flex-account administrator will need to process your claim, he says.
Without the Explanation of Benefits, Keyes says, the flex-account administrator has no way of knowing how much of the cost was covered by your insurer. Suppose, for example, that you went to your doctor in December, paid $200 for the service, and were reimbursed by your insurer for $100. Your Explanation of Benefits will show that your actual out-of-pocket cost was $100, Keyes says.
•Set up online bank and credit card accounts. These services, which are usually free, allow you to pay bills electronically, monitor your accounts for suspicious transactions, and create electronic alerts that will remind you when bills are due.
An online account will also help you check bank account balances — an important feature if money is tight. In 2008, the average bounced-check fee was $28.95, according to Bankrate.com.
•Avoid ATM fees. Can't bear the thought of giving up your daily cappuccino? Here's a resolution that's easier to keep: Use your own bank's ATMs.
The average ATM surcharge — the amount banks charge non-account holders for using their ATMs — was $1.97 in 2008, according to Bankrate.com. Most banks also charge a fee when their customers use another bank's ATM. The average amount of that fee was $1.46 in 2008. That works out to $3.43 — or the equivalent of paying nearly 7% interest on a $50 withdrawal.
Sandra Block covers personal finance for USA TODAY. Her Your Money column appears Tuesdays. Click here for an index of Your Money columns. E-mail her at: firstname.lastname@example.com.
Guidelines: You share in the USA TODAY community, so please keep your comments smart and civil. Don't attack other readers personally, and keep your language decent. Use the "Report Abuse" button to make a difference. Read more.


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|1970s 1990s
Pete Robinson (keyboards), John Gustafson (bass, vocals) and Mick Underwood (drums) were members of Episode Six during that renowned pop outfits final incarnation. They disbanded to form Quatermass in September 1969 and, over the ensuing 18 months, forged a career as a tight progressive power-trio. Their lone album was an impressive example of this genre, but failed to achieve a significant commercial breakthrough. Quatermass broke up in 1971. Gustafson, a former member of the Big Three and the Merseybeats, pursued a similar direction with Hard Stuff and later played with Roxy Music, while Robinson reverted to session work. Underwood joined several subsequent bands, including Strapps and Gillan. Underwood formed Quatermass II in the mid-90s, recording several demos with seasoned musicians Peter Taylor (vocals), Nick Simper (bass), and Bernie Tormé (guitar). A revamped line-up comprising Gary Davis (guitar), Don Airey (keyboards), Simper and Underwood...


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|Home > Bartha
♀ Bartha
What does Bartha mean?
Bartha [bartha] as a girls' name is of Old German origin, and the meaning of Bartha is "shining, brilliant". Variant of Bertha.
VARIANT Barta
OTHER FORMS VIA BERTHA Berrta, Berta▼, Berte, Berthe, Berti, Bertie▼, Bertina, Bertita, Berty, Birta, BirthaCREATIVE FORMS
(female) Badtha, Britha, ..
(male) MIDDLE NAME PAIRINGS
Bartha Ester (B.E.), ..
How popular is Bartha?
Bartha is a rare first name for females but a very popular last name for both adults and children (#35387 out of 150436, Top 24%). (2000 U.S. DEMOGRAPHICS)
Which version is better?with its variant form and related girl names. Berta, Bertha, Bertie, Bertina and Birtha are the popular alternative forms of Bartha (NOT RANKED). Other variants, like Berti, are seldom used. Usage of these forms of Bartha reached its most widespread a century ago (AVERAGE #708), but now, forms like Bertie have become less stylish. (TOP 2000 NAMES, 2015)
Similar NamesRecommended names are Artha, Bara, Barah, Barbra, Barra, Barucha, Basha, Bastia, Batya, Bert, Betha, Birte, Breda, Bretta, Brita, Britta, Caltha, Catha, Dortha▼, Eartha, Ertha, Hertha, Iantha, Katha, Laretha, Marsha▼, Marta▼, Marth, Martha▼, Marthe, Marthy, Martia, Martta, Matha, Mirtha, Sarha, Xantha and Zantha. These names tend to be more commonly used than Bartha.See names in meaning and etymology.


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|Cannondale R1000 Feminine
The Cannondale R1000 Feminine is a high-quality bike with 10 speeds and a lightweight frame. It features Shimano Ultegra hand brakes and top-quality wheels. This bike is available in a number of colours, including light blue, purple, and more. This bike is pricey, but it is made to last.
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K2 is a company known for making quality bikes and other sports equipment. The K2 Myste is a fairly inexpensive road bike designed specifically for women. It features eight speeds and a lightweight aluminium frame, as well as Tektro dual-pivot hand brakes. This bike has excellent manoeuvrability and it is comfortable to ride, even for extended periods.
|Pinnacle Dolomite
Pinnacle Dolomite women's road bikes are durable and sturdy, but still lightweight. They offer between 10 and 16 speeds and are very fast. This makes them ideal for racing or casual riding. They are available in a wide variety of colours, including white, red, blue, silver, and black. These bikes are expensive, but they feature superior-quality materials.
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The Amira road bike for women is an exceptional product. It features a carbon frame, which is strong but extremely lightweight. Though the Amira is a racing bike, it is also great for recreational riding. It has nine speeds and Axis DC hand brakes.
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The Ruby road bike for women features a carbon frame, so it is exceptionally lightweight. It offers nine speeds and vibration dampeners in the frame and fork to ensure a smooth ride. The handles feature gel pads covered in specialised grip tape for a secure yet comfortable grip. Ruby road bikes also have Axis DC hand brakes.
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The Ariel Sport Disk features Tektro Draco hydraulic disc hand brakes, with short, easy-to-reach levers. This means the bike can come to a complete stop very quickly, which is a great safety feature. It has a geometrically designed aluminium frame made to accommodate a woman's body. The Ariel Sport Disk comes in a variety of colours.
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The Dolce Elite compact is a small but powerful women's bike. It has 10 speeds and a lightweight aluminium frame. This bike is perfect for women with a smaller stature who are looking for a fast bike for racing or recreational riding. The Dolce Elite compact is highly manoeuvrable and available in many colours, including black, white, and silver.
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Trek Madone 5.0 bikes are great for racing. They have strong but extremely lightweight carbon frames, 10 speeds, and triple chain sets. Trek Madone 5.0 bikes are fairly expensive, but the high price tag means excellent quality. They are fast, durable, and easy to manoeuvre. Furthermore, the saddle and seat mast of these bikes are comfortable and stable during long rides.
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The Orbea Diva TPB road bike for women comes in black, pink and black, red and black, or white. It has 20 speeds, a carbon frame, and Shimano Zeus Pro hand brakes. The high number of speeds means that this bike is great for climbing steep hills and accelerating quickly on flat ground.


text:
Napoli continues tear as Red Sox down A'sApr 23, 2013 - 7:48 AM
The red-hot Napoli, who was named the American League Co-Player of the Week with teammate Andrew Bailey on Monday, has driven in 17 runs over a nine-game tear and is batting a cool .351 over the stretch.
"He's a very good hitter," Red Sox manager John Farrell said of Napoli. "He's got a long track record of extra-base and RBI production, but what he's done in this ballpark over the course of his career, those things are beacons when you look to select a player, and he's a very good fit here."
The Red Sox were swept in Sunday's doubleheader against Kansas City, but Felix Doubront (2-0) gave them a lift with a solid 6 2/3 innings of work. He struck out eight and gave up three runs on just three hits and five walks.
Overall, Boston starters are a combined 10-3 on the young season.
Will Middlebrooks cranked a three-run homer, Jarrod Saltalamacchia contributed an RBI double and Bailey notched his fourth save of season with a scoreless ninth after Oakland scored three runs in the eighth.
Oakland starter A.J. Griffin (2-1), who hadn't allowed more than two runs in any of his first three starts, was tagged for all nine runs -- seven earned -- on eight hits over four-plus innings.
The Athletics, losers in four straight, managed just five hits. Josh Reddick and Josh Donaldson had two RBI apiece and Chris Young drove in a run in the setback.
Oakland trailed 4-2 going into the fifth and had an excellent opportunity for a big inning, but managed just a single run.
Young pulled Oakland within one with a bases-loaded sac fly, but after Dubront loaded the bases again with a walk to Brandon Moss, Jed Lowrie to flew out to center to end the inning.
Napoli made them pay in the bottom half when he came to the plate with the bases full, smacking Griffin's offering over the Green Monster in left to give Boston an 8-3 advantage.
Griffin was then chased from the game and replaced by Chris Resop after first baseman Moss committed a fielding error on Daniel Nava's chopper. The Red Sox would tack on one more run later in the frame on a Saltalamacchia double before Dustin Pedroia left the bases loaded.
The Athletics were able to plate three runs in the eighth off back-to-back doubles from Donaldson and Reddick, but the rally was stifled when Boston reliever Junichi Tazawa set down both pinch-hitter Eric Sogard and Coco Crisp.
Oakland scored first with two runs in the second. A wild pitch from Doubront with runners on second and third plated the first run, and Reddick made it 2-0 with a base hit up the middle.
Boston got one back in the bottom half of the frame following back-to-back doubles from David Ortiz and Napoli, then jumped in front for good in the fourth on Middlebrooks' three-run homer.
Game Notes
Boston's Jacoby Ellsbury was 1-for-3 and has hit safely in 12 straight games. He also swiped his ninth base of the season ... Oakland is just 1-6 against teams not in the AL West ... The A's haven't lost four in a row since a nine- game slide between May 22-June 1 of last season ... Boston snapped an eight- game losing streak to Oakland.
Apr 22 10:08 PM
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Athletics 6, Red Sox 9 Top 9, 0 OutsApr 22 10:01 PM
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Athletics 6, Red Sox 9 Mid 8Apr 22 9:49 PM
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Athletics 6, Red Sox 9 Top 8, 1 OutApr 22 9:40 PM
- Josh Reddick doubled to left scoring Josh Donaldson. Runner on second with one out and Nate Freiman due up.
Athletics 6, Red Sox 9 Top 8, 1 OutApr 22 9:35 PM
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Athletics 6, Red Sox 9 Top 8, 1 OutApr 22 9:35 PM
- Josh Donaldson doubled to right-center scoring Brandon Moss and Derek Norris. Runner on second with one out and Josh Reddick due up.
Athletics 5, Red Sox 9 Top 8, 1 OutApr 22 9:34 PM
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Athletics 5, Red Sox 9 Top 8, 1 OutApr 22 9:34 PM
- Jarrod Saltalamacchia doubled to left-center scoring Will Middlebrooks. Runner on second with one out and Stephen Drew due up.
Athletics 3, Red Sox 9 Bot 5, 1 OutApr 22 8:27 PM
- Mike Napoli none-out, Grand Slam (4) to left-center scored Shane Victorino, Dustin Pedroia and David Ortiz. Daniel Nava due up.
Athletics 3, Red Sox 8 Bot 5, 0 OutsApr 22 8:18 PM
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Athletics 3, Red Sox 8 Bot 5, 0 OutsApr 22 8:18 PM
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Athletics 3, Red Sox 4 Mid 5Apr 22 8:09 PM
- Chris Young sacrifice fly to center scored Nate Freiman. On the play, Andy Parrino advances to third. Runners on first and third and with two outs and Brandon Moss due up.
Athletics 3, Red Sox 4 Top 5, 2 OutsApr 22 8:02 PM
- 41
rootsRUWTbot Added 17 roots
Athletics 2, Red Sox 4 Top 5, 1 OutApr 22 7:57 PM
- Will Middlebrooks one-out, three-run Home Run (5) to left scored Mike Napoli and Daniel Nava. Jarrod Saltalamacchia is due up.
Athletics 2, Red Sox 4 Bot 4, 1 OutApr 22 7:43 PM
- 24
rootsRUWTbot Added 11 roots
Athletics 2, Red Sox 1 Bot 4, 1 OutApr 22 7:41 PM
- Mike Napoli doubled to right scoring David Ortiz. Runner on second with none out and Daniel Nava due up.
Athletics 2, Red Sox 1 Bot 2, 0 OutsApr 22 7:09 PM
- Josh Reddick singled to center scoring Josh Donaldson. Runner on first with one out and Nate Freiman due up.
Athletics 2, Red Sox 0 Top 2, 1 OutApr 22 7:00 PM
- Felix Doubront's wild pitch scored Jed Lowrie. On the play, Josh Donaldson advances to third. Runner on third with one out and Josh Reddick at the plate.
Athletics 1, Red Sox 0 Top 2, 1 OutApr 22 7:00 PM
Top 1, 0 Outs
Apr 22 6:38 PM
- 13
rootsRUWTbot Added 13 roots
Athletics vs. Red SoxJan 10 8:15 AM
Related News
- Ventura: Relievers, media pose threats to Bradley's streak May 23
- Athletics-Mariners Preview May 23
- LEADING OFF: Kershaw tests streak vs Reds, Colon faces Nats May 23
- Ejections aplenty for Blue Jays lately May 23
- Baseball Capsules May 22
- Bad day gets worse for Athletics in 5-4 loss to Yankees May 22
- Pineda ends slump; Yanks top Oakland 5-4 to win 5th straight May 22
- Ortiz gets 4 hits, Bradley's streak at 27, Bosox top Indians May 22
- Boston's Bradley extends hitting streak to 27 games May 22
- Gray put on DL by A's because of strained muscle May 22
- Indians-Red Sox Preview May 21
- Yankees-Athletics Preview May 21


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Thousands fewer Sydney families are living in public housing, as demand grows and the government demolishes and sells homes just to keep the system afloat.
New figures reveal about 139,500 NSW households live in public housing, or 2300 fewer than two years ago.
The NSW government has demolished or sold about 6000 properties in the past four years, as its bill for repairs rises towards half a billion dollars.
In the same period, the number of new homes the government builds each year has more than halved, to 440 in the past financial year.
Hal Pawson, a Professor of Housing Research and Policy at UNSW, said the figures showed the state government was being forced to use more of the federal government's grants for building homes instead to pay for day-to-day repairs.
"The government is addicted to sales," he said. "Funding used to be used for investment in new housing, but over the past 10 years the growing deficit means none of that is available. They're having to use all of it just to balance the books.
"It means there's going to be fewer houses available."
Sale of public housing has been under way for a decade but reached a peak last year and will rise again in the coming two.
The housing waiting list grew about 3.5 per cent to nearly 60,000 this year; it will grow to more than 80,000 by 2016 according to projections.
And the shortage is hitting Sydney's major growth in the south-western suburbs the hardest, analysis by the state opposition shows.
About 670 more Bankstown families are seeking public housing now than three years ago but in that time the number of local houses available to them has fallen by about 220.
In Campbelltown there are 1400 fewer houses and about 130 additional families.
"Society's most vulnerable, the elderly, frail and low-income families are missing out," opposition housing spokeswoman Sophie Cotsis said.
Numbers on the waiting list represent about half those who need housing, according to the NSW Auditor-General's report in 2013.
Deputy CEO of the NSW Council of Social Services John Mikelsons is lobbying the state government to devote money from the privatisation of assets to new housing: "It is clear NSW is becoming less and less capable of meeting demand."
The sale of 206 homes at Millers Point alone is, on some estimates, likely to raise up to $500 million for government coffers.
Family and Community Services Minister Gabrielle Upton has refused to say specifically where it might be spent, except to guarantee it would go back into the social housing system. "We can reinvest the proceeds into more new homes," she said.
Forecasts by the NSW Auditor suggested a total of 4000 homes may be up for sale in 2014 and following three years.
Mrs Upton said the government inherited a housing system with deep structural problems.
The minister said next year's budget allocated $121 million for new homes, and $491 million for property maintenance and upgrades, a 30 per cent increase on previous years.
"The budget of $121 million in new supply will see the commencement of 759 public housing dwellings and the completion of 443 dwellings," she said.
But Professor Pawson said it was questionable how much would be used to buy more houses.
"It's very notable that no clear commitment has been made," he said. "You can only draw the conclusion that the priority is entirely about keeping the system afloat".
He said federal government funding was not enough to bridge the gap between what poor tenants could pay and the cost of running houses
The government's maintenance bill last year hit $220 million and a further $180 million was spent on upgrades.


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Raffaello and Toblerone Chocolates
Toblerone is that unique chocolate with a bold triangular shape, and a refined recipe of delicious Swiss milk chocolate with almond and honey nougat. Sweet and graceful, dainty and exquisite, a refined take of moist cake filled with whipped cream and coconut of rafaello. Product Includes: 1 Bar of Toblerone (Weight : 50 gms), 3 pcs of Raffaello Chocolates (Weight : 30 gms).


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America is in the throes of crisis. Polling as the least popular starting President in history, Donald Trump has taken the reins and intends to force his agenda on the world by brute force. Alongside him, outright white supremacists are coming out of the woodwork, convinced that their time has arrived. Millions who once counted on the Democratic Party are losing faith in the entire political system. Can anything be done to halt the rise of tyranny?...
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2 months of protest against the neoliberal assault on education in Mexico have finally resulted in talks between The Coordinadora Nacional de Trabajadores de la Educación (CNTE) and Interior Minister Miguel Angel Osorio Chong.Highways remain blockaded in 5 states and the teachers send us a video of protests from Monday, 11jul16...
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Tanks and barbed wire had surrounded Egypt's presidential palace and crowds of protesters were swarming around last week when President Obama placed a call to President Mohamed Morsi. Mr. Morsi and his allies in the Muslim Brotherhood stood accused of a sudden turn toward authoritarianism, as they fulminated about conspiracies, steamrollered over opponents, and sent their supporters into a confrontation with protesters the night before that call; the clash left seven people dead. But Mr. Obama did not reprimand Mr. Morsi, advisers to both leaders said.
Read Full Article »


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Running Time: 126 minutes
Released: 09/2008
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Eugene Edwards
"It's easy to see why Edwards has become a favorite on the L.A. pop scene - his revved up live shows and sterling guitar licks put muscle behind his melodies." - LA Times Elvis Costello. Squeeze. Nick Lowe. Foo Fighters. Tom Petty. Marshall Crenshaw. The Guess Who. Even the Beatles. Yeah, Eugene Edwards could hide from the comparisons he gets, from both fans and critics alike, but he's not like ...more


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The Purifiers were a group of mutants who believed that mutants were destined to rule the world.
Leader: Reverend Icarus (TIA)
Members: Anole (TIA), Dust (TIA), Dryad (TIA), Indra (TIA),
Loa (TIA), Pixie (TIA), Surge (TIA), and Wind Dancer (TIA).
Allies: Magneto (TIA) New X-Men(TIA), X-Men (TIA).
Enemies: Brotherhood of Mutants(TIA), Friends of Humanity (TIA), Hellions(TIA), Reverend Stryker (TIA)
Marvel Fanon at Wikia


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I am a definit expert on this.
My 2nd born was colic and peds. don't give you to much advise.I tried everything. To the point of tears and feeling like a failure.I took her back to the docs.4 different times.They gave her medicine that contained alcohol.I got a referral for a gastrointernologist.He was AMAZING!he had me feed her a bottle containing my breast milk and something that would show up in a x-ray.She had a common problem called GERD.To give you the best advise it's not nessacerally The formula or breast milk but what he recommened was a teaspoon of rice cereal for every 2oz of formula or breastmilk.Along with regular Maylox.It has calcium and simithicon(That's what is in gas drops)I had to put 1.0mls for 4oz.Keep your baby upright after feeding for 20min.to help digest and for sleep either cradle swing or raise the part of their mattress up with a blanket under it kinda like those beds you see on tv that that raise for elder people.You can even put your baby in a vibrating chair and put that in their crib for safty.Keep in mind my daughter was only 5wks.when I started.I used this method with my next baby,and so on.Don't wast your $ on changing bottles buying colic cures and formula changes.Best of luck