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Preamble

A two qubit/spin-1/2 system can be represented as

$$|\psi\rangle=\alpha|\uparrow\uparrow\rangle+\beta|\uparrow\downarrow\rangle+\gamma|\downarrow\uparrow\rangle+\delta|\downarrow\downarrow\rangle=\begin{pmatrix}\alpha\\\beta\\\gamma\\\delta\end{pmatrix}$$ where $\alpha,\beta,\gamma,\delta$ are complex numbers such that $|\alpha|^2+|\beta|^2+|\gamma|^2+|\delta|^2=1\tag{1}$ is the normalization condition. For convenience we can choose $\alpha=a$ strictly real as the global phase does not matter. This means that to specify the state, we need 7 real parameters. The equation (1) represents then the surface of a 7D Bloch sphere.

My question concerns non-entangled states in that sphere. For a state to be non-entangled it has to be of the form

$$|\psi\rangle=(\epsilon|\uparrow\rangle+\zeta|\downarrow\rangle)\otimes(\eta|\uparrow\rangle+\theta|\downarrow\rangle)=\begin{pmatrix}\epsilon\eta\\\epsilon\theta\\\zeta\eta\\\zeta\theta\end{pmatrix}$$ where $\epsilon,\eta,\zeta,\theta$ are complex numbers, such that $$|\epsilon|^2+|\zeta|^2=1=|\eta|^2+|\theta|^2.\tag{2}$$

Question

Equation (2) represents some kind of lower dimensional spheres surfaces embedded in the 7-sphere of eq. (1). This seems ok, the 7D-surface of the non-entangled states is null as there are more entangled states than non-entangled states (see Are there more entangled states or non-entangled ones? ).

How is the 7D Bloch sphere divided with respect to this non-entangled boundary? Does this boundary cut the surface of the Bloch-sphere on two regions?

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  • $\begingroup$ Simpler: If a general state is written $a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$, then the condition for the state to be separable with respect to the standard factorization of the Hilbert space is $ad=bc$. $\endgroup$
    – Chiral Anomaly
    Commented Sep 26, 2021 at 15:08
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    $\begingroup$ I haven't thought carefully enough about the topology to add anything to Zack's answer. I mentioned the $ad=bc$ approach because this is the condition for a $2\times 2$ complex matrix to have zero determinant, and I'm willing to bet that some good pure-math literature is available about the topology of the zero-determinant manifold, maybe under the heading "algebraic geometry." $\endgroup$
    – Chiral Anomaly
    Commented Sep 26, 2021 at 17:37
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    $\begingroup$ This means that to specify the state, we need 7 real parameters. The equation (1) represents then the surface of a 7D Bloch sphere: not quite. You also have to take into account the phase freedom. The overall space of pure 4dim states is $\mathbb{CP}^7$, which is not isomorphic to a (hyper)sphere. See e.g. Do pure qudit states lie on a hypersphere in the Bloch representation?. Regarding the overall question, you might want to look up en.wikipedia.org/wiki/Segre_embedding and en.wikipedia.org/wiki/Determinantal_variety $\endgroup$
    – glS
    Commented Sep 27, 2021 at 7:14
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    $\begingroup$ the number of free (real) parameters needed to specify an $n$-dimensional pure state is $2n-2$. So this would be $6$ for a four-dimensional system (which a two-qubit state is). But dimension aside, the topology itself is different from that of a hypersphere, so you cannot even say the space is isomorphic to $S^6$. Regarding the subset of non-entangled (and thus product) states, I agree with the answer below that this is $S^2\times S^2$. But note that it's an accident of this case. You get a nice sphere only for single qubits, because $\mathbb{CP}^1\simeq S^2$ $\endgroup$
    – glS
    Commented Sep 27, 2021 at 7:33
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    $\begingroup$ @glS no ok but dimensionality was never an issue. I assumed the surface of a 7D hypersphere (6 parameters) $S^6$ from normalization and global phase freedom. However, you are telling me that it is not a $S^6$, and I am guessing that’s because I assumed $\alpha$ real but that’s not sufficient to account for all possible global phases, specially when $\alpha=0$. $\endgroup$
    – Mauricio
    Commented Sep 27, 2021 at 8:38

1 Answer 1

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Two somewhat trivial observations:

  1. Topologically, the non-entangled states must be two spheres, ie $S^2 \times S^2$. This is because the product state form of your middle equation is specified by two separate and completely independent pure states, which each topologically comprise an ordinary Bloch sphere.

  2. These two Bloch spheres $S^2 \times S^2$ comprise a four-dimensional manifold; meanwhile, the total Bloch space is six (seven?) dimensional, as you observe. Thus, the non-entangled states cannot possibly cut the total Bloch space into two regions, in the same way a one-dimensional manifold cannot cut a three-dimensional manifold into two regions.

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  • $\begingroup$ Thanks! Minor detail, if you are defining the dimension by the minimal number of parameters that define your position in the hypersphere, I guess then is 6 dimensional ($S^6$). $\endgroup$
    – Mauricio
    Commented Sep 25, 2021 at 9:31
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    $\begingroup$ "the total Bloch hypersphere is six-dimensional" unless I'm misreading the context, this is not true. The space of quantum states of dimension larger than two is not a (hyper)sphere. The underlying reason being that ${\Bbb CP}^n\simeq S^m$ only holds for $n=1$ (and then $m=2$). See e.g. math.stackexchange.com/a/4182341/173147, quantumcomputing.stackexchange.com/q/6009/55, and quantumcomputing.stackexchange.com/q/8416/55 $\endgroup$
    – glS
    Commented Sep 27, 2021 at 7:18
  • $\begingroup$ I agree -- I was suspicious of this as well, but didn't know for certain. I'll edit the language. $\endgroup$
    – Zack
    Commented Sep 27, 2021 at 14:58
  • $\begingroup$ The space is still 6 dimensional, hypersphere surface or not $\endgroup$
    – Mauricio
    Commented Sep 27, 2021 at 15:34

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