The angular momentum operator along the $z$-axis $L_z$ satisfies the secular equation $$ L_z|m,l\rangle = \hbar m |m,l\rangle,$$ where $l$ is the corresponding (integer-valued) eigenvalue of the simultaneously diagonalizable operator $|\mathbf L|^2$ and $m$ is an integer between $-l$ and $+l$. Dropping the number $l$ from the state, we can write in angular coordinates the associated wavefunction $$\chi_m(\varphi):=\langle \varphi|m\rangle =\frac{1}{\sqrt{2\pi}}e^{im\varphi}.$$ Consider the (angular) position eigenket $|\varphi\rangle$. Since the eigenvectors of $L_z$ form a complete set of the Hilbert space, we should be able to write $$ |\varphi\rangle=\sum_m \langle m|\varphi\rangle|m\rangle=\sum_m \chi_m^*(\varphi)|m\rangle,$$ or more generally any wavefunction $\Psi(\varphi)$ as $$\Psi(\varphi)=\langle \varphi|\Psi\rangle=\sum_m \langle\varphi|m\rangle\langle m|\Psi\rangle=\sum_m c_m\chi_m(\varphi)=\frac{1}{\sqrt{2\pi}}\sum_m c_m e^{im\varphi} $$ which is just a Fourier expansion with coefficients given by $$c_m=\int d\varphi\langle m|\varphi\rangle\langle \varphi|\Psi\rangle=\int d\varphi \chi_m^*(\varphi)\Psi(\varphi).$$ My question, coming back to the original eigenket $|m,l\rangle$, is the following: on all of these sums indexed by $m$, does $m$ vary freely in $\mathbb Z$ or is it bound by the quantum numer $l$?

  • $\begingroup$ Let me get this straight: you are looking at even l spherical harmonics for θ =0? That is, are you truly considering SU(2), of just U(1), as your bottom manipulations assume? $\endgroup$ Apr 30, 2021 at 23:06
  • $\begingroup$ @CosmasZachos I'm basically considering just the $\varphi$-dependant part of $Y_{lm}(\theta,\varphi)$. If you will, the circle rather than the sphere... $\endgroup$ Apr 30, 2021 at 23:55
  • $\begingroup$ Indeed, for the circle you have only U(1) and unlimited range m s. $\endgroup$ May 1, 2021 at 0:56
  • $\begingroup$ Thank you @CosmasZachos. I have to admit that I am not very familiar with group theoretic language. Could you elaborate a little on that? $\endgroup$ May 1, 2021 at 14:16
  • $\begingroup$ No, not really. You have to study up on properties of spherical harmonics and why $|m|\leq l$ when the m s are not limited in a circle. $\endgroup$ May 1, 2021 at 14:33


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.