For a particle of spin 1, does there exist a rotation $D(R)$ such that $D(R)|1,0\rangle=|1,1\rangle$? Given a particle with spin $1$, let its states be $|1,1\rangle$, $|1,0\rangle$, $|1,-1\rangle$.


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*Question:


Does there exist a rotation $D(R)$ such that $D(R)|1,0\rangle=|1,1\rangle$?


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*My solution:


$D(R)$ is given by $\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)$ for some unit vector $\hat n$ and angle $\phi$.
Also, we see that $\langle 1,1| \hat{n}\cdot\vec{S} |1,1\rangle=n_z\hbar $ because:
$\langle s',m'|S_{z}|s,m\rangle=m\hbar\delta_{s',s}\delta_{m',m}$
$ \langle s',m'|S_{\pm}|s,m\rangle =\sqrt{(s\mp m)(s\pm m+1)}\hbar\delta_{s',s}\delta_{m',m\pm1}$
$S_{x}=\frac{1}{2}(S_{+}+S_{-})$
$S_{y}=\frac{1}{2i}(S_{+}-S_{-})$
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Now, let's assume there is such a $D(R)$. So there is also a $D'(R)$ fulfilling:
$D'(R)|1,1\rangle=|1,0\rangle$
Say $D'(R)=\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)$
So we have: $\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)|1,1\rangle=|1,0\rangle$.
Multiplying both sides by the bra $\langle 1,1|$, we get:
$\exp\left(-i\phi n_z\right)=0$
Which can never happen. $\blacksquare$


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*Does my proof seem all-right? Any comments on my way of reasoning? Any ideas of how to do it in other ways? What is the general approach to problems of this kind?

 A: The answer is negative. 
Suppose there is such rotation $R$ and let us denote its inverse by $R'$. In this case
$$S_z D(R') |1,1\rangle = S_z |1,0\rangle=0\:,$$
As a consequence,
$$D(R')^\dagger S_z D(R') |1,1\rangle =0$$
that is
$$\vec{n} \cdot \vec{S}|1,1 \rangle =0\tag{1}$$
where $\vec{n}= R' \vec{e}_z$. 
Using $S_z|1,1\rangle=|1,1\rangle$, (1) implies
$$n_x S_x|1,1 \rangle + n_y S_y |1,1 \rangle = - n_z |1,1 \rangle \tag{2}\:.$$
Appliyng $S_z$ to both sides, we also have
$$n_x S_zS_x|1,1 \rangle + n_y S_zS_y |1,1 \rangle =- n_z |1,1 \rangle\tag{3}\:,$$
which, together with
$$n_x S_xS_z|1,1 \rangle + n_y S_yS_z |1,1 \rangle =- n_z |1,1 \rangle \tag{4}\:,$$
implies
$$n_x [S_z,S_x]|1,1 \rangle + n_y [S_z,S_y] |1,1 \rangle =0 \tag{5}\:.$$
Using the commutation relations of the $S_k$ operators we have
$$n_x S_y|1,1 \rangle - n_y S_x |1,1 \rangle =0 \tag{6}\:.$$
Now focus on (2) and (6). Computing the determinant $d$ of the linear system made of this pair of equations (the unknowns being the two vectors $S_x |1,0 \rangle $ and $S_y |1,1 \rangle$)  we find $d= n_x^2 +n_y^2$. 
If $d=0$ that is $n_x=  n_y=0$, we have  that $\vec{n}= \vec{e}_z$ and thus $R$ is a rotation around $\vec{e}_z$. This is not possible because 
$$D(R_z(\theta))|1,0\rangle = e^0 |1,0\rangle =  |1,0\rangle \neq |1,1 \rangle\:.$$
If $d\neq 0$, the unique solution of the afore-mentioned system leads to either
$$S_x |1,1 \rangle =-\frac{n_xn_z}{1-n_z^2}|1,1 \rangle$$
or
$$S_y |1,1 \rangle =-\frac{n_yn_z}{1-n_z^2} |1,1 \rangle$$
where $n^2_z \neq 1$ otherwise $d=0$,
which are false by direct inspection, using the explicit expressions of the matrices $S_x$ and $S_y$ because $|1,1\rangle$ is not an eigenvector of $S_x$ and $S_y$. 
We conclude that there are no chances to find the wanted $R$.
A: I don't know if this follows from some general theorem, but I've found a simple proof. 
Let me denote with $\vert 0,\pm1\rangle$ the base of the $j=1$ representation, with $\vert \uparrow \downarrow\rangle$ that of the $j=\frac{1}{2}$ representation.  If there exists such an $R$: $$D^{(1)}(R)\vert 0\rangle =\vert 1\rangle,$$ then we must have, in the standard phase convention: $$D^{(1)}(R)\dfrac{\vert \uparrow \rangle\vert \downarrow \rangle +\vert \downarrow \rangle\vert \uparrow \rangle}{\sqrt 2}=\dfrac{D^{(\frac{1}{2})}(R)\vert \uparrow \rangle \otimes D^{(\frac{1}{2})}(R) \vert \downarrow \rangle +D^{(\frac{1}{2})}(R) \vert \downarrow \rangle \otimes D^{(\frac{1}{2})}(R) \vert \uparrow \rangle}{\sqrt 2}=\vert \uparrow \rangle\vert \uparrow \rangle. $$ Now, if we put $$D^{(\frac{1}{2})}(R)\vert \uparrow\rangle = \alpha \vert \uparrow\rangle +\beta \vert \downarrow \rangle \\D^{(\frac{1}{2})}(R)\vert \downarrow\rangle = \gamma \vert \uparrow\rangle +\delta \vert \downarrow \rangle, \\$$ we obtain the conditions: $$\alpha \gamma \neq 0\\ \alpha \delta+\gamma \beta =0\\ \beta \delta =0.$$ Taken together, these three equations imply $\beta = \delta =0$. So $D^{(\frac{1}{2})}(R)\vert \uparrow\rangle$ and $D^{(\frac{1}{2})}(R)\vert \downarrow\rangle$ are proportional to each other, which is impossible since $D(R)$ is invertible.
A: Consider the inverse rotation R'.
Compute the expectation value
$1 = \langle U^\dagger(R') L_z  U(R') \rangle_0 = \langle R_{3,i}  \cdot L_i \rangle_0 = 0$
The first equality is because the rotation sends you to the "highest" eigenvector, the second one is because L transform as a vector under rotation (it is a result of representation theory), the last one is because all of the three components of L have zero mean. 
