Wednesday 1 April 2020

quantum mechanics - Why do we get fake half-spin values for orbital angular momentum if we solve it algebraically?


It is a well-known fact that the values for the square of the orbital angular momentum of a particle $L^2$ and it's projection in the $z$-direction $L_z$ are $m\hbar$ and $l(l+1)\hbar$ and that $l$ and $m$ can take integer values ($...,-2,-1,0,1,2,...$) And they are the values you obtain from solving the wave equation for the observables.


Despite this, if you were to solve algebraically for the eigenvalues of $$\hat L_z|l,m\rangle$$ and $$\hat L^2|l,m\rangle$$ using ladder operators or outright solving for the individual matrix components (Like in Born's, Heisenberg's and Jordan's On Quantum Mechanics II) we obtain that the quantum number can have half-integer value ($...,1,-\frac{1}{2},0,\frac{1}{2},1,...$) and we can only reduce them to only integers by appealing to rotation operators or, once again, the wave equation.


I find the fact that we get an extra set of results extremely bizarre, since the wave equation can be derived from the matrices (like in Heisenberg's The Physical Principles of the Quantum Theory). I also find bizarre that, if one didn't know about rotation operators or the wave equation one could live their life thinking either orbital angular momentum can have half-integer values, or that Quantum Theory is wrong, as it disagrees with experiment, since those wrong values are eigenvalues of $L_z$ and $L^2$ The only explanation I can think of is that orbital angular momentum is actually not truly observable, only total angular momentum is, but this reason seems neither right,nor satisfactory.



Answer



If I understand your question correctly, then the answer is that not all properties of an operator are encoded in its commutation relations.


If you are given three operators called $L_1, L_2$, and $L_3$ and told only that they obey the commutation relations $[L_i,L_j]=(i\hbar) \epsilon_{ijk} L_k$, one might ask what (if anything) they can conclude about the operators or the states on which they act. It is a priori possible that you can't conclude anything about the operators without providing a concrete implementation of them on some Hilbert space, but of course this is not true; this information is sufficient to establish the existence of the Casimir operator $L^2$, as well as to derive all possible eigenvalues of $L^2$ and $L_3$ which would be consistent with the commutators.


But this isn't the whole story. A concrete realization of these operators contains more information than the commutation relations alone. When we define $L_i := (-i\hbar)\epsilon_{ijk}x_j \partial_k$ which acts on some subset of $L^2(\mathbb R^3)$, then not only do we recover the commutation relations we started with, we also find that only integer values of $l$ are allowed.


We should not be surprised by this. The spectra which are derived "algebraically" - which admit the possibility of half-integer values of $l$ - are based on the commutators and the commutators alone. In other words, we used the commutators to constrain (note: not exactly specify) the possible spectra of the operators. A concrete implementation of the operators contains the commutation relations, but also more detailed information about how the operator acts on the Hilbert space. Therefore, it's entirely possible that such an implementation would not exhaust the algebraically derived spectra, and indeed this turns out to be the case.





You don't need to resort to a wave equation to see this, by the way. The angular momentum operators which act on $\mathbb C^2$ (spin 1/2) and those which act on $\mathbb C^3$ (spin 1) obey precisely the same commutation relations but have entirely different spectra; it's therefore obvious that commutation relations are not sufficient to tell you everything you need to know about the associated operators.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...