# Probability and the Magnetic Quantum Number

I am currently self-studying quantum mechanics, and I'm working problems on angular momentum. The problem I'm currently working on asks one to consider a particle subjected to a spherically symmetric potential with wave function given by

$$\begin{equation*} \psi(\mathbf{x}) = (x+y+3z)f(r). \end{equation*}$$

It first asks whether or not $$\psi$$ is an eigenfunction of $$\mathbf{L}^{2}$$ and if so, for which value of $$\ell$$. By rewriting

$$\begin{equation*} \psi(\mathbf{x}) = \left[c_{-1}Y_{1}^{-1}(\theta,\phi) + c_{1}Y_{1}^{1}(\theta,\phi) + c_{0}Y_{1}^{0}(\theta,\phi)\right]rf(r) \end{equation*}$$

where

$$\begin{equation*} c_{-1} = \sqrt{\frac{2\pi}{3}}(1+i),\hspace{1pc}c_{1} = \sqrt{\frac{2\pi}{3}}(i-1),\hspace{1pc}\mbox{ and }\hspace{1pc}c_{0} = 2\sqrt{3\pi}, \end{equation*}$$

I have determined that $$\psi$$ is indeed an eigenfunction of $$\mathbf{L}^{2}$$ corresponding to $$\ell = 1$$, because for $$\mathbf{L}^{2}$$, the eigenvalue of $$Y_{\ell}^{m}(\theta,\phi)$$ is independent of $$m$$ and so any linear combination of $$Y_{\ell}^{m}$$ (with $$\ell$$ fixed) is an eigenfunction of $$\mathbf{L}^{2}$$.

The problem then asks what the probabilities are that the particle is found in its various $$m_{\ell}$$ states (i.e., $$m_{\ell} = -1,0,1$$). The solutions I have for the problem say tersely that the probability of $$\psi$$ being found in the state $$|1,m\rangle$$ is given by

$$\begin{equation*} P(m) = \frac{\left|c_{m}\right|^{2}}{\sum_{m=-1}^{1}{\left|c_{m}\right|^{2}}}. \end{equation*}$$

I'm not sure how to arrive at this conclusion, though. My thought process is as follows. If the particle is found in the $$m_{\ell} = 1$$ state, its wavefunction is, upon normalization,

$$\begin{equation*} \psi^{1}(\theta,\phi) = \tilde{c}_{1}Y_{1}^{1}(\theta,\phi) rf(r). \end{equation*}$$

Therefore, the probability of the particle being found in this state is

$$\begin{equation*} \int{\left[\psi^{1}(\theta,\phi)\right]^{\ast}\psi(\theta,\phi)\mathrm{d}^{3}x} = \int_{0}^{\infty}{\int_{0}^{\pi}{\int_{0}^{2\pi}{\tilde{c}_{1}^{\ast}\left[Y_{1}^{1}(\theta,\phi)\right]^{\ast}rf(r)\left[c_{-1}Y_{1}^{-1}(\theta,\phi) + c_{1}Y_{1}^{1}(\theta,\phi) + c_{0}Y_{1}^{0}(\theta,\phi)\right]rf(r)r^{2}\sin{(\theta)}\mathrm{d}\phi}\mathrm{d}\theta}\mathrm{d}r} \end{equation*}$$

and then to use the orthonormality of the spherical harmonics.

• I don't really understand the close votes. Sure, this is from a homework problem, but essentially the OP asks how to compute probabilities concerning observables which are defined only on a subsystem (here the angular part) - which is completely conceptual! Commented May 21, 2023 at 15:43
• @OP: If this is really what you want to ask, you could edit the question to make this more explicit... Commented May 22, 2023 at 7:01
• @TobiasFünke this is totally an assignment question. Indeed it is taken from a textbook: I know because I assigned it (although I can’t immediately recall the exact book). If you disagree it’s an assignment question, it qualifies as a check-my-work question (“what mistake am I making”?) Commented May 22, 2023 at 12:56
• @ZeroTheHero I disagree here in the sense that the wording of the question makes it sound as a check-my-work question, but I interpret the question differently. The "mistake" OP is doing (the question they are asking) in computing the probabilities is of a conceptual kind (at least how I read the question), so I think this is fine Commented May 22, 2023 at 12:59
• @OP Given the above comment, you should really consider to rephrase the question accordingly. I also have removed your last edit, because questions (and answers) should stand for their own and are not suitable for a discussion of this kind. That being said, I think you can safely cut-off the first half of the question by just stating the wave function and the problem ("given $\psi$, I want to compute the probabilities...") and then state the problem you encounter to fulfill this task. No need to mention the other parts of the task, for example. Commented May 22, 2023 at 13:02

From a more general point of view, we have $$L^2(\mathbb R^3)\cong L^2(\Omega)\otimes L^2(\mathbb R_+)$$ with $$\psi =\varphi\otimes g$$, which means $$\psi(\mathbf x):= \varphi(\phi,\theta)\, g(r)$$. Note that the $$\psi$$ in the question is indeed a function of this form.

The inner product between any two such (tensor) product functions is simply $$\langle \psi_1,\psi_2\rangle_{L^2(\mathbb R^3)}= \langle \varphi_1,\varphi_2\rangle_{L^2(\Omega)} \langle g_1,g_2\rangle_{L^2(\mathbb R_+)} \quad ,$$

with $$\langle \varphi_1,\varphi_2\rangle_{L^2(\Omega)}:=\int \int\mathrm d\phi\, \mathrm d\theta \sin(\theta)\, \varphi_1(\phi,\theta)\, \varphi_2(\phi,\theta)$$ and $$\langle g_1,g_2\rangle_{L^2(\mathbb R_+)}:=\int\limits_{0}^{\infty}\mathrm dr\, r^2 g_1(r)\, g_2(r) \quad .$$

Now you have to make use of the following general fact (axiom in some sense): Given a bipartite Hilbert space $$H=H_1\otimes H_2$$, the probability to measure in a normalized state $$\psi \in H$$ some value $$o$$ for some observable $$O$$ on $$H_1$$ is given by $$\langle \psi, P_o \otimes \mathbb I_2 \, \psi\rangle_H$$, where $$\mathbb I_2$$ is the identity on $$H_2$$ and $$P_o$$ the projector appearing in the spectral decomposition of $$O$$ corresponding to the eigenvalue $$o$$.

This should be explained in any textbook on e.g. quantum information or good book on QM - and after a little bit of thought it also seems rather intuitive.

Thus: The probability to measure the value $$m$$ (given $$\ell$$) is, in your case: $$\langle \psi, |\ell, m\rangle\langle \ell, m| \otimes \mathbb I_{L^2(\mathbb R_+)} \,\psi\rangle_{L^2(\mathbb R^3)} = \langle \varphi |\ell,m\rangle\langle \ell,m| \,\varphi\rangle_{L^2(\Omega)}\, \langle g,g\rangle_{L^2(\mathbb R_+)} \quad .$$

The orthonormality of the spherical harmonics now yields that $$\langle \varphi |\ell,m\rangle\langle \ell,m| \,\varphi\rangle_{L^2(\Omega)} = |c_m|^2$$.

It seems the task did not normalize $$\psi$$ before, so you have to divide $$\psi$$ by $$||\psi||=||\varphi|| ||g||$$ first, which then should yield the desired result.

• This answers my question and then some--it makes explicit what my current textbook seems nervous to mention for fear of scaring of physicists with too much math. Is it worth editing my question to try to re-open it? I guess the way to avoid getting closed for "Homework-like questions" is just to avoid any mention that your problem originated from a textbook. Also, do you have any recommendations for a quantum mechanics text that treats the mathematics involved more rigorously? Commented May 22, 2023 at 7:22
• Hi @kandb, hmhm whether or not it is worth is up to you. Glad my answer helped. Yes, you're right, many books are (sometimes) sloppy. Regarding the book: It really depends what you are looking for: From a conceptional point of view, I really (really) like Isham's book on Quantum Theory! He explains many things you cannot find in other books, but therefore skips the usual model systems, such as QHO, particle in a box etc. Also, he only refers to rigor at some points, but is is not rigorous himself, but still everything is very clear and formally written down. Commented May 22, 2023 at 7:25
• Other than that, I also enjoy Jochen Rau's book: it is really nice because he has a different way to explain QM, more from an operational point of view, and similarly to Isham he formally develops everything very nice, but not as rigorous as a mathematician would - and still there are the common models missing. A completely rigorous book could be one of Valter Moretti's book, but also the books from Hall (2013) or Browers (although I only skimmed through it). I also like the book by Heinosaari and Ziman... Commented May 22, 2023 at 7:29
• Long story short: For rigorous books: Moretti or Hall, for example. For conceptional books (but also at a formal mathematical level very clear): Isham. Let me know if you still have questions. Edit: Ballentine is very nice, too! I guess it is a perfect mix between the "standard courses" regarding the examples etc., but at the same time is clear at a mathematical and conceptual level! The same goes for Galindo and Pascual! Hope this helps. Commented May 22, 2023 at 7:30
• @kandb if you think people close homework question just because they are tagged as such you are sorely mistaken. I VtC and your question was reopened: good for you but it's still an assignment question (at least as phrased) and should be closed as such. Commented May 22, 2023 at 13:17