It began today just by reviewing for you the content of the opinion record there. But I see I have just one piece of piece of talk today. So let me just go over what it means. The opinion record concerns the matrix elements, the very reasonable tensor operators, TKQ, between two basis vectors of the standard angular momentum basis. It says that the dependence of the matrix element of the magnetic quantum numbers, of which there are three, m prime q and m, is captured by a Fletch-Gordon coefficient that you see over here, with m prime q and m, the quantum numbers same for fairly over here. When I say it's captured by the Fletch-Gordon coefficient, I mean the original matrix element is proportional to the Fletch-Gordon coefficient and the proportionality factor is something that does not depend on those magnetic quantum numbers. In general, it depends on everything else, however. It depends on the gambas and the J's and which tensor operator it is and the K value. This proportionality coefficient is conventional to denote it as I've indicated up here. It's called the reduced matrix element. It's got this double bars that indicate that it's not a ordinary matrix element. But basically it's just a way of listing everything else that the magnetic could depend on. And the definition of the reduced matrix element is that it's just the coefficient or proportionality factor that makes this equation work. Now last time I explained the two main applications of the bigger element right there. One is for selection rules and the other is to actually calculate a lot of these matrix elements within the magnetic quantum numbers over here of variables. I won't go through that again, but that's what it's used for. As far as selection rules are concerned, it's actually very easy to use the bigger element right there. All right. Now I'm not going to prove the bigger element right there. Mainly because it takes too long to write it all out. The proof is in the notes. I'll try to give you at least a general idea of what's involved in the proof. To get the general idea, let's first suppress this prime bra which is over here. And just look at this very useful, tens-properly accurate one in this case. So we've got T k q times the average a m. So we can describe this product as being an operator that has certain angular momentum quantum numbers, multiply times the ket to best certain angular momentum quantum numbers, namely k q on this side and j on the other side. If we let the magnetic quantum numbers be variable, the q would range from minus k up to plus k, and the n would range from minus j up to plus j. And so you'd end up with a total of 2k plus 1 times 2j plus 1 ket, that's if you did that. Now this is actually very similar to what we do when we're studying clutch-Gordon coefficients, when we combine angular momentum, for example, orbital and spin.  It doesn't matter. Then in that case, you might consider a product that would look like this. You'd have a k q-stake that would represent, let's say, an orbital, and a j m-stake that would represent spin. 2k is not a product that can operate a times a ket, but with the same quantum numbers. And again, you'd have the same number if you vary the magnetic quantum numbers, you'd have the same number of total states. And those total states that you get this way constitutes what are called the uncoupled basis in the process of coupling angular momentum. As we know, you can transform this into a couple basis, which would be eigenstates of total J squared and J z, and the expansion coefficients of the couple basis in terms of the uncoupled basis are the clutch burden coefficients. In fact, they're exactly the same as clutch burden coefficients in the Bureau of Europe. And so the idea is that, in a sense, the total angular momentum content, I mean, this is a set of vectors here, if I vary the magnetic numbers. This is a set of vectors in our Hilbert space. And one can think about expanding this as a linear combination of, in terms of angular momentum eigenstates, a standard angular momentum basis. If you wanted to pick out the coefficient of one of those basis vectors, you'd take the scalar product with exactly this broad here, so that J prime and M prime would play the role of the total angular momentum. That's just what the robot plays over here. And so the idea is that this set of states with this set, when they make magnetic quantum numbers run over their allowed values, this set of kets produced in this way has the same transformation properties under angular momentum rotations and so on. As a set of kets you get if I just multiply kets. Multiplying an operator times a ket is not so different from multiplying a ket times a ket. By the way, multiplying two operators, the irregissible tensor operators, is a very similar story. If I have a, let's say, a Tk1q1 times a Tk2q2, if you want an example of this, think of a product of the position operator. Remember that position operator is a k equals one irregissible tensor operator if you put it into the spherical basis. It's what we call RQ. Well, maybe I have that in RQ times an RQ prime. This is a tensor, like an xi, xj, but in a spherical basis. Exactly this thing occurs in the quadruple moment tensor, which I'll say something about later in the hour. Anyway, the point is that this product of operators, again, has the same angular momentum decomposition as a product of kets or a ket times an operator. The rules work in the same way in terms of angular momentum combined. And that's the basic idea that lies behind the proof of the theorem. It's just to exploit these facts. All right. There is one tricky thing, however, which is that looking at this, looking at these kets here, or that part of the kets here, which is the same, you'd think that what you'd be doing is probably k times j like this. But actually, if you look in the question of the other way around, it's j times k, which is being done here. The fact that these are swapped is purely a matter of convention, and it has nothing to do with the essential ideas, but it just makes it slightly harder to remember. Do you ever remember to do this in the right order? This could have been defined the other way around, and it would still be valid, but this is the conventional way of doing it. All right. Now, so that's the review of the ring of that card theorem, and at least some of the ideas involved with the proof. Now, the next thing I'd like to turn to is a special case of scalar operators. If k is a scalar operator, we know that it commutes with all rotations, but it's also an irreducible tensor operator with the values k equals 0 and q equals 0. There's only one value of q and k. And if we consider the major solves of the scalar operator in a standard angular momentum basis with the primes on the left side, the scalar operator in the middle of the unprime basis vector on the right side like this, the ring of that card theorem tells us that there's first of all reduced matrix settlement, which is gamma prime j prime double bond k double r gamma j times the Klebsch-Gordon coefficient, which I'll fill in at the moment, since the Klebsch-Gordon coefficient, since k and q are equal to 0 here, it turns into, it turns into j prime m prime, a scalar product with j 0 m 0 like this. In other words, the one what it's doing here is taking j across to 0, j for the space that the k is acting on, and 0 for the k itself. Well, if you combine an angular momentum with some value of spin, for example, with another angular momentum, which is 0, adding 0 to an angular momentum doesn't change it at all. In other words, if a product like this is trivial, it means that this Klebsch-Gordon coefficient is the new j. It's the same as the old j. There's only one value that comes out of this. And so this is actually a prime for delta j prime for delta j prime for delta j prime at any time. That's what the Klebsch-Gordon coefficient is. So the goal of this end of the theorem is to find delta j state prime delta m m prime like this. So I'm going to apply to scalars. The thing that I'm trying to say is that the matrix element of a scalar operator, I'll remind you that the most important scalar operators are Hamiltonians for isolated systems, of which we'll see quite a few in the course. The matrix elements of the scalar operator between basis vectors and the standard angular momentum basis are in the first place, they're diagonal in the j quantum number. Secondly, they're diagonal in the magnetic quantum number. And thirdly, the matrix elements themselves that are left over, which are on the diagonal, are actually independent of the magnetic quantum in the gamma. They do depend on j, as you see. j must equal j prime, and this will be zero. But they depend on j. They depend on gamma and gamma prime. They're not diagonal and gamma. There's no reason why they should be diagonal and gamma. Gamma is just an extra index that is introduced to resolve the degeneracies of j squared and jz. And in general, they have no relation to the operator k, for example, in Hamiltonian. So there's no reason why it should be diagonal and gamma. But it is got diagonal j's and m's, and as I say, it's independent of the magnetic quantum number, m. If you like, you can do this as a matrix in the magnetic quantum numbers. The same thing as proportional to chronic or delta means the matrix is diagonal. However, that doesn't mean the same matrix is diagonal, it doesn't mean that all the diagonal elements are equal. But that, in fact, is what happens here. All the diagonal elements are equal. So, in fact, it's a multiple of the identity matrix, as far as the magnetic quantum numbers are concerned. Now, physically, that means, for Hamiltonians, that means the energy doesn't depend on the orientation of the system. It's an easy fact to understand. But this is how it comes out. All right. Now, these results can be understood more simply from another argument, which I like to give. It doesn't go through all the big record term. What suppose we've got a scalar operator? We know that it commutes with jz and also with j squared. Suppose we're given an operator k, which has these commutational relations like this. Then let's work on the jz first. Let's say the sandwich of both sides of this jz commutator between the two basis vectors of the standard angle and the basis. And this thing gives us the prime gamma jm on one side. And then the commutator all right on its jz times k minus k times jz. And the other, then, we've got the unprime gamma jm on the other side. And this is equal to 0 because we've got 0 over to the right-hand side. Now, here we can let this one jz act to the left and the other jz act to the right. And the result is that we get m prime minus m times the matrix of prime gamma jm on one side k to the middle and then gamma jm on the other side. This is equal to 0. So what you see by just playing with the commutator is that either m equals m prime or else the matrix element is equal to 0. That's equivalent to saying that the matrix element is proportional to the particle delta. Likewise, if you did the similar game with j squared, you'd find the matrix element is diagonal and also the j component of the same argument. We'll see this argument again later on. However, these arguments by themselves don't tell you that the matrix element is independent of the magnetic component. That takes an extra argument, and in fact, it doesn't even follow from just these two commutation relations. It really requires a full rotational of variance of k that's became less commute not only with these two operators, but also with jx and jy. It doesn't appear just here. That means it commutes with all rotations. Well, in particular, if it commutes with jx and jy, it commutes with a raise in the Lorentz operators. So let me show you the effect of that. Let's see. Let's see here. Oh, I want to do that. Actually, I think, let's see. Yeah. Maybe what I'll do is, okay, here's what I want to do. Let's leave this discussion with general scalar operators. It would this for now, and just say that the bigger record there makes a stronger statement than just these two commutator arguments because it says that the matrix elements are independent of something else. If the Hamiltonian, I'll make this through the remark, but if the k were a Hamiltonian and you wanted to find energy eigenvalues, then what you needed is to diagonalize this matrix and the, since it's not in j, sorry, diagonal j, you need to diagonalize it with the gammas. Again, we've got some arbitrary basis. And if you do, you'll find the energy eigenvalues. But the result will be that the energy eigenvalues won't be dependent on the magnetic quantum numbers. Again, this is physically the independence of the fact that the energy is independent of the orientation. Now, we've seen a special case of this already in central force motion where the energy eigenfunctions have the importance of NL like this. And we derive this by working through the radial Schrodinger equation. What I propose to show you now is that a very similar conclusion applies to any isolated system, not just the central force problem. It could make it much more complicated. You could add spin orbit effects and other optimistic effects. You could include the magnetic interactions with the nucleus, which is the hyperfine effects. You could make it a multi-particle system, so you've got more than one electron. You could include the dynamics of the nucleus, so it's protons and neutrons that interact. You could make a very complicated problem. The conclusion still holds is that you can write the energy levels in terms of an angle of the quantum number and another one that labels the energies. But they don't depend on the magnetic quantum number. And that's a general conclusion. So now I'll show you how that general conclusion follows from the theory of development. So it works like this. If we have I'll speak up in Hamiltonian because that's the most important application. We will assume it's a scalar, so it can use all rotations in its isolated system, but possibly very complicated for it. That means in particular it can use with all three components of J. And as a result, it can use also J squared and JC.  And one of the ways to find those eigenstates, at least in principle, is to start by diagonalizing J squared and JC with what we call the quantum numbers J and M. This is the total eigenvalue for the system in a total JC. Let's call the eigenspace of J squared and JC with eigenvalues J and M. Let's call it EJM, like this. This is in general multi-dimensional. In fact, usually it's infinite-dimensional. This means the total element space is decomposed into orthogonal subspaces, each of which is characterized by the JM values. We write it this way as the sum of J and M where it's really a direct sum, but not an ordinary sum. This direct sum just means orthogonal subspaces. Then to find the energy eigenvalues, what we do is we just diagonalize the Hamiltonian in each of these subspaces. And if you do that, you're going to get some energy spectrum. Different energy spectrum in general Now, for simplicity, let's suppose that spectrum is discreet. Moreover, let's assume that it's non-generate. That's usually true, but let's make this an assumption. So then we can label the energy eigenvalues on a JM subspace by another quantum number called N goes 1, 2, 3. It just counts them and asks in the order. That's all it does. This means that the energy eigenspace then can be labeled by the three quantum numbers so that H acts on NJM brings out an energy and the energy depends on NJM and acts on a bulk-like NJM. The reason the energy depends on NJM is because we're diagonalizing the Hamiltonian in each of these subspaces separately and N is just the sequence number in that subspace. So based on everything I said so far, there's no reason why the energy shouldn't depend on it. However, I'm now going to show you that if you have full rotational invariance depending on it, it works this way. Let's apply a low-ring operator to this equation. Let's apply a J minus H NJM on the left and if H can use with all three components of J, then I can compute these two these two through and put the H on the left and the J minus on the right. J minus will act on the NJM and so it will bring out a square root, the 10 square root which will be J plus M times J minus N plus 1 and then we've got H adjacent to NJM like this that did somebody steal the eraser here? Let's see looks like they did. I can't believe I'm the eraser. Excuse me, I'll find the eraser. Okay. If you allow me to use my eraser, H has some NJM which brings up the eigenvalue which is the energy that we're assuming NJM and in multiple... excuse me let me go back a little bit. We run through it again. J can use with H and apply J minus to the left side of this equation I'll compute the J minus through the H let it act on NJM that will lower the value of M so I've got NJ comma M minus 1 this is being acted on by H so that brings out an energy NJ comma M minus 1 here we're having an open mind and allowing for the energy to depend on the magnetic quantum memory so that's J minus acting on the left-hand side now letting J minus acting on the right-hand side the energy is just the number so the J minus just goes into the ket and what we get is B NJM times the same square root times the same ket which is lowered by one magnetic quantum memory minus one and so if you look at these two equations the square roots cancel and the kets are the same and the only thing that's different are the energy values so what you see is that E NJ comma M minus 1 is equal to E NJM the result is is that the energies are the same for M N minus 1 in fact by induction it shows that all the energies don't depend on them so go back to this original statement here we can drop to him and you see we have a generalization of the fact that we learned for central force problems this is an opinion isolated system the energy levels are characterized by an angular magnetic quantum number and possibly generally other quantum numbers but they don't depend on the magnetic quantum number now what that means is that even if we make a very complicated system it means that the energy levels what you call the energy levels are actually the degenerate energy eigenspaces characterized by an angular momentum and they have a dimension of 2J plus 1 in other words there's a degeneracy this is the degeneracy of 2J plus 1 full of degeneracies that arise on the basis of rotational variance alright let me give you an example of this actually if you get an example it's just a hydrogen atom on the other side of the force problems but I'll give you a much more complicated example just so you can see how the idea carries through the more complicated systems let's talk about the higher 57 nucleus higher I think is 26 it's not as easy as 26 and so that means the number of neutrons is equal to 31 and iron 37 so this is a 57 body problem in nuclear physics with protons and neutrons and people don't even understand the forces very well so the Hamiltonian is complicated too what can we say about it well what we can say about it is is that the energy levels are going to be characterized by an angular momentum quantum number what we call the spin of the energy level but it's really the spin the sum of the ordinal and spin angular momentum of all the constituent particles in the nucleus and that energy level I call the spin I I is the usual number for the energy level will be 2I plus 24 will be generated and in particular this has got to be true for the ground state of the nucleus so here's the ground state of the iron 57 nucleus we just call it iron 57 like this and this actually has an I equals 1 half level iron 57 is used in the moss power effect and I'm not going to explain the moss power effect in detail it's not my purpose here let me give you at least some background on it there are some other levels here there's one level up here which has an excited state of the iron 57 and as I equals 3 has this is called let's call this 57 iron 57 iron let's call this with a single star like this the energy difference between here is about 14 kiloelectron volts there's another excited state let's call it 57 iron double star instead of a higher energy it makes transitions both down to the ground state and also down to this intermediate state all of these nuclei are in stable ground state the way one normally produces is iron 57 double star if you get a further decay up here coming from cobalt 57 and so if you have a sample of cobalt 57 and in other words this decay produces some for each period of time these excited states as it rattles down to the ground state it's this 14K transition it's the most important one in the case of the moss power effect this is a doubly excited state by the way let's see here I'll fill this in the first star in state has an I equals 3 has and the doubly double star in state has an I equals 5 has so these are the spins as we would say in these different states now if you think of the iron 57 nucleus as a unibody problem what's indicated there's just three balance states in the system but there's other balance states up there the huge enormous number of balance states the spectrum of nuclei I'm just talking about the balance states can be extremely complicated for a heavier nuclei and above that there's own balance states there's a continuum in which the protons and neutrons are separated so that's the Hilbert space for the iron 57 nucleus and it's an enormous space it has way functions of many variables nevertheless the ground state is a state that actually just consists of two different states there's the energy of the ground state there's the I the basis factors that span the ground state could be labeled by this if it's understood that I'm only talking about the ground state I don't need any other labels in it if I want to have another label in it I can put it in here so this is named the N equals 1 we call that N equals 2 but if we're only talking about the ground state we're going to suppress the N and just talk about the I M I now when we talk about when we talk about magnetic resonance and spins and magnetic fields we consider the spinning Hilbert space which we call the spin which was just basically the span of the basis factors looked like this I M I where I runs from minus I excuse me where M I runs from minus I plus I I think we call this S instead of I but it's conventional in nuclear physics to use I for this nuclear spin it just means the spin of a particle and this is a finite dimensional space and on this space we we wrote we wrote the magnetic moment as proportional to the spin operator with proportionality factors and then we took it from there minus U dot B we've noticed what we're doing so far as the iron nucleus is concerned is we're really only considering a very small subspace of the entire Hilbert space what allows us to neglect all the rest of this the answer is is that the energy differences are rather large the U dot B interaction of an iron 57 nucleus and any reasonable magnetic field is a very small energy compared to this 14 KV so you might as well just ignore this because this is the only subspace of the nuclear Hilbert space that's in effect what we did and this by the way is the explanation why for at least for composite particles like nuclei it's the reason why in dealing with magnetic resonance kinds of problems why we can just assume the Hilbert space consists of just a single irreducible subspace and the rotations characterized by a single incremental polynomial it's because it's the ground state of the system and because all the energy levels including the ground state are characterized by an angular momentum quantum number and they've got a 2 J plus 1 or i is the notation in this case degeneracy, pole degeneracy okay so those are some of the lessons that come out of this alright alright this fact that the standard changes I'm going from the 3 half to the 5 half this transition is an electromagnetic transition it means in the midst of photon and the photon carries off as you see one angular momentum one unit of angular momentum it's very similar to the bipolar transitions in atomic physics alright alright so this is some of the lessons that come from the fact that the Hamiltonian is a scalar operator okay now the next thing I'd like to discuss today is a somewhat different topic this is going to come up in this week's homework so I'm going to give you the preparation for it I want to talk about quadrupole moment operators so let me begin by reviewing some things from classical electric magnetism let's suppose we have a charge distribution which is localized let me call it rho r it's a charge density let's put a coordinate system x, y, z here somewhere in at the end or near the charge distribution let's denote positions inside the charge distribution by a prime position vector and I can call it r prime so in fact I'll call this rho r prime here like this similar out here we have a field point where we want to measure the potential it's called this r we want to measure the potential of the charge distribution of the field point so we see let's call it capital phi it's a potential capital phi r well by electrostatics this is the integral of the source distribution between r prime times the coolant denominator which is one of the distance between the source and the field point like this and if you like SI units you can put it in the ugly 4 pi s or not so I'll leave them out now the way I set this up we'll imagine the charge distribution is localized and the field point is far away from it so the magnitude of r is much greater than the magnitude of r prime r prime is the vertical integration but it's bounded by the dimension of the charge distribution so that means we can expand this coolant denominator in powers of r prime over r and if you do this here's the expansion you get the first term is just 1 over r and the next term is this it's r vector over r cube dotted into r prime vector the next term is this it's the factor of 1 over 2 r to the fifth this is sum on i and j we call it x i x j x i x j let's say r vector here is x 1 x 2 x 3 let's say r vector prime is the prime version of that x 1 prime x 2 prime x 3 prime so it's the sum on i j of x i x j and then this is the whole point x 3 x prime i x prime j minus delta i j times r prime squared and then this is the whole point x 3 squared and through the second order this is the expansion of the school on the denominator you get it just by straight for a Taylor series I won't do the algebra now this is under an integral so we integrate over r prime we need to do the r prime integral for the first term you just get an integral here's the integrals you get for the first term you get integral d cube r prime row of r prime that's just the total charge which I'll call the lowercase q you get d cube r prime row of r prime multiply times this r prime vector which appears here that's the charge weighted position vector which is also called diphenon vector I'll call it d that's just characteristic of the charge distribution second order you need to do the prime integral over this quantity of parentheses you get integral d cube r prime of row of r prime multiply times 3x prime ix prime j divide as delta ij times r prime square and this is conventionally called the quadruple moment tensor and it's called q ij right? now so I'll do the integral again what we get is this is 5r the field point is equal to the total charge q divided by the radius that's just the obvious coulomb term then we get the dipole moment dotted into our vector over r cube then we get 1 half 1 over r to the fifth times someone i and j x i and j times the quadruple moment tensor q ij that's your second order that's what it is now about this tensor p y j this is obviously a symmetric tensor and it's also traceless the tracelessness just comes from doing a Taylor series expansion of the coulomb denominator it's not an accident but it just comes out if you do it to say that it's traceless means that if I sum on i and j I get 0 so if I sum on i and j this term here we get 3 times r prime square because you say i equals j at sum if you sum the second term on i and j well that means you've taken the trace of the identity matrix which is 3 so it's minus 3 r prime square and we've solved it in 0 q ij is a symmetric and traceless tensor we also explained earlier that's actually a k equals 2 well in classical E&M this is not an operator so I can't really speak a very useful tensor but it corresponds to the k equals 2 or a 2 angular momentum a 1 in the angular momentum of 2 there's 5 independent components in a symmetric and traceless tensor and by doing linear combinations they can be put into the form of q value running from minus 2 up to plus 2 alright this is the Cartesian form of the tensor rather than the spherical form alright anyway this is the q ij now there's an interesting thing you can do with this expression for the potential when you move the q ij overload it makes space it allows you to multiply it so I put a 3 here and this becomes a 6 down here if I do that the interesting thing you can do is you can take this 3 xi xj and you can set it off the trace of that which is delta ij times r square and put it next to the term in there so it doesn't change the answer it doesn't change the answer because this delta ij term contracted against q gives the trace of q which is 0 so actually this term doesn't do anything do the answer, it doesn't change it however what it does do is it shows the second order of contribution to the potential and now it involves a complete contraction of one k equals to tensor with another k equals to tensor both are the same for both symmetric and faceless this would involve the source point this would involve the field point anyway so this is one of the ways that the, this is the way that the particle moment tensor arises from the classical now here's where it gets applied in for example nuclear physics again let's suppose we've got a localized charge distribution here I'm really thinking about the nucleus there's an ordinary system here x, y, and z and let's take a field let's take a point here r, r, and a charge distribution I won't try it at this time let's suppose we have also some potential coming from external charges somewhere out somewhere else there's potential by external now on this blackboard I'm just going to call this phi r because I get targeted by the external but this is not the same phi as in the blackboard above the blackboard above the phi is the potential produced by the charge distribution here it's an external potential which is interacting with the charge distribution the nuclear physics, the idea is that by external it comes from the electrons that are elsewhere and the rho of r is the nucleus alright so what we would like to find is the energy of interaction from the nucleus in the presence of this external electric field so to do this the energy in classical theory of a charge distribution in an external potential it's called a w it's the integral over space and the charge density rho of r times the potential that's all it is so what we do is to put this in more convenient form is to expand the potential about the origin if you think of the nucleus the range of the variable r is actually very small and so it's reasonable to expand the external potential presumably slowly varying over the extent of the nucleus around the center of the nucleus of some origin of the nucleus and if we do this and we just make this a Taylor series this becomes phi 0 plus the radius r dotted in the radius of the phi plus one half of right of this way sum of i and j of x by x j times the second derivative of the external potential with respect to x by x j evaluated at zero and I'll stop there because you can keep on going obviously and if we do this and we see the the integrals over the charge density can be done we first of all have just the total charge q itself so w is equal to q times phi evaluated at zero that's the first term as far as the second term we go we should have the charge weighted radius r that's the dipole moment so it becomes plus the dipole moment d dotted into the gradient of phi evaluated in the origin there's a quadrupole term we'll come to in a moment before I do that allow me to observe that the gradient of phi is of course minus the electric field evaluated at the origin and if I write this term in slightly different forms minus d dotted into the electric field this is the external electric field evaluated at the origin and now you see the energy of the dipole in the external field so now it's the electric dipole which minus d dotted into d as far as the second word of the term goes let me take this this expression before I do the integrals let me take this expression and massage it a little bit so I have three so I have one six of three x i x j let's also use the fact that we'll assume the charges are far away so the sources of the external potential are way too nucleus if that's the case I mean the plus is equationally the plus theta phi is zero and so that means that this tensor is actually traceless because the gradient is the trace of this tensor and so in this form here where I type the second derivative I can subtract off the trace with a one third factor delta i j times let's write it in positive five because that term is zero anyway but the result is that I have symmetric and traceless tensor now having done that I can use the same trick as before I won't find this by three x i x j since the thing in the square of that track is a symmetric and traceless I can subtract off the trace of this term so to make a long story short this whole thing if you look in this way is one sixth sum on i of j of three x i x j minus delta i j r squared times the whole thing in the square of the brackets and the point of this is is that I'm now going to integral over the density but it makes its appearance the same particle tensor that appeared in the other problem and so the net effect of this is the energy of interaction of the nucleus for the external field is q times five evaluated in the position of the nucleus minus the dot for the moment factor dotted in the electric field evaluated in the center of the nucleus plus one sixth of the dot for the moment the tensor of the nucleus multiplied times the second derivative potential evaluated in x i x j with respect to x i x j evaluated in zero like this so this becomes the expression for the energy of the nucleus and the presence of the external field this is almost all I'm going to say about this for now it's just a setup for a homework problem but I might add just one more thing which is that it can be shown in the example model of the nucleus vanishes and this comes from the conservation of parity and time reversal which we'll talk about later on but as a result this time disappears and this is why the particle moment term is so important because it's the first amount of energy correction in the coulomb interaction which is the obvious part and this gives rise to additional contributions to what's called the hyperfine structure of the time expected so I just want to say more about that now alright so that's all I want to say about maybe I'll write up some more I'll turn to a new subject now which is that parity this puts the end to our study of rotations and all this work we've been doing in rotations we've been talking exclusively about proper rotations proper rotations are rotations that belong to the group n sub 3 and that just means that they are plus 1 it means that as operations on space they preserve the handiness of the coordinate system they don't flip the right hand or the left hand but I'm going to introduce now a certain operation so the geometrical setup I wanted to invoke now is the same one that we used earlier in talking about classical rotations that's to say we'll posit some inertial coordinate system x, y, z and we'll refer operations to the coordinate system so before we use rotations we're rotations about the origin of the coordinate system now what I like to do is introduce an operation I'll call p which in words we'll define we'll call the spatial inversion of operation and what it does is it takes a vector it just maps it in minus itself spatial inversion because you start here with the vector r you just go up to the opposite side and you just reflect it through the origin to write this as a matrix p is that matrix with minus 1 is in the diagonal and 0 is in the off diagonal in other words it's just minus the identity you can see that the determinant is equal to minus 1 it's also an orthogonal matrix so this is an example of an improper rotation so we're talking about improper rotations now now previously we were talking about proper rotations when we switched to quantum mechanics we were interested in the unitary operators in our quantum mechanical systems which implement the classical rotations so we can write this this way it's that there's some mapping that takes a classical rotation and converts it into a unitary operator reacting on our quantum system depends on the system what the operator is so let's assume we have some quantum mechanical system around what I'd like to do now is do something similar for the case of the spatial inversion operation which exists at the level of classical geometry spatial geometry to convert this into an operator which we'll call pi that acts on our moment space the problem is to find the operator pi this is a similar task what we did earlier when we were looking for what the operators U and R were although this is not a much simpler task because there's only one P and there are lots of Rs so what we did before in front of the U of R was we said the U of R should reproduce the multiplication law of the classical rotations we're going to use a similar criterion for finding the operator pi so let's review what are the properties of the spatial inversion operator at the level of geometry in three-dimensional space here are the principal ones that we want first of all if you take the operator P and square it the identity has as obvious from reflecting from the origin twice the next thing is that P commutes from any rotation any proper rotation and this is also obvious because P is just minus the identity matrix so an identity matrix commutes with everything it actually commutes with all operations not just rotations so these are the principal properties let's borrow those into quantum mechanics and so we're loading up with requirements for postulates on the operator requirements for pi well the first requirement is that it should be unitary and we make that requirement because we expect symmetry operations to preserve probabilities we did something similar with the rotational operators secondly let's require that the square of pi be equally identity because that's what happens for the spatial rotation operation and then thirdly let's require that the pi commutes with rotation operators because it does so at the classical level as well this is for OR and SO3 all proper rotations these are our three requirements on the parity operator pi now these three requirements don't uniquely determine pi but they narrow the possibilities down there considerably so depending on the system it becomes possible to find an explicit formula for pi let's begin with the simplest case which is a spinless particle in 3D in which the wave function of psi of r as we know we know that if we apply a rotation to this this gets mapped into psi of rotation inverse multiplying by r what do you suppose should be the definition of the action of parity on the wave function well it's logical that we should use P inverse here's the P inverse acting on a radius factor but P inverse is the same as P as minus identity so this becomes psi of minus r so that's a guess for the action of parity as well as this guess as a matter of fact for the rotations there's another way of doing this instead of looking at wave functions let's look at kets rotation operator U of r acting on a ket acting on a position operator what it does is it just rotates the position of the ket and so likewise it's a good guess that pi acting on a position of eigenket would take it into minus the eigenket of the spatially inverted position these two statements here on the acting on a position of eigenket are completely equivalent to the two above the line which refer to wave functions guess for a definition of pi let's see if it satisfies our requirements first of all is it unitary the answer is yes an operator is unitary if and only if it preserves all scalar products so let's just look at the norm of the wave function of a parity inverted wave function of psi of minus r this would be integrating over all space this is the absolute value of psi of minus r of 1 e squared just by doing a change of variable taking r of minus r this is equal to the integral dq of r of psi plus r of y squared and so the norm of the wave function is preserved I think this is clear anyway just to make sure what this does so it is indeed a unitary operator secondly is it squared equal to 1 the answer to that is yes obviously if we did this operation twice we could flip the psi twice we could do this operation twice thirdly does it commute with all rotations the answer to this is again yes let's just put it in different orders let's take psi of r as an initial wave function let's apply first of all a a rotation operator called u of r then this turns into psi of r inverse r and then let's apply parity by this provision of definition this becomes psi it just takes r that places it by a minus r so we get psi of minus r inverse r like this and if you do this in the reverse order psi of r of parity first this goes into psi of minus r and if you find the rotation next this goes into psi of minus r inverse r and so they agree and we see that this parity operator commutes with all rotations of course if an operator commutes with all rotations that's our definition of a scalar operator and we can say that one of these requirements of postulates to the parity operator is that it should be scalar that means in particular commutes with all rotations it commutes with infinitesimal rotations therefore it has to commute with the angular momentum of the system j whenever it is it's just another way of writing a property 3 we call this 3a and 3d if you want but in a particular case that we're talking about spinless particles from 3d the angular momentum is of course the orbital angular momentum and so it follows we've probably done that for this particular case of a spinless particle in 3d that it commutes with orbital angular momentum this is the commutation relation you can check directly from the definition of how that's a wave function if you want to it's a necessary consequence of the commutativity of rotations alright so this will be we'll take this as our definition of parity then for spinless particles the general strategy here is to study the properties of the parity operator and then we'll turn to the question of what happens with Hamiltonian's commutative parity when they commute with parity when they don't when they do so right now we'll be working on just the properties of parity next question is one about spin let's take it one step at a time let's talk about a case in which it's only the spin of the particle we're interested in and we want the spatial part so in this case our Hilbert space E is the span of just the spin eigen space of the spin space that's the animal like this where the animal's remind us that's not the problem so another question is what does the parity do to one of these space SM well we use the fact that the parity has to commute with the eigen of the system which is S that's the consequence of 3D up here so let's take a look at first of all the case of S and Z so if we apply S and Z to pi acting on SM this is the same thing as pi times S and Z acting on SM which is times pi acting on SM so you see pi acting on SM is actually an eigenstate of S and Z with eigenvalue M but there is only one state in this Hilbert space that has such an eigenvalue of SMC so this must be equal to must be proportional to SM or C is some proportionality factor but in other words all of the basis factors in the standard angle of basis in the spin space are eigenstates of parity with some eigenvalue which is calling C here well the fact that the parity operator squared is equal to 1 means that the eigenvalue must be equal to plus or minus 1 so this C value here becomes plus or minus 1 what is called C here which equals to plus or minus 1 like this well I'm running out of time I am out of time so I want to stop I have to stop let me just say that it's too bad to stop right here it's just in the middle of this argument the next step of this argument is to show that these eigenvalues of the parity operator for these basis states are actually independent of them and thus there is a single eigenvalue which is plus or minus 1 that applies to the entire spin space that's the next conclusion in fact it's exactly the same conclusion that we made earlier when talking about Hamiltonians that the energy eigenvalues are independent of that magnetic formula here we say pi is the scalar operator so what's eigenvalue equal to 10 in here


Hello, everybody, welcome back to another Mid Journey, version five tutorial. In this one, I'm gonna show you how to create a logo in Mid Journey, and then more importantly, I'm gonna show you how to turn it into vector art and turning it into vector art for free. There's no charge on this. You literally, in the next two minutes, you're gonna become the best logo designer, the best logo creator you know. Let's get started. First step, go into Mid Journey. As you see, I'm in here in the Mid Journey bot. Forward slash imagine, and then copy in what you want your logo to be. It can be anything. The only thing I recommend is using white background, and the reason why is easier to remove the background when it's in white as opposed to a gradient. Anyways, I've published that. I punched that in. I'm gonna let it run. When it's done, we're gonna come back and we're gonna take it to the next step. All right, it's done. Welcome back. So if I just click on this image here, this grid of four, you'll see here that it gave me four different cracks at it. I quite like all of them, but the top right one really appeals to me. So what I'm going to do is I'm just gonna click on this U2, which means Upsample or Upscale 2. And by the way, if I wanna create variations, I can click on V2 or V Anything, and it allows me to go ahead and remix it. So if I wanted it to have a, instead of flat 2D, I could do flat 3D, for example, and click Submit, and then it will rerun it for me. So it's a little bit outside the scope of this tutorial, but I wanted to show you that technique. Okay, so we've got the image or we've got what we want here, and we wanna turn it into a vector art logo. How do I do it?  When I save it, I'm just gonna save it to my desktop, and then bang, there we go, I've saved it. The next step is I'm gonna go to my browser, and you're gonna already see here that I've done it once already, but I'll do it again. So what we're gonna do is we're gonna go to vectorizer.ai, the link will be in the description below, and I'm gonna go into Finder now, and I'm gonna grab that image we just made and downloaded from Mid Journey. I'm going to drag and drop it right here, and then it's in vectorizer.ai, and then very simple. Here's the original image. Here is the vectorized result. So this is the vector. Now I'm gonna go ahead and click on Download, and when I do that, you're gonna see here at the bottom, then it says dot, dot, dot SVG. That means Scalable Vector Graphic. Now I'm gonna go ahead and show it in Finder, and I'm gonna go ahead now. I'm just gonna right click on it, and I'm gonna open it with Adobe Illustrator, and the reason why I did that is because this way, when I click on it, and I'm just gonna zoom in here, pardon me, when I click on it, you'll see here that this is vector. So there you go, this is how we create amazing vector art vector logos in under two minutes. Thanks for watching.


Hey everyone, welcome back for more ExoZero's content. In today's video, we're going to discuss Unleash potential of Faded Heroes. So this is a part one, I think, of a five part series or a four part series, I'm not sure yet. But before that, if you want to see more ExoZero's content, please hit the subscribe button down below. Hey guys, welcome to our second Unleash potential video. The first one that we had was for Generals. I'll put the link up there so that you may take a look at it later. But for this one, this will be an Unleash potential video for Faded Heroes. These are technically your lieutenants. So this video will feature the following Faded Heroes. So you have Rudley, Rera, Valentina, Shakan, Jin, Naomi, and Sewa. So they are your lieutenants and they are the first batch of Faded Heroes that will feature for Unleash potential. So just take note first before you Unleash the potentials of any hero because again, it's really hard to find the materials for them. Number one is always check for who you want to Unleash first. Okay, in the hierarchy of things, it's preferable that you have that you need to Unleash first Generals before the lieutenants. So this batch because Generals affect a lot of people in terms of their passive skills. And also you need to address as well on what skills or stats do you need to increase and to address first. So first is who and what skills and stats. Also take note of their fate course because some fate course have new stats that you need to take note before you do Unleash potential. Like for example, I'll start off with Rudley. So for Rudley, let's take a look at his skills. So he needs to be increased in health and also in yeah, in health. But when you look at his fate core, I don't have FC Rudley's fate core, I'm sorry. So I need to take a look at where he's at. So he's here. He's here in his fate core, just a preparation if you want to get him. So you need also to increase his, so again for current health. But the other one is defense. So you also need to increase his defense stat to also make sure that you give an increased defense based on your defense percentage and defense on own defense, yeah, on Rudley. This will be based on Rudley's defense. So this is very important for Rudley. So to start with Rudley, let's go back to his, to his, what do you call this, to his Unleash potential. So let's go here. So my recommendation is you unleash his potential under stamina first because you need to increase his health here and his health here. And you eventually will need, sorry, this one. So grants man to self when the round starts. So this is actually good for him. The second one that I would highly recommend is his defense because after you've seen in his fate core, you will need defense.  And ultimately for his Unleash potential skill, effects, sorry, resets turn one time per round if old mana is under three after skill hits. I think that's the same for this one, yes. So for Rudley, stamina first, then defense. Next would be Rera. So Rera is the lieutenant of Bathory. So for Rera, I've advanced checked her FC, no other implication in terms of her FC. So I'll go straight to her, to her skills. So also for her skills, I've also checked it. There are no stats that you need to increase. For her, you should focus on stamina as well because you will need to, you will need to basically reach this one, the Unleash potential effect for additional 1,000 pierce damage. This is actually good. And you also, you go through health as well. And I think there's also a speed here, yeah, speed increase for Rera as well. So go through the stamina. And the other one is you go to attack because she's an attack based hero. You will need this one as well. Bind for two turns and it also is here. But I would suggest you prioritize attack because she's an attack based character and you increase her attack here as well. Okay, so next would be Valentina. So Valentina is the lieutenant of Shufraken. Actually, she's the daughter of Shufraken, so sorry. She's the daughter lieutenant. Okay, so for Valentina, you don't have anything of a stat that you should need increased. So for her kit, her Unleash potential, you start with stamina because you'll need additional, number one stamina for additional health for survivability. And you have to reach this one for additional damage for pierce. The second one would be defense tree because Valentina doesn't need really to deal a lot of damage.  So you need her to survive because she has, her attack speed is pretty much very low and she's the last to go usually. So she needs to survive the round for her to apply either her S1 or her S2 bind and mana burn. So for Unleash potential for defense, you'll get another bind as well. But I think for her, this is not needed, but you can get this anyway. But you have the option not to get this one anymore. So stamina and defense because for her survivability. Next lieutenant that I'm talking about is Siakan. So he is the lieutenant of Rachel. So again, nothing, nothing important for him or his FC to what do you call this to, to enhance in terms of stat. So we'll go to his Unleash potential. So Unleash potential for Siakan is you start with stamina because number one, he is a, he is a defensive character. He needs to block and he needs to apply block to the other heroes. So health as well. And you have to get this one. This is a good passive for Unleash potential reduces all allies damage by 15%. So can't be stacked with other similar, similar abilities.  So you need to increase his defense. And for his addition potential deals additional 45% of own defense when that attack hits again, you will need to make sure that you, you go through the defense tree. And also once you go to the attack, um, so deals additional 15% of own defense when attack hits similar to your defense tree. Okay. Next is Jin. So Jin is also the lieutenant of, of Jinai. So again, I checked his skills for his regular form and for FC, no implications in terms of stats. So we move on to his Unleash potential. So again, for Jin, you need to go through stamina because he also needs survivability in terms of health, but you would, would you would want to need, you would want this one, the deals 1000 piercing damage when attacking because he needs a little bit of attack damage in a way compared to Baraka. He has less damage that he deals compared to Baraka. So and also once you're done with the stamina, you could either go defense or attack, but I would prefer you go with attack because he's an attack based character or hero. So you go to this one, increase attack, increase speed, but you will have to get this one eventually for the bind. Bind is they, they don't gain mana for two turns. So that is the skill that is what it's all about. So you can actually spam this one if you have enough mana. It's only one mana cost for this spell. Okay, so let's move on to Naomi. Naomi is the, where's Naomi? Naomi is the lieutenant of Dorca, correct Dorca, sorry, but actually, but actually Naomi is the queen. So I'm not sure if she can be the lieutenant, but anyway, she's the second in command in terms of hierarchy among faded. That's what I want to tell you guys. So again, for her, for her skills, so S1, none, but for her S2, you need to increase her health because of this one. There's a multiplier for, to increase the health, own maximum health by 46%. So also increase own defense by 51%. So this is in terms of shared health. So you know where I'm going at. So when you take a look at her, her unleash potential, you go first with, don't go first with defense, go with stamina because you need health, of course, because of her S2, but also you need this one. The reduces all allies damage by 15% can't be stacked. You need this one first, then you go next to your defense tree, which increases all defense by, all defense for six turns by 15% of own defense. So I think she has this as well here. But for additional damage, these additional 30% of own defense when attack hits.  The last lieutenant that I wanted to feature here would be Seowa. I hope I'm pronouncing her name right. It's Seowa or say S-A-S-A-O-A or no, it's Seowa. Anyway, for Seowa, for her skills, again, you need to increase her health. And yeah, that's it, health. I actually, Seowa is one unique passive actually, not really unique, but there are only a few in this, in this, what do you call this, in the Fated Heroes. She needs to increase her health, right? So you start with the stamina tree, health, health here. But if you look at her Andish potential, she reduces damage against all allies by 30%. I've seen other characters have this one. I think it's only Seowa, Ramji and Shell that has this one, the 30%, reduced damage by 30%. And this is big. If you intend to bring Seowa in terms of, I'm interested in bringing her in PvP tag. So please make sure that you have this on her. You finish the tree first, then you go next to this one. The defense tree for additional defense, so maximum health, reduced additional damage. And this one is grand status effect immunity, one-time self can't be stacked. Okay, so anyway, guys, that is all for this video. Please put some comments below, any suggestion, comments on what they call is on the Anish potential part one. Also, don't forget to subscribe to my channel. This really helps me put out good content and also put out regular content as well. Anyway, guys, this is it for me. So this is The Warden and I'm out.


Okay, very good morning to you. Hope you're doing well. Happy Friday, 5th of February. Just a quick shout out to the Amplify Live community. It's been a really great week just for the guys trading but also in terms of content. I had a really great interview with one of the traders who's going on to bigger and better things, joining one of the biggest remaining kind of prop firms, Geneva Trading, based in Dublin and Chicago. So good luck to him. And then we've had a really great session and an update in depth about the vaccine situation and the ongoing global rollout program, which is really critical to understand and the shape of what the future looks like in the coming months, particularly the second half of the year. And then we've had the live streaming sessions from the Bank of England yesterday. We'll cover the NFP in full later on this afternoon. So remember, if you are watching this on YouTube, this is delayed because the Amplify Live guys get it early. So remember to check out the link below if you're interested. But otherwise, let's get straight into the market and talk about what's going on today and a little bit of a look ahead to non-farm pay rise and what we can expect there. But really, I just wanted to wrap up the week with a quote. And don't worry, I'm not going to start spouting high coups at you. I'm just going to talk about someone, a piece I was reading and they said, quote, the virus is getting incrementally better at the very same time the earning season economic data seems to be showing improvement. I think that's a really nice summary generally of what's going on in markets right now. Remember, this week, what we've had, and this is this week alone, you've had ISM non-manufacturing PMI the highest since February of 2018. In terms of the manufacturing, the highest since February of 2019 ADP 174,000 against expectations of 49,000 weekly jobless claims to third straight week of declines in the lowest number since the last week of November, Amazon, Alphabet, smashing 100 billion or 100 billion, I should say revenues in a quarter. So overall then, all of that has come in the context of federal reserve officials continuing to push against any notion of tightening irrespective of the economy showing some signs of heating up. Now, one of the latest was Esther George. She is a non-voter, but she spoke last night and she said the Fed are still, quote, far away from achieving its goals and will be premature to start a debate on scaling back its massive bond buying program. Again, very similar to what we heard with Evans the previous day and Bullard and others. And someone like Esther George typically is defined as being a hawk and yet she herself is saying it's far too early to start calling it the end of the bond buying program. So I think overall you've got a bit of a perfect storm here when it comes specifically for why equity markets are moving back to all times high, all times high. You've got a bit of momentum picking up economically supported by the lower for longer mantra, which is definitely here to stay. And the idea being that the people who are pulling these levers making these decisions from Treasury and Yellen to Powell at the Fed, everyone is there to provide this supportive environment going forward. So let's have a look at some of these charts and really encapsulate that mentality because here is the S&P 500 and technically then an important breakout we had in price activity going into the close. So that usual kind of flush in volume going into the close on the night last night just saw us push through that double top and fresh all time highs then. And then as we drifted in the futures in the overnight Asia Pacific session, build that momentum just fatigue after the close and then Asia just took it on as well. Again, in terms of the kind of geographic sentiment, it's definitely being Asia is definitely has its own unique domestic narratives, but definitely not moving the global kind of sense of sentiment. And that's definitely been derived from generally the high close on Wall Street, Asia generally following suit. And in Asia, we've drifted up futures still pushing on up. And if we look on a daily on the S&P, you can see then quite clearly that 3860, which was that kind of obstacle that restricted price at the end of January broke that last night. And that's now providing a degree of support to price. And, you know, looking on the upside today, whether it does or not, you know, the idea is here, this market, I still feel for 2021, it's not just going to 3900. It's not just going to 4000. It's going higher. So all things remaining equal, I think there's enough of the groundwork set on the just general summary of what I was just describing to see equities just continue the upside trend. That doesn't mean, though, that as we go up, we don't have the odd afternoon or day of pullback. But that doesn't detract from the trend, which is, you know, this is looking on a daily since November that this kind of trend just continues. Perhaps the acceleration of the upward gain starts to decrease slightly. But the market's still in a bullish mindset, I think going forward. And I think US equities have got plenty left under the bonnet. Otherwise, elsewhere, what was really interesting about yesterday's session was a lot of the dollar based movement. And this is a look at the Dixie looking back to really encapsulate the story of the pandemic. So here you can see the initial volatility, the push up on the kind of flight to quality bid in the dollar at the depths of the sell off that we saw in March and global equities. And then generally, the response obviously coming heavy handed from the Federal Reserve to instill ultra loose monetary policy. And that's just seen the continuation of dollar weakness. But taking the reference point from the May high and a retest in November, we had yesterday, this is something I was talking about a lot as a trigger point then for subsequent movement elsewhere. And I'll explain this now is that the dollar broke out after two attempts on this trend line yesterday. And that coincided then with your dollar breaking in a very meaningful area, which was 120. So we did looking at the your dollar futures here, close below 120. And that is meaningful from a technical point of view. It's a big level. Psychologically is one in the past of which the ECB is used as a bit of a reference point in terms of their uncomfortableness in terms of the euro level. And it's acted as resistance before. And that was a real severe about turn previously back in September, when the ECB originally voiced their concern. And you can see there, it was strong resistance. And here, how meaningful it was when it broke, because that really saw the euro fire up. And we eventually moved up even four points higher than that in the following weeks thereafter. So we're back below that point. And that was what was interesting. The initial Dixie trend line break that then led to the euro was consolidating at the time around this 120. But when it broke down, that further lifted the dollar. And then subsequently, that started to weigh on some of the precious metals. So here, we saw a nice breakdown in the gold price. And you can see we dropped down to around 1811. You can see how technically driven the gold market is and momentum based when it breaks these levels. So under the prevailing strength of the dollar, which has been persistent this week, you've had gold consolidate 1833 was a big level that breaks, the price comes down pretty quickly. And then again, the 1811 was a big area and it dropped down. And this was yesterday afternoon at mid some of those currency moves. So we've recovered a little bit here. Just breaking out now some of the Asia pack range. It's got a bit of room to the upside 1800 of course is psychological. And 1800 is noteworthy given that that was the low we had back on the 19th of January, as you can see here. So these were a lot of the technical points that were helping assist some of the downsides sell off yesterday. And similarly, silver was a was a similar price pattern in terms of it broke through a lower bound of its trading range for the time being. So at the moment, the Dixie is flat pretty much. So not having too much of an impact on markets right now. The point is, it's held on to its relative gains throughout the week, which is which has been pretty, pretty decent move in the greenback. For cable, of course, yeah, really powerful move on the back of the admission really of any great detail in the statement or their policy report about negative rates. So it's definitely the idea I think that the Bank of England wanted to convey yesterday was that look, we're talking to banks, we're getting them to prepare for the what if scenario if negative rates do need to be adopted in the future. But the idea being is that, yes, Q one, perhaps growth wise looks a little bit more tepid than perhaps what they previously saw back in November's projections, just given the variant of the virus and the subsequent lockdowns we've seen. But the second half of the year looks promising for a strong return and that therefore rendering further stimulus in a policy sense like negative rates not really appropriate. If all things remain equal, and that is the deployment of the vaccine vaccination program goes as it has been, which is fairly successfully. And so we had a really big about turn in sterling. And yeah, we've got a couple of speakers today, Bailey himself, you've got the the deputy, also speaking, Ben Broadbent later on this morning. So Broadbent 10, Bailey 130, I wouldn't expect them to say anything new. I wouldn't be looking for them to really shift the communication from what the Bank of England said yesterday. But again, another, I guess, meaningful cushion with the vaccination situation has helped sterling somewhat outperform a little bit the euro, which technically now also is is going to trade perhaps little heavy now it's through that 120 for the time being. And then following on from your time highs and equities that that kind of narrative then of this kind of semi Goldilocks scenario means that oil prices are still just loving life for the time being. And again, we've had a couple of meaningful kind of technical platforms for price to work off, which originally was around this 5625. We broke through that yesterday evening. And then we've just accelerated up and then in Asia Pact session, had a bit of a floor for price of which it is just now testing.  And, you know, yesterday, we found a bit of hiatus at around that $56 handle, which was what we're looking at for a target to book some of the longs on the break break of that Feb 20 2020 high. And now we had pulled back decent pullback to 5530. But here we are back trading a dollar and 20 back above that area. And having got above that zone of resistance really upside technically on the daily chart, I don't really see a great deal until we get further up to 5736. So how that plays out, I think a solid payrolls. And we'll talk about payrolls. Let's just talk about payrolls now. So a solid payrolls, I think is very much kind of priced into the market. I think if you did get a good payroll report, perhaps you just see a bit of a continuation, but in a moderate sense to the general directional trends that we've had, which is kind of equity strengths, dollar maintains, it's kind of bid from the week. And we have yesterday seen a selloff in 30 year Treasury is pitching the US yield curve to its steepest in more than five years. You know, don't forget as well, there's still this idea that stimulus, a large additional injection of economic stimulus is still forthcoming, albeit it might be watered down as it goes through its various kind of procedural passage in Congress. That with the global growth expected given vaccination drives gathering pace. And with payrolls kind of feeding that narrative just is why we've seen the moves that we've had in markets this week.  And as we were saying at the time, not one that global markets are really looking at. It's a very concentrated small catchment of people that are looking at those things that doesn't reverberate out to the broader market. But as far as non non fun payrolls is concerned, then, I guess for me, it's about appreciating the markets positioning. We've had a slew of good data, as I just mentioned, from the US. And that does mean then that expectations are for an upside print, because the actual consensus is kind of like with ADP was around 50 k for the headline, but definitely markets are sitting on the higher end of that. ADP in itself was 174,000. So yeah, looking at the employment constituent breakdowns of the various different things that we've had this week, they would all be indicative of a positive report. So yeah, the disappointment then is, what if it's not great? And just given that market positioning, do we get a little bit of a pullback? In the in summary, it doesn't really matter non farm payrolls because that might create an intraday interesting movement. If it is weak, do we get equities and just book profit going into the weekend? We see a bit of a pullback that feels heavy. But in the in the context of the week's gains, it's really quite marginal. And then with the dollar longs as well, profit taking just looking to just come out of some of that and the euro just drips back up to 120. That could well be a scenario today. And maybe you get some bottom feeders coming down in some of the precious space just to lift things up into the close under those conditions. But does it really change the the idea of what the Fed are going to do and the tactical approach of just being far away from anywhere near to discussions about ending bond buying? No, it doesn't. So beyond today, I don't think it really changes it. And hence the reason why some of these patterns that we've seen form throughout the week might just continue thereafter. But look, a few other things to briefly mention away from that, that's kind of the overall take and generally the most important thing. So a couple of side notes on the vaccine front, Johnson and Johnson have asked us drug regulators to clear its COVID-19 vaccine for emergency use. If cleared, it would be the US a third vaccine for the US after Pfizer by Nteca Moderna, of which their kind of rollout strategy has been heavily dependent on. Again, we did a really great chat with one of the guys not me, the content really coming from him, one of the guys in the community and amplify live. If you would like me to share that, then I'm happy to do so. Just drop me a comment in the link and perhaps I can put out an edited version over the weekend on the YouTube channel. But for me, understanding really the situation of viruses, although it's not like we're seeing these headlines driving markets, but from a macro perspective over the medium term, that is the critical key to understand where we're at there, and with the vaccine and distribution and rollout as to then timings and what the geographic timings would be on the economic recovery that we could see through the second half of this year. So it's really important to understand that stuff. So J&J looking to get that US approval separately, the FDA and the US are drawing up new rules to make it faster to approve treatments and vaccines for the new variants of COVID-19. And then the last thing is, I thought just an update generally, because people don't really mention too much at the moment, the numbers and cases and hospitalizations and so on in the US. What does that look like right now? Well, hospitalizations yesterday fell below 90,000. And as a marker, that's the first time that's happened since late November. So a couple of weeks, while cases they have risen by the most in five days. So this is one of those. This is another key reason why this kind of Goldilocks scenario exists. So the economy is is heating up through macroeconomic data points, as we can see this week, earnings looking generally robust. But COVID-19 in itself is still a long way from going away. And so the Fed and policy officials, whether government led with their federal state level strategies, or from a monetary policy perspective, don't have a choice, they've got to keep, they've got to keep the stimulus checks coming, they've got to keep a low lending environment coming. And that's why you're getting this kind of perfect storm almost for equities just to continue back to all time highs. The other thing I just wanted to mention was Italy, just very quickly. Yeah, definitely got the call wrong yesterday thinking that it would take a bit of time perhaps for Draghi to pull the relevant strings in order to form a government. The latest is that Italy's PD Senate Chief expects Draghi has an 80 to 90% chance of forming a government and without getting too bogged down into the details, basically, if you look at the the kind of strategy of who we can talk to to acquire how many votes to form a government and it's basically in any shape or form depending on which way he goes is looking likely he will be at a former government. And as we know Draghi an incredibly well respected individual, both from a political standpoint, but also from a market's perception. And the yields in Italy have just continued to fall and we've broken out of that important technical range in BTP futures at 52-14. You can see here the high on the 20th and this air of resistance we've had throughout the week since that aggressive gap up on the confirmation of Draghi towards the beginning of the week. And BTPs have continued to rally into the close yesterday. So upside targets, you've got the previous high we had, which is the year to date high in BTPs at 52-66. Alright, quick look at the calendar. So aside from non farms, what else is going on? Well, the UK and European morning is very quiet. There's nothing really of magnitude to be aware of. So all things being equal, it could be quite quiet and the run into the jobs data payroll hits at 130. So I'll be on the live stream with the team 130 on Amplify live. And then going into the afternoon that really is the main thing to look out for. So with that, I'm going to wish you a great session ahead. Have a great weekend, take care. And if you're watching this on YouTube, as I always say, really appreciate it. If you could like and subscribe, help grow the channel. And hopefully other people can benefit from this this macro fundamental run down every morning. Alright, guys, take care.


You may remember this distressing video going viral in 2018. Now in that clip you saw Jamar Hijazi then 15 being pushed to the ground and threatened with drowning. Hijazi and his family had come to the UK as refugees from Oms in Syria. It was obviously a horrible welcome for a schoolboy fleeing war. It was made worse when Stephen Yaxley Lennon, otherwise known as Tommy Robinson, made a Facebook video claiming Jamal was not innocent and violently attacks young English girls in his school and that he had, quote, beat a girl black and blue, unquote. Hijazi claimed that Yaxley Lennon's video led him to be subject to death threats and extremist agitation. And this week he successfully sued the far right activist for libel. Mr Justice Nicklin found that as was entirely predictable, the claimant became the target of abuse, which ultimately led to him and his family having to leave their home and the claimant to have to abandon his education. The defendant is responsible for this harm, some of the scars of which particularly the impact on the claimant's education are likely to last for many years, if not a lifetime. Yaxley Lennon had fought the case on the grounds that what he had said was true or his claim was that what he said was true. This was rejected by the judge. The judge said, I am satisfied that if the claimant had behaved so repeatedly in the abusive manner alleged, including to members of staff, then this would have been recorded in the claimant school records. There is no trace of any such behavior by the claimant in these records. On the contrary, his behavioral record is overwhelmingly positive. So he's saying Tommy Robinson's claims do not stack up. And they also caused an incredible amount of harm. He said the language used in the videos was calculated to inflame the situation. Robinson has been ordered to pay damages to Hijazi of £100,000. He is also liable to pay his legal fees, which the BBC reports were around half a million pounds. Tommy Robinson has since declared bankruptcy. A horrible story. The consequences of Tommy Robinson sharing those lies was incredibly horrible.  You've already been bullied by people in a country you've just moved to after fleeing civil war. Then you get death threats from far-right activists because Tommy Robinson has made up stuff about you on Facebook. The payout to the families of £100,000, I don't think that touches what's happened to them. I don't think that goes anywhere near compensating them. On the other hand, if Tommy Robinson hasn't even got that, then what difference does it make? As I understand it, their legal costs are also around half a million pounds. So Robinson also would have to pay that. He was already in financial trouble before this. Again, as I understand it, I don't want to say things aren't true because I don't want to do what Tommy Robinson has done. But from what has been said in the media, what's been written in the media, marital problems and so on and so forth. And what really, like you say, what really hammered him was before this, it was changes to the Facebook algorithm because that's what really took away his prominence and his profile, the fact that he was blacklisted effectively from Twitter, from Facebook, from Instagram. Without social media, he hasn't got any means to sell his product, which is hate, which is himself. And so he was a busted flush before this. But I do really feel sorry for the family because I don't think 100,000 pounds is such a size. I really don't. This young kid had to leave school.  This has been going on for more than a year. His family's already probably, they've probably all got PTSD from leaving Syria, fleeing civil war. I think it's appalling. I think it's really, really appalling. I think Tommy Robinson should be doing prison time personally. And I'm a prison abolitionist. I have no problem with Tommy Robinson going to prison, believe me. So I don't think it's a sufficient punishment. I really, really don't, Michael. And maybe some parts of the left might disagree with me for that. I really don't.  For me, this seems like inciting racial hatred. And clearly that's what happened. And I would have thought, especially this is about a minor, this is about a 15-year-old. It seems surprising to me that it's just a financial response. I think at this time, probably Tommy Robinson has less financial backers than he once did. But a couple of years ago, three years ago, he could have presumably called in money from, he had far-right backers in North America. Potentially that moment is over. And he has declared bankruptcy. But I think you're right. It does seem like he's, this seems like a light punishment from my perspective. Although a victory for the kid involved. Aaron, your final thoughts on this story? Yeah. I mean, his big financial backer came from Canada.  And he was expanding over here. I suppose one sort of one sort of counter argument would be, yes, Tommy Robinson was huge in 2017, 2018. He was finding a huge audience. I remember in 2018, before he got banned, some of the analytics were just deeply disturbing.  So on the one hand, you know, that yes, the algorithm screwed him. But at the same time, like we have to remember that in this country in 2017, we had a high watermark for socialist politics. It just goes to show how politically volatile that moment was, really, both on the left and the right. And it was very much a case of, you know, there's the center ground of politics. The common sense was up for grabs. You know, has that changed significantly? I do think it's, I do think it's changed. But I also think many of those questions are settled. I think many of the parts of the sort of economic agenda of Corbyn and McDonald are now broadly the consensus of the public. I think the Tories have to at least acknowledge that. And I think that ultimately on the immigration stuff, you know, the left lost big time, big time, big time. And it expended huge amounts of capital because of chasing a second referendum. So Tommy Robinson didn't win. But I think many of his viewers, his supporters, you know, we did end up with the government and a set of policy outcomes that they are absolutely overwhelmingly pleased with.


Good morning traders, Patrick Munder here with another tickmail chart. I just want to draw your attention today to some potential two-way opportunities developing in the sterling Swiss on the four-hour timeframe. You can see that we have prices of extending here and what could be a leading diagonal. We have 1.2, we appear to be in the third wave extension here, and two areas of interest for me. I'm looking for a pop hire here into daily and weekly range resistance, 1.2350, 1.2330, but to potentially fade strength here, looking for a pullback into the trend line support at 1.22 for the fifth wave to target 1.25. So as always, traders, watch for reversal patterns to confirm the setup and the trade. Trade the plant, most importantly manage your risk. Until next time, thanks very much.


This study demonstrated that a passive exoskeleton can be used to enhance vertical jump performance by storing and releasing energy during the negative and positive phases of knee flexion, respectively. The exoskeleton increased the average jump height by 6.4 inches, suggesting that this technology could potentially be used to improve athletic performance. This article was authored by Coral Bendavid, Baraka Straitch, and Roziel Riemer.


Hello, everyone. My name is Muhammad Eid. I'm from New York University, Abu Dhabi, and I would like to present our work on a comparison of Vibrotactile feedback and EMS for motor response during active hand movement. Wearable haptic technologies have garnered widespread attention due to increased accessibility, functionality, and affordance. Vibrotactile stimulation is probably the most popular since Vibrotactile actuators are small in size, easy to control, and, more importantly, affordable and expressive. On the other hand, EMS, a technique that applies electrical pulses to the user's muscles in order to involuntarily contract them, is becoming a hot topic as it provides a similar set of advantages in addition to its ability to provide force feedback. Both technologies can be applied at various body parts as shown in these many examples. The purpose of this study is to compare Vibrotactile stimulation and EMS to elicit a motor response during active hand movement. We created an experimental setup for this purpose that included a strip of five LEDs for visual feedback, a vibration motor in 9 degrees of freedom IMU for tracking the hand movement, and an EMS system. The stimulation cues were all set to 500 milliseconds with a 95% plus detectability. We opted for touching or avoid touching the face as the experimental task since it's an everyday motor task that goes often without thinking. Participants were asked to wear the wristband and keep their hand on a table in front of them. They were then prompted to touch their face during which sensory feedback was provided in the form of visual, Vibrotactile, or EMS. We recruited a total of 45 participants, each completed 100 trials. 70% of these trials were for the control condition where no feedback was provided, and 30% were divided equally among the three modalities. We calculated the response time and error rate. On the left-hand side, you see that EMS and Vibrotactile feedback have a significantly shorter response time compared to visual feedback. On the right-hand side, you can see that EMS has significantly smaller error rate compared to visual feedback. We also conducted a survey to capture the user experience where all participants reported no pain or fatigue for all the modalities. It's worth noting that 93% of participants reported Vibrotactile feedback as pleasant, whereas only 60% of participants reported EMS as pleasant. In summary, EMS feedback can be more favorable in critical applications where errors must be avoided, whereas Vibrotactile feedback is more convenient for non-critical applications due to its lower coast and higher usability.


do anything with it the shot was deflected I should say but outside it up really hasn't really hasn't been a police FC but here they come to change here Jones on his right foot so that's it well they found the net finally deadlock is broken and is that man Jabari Mitchell who's at the target that's the first goal of the new season a bit fortuitous with the deflection but nevertheless it's the first goal of this new season ball laid out to Jones and selfishly lays it on for Mitchell who strikes it defender has gone about blocking a shot in the wrong way charged out he just needed to make his body big the defender instead of turning his back to the ball and for all you young defenders looking out there it's exactly