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Adding Angular Momenta

Michael Fowler, UVa. 11/28/07

Introduction

Consider a system having two angular momenta, for example an electron in a hydrogen atom having both orbital angular momentum and spin.  The ket space for a single angular momentum has an orthonormal basis  so for two angular momenta an obvious orthonormal basis is the set of direct product kets   What does this mean, exactly?  Suppose the first angular momentum  has magnitude , and is in the state , and similarly the second angular momentum  is in the state .  Evidently the probability amplitude for finding the first spin in state m1 and at the same time the second in m2 is , and we denote that state by   How to handle these direct product spaces will become clear on examining specific examples, as we do below, beginning with two spins one-half.

 

Now the sum of two angular momenta

 

 

is itself an angular momentum, operating in a space with a complete basis   

 

This is easy to prove: the components of  satisfy , and similarly for the components of .  The components of  commute with the components of , of course, from which it follows immediately that the vector components of  do indeed obey the angular momentum commutation relations: and recall that the commutation relations were sufficient to determine the allowed sets of eigenvalues.

 

We shall prove later that the eigenstates  of  are a complete basis for the product space of the eigenkets of —to establish this, we must first find the possible allowed values of the total angular momentum quantum number j.

 

Here we have, then, two different orthonormal bases for what is evidently the same vector space.  In practical applications, it often turns out that we have to translate from one of these bases to the other.  Our present task is to construct the appropriate transformation: we accomplish this by finding the coefficients of any in the  basis.  (These are called the Clebsch-Gordan coefficients.)

 

We shall build gradually, beginning with adding two spins one-half, then a spin one-half with an orbital angular momentum, finally two general angular momenta.  This is a very important part of quantum mechanics: we give every detail.  Readers already somewhat familiar with the subject may find this a bit tedious, they can glance over the introductory examples and go to the general case.

Adding Two Spins: the Basis States and Spin Operators

The most elementary example of a system having two angular momenta is the hydrogen atom in its ground state.  The orbital angular momentum is zero, the electron has spin angular momentum , and the proton has spin .

 

The space of possible states of the electron spin has the two basis kets , (also variously written as !)  the basis proton spin kets are , so the possible states of the combined system are kets in the direct product space which has a basis of four kets:

 

 

using  as shorthand for .

 

Note here that we’ve written the kets in “alphabetical order” with as the first letter, as the second.  That is to say, we’ve first written all the kets having  as the first letter, etc. 

 

For the more general case of adding j1 to j2, to be considered shortly, we’ll order the kets in the same “alphabetical” way, writing first all the kets having m1 = j1, and so on down to m1 =  –j1, so the possible sets m1m2 are:

 

 

The dimensionality of this space is then (2j1 +1)×(2j2 + 1).

 

Now the first block of 2j2 +1 elements all have the same m-component of j1, that is, m1 = j1, the next block has m1 = j1­ –1, and so on.  Think about what this means for constructing a rotation operator acting on the kets in this space:  if it operates only on the angular momentum j1, it will change the factors mi multiplying the blocks, if the operator rotates only  j2, it will operate within each block, all the blocks being changed in the same way.

 

To get a feeling for how this works in practice, we go back to the simplest case, two spins one-half.

 

The space is four-dimensional, having basis

 

.

 

Any operator acting on the spins will be represented by a 4×4 matrix, best thought of as a 2×2 matrix made up of 2×2 blocks: an operator acting on the proton spin acts within the blocks, one operating on the electron spin acts on the blocks themselves, regarded as single entities.

 

Let’s look at a few examples.  Recall that the raising operator for a single spin is the 2×2 matrix   So what is the raising operator for the electron spin? 

 

 

We use bold to denote 2×2 matrices.

The pattern is clear: the big structure (in bold above), that of the four 2×2 blocks, reflect the structure of the electron spin operator  within those blocks (of which only one survives) the identity operator  acts on the proton spin. 

 

The operator that raises the proton spin is:

 

 

What about the operator that raises both electron and proton spin? In this case, the pattern of blocks, and the pattern within each block, must both be , so

 

 

There is only one nonzero matrix element because only one member of the base survives this operation. 

 

If two spins interact (via their magnetic moments, for example) in a way that preserves total angular momentum, a possible term in the Hamiltonian would be  represented by:

 

 

Representing the Rotation Operator for Two Spins

Recall from the lecture on spin that the rotation operator on a single spin one-half is

 

 

in the 2×2 spinor space.  As we established, this matrix operator has the form

 

 

with   

 

This set of unitary 2×2 matrices form a representation of the rotation group in the sense that the total resulting from two successive rotations is given by the matrix which is the matrix product of those corresponding to the two rotations.

 

From the discussion in the previous section, it should be clear that in the product space of the two spins, the representation of the rotation operator—both spins of course undergoing the same rotation—is:

 

 

This set of 4×4 matrices, again with  must also form a representation of the rotation group over the four-dimensional space.  We shall shortly discover that this representation can be simplified, but to achieve that we need to analyze the states in terms of total angular momentum.

Representing States of Two Spins in Terms of Total Angular Momentum

We’re now ready to look at total spin states for the ground-state (zero orbital angular momentum) hydrogen atom. 

 

Consider first the state with both electron and proton spin pointing upwards,  .  The z-component of the total spin is , so .  Labeling the total spin state , we have a state with m = 1, so s = 1.  (To confirm that this state indeed has s = 1 we can apply the total-spin raising operator .  Since both component spins have maximum m value, , but  only gives zero when acting on the  member of a multiplet. )

 

We find, then, that  where we’ve added the suffix sm to make clear that the numbers in the last ket signify  for the total spin.  The total spin s = 1, being a total angular momentum eigenstate, has a triplet of m values, ,   being the top member.  The m = 0 member is found by applying the lowering operator to :

 

which together with

 

gives

 

 

Obviously, the third member of the triplet, . 

 

But this triplet only accounts for three basis states in the  total angular momentum representation.  A fourth state, orthogonal to these three and normalized, is .  This has m = 0, and also has , easily checked by noting that the total spin raising operator  acting on this state gives zero, so the state has the maximum allowed m for its  s value.

 

To summarize: in the total angular momentum representation for two spins one-half, the four basis states are .  This orthonormal basis spans the same space as the other orthonormal set .  Our construction of the  states above amounts to finding one set of basis kets in terms of the others.

 

 

Note that since both sets of basis kets are orthonormal, mapping a vector from one set to the other is a unitary transformation.  But there’s more: the coefficients we found expressing one basis ket in the other basis are all real.  This means that if any ket has real coefficients in one basis, it does in the other.  For this special case of all real coefficients, a unitary transformation is termed orthogonal.

 

The orthogonal transformation expressing one base in terms of the other is easy to construct:

 

 

The matrix is orthogonal and symmetric, so is its own inverse.

 

Geometrically, s = 1 means the component spins are parallel, for s = 0 they are antiparallel.  This can be stated more precisely: , so for s = 1, , and for s = 0   .  This makes it easy to construct projection operators into the s = 0 and s = 1 subspaces: .

Representing the Rotation Operator in the Total Angular Momentum Basis

We’ve already established that the rotation operator, acting on the two spin system, can be represented by a 4×4 matrix, and that the new (total angular momentum) basis can be reached from the original (two separate spin) basis by the orthogonal transformation given explicitly above.  Therefore, pre-and post-multiplying the two-spin rotation operator will in fact give a 4×4 matrix representation of the rotation operator in the new total angular momentum basis.

 

However, that approach misses the point: first, the singlet state  has zero angular momentum, and so is not changed by rotation. 

 

Second, the triplet state has angular momentum one, so rotation operators must act on it just as we found earlier for an angular momentum one:

 

 

This means that, as far as rotations are concerned, the space spanned by the four kets is actually a sum of two separate subspaces, the one-dimensional space , and the three-dimensional space having basis .  Under rotation, a vector in one of these subspaces stays there: there are no cross terms in the matrix mixing the spaces.

 

This means that the rotation matrix has the form  where R3 is the 3×3 matrix for spin one, I is just the 1×1 trivial matrix in the singlet subspace, in other words 1, and the O’s are 1×3 and 3×1 sets of zeroes.

 

A state of the spins can of course be a sum of components in the two subspaces, for example

 

Reducible and Irreducible Group Representations

We began our discussion of two spins one-half by examining properties of spin operators in the four-dimensional product space of the two two-dimensional spin spaces, and went on to construct a four-dimensional representation of the general rotation operator in that space: a matrix representation of the rotation group.  But when the two-spin system is labeled in terms of total angular momentum, we find that in fact this four-dimensional rotation operator is a sum of a three-dimensional rotation, and a trivial identity rotation for an angular momentum zero state.  The four-dimensional operator can be “diagonalized”: the space split into a three dimensional space and a one-dimensional space that don’t mix under rotation, and any state of the system is a sum of kets from the two spaces.

 

This is often expressed by saying the product space of two spins one-half is the sum of a spin one space and a spin zero space, and written

 

 

Putting in the dimensionalities of the spaces in this equation,

 

 

This simple check on total dimensionality sets the pattern for more complicated product spaces examined below.

 

The 4×4 representation of the rotation operator is said to be a reducible representation: it can be reduced to a sum of smaller dimensional representations.  An irreducible representation is one in which there are no subspaces invariant under all rotations.

 

Recall that we constructed the reducible 4×4 representation by taking a direct product of the 2×2 spin one-half representations of the rotation group.  The equation  we used above to describe the ket spaces equivalently describes the rotation group representations within those subspaces.

 

One might wonder why we would bother to build two different bases for the same vector space. The reason is that different problems need different bases. For a system of two spins in an external magnetic field, not interacting with each other, the independent spins basis , etc., is natural.  On the other hand, for a hydrogen atom in no external field, but including an interaction between the spins (which are aligned with the magnetic dipole moments of the particles) the  basis is the right one: the interaction Hamiltonian is proportional to , which can be written , where we recognize the raising and lowering operators for the individual spins.  This means that the state , for example, cannot be an eigenstate if the spin term in the Hamiltonian is , but the states  are eigenstates because  commutes with the total angular momentum and its components.

 

But what would be a good basis for a hydrogen atom, including the  term, and in an external magnetic field?  That is a nice exercise for the reader.

Adding a Spin to an Orbital Angular Momentum

In this section, we consider a hydrogen atom in a state with nonzero orbital angular momentum, .  Such orbital motion is equivalent to an electric current loop and generates a magnetic field. The magnetic dipole moment associated with the electron spin interacts with this field, the appropriate Hamiltonian having a term proportional to , and is termed the spin-orbit interaction.  The proton also has a magnetic moment, but that is three orders of magnitude smaller than the electron’s, so we’ll neglect it for now.

 

The spin-orbit interaction  is most naturally analyzed in the basis states of total angular momentum, , where  (see the analogous discussion of the spin-spin interaction above).  Write the orbital angular momentum eigenstates  and the spin states  where  and .  The product space is  dimensional: a single ket in this product space would be fully described by , but since both l, s are constant throughout the problem, the only actual variables are  so we’ll write the ket in the more compact form , for example .

 

The maximum possible angular momentum component in the z-direction is clearly, for the state .  In the total angular momentum representation, this must be the state .  So the two different bases have a common member:

 

.

 

In the total angular momentum  representation, is the top m state of a multiplet having  members.  The proof is the same as that for adding two spins one-half: the total spin raising operator gives zero acting on this state.  And, just as for the spin-spin case, the next member down of the multiplet is generated by applying the lowering operator:

 

Therefore

 

 

This state  lies in the  subspace, which is two-dimensional, having basis vectors and in the  representation.  So it must have two basis vectors in the  representation as well.  The other  ket must be orthogonal to  and normalized: it can only be

 

 

We’ve represented this new ket in  as the top state of a  multiplet.  It’s easy to check that this is indeed the case: it has , andacting on it gives zero, so it has to be the top member of its multiplet. 

 

The only ambiguity is an overall phase:  the Condon-Shortley convention is that the highest m-state of the larger component angular momentum is assigned a positive coefficient. 

 

So is the top state of a new multiplet having  members. The two multiplets and  taken together have members, and therefore span the whole dimensional space. The rest of the   basis vectors are generated by repeated application of the lowering operator in the two multiplets.

 

The reason there are only two multiplets in this problem is that there are only two ways the spin one-half can point relative to the orbital angular momentum.  Recalling that for the two spins we expressed the product space a sum of a spin 1 space and a spin 0 space, , the analogous equation here is

 

.

 

For the general case of adding angular momenta ,   multiplets are generated, corresponding to the number of possible relative orientations of the two angular momenta.

Adding Two Angular Momenta: the General Case

The space of kets describing two angular momenta   is the direct product of two spaces each for a single angular momentum, but the direct product nature of the kets is usually not made explicit, for example  is usually written as a single ket .  Just as in  the examples above, since  are fixed throughout, they don’t need to be written into every ket, we’ll just denote the ket by , or, when dealing with numerical values, append  as a suffix, for example . 

 

The kets  form a complete orthonormal basis of the  dimensional product space of the two angular momenta: they are the eigenstates of the complete set of commuting variables .

Total Angular Momentum Basis States

There is of course an alternative complete orthogonal basis of the space of the two angular momenta:  for total angular momentum , a different set of complete commuting variables is:  .  (This is not the same set of states as in the previous paragraph: for example,  does not commute with .  Check it out!)   We shall establish later in the lecture that the allowed values of total angular momentum range from  to , just as one would naïvely expect.

 

This alternative set is a better basis set for two angular momenta interacting with each other—an interaction term like  can change  but not .

 

As always, we’re taking