10.3: Total Angular Momentum
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In 3D space, if you have three components of a vector →v, then the magnitude of that vector squared is v2=v2x+v2y+v2z. Angular momentum is a vector, and so this rule would apply to angular momentum as well. However, in quantum mechanics, we see that angular momentum behaves very differently from how it does in classical physics. In particular, if an object has a definite z component of angular momentum, then it has an indefinite x component of angular momentum. Does that mean that total angular momentum must also be indefinite? In order to answer this question, we must ask it in a proper quantum manner.
In quantum mechanics, we associate each observable quantity with an operator. We can then use that operator on one of its eigenstates (i.e. a state where the observable has a definite value) to pull out the value of the observable as the eigenvalue a in the equation ˆA|ϕ⟩=a|ϕ⟩. If the system is not in an eigenstate, we can figure out the “expectation value” ⟨a⟩ (i.e. the weighted average of all values that could be observed) using the operator in the equation
⟨a⟩=⟨ψ|ˆA|ψ⟩
What then is the operator that corresponds to total angular momentum? By analogy to classical physics, we can guess that the operator for total angular momentum squared is:
ˆS2=ˆS2x+ˆS2y+ˆS2z
This begs the question as to what ˆS2x means; we know how to square numbers and variables, but how do you square an operator? For ˆS2, we will just treat that as an operator itself, the “total spin squared” operator. We’ll not treat the superscript 2 as squaring, rather we’ll just consider it part of the name. As for the others, what really matters about an operator is what it does to a vector representing a quantum state that it’s supposed to operate on. In order to make this definition of the spin angular momentum squared operator to work, we need to interpret them as follows:
ˆS2x|ψ⟩=ˆSxˆSx|ψ⟩
In other words, first apply the ˆSx operator to the state |ψ⟩, and then apply the ˆSx operator again to the vector that resulted from the first application. As an example, let’s consider the ˆS2x operator on the state |+z⟩:
ˆS2x|+z⟩=ℏ2[0110]ℏ2[0110][10]=ℏ24[0110][0110][10]
All we’ve done in the first step is pulled the scalar constants out front. To perform this matrix multiplication, first we must multiply the rightmost matrix by the vector, and then we can multiply the first matrix by the result.1
ˆS2x|+z⟩=ℏ24[0110][01]=ℏ24[10]=ℏ24|+z⟩
Interestingly, it seems that |+z⟩ is in fact an eigenstate of ˆS2x, even though it’s not an eigenstate of ˆSx!
Armed with these techniques, it is possible to show that any properly normalized spin-1/2 state |ψ⟩ is an eigenstate of ˆS2 with eigenvalue 34ℏ2. Although it may be surprising that |+z⟩ is an eigenstate of ˆS2x, in retrospect it should not be surprising that all states are eigenstates of the total angular momentum operator. We’ve been saying all along that the total angular momentum of an electron is √32ℏ; what can be in an indefinite state is the components of that angular momentum along various axes.
1If you know how to multiply 2×2 matrices, you can do the matrix multiplication first if you wish. As we will see, the commutative property does not apply to matrix multiplication, but the associative property does.