Rotation of Entangled States $(R \otimes R)|\phi\rangle = e^{i\delta}|\phi\rangle$

Question

Maximally entangled states are said to be rotational invariant should a rotation be applied to both particles. That is (for some rotation operator $R$):

$$(R \otimes R)|\phi\rangle = e^{i\delta}|\phi\rangle$$

Let us take the EPR state:

$$|\phi\rangle = \frac{1}{\sqrt{2}}(|+\rangle|-\rangle - |-\rangle|+\rangle)$$

$$(R \otimes R)|\phi\rangle = \frac{1}{\sqrt{2}}(|n;+\rangle|n;-\rangle - |n;-\rangle|n;+\rangle)$$

We know that $|n;\pm\rangle = \cos(\theta/2)|+\rangle \pm e^{i\psi}\sin(\theta/2)|-\rangle$. When you work out this calculation you get:

$$(R \otimes R)|\phi\rangle = \frac{1}{\sqrt{2}}(\cos(\theta/2)|+\rangle + e^{i\psi}\sin(\theta/2)|-\rangle \otimes (\cos(\theta/2)|+\rangle - e^{i\psi}\sin(\theta/2)|-\rangle) - (\cos(\theta/2)|+\rangle + e^{i\psi}\sin(\theta/2)|-\rangle) \otimes (\cos(\theta/2)|+\rangle - e^{i\psi}\sin(\theta/2)|-\rangle)$$ $$= - \sin(\theta)e^{i\phi} \left( \frac{|+\rangle|-\rangle - |-\rangle|+\rangle}{\sqrt{2}}\right)$$

The $\sin(\theta)$ is not a phase term, is is a re-normalization term. Granted, even if you renormalize the system you get an equivalent state, but then $(R \otimes R)|\phi\rangle = e^{i\delta}|\phi\rangle$ should be $(R \otimes R)|\phi\rangle \propto e^{i\delta}|\phi\rangle$. Is there something that I am misunderstanding in this rotation operation?

Answer 1

There are a few things to be mentioned: (while admittedly digressing a bit)

1. Maximally entangled states are not invariant under identical rotations applied to both particles. That is only true for the "singlet state" in your example, but not for the other three Bell states, even though they are all maximally entangled. For instance $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|+\rangle|+\rangle + |-\rangle|-\rangle)$ is only invariant under $z$-axis rotations.
2. You say that $|n;\pm\rangle = \cos(\theta/2)|+\rangle \pm e^{i\psi}\sin(\theta/2)|-\rangle$, but that must be a typo, since that would mean for rotations with small $\theta$ that $|n;+\rangle$ and $|n;-\rangle$ are almost identical (whereas they should be orthogonal). In particular, when $\theta=0$ (the identical rotation) we would have $|n;+\rangle=|+\rangle$ and also $|n;-\rangle=|+\rangle$.
3. You say that if you work this out further you still get the original singlet state, albeit renormalized with $\sin\theta$. That would seem to be true at least for the case $\theta=0$ mentioned above, since $|\phi\rangle$ is then mapped to $\frac{1}{\sqrt{2}}(|+\rangle|+\rangle - |+\rangle|+\rangle) = 0$.

The main error is of course point 2), where the rotated states are incorrect. But if you fix that problem you still cannot prove it for all maximally entangled states. They form a singlet and a triplet if you see them as angular momentum states. Only the singlet is invariant, the other three transform among each other under the spin-1 representation, i.e. with ordinary 3D rotations. At least if the same rotation is applied to both particles.

You can also apply different rotations to each particle, and then the state still will remain maximally entangled (you cannot break entanglement if the particles are not interacting but only individually manipulated). What happens in that case is that you have transformations from the group $SU(2) \times SU(2)$. That group is isomorphic to $SO(4)$ and it acts in the real (not complex) 4-dimensional subspace of all maximally entangled states. The fact that it is a real vector space is also visible when you make a linear combination of the Bell states to obtain all possible maximally entangled states. That doesn't work for arbitrary complex coefficients (or else you could create any state!) but only for: $$ |\psi_\text{ max entang.}\rangle =e^{i\varphi}\ \big( \ a\ |\Phi^+\rangle+b\ \ i|\Phi^-\rangle+c\ |\Psi^+\rangle+d\ \ i|\Psi^-\rangle \ \big) $$ where $a,b,c,d$ are real coefficients and we use the standard [Bell basis], except that the second and fourth state are redefined with a factor $i$. There is an extra overall phase factor $e^{i\varphi}$ allowed, so that would make the maximally entangled states a 5 real-parameter family of states, but if we normalize them with $a^2+b^2+c^2+d^2=1$ it is only a 4 real-parameter family and if we ignore the phase $\varphi$ only 3 are left.

So they can be represented on the 3-sphere and after 4D stereographic projection visualized in ${\mathbb R}^3$ (e.g. the singlet state in the origin, triplet states on the ${\mathbb R}^3$ basis vectors). All possible maximal entanglement preserving transformations can then be visualized as $SO(4)$ single or [double rotations] in this projection. The case of an identical rotation on both particles actually becomes a single rotation in this ${\mathbb R}^3$ space, but the more arbitrary $SU(2) \times SU(2)$ transformations will become $SO(4)$ double rotations where each state in the ${\mathbb R}^3$ projection move in a spiral over a Clifford torus, as in [this drawing:]

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