My professor when talking about the EPR paradox said that the singlet spin state,
$$ |00\rangle=\frac{1}{\sqrt 2}(|+-\rangle-|-+\rangle) $$
is symmetric under rotation because its density matrix is
$$\rho=\frac{1}{4}(1-\vec{\sigma_1}\cdot\vec{\sigma_2}),$$
where $\vec{\sigma_1}$ and $\vec{\sigma_2}$ are the Pauli matrices for the first and second particle.
This is because for example a rotation around the $y$ axis that sends $\vec{e_z}$ to $\vec{e_x}$ and $\vec{e_x}$ to $-\vec{e_z}$, sends ${\sigma_i^z}$ to ${\sigma_i^x}$ and ${\sigma_i^x}$ to $-{\sigma_i^z}$ and so leaves $\rho$ unchanged.
But my professor said that $\rho$ for the state
$$ |10\rangle=\frac{1}{\sqrt 2}(|+-\rangle+|-+\rangle) $$
is
$$\rho=\frac{1}{2}(1+\vec{\sigma_1}\cdot\vec{\sigma_2}).$$
So also this state should be symmetric under rotations because there is a scalar product.
But this state isn't symmetric because, while the uncertainty of total spin third component $S^z$ is $0$ (the states is an eigenstate), the uncertainties of $S_x$ and $S_y$ are not $0$. This for example follows from the fact that $\langle S_x \rangle=\langle S_y \rangle=0$ and $\langle S_x^2 \rangle+\langle S_y^2 \rangle=\langle S^2 \rangle = 2\hbar$.
How to solve this contraddiction?
EDIT:
I think I figured out what happened: the $\rho$ matrix for $|10\rangle$ state can't be the one in the OP as shown in this calculation:
$$\vec{\sigma_1}\cdot\vec{\sigma_2}=
\frac{1}{2}[(\vec{\sigma_1}^2+\vec{\sigma_1})^2-\vec{\sigma_1}^2-\vec{\sigma_2}^2]=
\frac{1}{2}(\frac{2}{\hbar})^2 S_{tot}^2-3 =
\frac{2}{\hbar^2}S_{tot}^2-3$$
$$\rho=\frac{1}{\hbar^2}S_{tot}^2-1$$
$$\rho|10\rangle=|10\rangle$$
$$\rho|11\rangle=|11\rangle$$
$$\rho|1-1\rangle=|1-1\rangle$$
So $\rho$ can't be the projector onto the $|10\rangle$ state.
Can someone give me a confirmation about this?
Best Answer
Yes. This is a total-spin zero ($S^2=0$) state and also has $S_z = 0$. This means it transforms trivially under rotations.
No. No it should not be symmetric. The state you wrote above is the $S_z=0$ part of the triplet. It transforms like a vector component into the other components of the triplet.
Yes. The components of the triplet do transform under rotations.
There is no contradiction. Also, as discussed below, one of the expressions for one of the density matrices you wrote above is incorrect.
Update:
Yes. I agree. The density matrix for $|10\rangle$ is: $$ |10\rangle\langle 10| \propto \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right) $$ whereas the expression $1 + \vec{\sigma_1}\cdot\vec{\sigma_2}$ is: $$ 1 + \vec{\sigma_1}\cdot\vec{\sigma_2} \propto \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right)\;, $$ so they are not the same.
However, the expression for $|00\rangle\langle 00| = \frac{1}{4}\left(1 - \sigma_1\cdot\sigma_2\right)$ is fine.
I can confirm in the sense provided in the update above.