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Let me rephrase this question:

Let us assume we know that symmetry transformations always look like this:

$$U(s)=e^{iKs} $$

with a hermitian Operator K. This tells us that for very small $s$:

$$U(ds)=1+iKs$$.

Sakurai for example then defines the general angular momentum operator $J_i$ to be the one defining the Hilbert space version of a rotation of the system in real space i.e. $K \to { J_i \over \hbar}$ and $s \to d\varphi$. Bute here comes the point. Why then is a rotation around am arbitrary axis then: $$U(\vec n , d\varphi)=1 + i { \vec J \vec n \over \hbar}d\varphi$$

If you start with the orbital angular momentum it is easy so see why we can write: $U=e^{{i \over \hbar} \vec \varphi \vec L}$ with $\vec \varphi = \varphi \vec n$ for a rotation of $\varphi$ around the axis $ \vec n$ by using :

$$ e^{{i\over \hbar} \vec \varphi \underbrace{\vec L}_{\vec x \times \vec p}}\Psi(\vec x) =e^{{i\over \hbar} (\vec \varphi \times \vec x) \vec p}\Psi(\vec x)= \Psi(\vec x+\vec \varphi \times \vec x)$$

witch is for small $\varphi$ exactly what you need since then a rotation-matrix $D$ around $\vec \varphi$ has this property $D \vec x=\vec x + \vec \varphi \times \vec x$.

But while in the latter case it is obvious that we are dealing with a rotation in a real 3d space in the first case it is not clear why the operators $J_i$ representing the real rotation in Hilbert space should obay the same rule that $1 + i { \vec J \vec n \over \hbar}d\varphi$ gives the rotation operator for the apropriate Hilbertspace version of a real rotation around $\vec n$.

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An aside first: We do not need to "assume" that we know that one-parameter symmetry transformations are unitary operators $$ U(\phi) = \mathrm{e}^{\mathrm{i}\phi T}$$ for some self-adjoint operator $T$, since Wigner's theorem guarantees that symmetries are (anti-)unitary, and Stone's theorem guarantees (strongly continuous) one-parameter unitary operators are generated as exponentials of a self-adjoint operator.

Now, for the actual question: The case we are looking at is that of a projective representation of the spatial rotation group $\mathrm{SO}(3)$ onto our Hilbert space. Given an axis unit vector $\hat n$, we know that we may write the element of $\mathrm{SO}(3)$ corresponing to a rotation by $\phi$ around $\hat n$ as $$ R_\hat{n}(\phi) = \mathrm{e}^{\mathrm{i}\phi\hat n \cdot \vec L}$$ where $\vec L$ is the 3D angular momentum operator, or equivalently a vector of generators of the Lie algebra $\mathfrak{su}(3)$. One can see this by choosing an explicit realization of the $T^1,T^2,T^3$ (e.g. $L_x,L_y,L_z$ here) and indeed computing the exponential.

A projective representation of the rotation group $\mathrm{SO}(3)$ is induced by any representation of its Lie algebra, which in turn is determined if one gives the images of the generators. Giving the image of the generators by $\vec L \to \vec J$ (with $\vec J$ self-adjoint for unitarity of the representation), the induced representation is $$ U_\hat{n}(\phi) = \mathrm{e}^{\mathrm{i}\phi\hat{n}\cdot \vec J}$$ or, infinitesimally, $$ U_\hat{n}(\delta\phi) = 1 + \mathrm{i}\phi\hat{n}\cdot \vec J + \mathcal{O}(\delta\phi^2)$$ by substitution of the defining series.

tl;dr:: The angular momentum operators on the Hilbert space obey the same relations as the original $\mathrm{SO}(3)$ matrices simply because they are a representation.

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  • $\begingroup$ So because we know for the rotations af small angles we know $D_x(n_1 \varphi)D_y(n_2 \varphi)D_z(n_3 \varphi)=D_{\vec n}(\varphi)$ and we know there is some homomorphism $h$ going from the rotation group to these $U(\Phi)=e^{i\phi T}$ we know the property i mention in the beginning carries over by something like this: $h(D_x(n_1 \varphi)D_y(n_2 \varphi)D_z(n_3 \varphi))=h(D_x(n_1 \varphi))h(D_y(n_2 \varphi))h(D_z(n_3 \varphi))=e^{i n_1 J_x}e^{i n_2 J_y}e^{i n_3 J_z}=e^{\vec n \vec J}$ is that about right? $\endgroup$ – pindakaas May 28 '15 at 17:17
  • $\begingroup$ So am i totally of with this? $\endgroup$ – pindakaas Jun 1 '15 at 8:48
  • $\begingroup$ @pindakaas: Sorry for the late reply (I somehow managed to overlook your first comment). I think your summary captures the essential idea, you're not totally off with this. $\endgroup$ – ACuriousMind Jun 1 '15 at 14:49
  • $\begingroup$ Okay thanks for the anwer more details would have to wait anyway. $\endgroup$ – pindakaas Jun 1 '15 at 16:48

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