Lecture 5 Spin Hamiltonians Angular momentum. Angular momentum operators in matrix representation. Angular momentum as a spin of higher size. Spin Hamiltonians. Addition of angular momenta. Clebsh‐Gordun coefficients.

1) Angular momentum Hydrogen atom and its es (eigenstates)

R 𝑟 radial part of wave function. 𝑌 θ, ϕ – spherical harmonic, angular part of w.f. 𝑛 – principal quantum number, defines energy level. ℓ ‐ angular momentum quantum number, defines magnitude of the angular momentum. 𝑚 projection quantum number, defines projection of angular momentum on axis z (taken as a convention). 𝑔 – orbital degeneracy number.

Classical angular momentum 𝐿⃗ 𝑟⃗𝑝⃗. Quantum angular momentum is a vector operator with three components

All these operators can be represented in spherical coordinates θ, ϕ. See textbook. Spherical harmonics are the eigenstates of operators 𝐿 and 𝐿.

From definition of angular momentum (eq.1) it can be shown that they satisfy the following relation.

For a given ℓ , 𝑚 takes the numbers 𝑚 ℓ,ℓ1,..0,..,ℓ resulting in degeneracy 𝑔2ℓ1.

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We can also introduce rising and lowering operator, which are defined and act as following: (see textbook for details)

From the above relation one can see that the set of angular wave functions, 𝑌 ,…,𝑌 ,….,𝑌 , is a closed set, meaning that result of action of any angular momentum operator on 𝑌 can be expressed as a linear superposition of spherical harmonics from this set. So it is customary to speak about such objects as higher spins.

We can create the corresponding matrix description of angular momentum (spin) rd operators in the basis of 𝑌 . See problems in Griffiths (3 ed. 4.34, 4.61, 4.62) . Let’s do this for a particle with the spin 1, that is with ℓ1. First I switch, for consistency, from spherical harmonics to z‐components of spin 1. and then use them as basis vectors

The matrix for 𝐿 operator can be easily constructed. You can check that it acts on the basis vectors in accordance with eq.5.

In the next step we find how rising and lowering operator acts on the basis vectors using eq. (8,9)

Next we express 𝐿,𝐿 via 𝐿,𝐿 and construct the matrices.

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Continuing this procedure, we get

At this point we can equate 𝐿𝑆 meaning by 𝑆 a higher spin

Generalization: 1) From atomic orbitals (spherical harmonics) we can construct matrices for operators with rank 0,3,5,7.. corresponding to operators with 𝐿 0,1,2, …. Purely .. mathematically, using expression for raising and lowering operators we can get matrices for 𝑆 1/2, 3/2,5/2,.. Most importantly we get op erators with the lowest non‐zero rank 𝑆1/2, that is starting from 𝑆, get matrices for 𝑆,𝑆. This is remarkable, because we do not know how to write the wave function corresponding to say 𝑍/ 2) The same algebra of matrix operators works when we deal with elementary particles in high energy physics and in atomic physics.

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2. Spin Hamiltonians s

2.1. Spin in magnetic field

In general, μ⃗ γ𝑆⃗ , where μ⃗ magnetic moment, γ gyromagnetic ratio, 𝑆⃗ spin vector operator. || For spin ½ of electron μ⃗ g 𝑆⃗, where 𝑔2 so‐called g‐factor.

Another popular notation convention used both for spin ½ and higher spin J. μ𝑔μ𝐽 , where 𝐽 is the total angular momentum or spin measured in integer of half‐integer ||ℏ numbers : ‐1,0,+1 or ‐1/2 , +1/2. And μ is Bohr magneton. Notice that ℏ is moved from spin to μ. The minus sign in eq.2 and gyromagnetic ratio appears because magnetic moment is antiparallel to spin angular momentum for an electron. (electron has negative charge)

The energy of magnetic moment placed in magnetic field in classical EM is 𝐸μ⃗𝐵⃗ In QM, it is replaced with operator Hamiltonian (and this is our first spin Hamiltonian)

Problem: 𝐵⃗ 0,0, 𝐵, at 𝑡0, spin is in the state 𝑋, that is it point into positive x‐ direction. Q.1 Describe time evolution of the spin.

Here Ωγ𝐵 is Larmor frequency.

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The state of the spin at 𝑡0, can be represented as

Then its time evolution is given by the time evolution of the eigenstates,

Q.2 Find ⟨𝑆𝑡⟩ average value of 𝑆 as a function of time.

ℏ You can further show that 𝑆 𝑡 𝑐𝑜𝑠Ω𝑡. So the physical picture is similar to the classical behavior. Spin rotates (precesses) around z axis in x‐y plane.

Energy level splitting picture #1: Arrows indicate direction of magnetic moment. This picture is used in literature all the time.

Energy level splitting picture #2. Arrows indicate direction of the spin. It is different for electron (negative particle) and proton (positive particle)

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2.2 Spin Hamiltonians with the product term. Spin‐orbit, Heisenberg …..

The are many examples of physical systems where Hamiltonians depend on interaction between spins. Let’s give some examples.

1) Spin‐orbit coupling. Fine structure of hydrogen and other atoms. Griffiths p.299. I do not give all details just basics facts that allow us to write the Hamiltonian.

Static charge. Electrical field acts on the charge, magnetic field on magnetic moment.

However, moving electron experience Lorentz force 𝐹 𝑒𝑣⃗𝐵⃗. There is a mirror effect. A particle moving in electrical field “sees” magnetic field 𝐵⃗ 𝑣⃗𝐸⃗. Handwave explanation: In our usual frame, electron rotates around proton. However, in the reference frame of electron, proton rotates around it and produces magnetic field in the electron position.

The orbital motion of electron is related to its angular momentum. The Hamiltonian has the following form (actually it is a small correction to Coulomb potential). 𝑆⃗ and 𝐿⃗ characterize the same electron.

2) Hyperfine interaction (splitting) – interaction between spin of electron and spin of proton (nuclei)

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Here we have two magnetic moments interacting with each other via dipole‐dipole magnetic interaction. Additional “contact” interaction occurs for s electrons, because they have finite probability to be found at the origin, where the proton (nuclei) is located. When ℓ0 one can find:

3) Heisenberg model

In some approximation, the Hamiltonian of two interacting electrons in, say, diatomic molecule can be represented as: (Comm: we will derive this result)

𝐻 𝐽𝑆⃗𝑆⃗ This Hamiltonian is known as the Heisenberg model. 𝐽is the exchange constant. It can be both positive and negative. The Heisenberg model can be extended to the case of many spins in 1,2,3 dimension. Spin can be ½, 1, …. The model plays important role in the theory of magnetism.

To summarize, we have a rather generic model Hamiltonian, which has the product of two vector operators. Let’s use the matrix method to solve it for the simplest case of two interacting ½ spins. Importantly, here we start dealing with many‐body QM, so the approach has many generic features. First, we simplify Hamiltonian using Pauli matrices. Plank constant are absorbed in 𝐽.

The basis vectors are formed as direct product of states of the first spin and the second spin. Notation with arrows are often used in literature. I prefer more explicit notation which specify the state and particle. For completeness I also list the number of the state in the Hamiltonian basis and corresponding column.

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To compute matrix elements of H such as 𝑍 𝑍 𝐻𝑍 𝑍 we need to find an action of individual Pauli matrices on individual spin states

Let’s demonstrate how we find matrix element for Heisenberg Hamiltonian. The rule is each operator acts on its own spin sate 1 on 1, 2 on 2.

Some intermediate results needed for computation of matrix elements.

∗ Matrix is Hermitian so 𝐻 𝐻. Here is what we get

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We see that our matrix of rank 4 splits into three square matrices A, B, C of rank 1,2, and 1, respectively. The procedure of finding eigenstates and eigenvalues for these matrices can be done independently.

We see that the eigenstates of the Hamiltonian can be split into two groups.

The group with 𝐸𝐽 form multiplet corresponding to the total spin equal 1 (in ℏ units) and projections m=+1 for 𝑍 𝑍 , m=‐1 for 𝑍 𝑍 , and m=0 for 𝑍 𝑍 𝑍 𝑍 /√2. (You will be asked to prove the last statement in the homework problem. ) This multiplet has the special name triplet. The state with 𝐸3𝐽 is called singlet

The picture of the energy splitting when (positive) exchange interaction is present can be summarized as following.

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3. Addition of angular momentums. The result that we have obtained for the Heisenberg model can be obtain in a simpler, as some think, method of the addition of angular momentums. I still believe the matrix method is much superior – it uncovers all details, allows to compute all parameters and does not need analogy with classical physics case often used in the textbooks. Still for the sake of completeness I would like to consider the addition method as well.

We consider the spin‐orbit coupling term, where 𝐿⃗ is operator of the electron angular momentum and 𝑆⃗ its spin operator

Next we introduce total angular momentum operator and make some manipulations.

Our starting choice of the basis vectors is all combinations of angular momentum and spin component such as for example: 𝑌 𝑍 ⃗ ⃗ The 𝑌 𝑍 is not the eigenstate of the operator 𝐿𝑆.

However, we can go around this problem if we notice the following relations

So, we found that operators 𝐽 ,𝐿 ,𝑆 ,𝐽 commute with 𝑆⃗ 𝐿⃗ and therefore they have the same eigenstates and eigenvalues. The Hamiltonian in the basis |𝐽 ,𝐽,𝐿 ,𝑆 ⟩ takes diagonal form. Eigenvalues can be computed using relations

Which can be generalized for total angular momentum 𝐽

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So the eigenvalues of the 𝐿⃗𝑠⃗ take the form

Let’s see how it works for the case we just have solved using the matrix method, 𝐿 𝑆 1/2, 𝑆𝑆 1/2. Here we have two options for total angular momentum 𝐽 1, 𝐽 0.

We some dropped common factor we have got the same results from the matrix method.

For the higher orbital angular momentum and higher spin angular momentum the method works as following. The maximum and minimum values of the total angular momenta are determined as

And the sets of allowed total 𝐽goes as 𝐽,𝐽 1,….,𝐽

For example, if L=2 and S=1, we have 𝐽 3, 𝐽 1. The total number of states is ℓℓ1𝑠𝑠1 15 states. These 15 states are split into three multiplets.

Within each multiplet, the states are degenerate. The energy levels of multiplets can be found using formulas listed above.

The method of addition of angular momentum does not tell us right away in what combination the old basis components go into the eigenstates. In practice this is usually done using the table of Clebsh‐Gordan coefficients

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Let’s see how it works. The section 1/2x1/2 shows the case of addition of to spins. Going down along indicated columns we have

Couple of examples from the section 2x1

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