Law of transformation of vectors due to rotations

[LagrangeEuler]

I currently styding applications of Lie groups and algebras in quantum mechanics.
$$U^{\dagger}(R)V_{\alpha}U(R)=\sum_{\beta}R_{\alpha \beta}V_{\beta}$$
Where $U(R)$ represents rotation. Letter $U$ is used because it is unitary transformation and $R_{\alpha \beta}$ matrix elements of matrix of rotations. Why this is the way for vector transformation? Is there any explanation?
Also for me is interesting that
$$R_{\alpha \beta}=\delta_{\alpha \beta}+\omega_{\alpha \beta}$$
And from that
$$U(R)=I+\frac{i}{2}\sum_{\mu \nu}\omega_{\mu \nu}J_{\mu \nu}$$
$$U^{\dagger}(R)=I-\frac{i}{2}\sum_{\mu \nu}\omega_{\mu \nu}J_{\mu \nu}$$
where $\omega$ is parameter and $J$ is generator of rotation. Second question. How to now what to take for $U^{\dagger}(R)$ and what for $U(R)$? + or - sign.

 

[samalkhaiat]

Why this is the way for vector transformation? Is there any explanation?

As a classical vector on the relevant index space (in your case vector on $\mathbb{R}^{3}$), $V^{\mu}$ transforms exactly like the coordinates do, i.e., (with repeated indices are summed over) $$V^{\mu} \to R^{\mu}{}_{\nu} \ V^{\nu} , \ \ \mbox{where} \ \ R^{\mu}{}_{\nu} = \frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} \ .$$ But, as an operator $V^{\mu}$ transforms (for each value of the index $\mu$) like quantum-mechanical operators do, i.e., by a unitary transformation $$V^{\mu} \to U^{\dagger} V^{\mu} U \ .$$ This means that the expectation value of each component of $V^{\mu}$ in the state $|\psi \rangle$ transforms as $$\langle \psi |V^{\mu} | \psi \rangle \to \langle \psi |U^{\dagger} V^{\mu} U |\psi \rangle \ . \ \ \ \ (1)$$ But, for each $\mu$ the expectation value is a c-number (classical quantity). So, $\langle \psi |V^{\mu} | \psi \rangle$ should also transform like a classical vector on our index space, i.e., $$\langle \psi |V^{\mu} | \psi \rangle \to R^{\mu}{}_{\nu} \langle \psi |V^{\nu} | \psi \rangle \ . \ \ \ (2)$$ Now, comparing (1) with (2) and knowing that $|\psi \rangle$ is an arbitrary state, we obtain the transformation law for the vector operator (i.e., q-number) $$U^{\dagger}(R) \ V^{\mu} U(R) = R^{\mu}{}_{\nu} \ V^{\nu} \ . \ \ \ (3)$$ Group-theoretically speaking, this equation tells you how the (finite-dimensional) matrix representation $R$ of the group (in question) is related to the (not necessarily finite-dimensional) unitary (operator) representation of the same group on the Hilbert space of the states.

How to now what to take for $U^{\dagger}(R)$ and what for $U(R)$? + or - sign.

The sign is completely arbitrary. However, you need to be careful when $V^{\mu} = V^{\mu}(x)$ is a vector field. In this case, Eq(3) should read $$U^{\dagger}(R) \ V^{\mu}(x)\ U(R) = R^{\mu}{}_{\nu} \ V^{\nu} \left(R^{-1} x \right) \ .$$

 

[vanhees71]

Also for me is interesting that
$$R_{\alpha \beta}=\delta_{\alpha \beta}+\omega_{\alpha \beta}$$
And from that
$$U(R)=I+\frac{i}{2}\sum_{\mu \nu}\omega_{\mu \nu}J_{\mu \nu}$$
$$U^{\dagger}(R)=I-\frac{i}{2}\sum_{\mu \nu}\omega_{\mu \nu}J_{\mu \nu}$$
where $\omega$ is parameter and $J$ is generator of rotation. Second question. How to now what to take for $U^{\dagger}(R)$ and what for $U(R)$? + or - sign.

In addition to what @samalkhaiat already has written in #2 you should note that these are only "infinitesimal transformations". What you have here are the generators $\hat{J}_{\mu \nu}$ of rotations, i.e., a basis of the Lie algebra of the rotation group. You have $\hat{J}_{\mu \nu}=-\hat{J}_{\nu \mu}$ so that you as well can use the usual angular-momentum operators,
$$J_{\rho}=\frac{1}{2} \epsilon_{\rho \mu \nu} \hat{J}_{\mu \nu}.$$
Here and in the following I use the Einstein summation convention according to which over repeated indices one has to sum. I also do not distinguish between co- and contravariant components for this Euclidean case using a Cartesian basis.

You come from the infinitesimal transformations to finite transformations by exponentiation. In this case the rotation around an axis in direction $\vec{n}$ with rotation angle $\phi$ (in the sense of the right-hand rule), the corresponding unitary operation is given as
$$\hat{U}(\vec{n},\phi)=\exp(\mathrm{i} \phi \vec{n} \cdot \hat{\vec{J}}).$$
It's easy to see that this is a unitary operator, if all the $\hat{\vec{J}}$ are self-adjoint, because then
$$\hat{U}^{\dagger}(\vec{n},\phi)=\exp(-\mathrm{i} \phi \vec{n} \cdot \hat{\vec{J}})=\hat{U}^{-1}(\vec{n},\phi).$$
You can also show by analysis of how rotations compose that the generators fulfill fill the angular-momentum algebra,
$$[\hat{J}_{\mu},\hat{J}_{\nu}]=\mathrm{i} \epsilon_{\mu \nu \rho} J_{\rho},$$
and from these commutation relations you can construct all irreducible representations of the rotation group (or rather its covering group SU(2) which then introduces half-integer spin representations, a notion that is a pure quantum concept, not known within classical physics).

 

[LagrangeEuler]

Thanks a lot. I have one more question.
In calculations
$$\frac{i}{2}\sum_{\mu,\nu}\omega_{\mu,\nu}[V_{\alpha},J_{\mu,\nu}]=\sum_{\mu}\omega_{\alpha \mu}V_{\mu}$$
I one want to obtain commutator$$[V_{\alpha},J_{\mu,\nu}]$$, why you can not transform right hand side with
$$\sum_{\mu}\omega_{\alpha \mu}V_{\mu}=\sum_{\mu \nu}\omega_{\nu \mu}\delta_{\alpha \nu}V_{\mu}=-\sum_{\mu \nu}\omega_{\mu \nu}\delta_{\alpha \nu}V_{\mu}$$
and then
$$\frac{i}{2}[V_{\alpha},J_{\mu\nu}]=-\delta_{\alpha \nu}V_{\mu}$$
Is it because $\omega_{\alpha \beta}$ is antysimetric and $\delta_{\alpha \beta}$ is symetric?

 

[vanhees71]

How do you come to this conclusion? You have to consider the $\omega_{\mu \nu}$ as varying independently (of course within the constraints that they are antisymmetric under interchange of their indices, because any symmetric part within the $\omega_{\mu \nu}$ cannot contribute to the left-hand side, because the angular-momentum operators are antisymmetric under exchange of their indices). So the logic is rather
$$\frac{\mathrm{i}}{2} \sum_{\mu \nu} \omega_{\mu \nu} [V_\alpha,J_{\mu \nu}] = \sum_{\mu \nu} \omega_{\nu \mu} \delta_{\nu \alpha} V_{\mu}= \frac{1}{2} \sum_{\mu \nu} \omega_{\nu \mu} (\delta_{\nu \alpha} V_{\mu}-\delta_{\mu \alpha} V_{\nu}) ,$$
from which you get (using the antisymmetry of the $\omega_{\mu \nu}$)
$$\mathrm{i} [V_{\alpha}, J_{\mu \nu}] = (\delta_{\mu \alpha} V_{\nu} - \delta_{\nu \alpha} V_{\mu}).$$

 

[samalkhaiat]

why you can not transform
and then
$$\frac{i}{2}[V_{\alpha},J_{\mu\nu}]=-\delta_{\alpha \nu}V_{\mu}$$

Because you end up with a wrong equation if you factor out $\omega_{\mu\nu}$ before taking into the consideration the antisymmetric property of $\omega_{\mu\nu}$, as explained to you by vanhees71. As an advice for you, you should always check that your equations do make sense. And, it is very clear that your equation is very wrong because (1) the left-hand-side is antisymmetric in $\mu\nu$, while the right-hand-side has no definite symmetry with respect to $\mu\nu$, and that is a wrong thing to have in a tensorial equation, and (2) as a consequence of (1), if you multiply both-sides of your equation by the symmetric ($\mathbb{R}^{3}$-metric) tensor $\delta_{\mu\nu}$ and sum over $\mu\nu$, you obtain $0 = V_{\alpha}$ for all values of $\alpha$, i.e., your equation is “correct” only for the zero vector so there is nothing good about it. So, if you encounter equation of the form $$\sum_{\mu\nu} \omega_{\mu\nu} J_{\mu\nu} = \sum_{\mu\nu} \omega_{\mu\nu} A_{\mu\nu} \ , \ \ \ \ (1)$$ with $\omega_{\mu\nu} = - \omega_{\nu\mu}$ and $J_{\mu\nu} = - J_{\nu\mu}$, and you want to factor-out the $\omega$, what do you do? If you know that $A_{\mu\nu} = - A_{\nu\mu}$, then you simply get $J_{\mu\nu} = A_{\mu\nu}$, but if $A_{\mu\nu} \neq - A_{\nu\mu}$, then you have to anti-symmetrise $A_{\mu\nu}$ before factoring out the $\omega$, because $\omega_{\mu\nu}$ (being anti-symmetric) kills the symmetric part of $A_{\mu\nu}$: any tensor $A_{\mu\nu}$ can be uniquely decomposed into symmetric and anti-symmetric parts $$A_{\mu\nu} = \frac{1}{2} \left( A_{\mu\nu} + A_{\nu\mu}\right) + \frac{1}{2} \left( A_{\mu\nu} - A_{\nu\mu}\right) \ .$$ Now if you contract this with $\omega_{\mu\nu}$, the first term on the right-hand-side vanishes because it is symmetric, and you get $$\sum_{\mu\nu} \omega_{\mu\nu} A_{\mu\nu} = \frac{1}{2} \sum_{\mu\nu} \omega_{\mu\nu} \left( A_{\mu\nu} - A_{\nu\mu}\right) \ , \ \ \ (2)$$ Now, and only now, you can factor out $\omega_{\mu\nu}$ from (1) and (2) to obtain a valid tensor equation $$J_{\mu\nu} = \frac{1}{2} \left( A_{\mu\nu} - A_{\nu\mu} \right) .$$