quantum mechanics - What is "Rotational Invariance" in the context of qubits

In this question the state, $\frac{1}{\sqrt{3}}\left|00\right\rangle +\frac{1}{\sqrt{3}}\left|01\right\rangle +\frac{1}{\sqrt{3}}\left|11\right\rangle$, has been said in the answers to not be able to collapse into $\frac{1}{2}|11\rangle \langle11| + \frac{1}{2}|00\rangle \langle00|$ after measuring a $|0\rangle$ in the first qubit because this would break "rotational invariance".

I understand mathematically that we project the state into $|0\rangle\langle0|\otimes1\!\!1$ and it becomes $|0\rangle\otimes(|0\rangle+|1\rangle)/\sqrt{2}$.

(Phrasing the question in a way suggested by the comments:) If I consider the situation in which a measurement the 1st qubit in the 0,1 basis causes the second qubit to collapses in the 0,1 basis - why would this break rotational invariance? (And is this method of evauluation a useful way of investigating measuring qubits in general?)

Answer

Let's say we are using the -- incorrect -- rule that upon measuring the first qubit in the $\{|0\rangle,|1\rangle\}$ basis and obtaining the $|0\rangle$ outcome, the second qubit collapses in the $\{|0\rangle,|1\rangle\}$ basis, i.e. on $|00\rangle$ and $|01\rangle$.

Our test state is $$ |\Psi\rangle=\frac{1}{\sqrt{3}}\left|00\right\rangle +\frac{1}{\sqrt{3}}\left|01\right\rangle +\frac{1}{\sqrt{3}}\left|11\right\rangle\ . $$ By applying a Hadamard transformation to the second qubit, this becomes $$ |\Phi\rangle=\frac{1}{\sqrt{3}}\left|00\right\rangle +\frac{1}{\sqrt{3}}\left|10\right\rangle -\frac{1}{\sqrt{3}}\left|11\right\rangle\ . $$ Now let's apply our rule: In the first case, we get $|00\rangle$ and $|01\rangle$, each with probability $1/2$. In the second case, we get $|00\rangle$ with probability $1$ -- and if we undo the rotation, we get $|0\rangle(|0\rangle+|1\rangle)/\sqrt{2}$ with unit probability, rather than the 50/50 mixture we got in the first case.

Thus, this scheme is not invariant under a basis transformation of the $B$ system -- depending in which basis I express my $B$ system (this is what the Hadamard does), I get different results. So if you want a measurement scheme which gives the same result independent of the basis you choose for $B$, the proposed scheme does not work.