Spin states and exchange interaction

hello everyone and welcome to this virtual lecture course on quantum condensed metaphysics i'm dr andrew mitchell and in this lecture we'll be studying models of simple spin systems this will build upon the work we did in the previous lecture developing the formalism for spin systems in the first part we're going to look at a simple system of two spin half particles and we'll actually see that these can combine to make a triplet a spin 1 object we'll study this in detail and we'll actually see that the triplet state of two spin half particles is actually identical mathematically to a genuine spin one particle the operator algebra is the same the observables are exactly the same everything is the same so we can actually in some sense regard a spin one particle as being made up out of two spin half particles in a triplet combination and actually um this will come out as something interesting and useful to us later on in the course when we talk about fractionalization and the whole damn spin gap in this lecture we're going to talk quite a lot about quantum numbers we'll see that quantum numbers are conserved quantities that are not changed by a hamiltonian when the hamiltonian acts upon an eigen states labelled by particular quantum numbers um they are not changed by the action of the hamiltonian they are conserved through the dynamics of the system's evolution according to the hamiltonian there is a corresponding operator for each quantum number and um a conserved quantity uh corresponding to such a quantum number is one where this operator commutes with the hamiltonian and we'll be discussing that in this lecture specifically we'll be talking about the spin and spin projection quantum numbers individually for each uh each spin in our system and also the global spin and the uh the global spin projection we'll talk about simple models of spin systems especially the heisenberg exchange model we'll talk about the microscopic origins of the exchange interaction and we'll see where that comes from and what role it plays for a simple system involving two spin half particles we'll see that the exchange interaction can actually be written in terms of the total spin quantum number and therefore we can trivially write down our eigenstates and eigen energies of this system in terms of those quantum numbers we'll see that the energies of the three components of our spin one triplet are degenerate but these can be lifted by applying a magnetic field and we'll talk about simple systems like this in this lecture towards the end we'll talk about the more general case of heisenberg spin systems involving many spins and we'll see that life is not so simple in that case and that we need a more direct way of uh of solving the schrodinger equation for such systems and that will actually come in the next lecture so let's get down to work in this lecture we will continue with the analysis of the two-spin system but first let's recap some of the formalism that we developed in the previous lecture first we consider two spin half particles labeled a and b each of these spin half particles can have a z projection of the spin which is either up or down denoted here by these cartoon arrows up and down we label the z projection of the spin of particle i by s said i and it can take values of either plus a half or minus a half in this case i labels either the a or b spins now we consider the combined a b system which has a larger hilbert space it has a hilbert space now of dimension four because we have four different possibilities for combining the two spins we can enumerate a suitable basis from product states of the complex vector spaces from a and b spins what i mean by that is that i can write a single cat for the combined system with up up denoting that there is an up spin in a an upspin in b and this is just obtained from taking the direct product of the state of a with an upspin with the state of b also with an upspin similarly we have states of our combined a b system corresponding to up down down up and down down states this is the full space of possibilities of our combined a b system for two spin half particles in the product state basis we call this the product state basis because the states of the combined system are simply obtained by taking the product of the states of the individual subsystems a and b but of course in quantum mechanics this is not the only option we can also have as linear combinations or superpositions of these basis states a given system described by a given hamiltonian may have eigenstates which are linear combinations of these basis states the product states that i've written down here constituting this basis are actually kind of like classical states because they are not entangled as we will see quantum mechanical entanglement arises from the superposition of states we will quantify this in the next lecture in terms of the entanglement entropy which will make this concept more concrete now for the combined a b system we can also define some spin operators we can identify for example an operator for the z component of a given spin i which is either a or b and indeed the raising and lowering operators for that spin here we're applying these operators to the combined state one of these basis states that we've just enumerated but the operator affects only a component of the combined system for example sa acts only on spin a of our combined system independently of what's going on on spin b we can then write down total spin operators s total z and s total plus minus which are simply the sum over the a and b spins of the individual operators finally an important object will be the total spin operator squared s total squared this is simply the dot product of the vector operators for s total and because s total is the sum of a and b we can write that as s a vector operator squared plus s b vector operator squared plus twice of the dot product between the spin on a and the spin on b now since the individual particles are spin half particles and s squared acting on a spin uh half gives us an eigenvalue of s into s plus one which in this case is three quarters these first two terms here just give us a constant the first one gives us three quarters the second one gives us three quarters because each of them is a spin half particle however this third term here s a dot sb is non-trivial this is an interaction between the two spins i can write this in terms of the z components of the operators on a and b spins and the raising and lowering operators we derived this equation in the last lecture now we can apply these operators to our basis states and we would find that although these basis states have a definite value of s said total meaning that um these basis states here are actually eigenstates of the total s said operator with an eigenvalue of the total s said they are not icon states of the total spin if i apply s total squared to these basis states i would find that in particular the up down and down up states are not eigen states of this operator however we can define a new bases which are simultaneously eigenstates of the s total squared operator and the s total z operator this means that each of the basis states will have a definite eigenvalue for the s total squared operator and the s total z operator therefore we can label these basis states according to the total spin and total z projection these are therefore quantum numbers here are the states we can label them by s total and s total z you see here on the right hand side in the definition in terms of our original basis that we need to take superpositions or linear combinations of our product state bases so that these states are eigenstates of the spin operator with this definition it is then easy to check that when we apply s total squared to these states we get correctly the eigenvalue s into s plus 1 times the state back again and if we apply s total z to one of these states then we get s said times the state back again these are the usual rules for the angular momentum operators so indeed it makes sense to label these states by the total spin and total spin projection quantum numbers in particular we see that we have a multiplet structure we have a single state with spin zero this is referred to as a singlet and then we have a collection of three basis states all with spin one with different z projections and this is referred to as the triplet state this all makes good sense from the perspective of addition of angular momentum you know from the klebs gordon series that if i take two spins one half and i combine them then i can obtain a spin zero and a spin one and that's exactly what we're seeing here let's just quickly remind ourselves of the rules of angular momentum addition in quantum mechanics the sprint projection s said is an abelian quantum number this means that it is additive if i want to know the total s said i simply add up all the s sets for the individual contributions for example imagine that i act on an individual a spin up with the sa z operator then that's of course an eigenstate of the saz operator and it has an eigenvalue of plus a half if i act on a down state on the b particle with sbz then of course i get the state back again with an eigenvalue of minus a half if i now consider the combined a b system with the state up down and act on it with the total s z operator which i can decompose as the sum of the s a z and s b z operators then i have an eigenvalue of zero so this tells us that the total s said is zero it's simply the sum of the individual components as another example let's consider this superposition state on the previous slide this is the one with a total spin equals to one and a total said projection equals to zero we previously asserted i'm going to label this with a shorthand notation as the ket 1 0 for the spin and z projection if i apply s total z operator to this state i can write that in terms of the individual operators acting on these individual cats and i will see that because each of them gives me s said zero in total um by the analogous argument from the line above then the overall s said for this combined state will be zero so here the total s said is also equal to zero but it's slightly more subtle because each of the basis states comprising this linear combination here has an s total z is equal to zero but you can see that for example in this first term the s said a is plus a half an s s said b is minus a half whereas in the second ket ester a is minus a half and said b is plus a half so the state 1 0 has a definite total s z equals to zero however the s said of the individual spins a and b uh is not well defined if we focus on the a spin in the first component is plus a half but in the second component is minus a half so the punch line is that this state is an eigenstate of the total s z but it is not an eigenstate of the individual s seds for the a and b spins by contrast the spin s is a non-abelian quantum number it's not simply additive you might remember from the clerch gordon series that s1 plus s2 can take different values ranging from the absolute value of s1 minus s2 in integer steps all the way up to s1 plus s2 and for two spin half particles of course this implies that we have an s total of either 0 or 1 the singlet or triplet which is the result that we found directly in the last lecture so for example if i take an up spin on site a and apply to it the sa squared operator then i get three quarters times the state back again this state is an eigenstate of the sa squared operator with an eigenvalue three-quarters that's because we have a spin half particle and the eigenvalue is s into s plus one giving three-quarters likewise uh for a downspin in b if i apply sb squared i get quarters times the state back again it's the same eigenvalue because the particle has the same spin independently of the spin projection on the other hand if we now apply the s total squared operator to this state which we introduced on the previous slide then we get zero that tells us that we have a total spin of zero so we have two spin half particles but when we combine them we can get spin zero this is a feature of spin being a non-abelian symmetry and s being a non-abelian quantum number it's not simply additive let's now talk a little more about the underlying symmetries here if we just consider the s total squared operator then we see that the three multiple components of s equals one can actually be interchanged by rotations we can understand this because if we apply the s total squared operator to a state with a definite spin s equals 1 then we will find an eigenvalue of s into s plus 1 is equal to 2 in this case times the state back again and that is independent of the s total z so whether s total z is plus one zero or minus one this relation holds so we see that all of the three multiple components of s equals one are eigenstates of the total spin operator squared and indeed they have degenerate eigenvalues independent of the z projection you might recall that if you have degenerate eigenstates of a given operator then any linear combination of those states will also be an eigenstate of that operator suppose for example that we define a particular linear combination denoted psi which is going to be a normalized combination of the 1 1 the 1 0 and 1 -1 states here i'm again labeling the states by their spin and z projection um with arbitrary complex coefficients a b and c because all of these three uh components of this linear combination here all have definite spin we will find that this wave function psi as defined is an eigenstate of the total s squared operator you can try it out yourself to confirm that that's the case this works of course because the s-squared operator doesn't care about the z-projection which is different in these three kets it only cares about the total s which is the same in all three components on the other hand if we were to apply the total said operator to this state psi we would find that it is not an eigenstate and that's simply because this state does not have a definite s said as you can see here explicitly it's made up from components with different sets in fact as we'll see doing this linear combination of these basis states is equivalent to doing a rotation of our coordinate system in the new coordinate system we could measure the spin projection as said and we will find that it would have a definite value when we talk about the spin projection as said of course we have some coordinate system in mind and we can of course do a rotation of our coordinate system to define a new z-axis suppose we define a state with definite spin and spin projection labeled by this ket s with s said and in this case i'm going to take the triplet of states with s equals one meaning that each of these three basis states here has an s equals one but an s said of minus one zero and plus one respectively so all three of these basis states are eigenstates of uh the s squared operator with the same eigenvalue s and s plus 1 which in this case is equal to 2. now suppose we take some arbitrary linear combination as on the previous slide where a b and c are complex numbers and this factor at the beginning here is just to properly normalize the state as i argued on the previous slide if i apply the s squared operator to this linear combination i will see that it is an eigenstate with the eigenvalue of 2. it is a state with a definite spin s equals one however we started off with three orthogonal states the object that i've denoted psi one here is just a specific linear combination of those states there must actually in fact be two other states psi 2 and psi 3 which are mutually orthogonal to be more precise what i mean is that since we can take the inner product or overlap of these three basis states and expressed that in terms of this kronecker delta meaning that the states are individually normalized but orthogonal to each other that implies that if we define this psi 1 as this linear combination there must be two other states let's call them psi 2 and psi 3 which are also mutually orthogonal so is this really the case and what do these states look like for example let's say we take as our psi 1 the symmetric linear combination of our basis states then i can define explicitly a psi 2 and a psi 3 to find in the following way which satisfy this property of orthonormality i start from a given orthonormal basis of these states and i end up with a new basis which is still orthonormal you should check as an exercise yourself that these three states are normalized and that they're orthogonal to each other and therefore satisfy this relation so we can set up this problem in a more general way by considering it as a rotation in multiple space let's define this vector psi vector here as a row matrix containing these three components of our spin 1 multiplier i can do now a rotation within this multiple space by simply defining a new vector psi primed which is related to the old vector psi by multiplying it by this unitary matrix as i explained in the previous lecture we want this rotation matrix u to be unitary so that if psi dagger psi is equal to the identity matrix in the original basis which is equivalent to guaranteeing the orthonormality condition then psi primed dagger psi primed is equal to the identity matrix in the new basis and that just follows from the property of the unitary matrix that u dagger u is equal to the identity so for any arbitrary three by three complex unitary matrix we can define such a rotation importantly this new basis the one of the psi primed which will just contained linear combinations of the old basis states by the usual linear algebra of this equation these will be eigenstates of the spin operator they will still all have the same spin quantum number s equals one however they will no longer have definite s said because we're taking linear combinations of these states with different s said the point is that this can be viewed as a rotation of our coordinate system for measuring the z projection we can rotate our operators in a similar fashion so that when we measure the z projection in our new basis we again get plus one zero and minus one of course our choice of the z-axis in three-dimensional space is arbitrary this rotation is basically equivalent to rotating the coordinate system and picking a new z direction this is an example of a basis transformation which we began talking about in the previous lecture let's now consider the spin raising and lowering operators on these s equals one states first of all let's denote our states in terms of these kets labeled by the quantum number for the total spin and total z projection of the spin for s equals one we have s said of minus 1 0 and plus 1. as we saw these can be written in terms of our composite a b system in the following way now the raising and lowering operators on these states can be thought of as a sum of the raising and lowering operators for the a and the b subsystems so let's see how these operators act on these basis states let's consider for example the total spin raising operator s total plus acting on the state 1 0. writing out s total plus in terms of s a plus plus s b plus and writing out the state 1 0 as 1 over root 2 of up down plus down up we have the following expression now some of these terms we can see are immediately zero because for example in this first term we cannot further raise the a spin because it is already an upspin and in the second term we cannot further raise the b spin because it is already an upspin so these two terms can be immediately crossed out this therefore gives us 1 over root two of um the term sb plus acting on the up down state which gives us the up up state and s a plus acting on the down up state which again gives us the up up state overall we have 2 over root 2 which is root 2 times up up this is of course root two times the spin one s z equals one state so we've seen correctly that when we apply the total spin raising operator to our state 1 0 we generate 1 1. we get this coefficient here of square root 2. that's actually consistent and correct in terms of the familiar rules of raising and lowering operators on generalized angular momentum states suppose i have an angular momentum raising and lowering operators l plus or minus if they act upon a state l m then i get a state l m plus or minus one back again multiplied by this coefficient we discussed this in the second lecture and probably you've seen this from introductory quantum mechanics courses so in the case of a spin 1 object with ssd components zero i can raise the s said component of this to s said equals plus one and if you work this out the coefficient is indeed square root two so all of this algebra actually works out this tells us something very important which is that we can think of a spin one as really being built up out of two spin half particles in a symmetric combination so the triplet state of a combined two spin half particle system is actually indistinguishable from a genuine spin-one particle conversely we can say that a spin one particle might have a hidden internal structure made up of two spin half particles the algebra the way in which these things transform is exactly the same these are mathematically identical furthermore the symmetry of a spin one particle is indeed the full su-2 symmetry this is because the spin can be pointing in any direction in three-dimensional space we saw this explicitly with our rotation transformation within the multiplet space equivalent to a a rotation of our coordinate system also we have the symmetry of s said going to minus s said together this gives us the su-2 group described by the set of untree matrices encoding the rotations in three-dimensional space in this case we have the spin 1 representation of the su-2 group so what we've seen here is that certain states of a many particle system comprising entities with particular quantum numbers for example charge and spin can behave like particles with different quantum numbers we start off with two spins a half and we obtain an object that is identically mathematically to a spin one particle this is a general feature of condensed matter physics one can obtain in many body systems effective or collective states which behave differently from their individual constituents in fact as i hinted this works both ways as i'll show later in the lectures one can start off with particles of spin-1 type and see that in certain conditions effective spin half particles can be realized not as bare particles in the system but as emergent excitations this is a simple example of a phenomenon known as fractionalization this was first discussed by the famous physicist duncan haldane in relation to the so-called spin gap problem in fact holden won the nobel prize in physics for his contributions to quantum topology and fractionalization as we'll see we can have even more exotic examples of this kind of fractionalization in fermionic systems comprising electrons in a superconductor one can obtain so-called majorana quasi-particles these objects which are their own anti-particle they're kind of like half of a real fermion these are not the fundamental particles of the system but they are emergent or collective modes of the system they are quasi-particles of the system with fractionalized quantum numbers all of that's to come in this lecture course now i want to move on to discussing models of spin systems what kind of hamiltonians can we construct for example for our system comprising two spins one half the simplest kind of interaction between two spins is the so-called exchange interaction the exchange interaction between two spins a and b is simply sa dot sb there is a correction to the energy of the system due to the interaction between the spins a and b following from this term the parameter j here is just a coupling constant it's a scalar number which parameterizes the strength of the interaction it can either be positive or negative depending on the situation and j has units of energy before i talk about solving the schrodinger equation for such a hamiltonian first let me explain a bit about where this exchange interaction between spins comes from the microscopic origin of the exchange interaction is basically the pali principle and the fermionic wave function anti-symmetry it is called the exchange interaction because it's basically relating to the exchange statistics for fermions the heuristic argument for the form of the exchange interaction that i've written here goes as follows the total wave function of a system comprising two electrons must be anti-symmetric with respect to exchange of those particles however the wave function involves a product of a spatial part and a spin part therefore the spatial part must be symmetric if the spin part is anti-symmetric and the spatial part must be anti-symmetric if the spin part is symmetric if the spatial part of the wave function is anti-symmetric then on average the particles are further apart from each other than they would be if the spatial part of the wave function was symmetric this is basically due to the powly principle as we discussed in the first lecture and is sometimes called pauly repulsion because the particles have different average separations depending on the symmetry of their spin states we get different energies for states with different spin symmetry exactly how this plays out depends on the details of the microscopic hamiltonian for example the quantum tunneling between particles in different neighboring orbitals is enhanced by a closer separation but on the other hand the electrostatic coulomb repulsion between the two electrons is also larger when the separation is reduced the important thing to note is that the energy of the system is affected by the spin symmetry in particular whether or not the spin wave function is symmetric or anti-symmetric with respect to exchange of those particles therefore we will now study the spin symmetry of wave functions involving a system of two spin half particles we will do this using this permutation operator p a b which will swap the labels of a and b we then want to ask whether a given wave function is symmetric or anti-symmetric under this operation if i were to operate with this operator p a b on a basis state up up i would like this to return the state up up back again if i were to act with p a b on down down i would like this to return the down down state i'm swapping the spin labels of a and b in these two examples these spin labels are the same so i don't see any difference however if i act with pab on the up down state this would swap over the labels and i would obtain the down up state likewise if i act with pab on down up it should swap the labels and i should get the up down state this is what i mean by the permutation operator that swaps the spin labels of a and b written out explicitly the question is what is this operator p a b it turns out that this operator yields exactly the right effect when applied to these basis states two times s a dot sb plus one half i will leave it as an exercise for you to confirm that this really works try for example p a b acting on the up down state and you will get something like this which indeed gives us the correct result so the conclusion of all of that is that the permutation operator that swaps the spin labels is basically this spin spin interaction s a dot sb in fact that's why we call sa.sb an exchange interaction because it's related to the permutation operator that exchanges the spin labels of two particles the correction to the energy due to the spin state is basically coming from the exchange statistics of the fermions we've seen that when the permutation operator acts on these basis states we swap the a and b labels for the up up and down down states we see that these are eigen states of the permutation operator with eigenvalue plus one they are symmetric with respect to swapping the spin labels of a and b however the up down and down up states are not eigen states of this operator so let's now instead consider these linear combinations which i introduced earlier the plus minus states these are respectively the even and odd combination of the up down and down up basis states it is then easy to show from these simple properties that if we apply the permutation operator to the plus state then we get the plus state back again with an eigenvalue plus one however if we apply the permutation operator p a b to the minus state then we get the minus state back again with an eigenvalue of minus one so the plus state is symmetric with respect to exchange of spin labels a and b whereas the minor state is anti-symmetric so a hamiltonian of this form which is referred to as the heisenberg exchange hamiltonian would give different energies for the plus and minus states because they have different symmetries under exchange of particles that's because a heisenberg hamiltonian here is related to the permutation operator states that behave differently under the permutation p a b will pick up different energies in this hamiltonian finally note that the plus and minus states are actually related to the spin eigenstates of our bases the plus state is the spin 1 s s naught component whereas the minor state is the spin zero a sedinal component so this heisenberg hamiltonian is changing the energy depending on the total spin of the system so the heisenberg exchange hamiltonian for two spins a and b takes this very simple form and we'll just take j here as some phenomenological parameter it is determined by microscopic details it can be positive or negative but we won't assume anything about it here now earlier in the lecture we showed that s a dot sb can actually be written in terms of the total spin of the system when we do that we obtain the following expression we see one half times j times s total squared minus the constant three quarters j now the constant here of minus three quarters j doesn't really matter it doesn't affect the dynamics of the system because it just shifts up all of the eigenstates by a constant amount often this term would just be neglected from the hamiltonian altogether however i'll keep it in the analysis on this page just so we can compare to the original hamiltonian involving sa.sb this alternative expression for the exchange hamiltonian for two particles is useful because it allows us to immediately write down the eigenstates and eigen energies of this hamiltonian we can do that if we use a basis of states with a definite s total we spent quite some time in the last lecture developing that basis so we know what those states are you've seen these many times by now i'm simply writing a basis with a definite s total and s said in terms of the product state basis that we started with it has this expression this is a very nice basis to use with this hamiltonian because the hamiltonian only involves the operator s total squared and we know how s total squared acts on these states specifically if our exchange hamiltonian acts on a state with quantum numbers s total and s total z then we know we will get this state back again and therefore it is an eigenstate the eigenvalue is one half j times the eigenvalue of s total squared which is s into s plus one then we have this correction minus three quarters j these states are therefore eigen states of the hamiltonian and solve the schrodinger equation the corresponding energies are precisely these numbers specifically i obtain the same energy plus one quarter j for all of the states with s equals one independently of their z projection and i obtain minus three quarters j is the energy of the spin singlet state with s total equals zero we see that the three components of the spin one state are degenerate they have the same energy that is because the hamiltonian only depends on the total spin s not on its said projection of course we can lift the degeneracy of the triplet by introducing a magnetic field the magnetic field couples to the electron spin then we would obtain something like this the hamiltonian has the same exchange coupling term but now it is supplemented by two new terms this b vector here is a vector field corresponding to the magnetic field and we evaluate it at the position of particle a and then take the dot product of the vector field with the spin operator we can do the same likewise for particle b note that the b vector here is not an operator it is simply a vector field corresponding to the strength and direction of the magnetic field at a given point in space let's now make a couple of simplifications let's assume that the magnetic field is homogeneous and pointing in a particular direction let's also say that that direction is the z axis of course as emphasized earlier our choice of z-axis is arbitrary so without loss of generality we can say that the magnetic field is pointing along the z-direction if the magnetic field is homogeneous it means it couples with the same strength to the z-components of spins a and b i then obtain the simpler form for the coupling of the magnetic field to our spins furthermore notice that s a said plus sbz is actually the total spin projection s total z and indeed as we saw earlier we can write the exchange coupling term in terms of s total squared and so the hamiltonian can be written in this simple form involving as it does only the total spin operators s total squared and s total z this is particularly convenient because we have a basis of states which are labeled by s total and s total z our basis states are therefore eigenstates of this hamiltonian they solve the schrodinger equation and the eigenvalue is of course the energy which is simply related to the quantum numbers s total and s total z we see that the coupling to the magnetic field introduces this new term in the energy and this actually lifts the degeneracy of the three components of the spin one triplet the s said one component of the triplet now has a correction minus b the s0 component of the triplet doesn't couple to the magnetic field whereas the s said minus one component picks up a contribution to the energy of plus b the spin singlet state with which has s total z equals zero again does not pick up the corrections the energy from the magnetic field as a function of magnetic field therefore we can plot the energies of these states at b equals zero we have the spin singlet state with an energy of minus three quarters j and then triply degenerate uh states corresponding to the spin triplet the s z plus one zero and minus one components are all equal in energy at b equals zero however as we switch on the magnetic field these three components of our spin triplets become non-degenerate they change in energy in different ways depending on the magnetic field the said zero component does not couple to the field whereas the s said plus and minus one components uh have a correction c energy with an opposite sign in particular if i make the uh the magnetic field large enough i can have a crossing of the ground state from the spin singlet to the sn1 component of the spin triplet the change in the ground state of the system happens when the magnetic field is equal to the exchange interaction b equals j in the next lecture we'll actually look at the consequences of this in terms of the physical properties and thermodynamic observables of this system the previous example of our two-spin system was so easy to solve because we labeled our states by certain quantum numbers but our hamiltonian only involved operators for those quantum numbers in this case the quantum numbers were s and s said of course this is not the most general scenario but it does behoove us to think more generally about quantum numbers and how they can be used to help us we can label an arbitrary basis state phi by a quantum number x if x hat acting on phi gives us the state back again phi multiplied by the quantum number x and this holds for some generalized quantum mechanical operator x hat we see that the state phi is therefore an eigenstate of the operator x hat with eigenvalue x the example given previously in this lecture was s z total which is built out of s a z and s b z acting on for example the up up or up down states these are both eigenstates of s total z with eigenvalues one or zero importantly we can label eigenstate's psi of a given hamiltonian h by the same quantum numbers if and only if we can form a simultaneous eigen basis of h hat and x hat this can be done if h and x commute that is if the commutator of h with x is equal to zero only then are the wave functions that solve the schrodinger equation for a given hamiltonian h also eigenstates of our quantum number operator x hat and therefore only in this case can we label our eigenstates psi by the x quantum number as a concrete example used already in this lecture let's say that we have some states psi labeled by the quantum numbers s total and s total z these have corresponding operators which are s total squared and s total z now for a given hamiltonian for example the one that i've written down here from earlier in the lecture we can check to see whether our quantum number operators commute with this hamiltonian in the case of the quantum number corresponding to the total spin as total squared the commutator with the hamiltonian takes the following form we can write it in terms of the two separate commutators one involving the commutator of s total squared with itself which of course is equal to zero and the other one uh involving s total z with s total squared and we know from the rules of uh angular momentum that actually these again do commute so this whole thing is equal to zero this tells us that we can label the states of our hamiltonian according to s total squared likewise the operator for s total z also commutes with the hamiltonian as we see explicitly here this tells us that we can label our eigenstates of the hamiltonian psi in terms of our quantum numbers s total and s total said indeed we showed explicitly earlier in the lecture that this is the case the wave functions that solve the schrodinger equation for this hamiltonian can indeed be labeled in terms of their total spin and total spin projection this is because quantum numbers are conserved quantities they are not changed by the hamiltonian the quantum numbers do not change under the dynamical evolution of the system this follows from the property that the total time derivative of the quantum number operator x hat is equal to zero this would therefore imply that the corresponding quantum number x is a conserved quantity that does not change in time this is important because it implies that we can label our states according to this quantum number the hamiltonian does not change the quantum number and it remains fixed and constant throughout the dynamical evolution of the system in quantum mechanics you might recall that the time derivative of an operator is actually given by i upon h bar of the commutator of that operator with the hamiltonian this is an important result for what i'll say in the following so let me briefly prove this result in the heisenberg picture we have a time dependent operator x hat of t which is given in terms of these time evolution operators e to the i h t upon h bar and e to the minus i h t upon h bar this is for of course the case where the hamiltonian h is time independent we can now simply go ahead and take the time derivative d by dt of this operator x of t using the product rule we have to be careful however with the order of these operations for example in this first term i am taking the time derivative of the factor e to the i h t upon h bar and i can pull out to the front the constant i h upon h bar but in the second term when i pull this out i have to keep the order of these operators intact with a x coming before h but h and e to the iht upon h bar do commute this is because i can do a taylor series expansion of this exponential here i would then obtain 1 minus i ht upon h bar plus 1 over 2 factorial of i h t upon h bar squared and so on now h obviously commutes with each of these terms because we just have powers of h as the only operator featuring in this series that implies that i can can commute the h through this exponential factor and obtain this as the second term now i can identify this uh cluster of terms e to the i h t x e to the minus i h t in both of these two terms as simply the time evolved x of t from our original definition this gives me i upon h bar of h times x of t minus x of t times h and of course this gives i upon h bar of the commutator of the hamiltonian with the operator x as originally advertised okay so let's use this important result to say something about conserved quantities and quantum numbers since the time dependence of an operator x involves the commutator of the operator x hat with the hamiltonian we can see immediately that quantum numbers must commute with the hamiltonian that's because as defined here quantum numbers correspond to conserved quantities the quantum numbers do not change with time and therefore the corresponding operator must commute with h this tells us that we can label the eigenstates of h according to the quantum number x if h and x commute then d by dt of x is equal to zero meaning that there is no time dependence of the corresponding eigenvalue x it is constant in time and therefore a conserved quantity the hamiltonian operator does not change the quantum number x when it acts on an eigenstate therefore we can use the eigenvalue x to label these states indeed we can label a given state by several quantum numbers if their operators all mutually commute and of course commute with the hamiltonian we also saw an example of this in our two-spin system because we were able to label our eigenstates not just according to the total spin but also according to the total spin projection we had two quantum numbers with which to label the states these quantum number operators commuted with each other as well as with the hamiltonian as we will see in the next lecture we can exploit quantum numbers to help us to solve the schrodinger equation these quantum numbers also correspond to the conserved quantities in the system and are therefore physically important objects in their own right finally in this lecture let me briefly touch upon the heisenberg model for spin systems here i'm writing down the general form of an exchange hamiltonian between a pair of spins i and j with an exchange coupling j i j and now i'm summing over all pairs of spins in the system i and j here i could imagine having a large system of spins not just one or two spins as we've considered so far but many spins indeed maybe even in the thermodynamic limit of an infinite number of spins as before i can now introduce the total spin operator as just the sum over all of the individual spin operators for all of the individual sides labeled by i and likewise for the z projection of the total spin i can just express that as the sum over all the individual z components of the spins as i said it is now easy to show that the operator for s total squared and the operator for s total z both commute with this hamiltonian these commutators are equal to zero i will leave that as an exercise for you to check yourselves that tells us that the total spin and the total spin projection are again good quantum numbers these are conserved quantities that are not changed by the hamiltonian this means that we can label the eigenstates of our hamiltonian according to s total and s total z the problem is that the hamiltonian h is not written simply in terms of these quantum number operators therefore an arbitrary basis labeled by the total spin and the total spin projection will not in general be um eigenstates of this hamiltonian we have to find a systematic way of solving the schrodinger equation and working out what the eigenstates are we know that these eigenstates can be labeled by s and s said but what are these eigenstates how does labeling our basis states according to s and s said help us in this task this is going to be the topic of the next lecture

Channel: Dr Mitchell's physics channel

Views: 957

Rating: 5 out of 5

Keywords:

Id: FTTK1Ol31DA

Length: 50min 8sec (3008 seconds)

Published: Wed Jan 27 2021