Probability and the Magnetic Quantum Number

Question

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.

Is this a correct way to think about this problem?

Answer 1

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.

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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.