Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

For a state with non-zero angular momentum, why is it that it cannot be described by the spherically symmetric spherical harmonic?

share|improve this question
    
maybe this answer will help: physics.stackexchange.com/questions/90173/… –  Adam May 28 at 14:48

1 Answer 1

up vote 2 down vote accepted

This is because each spherically invariant state $\psi$ must have zero angular momentum.

Indeed, by hypotheses, the state $\psi$ verifies $$\psi(R_{\vec n}(\theta)\vec{x} ) = (e^{i \theta \vec{n}\cdot \vec{\hat{J}}} \psi)(\vec{x}) = \psi(\vec{x})\tag{1}$$ where $\vec{n}\cdot \vec{\hat{J}}$ is the self-adjoint generator of rotations $R_{\vec n}(\theta)$ around $\vec{n}$, i.e. it is the angular momentum along $\vec{n}$. Taking the $\theta$ derivative of (1) for $\theta=0$ we have $$\vec{n}\cdot \vec{\hat{J}} \psi =0$$ in particular, for $k=x,y,z$, $$\hat{J}_k \psi=0\:,$$ so that $$\hat{J}^2 \psi = \hat{J}^2_x\psi + \hat{J}^2\psi + \hat{J}_z^2=0\:.$$

ADDENDUM. Actually a state is represented by a normalized vector up to a phase. A spherically symmetric state is therefore represented by a vector satisfying version of (1) weaker than the one presented above: $$\psi(R_{\vec n}(\theta)\vec{x} ) = (e^{i \theta \vec{n}\cdot \vec{\hat{J}}} \psi)(\vec{x}) = \chi(\theta, \vec{n})\psi(\vec{x})\tag{2}$$ where $|\chi(\theta, \vec{n})|=1$. Taking the $\theta$ derivative for $\theta=1$ we find $$\vec{n}\cdot \vec{\hat{J}} \psi = \alpha(\vec{n}) \psi$$ where the eigenvalue is $$\alpha(\vec{n}) = \frac{d\chi(\theta, \vec{n})}{d\theta}|_{\theta=0}$$ which is a real number as easily follows from $|\chi(\theta, \vec{n})|=1$. The common eigenvectors $\psi \neq 0$ of $\hat{J}_x,\hat{J}_y,\hat{J}_z$ have the common eigenvalue $0$ as it can be proved by direct inspection (or by means of some straightforward theoretical argument exploiting the commutation relations $[\hat{J}_x,\hat{J}_y]= i\hat{J}_z$). We conclude that this more complete way leads to the same result as before.

share|improve this answer
    
Is there a more qualitative answer in physical terms? It was more like a 2 mark question.. –  user44840 May 28 at 13:22
    
Well, a spherically symmetric wavefunction must be constant along all angular directions. Fixing the $z$ axis in a arbitrary direction $L_z =-i\frac{\partial}{\partial \phi}$ and thus $L_z\psi=0$ since $\psi$ is constant in $\phi$. As $z$ can be fixed in an arbitrary direction, $L_x\psi=L_y\psi = L_z\psi =0$ and thus $L^2 \psi =L_xL_x\psi + L_yL_y\psi +L_zL_z\psi=0$. –  Valter Moretti May 28 at 15:53
    
I think they didn't say that the wavefunction was spherically symmetric.. –  user44840 May 29 at 20:51
    
You actually said spherically symmetric spherical harmonic. However, the complete wavefunction is the product of such symmetric spherical harmonic and a function of $r$, so it is a spherically symmetric wavefunction. –  Valter Moretti May 30 at 6:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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