One of the most important concepts in classical mechanics is that of angular momentum, since all isolated systems possess the fundamental invariance property that their angular momentum is conserved
In this respect, angular momentum can be compared with the total energy and total linear momentum, which are also conserved quantities for isolated system
The main difference from classical mechanics arises from the fact that in quantum mechanics the angular momentum is not an ordinary vector, but a vector operator, whose three components are operators that do not commute with each other
Why are we talking about angular momentum in a chapter about spin? The reason is because spin is a type of angular momentum
We know this from charged particles experiencing a magnetic force
We can usually explain this phenomenon by arguing that rotating charges produce a magnetic field which leads to the interaction. However, the same thing happens in the absence of any spatial rotations
This means that charged particles somehow possess an intrinsic angular momentum that exists independently from their spatial motion. We call this intrinsic angular momentum, spin
We must emphasize that the name, spin, is a misnomer, for it has nothing to do with rotation about an internal axis
Spin is a purely quantum mechanical observable with no classical analogue
It is an extra degree of freedom attached to elementary particles and must be prescribed together with the values of all other comparable properties of a particle in order to designate its state
Operators exist independently from the basis chosen, and the same holds true for the relations between operators
Hence, the commutator relations between operators are also not basis-dependent. It is therefore reasonable to define an operator using a set of commutator relations
Our goal is to find the set of commutator relations that belong to the orbital angular momentum operator using their position-basis representation
Eigenvalues of angular momentum stemming from these commutator relations are obtained, and it is at this point that a distinction between orbital and spin angular momentum first emerges
The reason why we start with orbital angular momentum is because it is closely akin to angular momentum in classical physics
In both cases, their angular momentum stems from rotation
We will later learn that this is not the case, for the angular momentum caused by spin is not associated with motion through space
Classical mechanics tells us that angular momentum can be defined as
An important property of angular moentum in quantum mechanics is that the angular momentum is not an ordinary vector, but a vector operator, whos three components do not commute with each other
Hence only one of the three Cartesian components of angular momentum may be specified in a quantum mechanical state
We are also concerned with the total magnitude of the angular momentum
Classically we can relate the magnitude of the angular moemntum and the correspodning components by the Pythagoras theorem
L=Lx2+Ly2+Lz2⟹L2=Lx2+Ly2+Lz2
Hence, the operator representing the square of the magnitude of the orbital angular momentum is defined as
L^2=Lx2+Ly2+Lz2
Although any two components of the orbital angular momentum operator do not commute, the square of the total orbital angular momentum operator does commute with each of the three components
The commutator between the square of the magnitude of the orbital angular momentum and any component is therefore
After cancelling terms, we will see that all three components commute with the square of the square of the magnitude of the orbital angular momentum operator
[L^2,Lx]=[L^2,Ly]=[L^2,Lz]=0
Hence both the magnitude of the orbital angular momentum and the definite value of any one of its component can be determined simultaneously
Although the commutator relations are consistent with the differential and coordinate operator relations, they may be taken independently of these and assumed to be the defining relations for quantum mechanical angular momentum
When such is the case, angular momentum need not refer to the spatial coordinates or linear momentum components of a particle since the relations by themselves do not
To avoid confusion, we shall use the general angular momentum operator as J
The general angular momentum must follow a set of commutator relations
The components of the general angular momentum must not commute with each other
The consequence of the above commutator relations is that any component of the general angular momentum operator must commute with the square of the total angular momentum operator
A ladder operator is defined as an operator that increases or decreases the eigenvalue of another operator by a unit amount
The operator whose eigenvalues we wish to raise or lower here is the angular momentum operator along the z axis
There are two ladder operators, the raising operator and the lowering oeperator, which increase or decrease the eigenvalues of the angular momentum operator along the z-axis by a unitary amount
It should be obvious that upon operating on the eigenvector with the ladder operator, we will obtain another eigenvector
Now that we have established the ction of the ladder operators, we should consider their commutator relations with the angular momentum operator along the z-axis
[Jz,J+][Jz,J−]
We can do so by considering the action of the commutators on the eigenvector of the angular momentum operator along the z-axis
Recall the relation between the four angular momentum operators
J2=Jx2+Jy2+Jz2
We are only concerned with the eigenvalues
of the square of the total angular momentum operator and the angular momentum operator along the z-axis, so we shall make them the subject
J2−Jz2=Jx2+Jy2
To see what the relation between the eigenvalues of the total angular momentum operator and the angular momentum operator along the z-axis is, we simply have to have them act on their common eigenvector
(J2−Jz2)∣α,β⟩
By definition we will obtain their respective eigenvalues
(J2−Jz2)∣α,β⟩=(α−β2)∣α,β⟩
It is clear that the vector here is also an eigenvector of the sum of the square of the angular momentum operators along the x and y axes
Jx2+Jy∣α,β⟩=(α−β2)∣α,β⟩
Since the angular momentum operators along the x and y axes are Hermitian, the sum of their square must also be Hermitian. Hence, their eigenvalues must be real and positive
α−β2≥0⟹a≥β2
The implication that β is bounded for a given α menas there are eigenstates that cannot be raised or lowered.
If were were to use the ladder opeartors on the highest or lowest eigenstate, we get
These two expressions seem to contradict the statement that βmax and βmin are the largest and smallest eigenvalues respectively. The only way out of this contradiction is to have the operated eigenvector go to zero.
The second root is rejected since it suggests that βmax is less than βmin
The raising and lowering operators suggest the neighboring values of β are separated by ℏ, therefore:
βmax−βminβmax+βmax2βmaxβmax=nℏ=nℏ=nℏ=2nℏ
And βmin is just the negative of this value. n can be any non-negative intger j. For an even n, i.e. n=2j
βmax=jℏ
Such that
α=βmax2+ℏβmax=βmax(βmax+ℏ)=jℏ(jℏ+ℏ)=j(j+1)ℏ2
β=mjℏ,mj=0,±1,±2,⋯,±j
The values of this quantum number, the magnetic quantum number, are restricted to −j≤mj≤j
The whole point of using the ladder operator is because we can increase the eigenvalue in a unitary step fashion. This means that it should not be surprising that the general angular momentum operator shares the same spectrum of eigenvalues as the orbital angular momentum operator. In fact, the eigenvalues of the orbital angular momentum operator is only a subset of the spectrum of eigenvalues of its general counterpart.
The general angular momentum operator has the same eigenvector expression
However, without the restrictions imposed on the case of the orbital angular momentum, the quantum numbers can take on more than just integer values
ifmj=0is included in the sequence ofmjvalues⟹j=integerifmj=0is not included in the sequence ofmjvalues⟹j=21×odd integer
Since we are dealing with a case of a generalized angular momentum Jz, where there are no specific boundary conditions, whether n is odd or even does not matter. In the case of the orbital angular momentum L^z, where we are constrained in ϕ (and θ in 3D), the eigenstates are:
We can find a form for the eigenstates of the constrained orbital angular momentum
This is useful, as it forms part of the solution for the time-independent Schrodinger equation, by separation of variables into the angular and radial parts of the wavefunction
We will construct the solution using the L^2 and L^z operators, and will arrive at an identical solution of spherical harmonics to the one in the subsequent 'Hydrogen Atom' Section
Spin can be though of as a quantum system in its own right, independent of any particle. As we will see, it has an operator, a set of eigenstates and eigenvalues, and forms a complete orthonormal basis in Hilbert Space
However, it's convenient to build an understanding of it as an 'intrinsic' angular momentum of an electron, in addition to its obrital angular momentum, and an extra degree of freedom
This is backed by two important experimental evidences involving atoms in magnetic fields. The Zeeman effect, and the Stern-Gerlach experiment
Spin does not arise from any overall physical motion, so we cannot construct a quantum mechanical spin operator in terms of position and momentum operators, like we have done for angular momentum.
However, spin must be additive to the orbital angular momentum for our purposes, so imposing the same commutation relations as orbital angular momentum [S^i,S^j]=iℏϵijkS^k, allows us to derive its eigenvalues using the ladder operator analogous to the general/orbital angular momentum case, and demonstrate a complete orthonormal basis of eigenstates.
Where due to the quantization of ms=±1/2 and s=1/2, the eigenstate can be represented by a 2D-vector, which takes different form whether we are in a case of 'spin-up' or 'spin-down', along the z-axis