# How would you build a device to measure $L^2$ (angular momentum squared) of a particle?

The formalism of Quantum Mechanics uses angular momentum operators such as $$L_x, L_y, L_z$$, and $$L^2$$.

The quantities corresponding to $$L_x, L_y, L_z$$ can be measured using a Stern-Gerlach apparatus, but how do you build a machine that measures $$L^2$$? Are there any examples of devices that do so?

• Commented Apr 26, 2022 at 16:46

When it comes to quantum mechanical properties, the best way to typically measure them is indirectly. Take for example the Stern-Gerlach experiment, we don't measure the spin of the electron directly, but we measure the deflection in a magnetic field due to the intrinsic magnetic moment $$\hat \mu = -g_s\frac{e}{2m_e}\hat S$$. Now, your question asks specifically about $$L^2$$, so, I invite you to consider spin-orbit coupling. The spin-orbit coupling describes the interaction of the electron's spin magnetic moment with the magnetic field produced by the proton, when seen from the electron's frame of reference. It provides a shift in the energy levels of hydrogen given by the following number: $$\Delta E = \frac {\beta (n, \ell)} {2} (j(j+1) -\ell(\ell+1) -s(s+1))$$ where $$\beta (n, \ell) = \frac {\mu_0} {4\pi} g_s \mu _B ^2 \frac 1{(n^3 a_0 ^3 \ell (\ell + \frac 12)(\ell +1)}.$$ Here $$\ell$$, the quantum number for total angular momentum, apperars explicitly. Now all that's left to do is to find an explicit relationship between $$\ell$$ and $$j$$, because $$s =\frac 12$$ is already known. It is a result from the theory of the addition of angular momentum that $$j=\ell \pm \frac 12$$. Thus, for a given state $$|nj\ell m_j\rangle$$ (We must use the coupled basis here, where I've suppressed the s, since it is constant), we find that the energy shift is (I'll let you do the calculations yourself, simply plug $$j =\ell +\frac 12$$ and $$j=\ell -\frac 12$$ into the equation for $$\Delta E$$) $$\Delta E^- = -\frac {\beta (n, \ell)} 2 (\ell +1)$$ and $$\Delta E^+ = \frac {\beta (n, \ell)}2 \ell.$$ We can measure both of these quantities by looking at the emission spectrum of the hydrogen atom, thus, measuring the energy shift allows us, in principle, to calculate $$\ell$$, and thus the eigenvalues of $$L^2$$, given only the energy levels, which can be found using the emission spectrum of hydrogen. No other experimental setup is really needed. I hope this helps. If you aren't familiar with spin-orbit coupling and the addition of angular momentum, check out Barton Zweibach's lectures on quantum mechanics 2 and 3, you can find them here and here.
• Hi Daniel, Unfortunately, this analysis ignores the relativistic correction that is mixed in experimentally. When both perturbations are included the $l$ and $s$ are no longer good quantum numbers, and the equation loses the desired form. In addition, in case there were any doubts about using an approximation, Griffiths (3rd ed., p. 304) gives the exact energy via the Dirac equation which also shows the desired $\hbar^2 l(l+1)$ form is missing too. Commented Mar 29 at 17:16