I learnt from my quantum mechanics courses that angular momentum operator is the generator of rotation, and applying angular momentum operator is like performing an infinitesimal rotation in 3D space.

Here is my question: why would a wavefunction "lengthen" under infinitesimal rotation? For instance, let's take a wavefunction $\psi$ with $l=2$ as an example. Obviously, it is invariant under rotation in z-axis and $l_{z}\psi=2\hbar\psi$, but I don't understand why a function will be scaled by a factor of $2\hbar$ when it is rotated. How can rotation reduce (or enhance) the amplitude of a function?

I understand that in this case it doesn't matter because wavefunction must be normalized and a constant factor in front won't change any observables. However, if I understand correctly, the use of rotational generator should be something universal, i.e. I could just as well apply the generators of rotation to any scalar field that maybe changing the amplitude do carry physical significance. This brings back to may question, how can an infinitesimal rotation change the amplitude of a function?

I am guessing that my interpretation of angular momentum operator is wrong. I can answer my homework's questions just fine as they mostly ask for mathematical calculations, which I have less trouble with, but I am having a hard time understanding intuitively how different operators operate. Thank you very much for your help.

  • $\begingroup$ Hi. The transformation, that is the rotation, comes from the action of the group element, that is the exponential of the generator $\exp ^{-iεL_z}$ $\endgroup$ – Constantine Black Jan 16 '17 at 17:03

The rotation operator is $$R(\hat{n},\varphi)=e^{-\frac{i}{\hbar}\varphi\hat{n} \cdot \vec{L}}$$, set $\hat{n}=\hat{z}$,so rotate the wave-function $\psi_{lm}$, we obtain $$R(\hat{n},\varphi)\psi_{lm}=e^{-i\varphi m}\psi_{lm}$$ the "length" is invariant. Considering an infinitesimal rotation $\hat{n}=\hat{z},\varphi \to \delta \varphi$, the rotation operator is approximate to $$1-\frac{i}{\hbar}\delta \varphi \hat{n}\cdot\vec{L}$$ then $$R(\hat{n},\delta\varphi)\psi_{lm}=\psi_{lm}-i\,\delta \varphi \,m\psi_{lm} \approx\psi_{lm}$$ the "length" is also invariant.


What a great question! The answer, I think, is that $l_z$ is not a generator of rotations per se. So for example your typical spin-$\frac12$ system the angular momentum operator in the $z$-direction is$$\frac\hbar 2 \begin{bmatrix}1&0\\0&-1\end{bmatrix}$$and you can just "see" that this is not a rotation matrix (and it's not just the scale factor, though that helps).

Note especially that this "length" is actually a probability to observe the system in any state therefore it better not change, even for simpler systems. What gives?

I think that there is a misunderstanding deep inside this, that Hermitian observables are directly affecting the wavefunctions in quantum mechanics. They never do this directly, they always have to go through something else, usually the Schrödinger equation. In the latter case, if $\hat H = \hbar\omega\sigma_z$, then the actual evolution operator is $U = e^{-i\omega t\sigma_z}=\cos(\omega t) I -i\sin(\omega t)\sigma_z$, which looks much more familiar as some sort of generated rotation.


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