1
$\begingroup$

What is the motivation for the values of the magnetic quantum number $m_l$ to take values of $ -l, -l+1, \cdots , l $ where $l$ is the angular momentum number? The ladder operators for angular momentum do imply that is some value $a\hbar$ is an eigenvalue of $L_z$ then so too is $(a\pm 1)\hbar$ provided the limiting values are not reached. Some constraint for these limiting values can be found from considering that $\langle L^2\rangle \geq \langle L_z\rangle$; that is, the magnitude of $m_l$ is bounded by $\sqrt{l(l+1)}$. However none of these considerations eliminate the possibility of $L_z$ having intermediate eigenvalues as well as the integer incremental steps suggested from the ladder operators. They also do not motivate why $l$ is integral and why the maximum values of $m_l$ ate given by $l$ and not, say, $\sqrt{l(l+1)}$ exactly.

$\endgroup$
1
$\begingroup$

This happens because of the commutator between $L_z$ and $L_{\pm}$.

$$[L_z,L_{\pm}]=\pm\hbar L_{\pm}\tag{1}$$

The above can be easily shown from the angular momentum algebra $[L_i,L_j]=\epsilon_{ijk}L_jL_k$ and from the definition of $L_{\pm}$ operators (that is, $L_{\pm}=L_x\pm iL_y$).

Here's how to conclude about the $\hbar$ separated spectrum of $L_z$:

Let $|l,m_{l}\rangle$ be an eigenket of $L_z$ with eigenvalue $m_l\hbar$. One can use (1) to show that , $L_{\pm}|l,m_{l}\rangle$ is an eigenket of $L_z$ as well.

$$L_z(L_{\pm}|l,m_l\rangle)=\underbrace{(L_zL_{\pm}-L_{\pm}L_z)}_{\pm \hbar L_{\pm}}|l,m_l\rangle +L_{\pm}\underbrace{L_z|l,m_l\rangle}_{m_l|l,m_l\rangle}=(m_l\pm \hbar)L_{\pm}|l,m_l\rangle$$

Thus, we find that $L_{\pm}$ creates a ladder of eigenstates for $L_{z}$ where the "rung" separation is exactly $\hbar$. One can further show that there exists a bottom rung and a top rung and thus they get bounded.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.