Please can you help me to understand how the dimension of the set of separable states is $\dim \cal H_1 + \dim \cal H_2$?

This is the relevant passage:

So far, we have assumed implicitly that the system is made of a single component. Suppose a system is made of two components; one lives in a Hilbert space $\cal H_1$ and the other in another Hilbert space $\cal H_2$. A system composed of two separate components is called bipartite. Then the system as a whole lives in a Hilbert space $\cal H = \cal H_1 \otimes \cal H_2$, whose general vector is written as $$\left|\, \psi \right\rangle = \sum_{i,j} c_{ij} \left|\,e_{1,i}\right \rangle \otimes \left|\,e_{2,j}\right\rangle, \tag{2.29}$$ where $\{|\,e_{a,i}\rangle\}$ ($a=1,2$) is an orthonormal basis in $\cal H_a$ and $\sum_{i,j} |c_{ij}|^2 = 1$.

A state $|\,\psi \rangle \in \cal H$ written as a tensor product of two vectors as $|\,\psi \rangle = |\,\psi_1 \rangle \otimes |\,\psi_2\rangle$, ($|\,\psi_a\rangle \in \cal H_a$) is called a separable state or a tensor product state. A separable state admits a classical interpretation such as “The first system is in the state $|\,\psi_1\rangle$, while the second system is in $|\,\psi_2\rangle$.” It is clear that the set of separable states has dimension $\dim \cal H_1 + \dim \cal H_2$.

  • $\begingroup$ From Wikipedia: If $\mathscr{H}_1$ and $\mathscr{H}_2$ have orthonormal bases $\{\varphi_k\}$ and $\{\psi_l\}$, respectively, then $\{\varphi_k \otimes \psi_l\}$ is an orthonormal basis for $\mathscr{H}_1\otimes\mathscr{H}_2$. In particular, the Hilbert dimension of the tensor product is the product (as cardinal numbers) of the Hilbert dimensions. $\endgroup$
    – yuggib
    Sep 1, 2014 at 13:21

2 Answers 2


Note that the space of separable states is not a vector space, and in particular not a subspace of the total Hilbert space: the sum of two separable states is unlikely to be separable. So dimension here means something more general than vector space dimension.

Having said that, I would disagree with the author on his dimension! I would say that the space of (nonzero) separable states has dimension $\dim \mathcal{H_1}+\dim\mathcal{H_2}-1$.

To specify a separable state, we can supply an element of each of $\mathcal{H_1}$ and $\mathcal{H_2}$, which means $\dim \mathcal{H_1}+\dim\mathcal{H_2}$ complex numbers. However, there is a redundancy here, because we can change each by an overall scaling ($|\psi_1\rangle\mapsto\lambda|\psi_1\rangle, |\psi_2\rangle\mapsto\lambda^{-1}|\psi_2\rangle$) without changing the product state, which reduces the dimension by 1.

A couple of simple examples:

1) If $\mathcal{H_1}$ is 1-dimensional (completely trivial!), then all states are separable, and $\mathcal{H_1}\otimes\mathcal{H_2}\simeq\mathcal{H_2}$.

2) If both $\mathcal{H_1}$ and $\mathcal{H_2}$ are two-dimensional, we can write a state of $\mathcal{H_1}\otimes\mathcal{H_2}$ as a 2x2 matrix. The separable states have proportional columns/rows, so are exactly the same as matrices of determinant zero. If we exclude 0, this is a 3-dimensional submanifold.

  • $\begingroup$ your scaling $|\psi_1\rangle\mapsto\lambda|\psi_1\rangle$ doesn't preserve the normalisation though. If $\dim\mathcal H_1$ is (as it should) $2(N-1)$ with $N$ the dimension of the space, then the phase and normalisation ambiguities have already been taken care of, so I don't understand why should there be one less parameter. For example, two-qubit separable states are determined by $2+2=4$ parameters, not three: $(\cos\theta|0\rangle+\sin\theta e^{i\phi}|1\rangle)\otimes(\cos\theta'|0\rangle+\sin\theta' e^{i\phi'}|1\rangle)$ $\endgroup$
    – glS
    Sep 13, 2019 at 1:35

This might not be what Nakahara has in mind, but one can make sense of this using the idea of projective Hilbert spaces. Let $\mathcal{P}(\mathcal{H})$ denote the projective space associated to the "normal" space $\mathcal{H}$.

The subset of separable states is not a subvectorspace in the proper sense, as Holographer notes. Yet it can be understood as a projective subvariety of the projective space associated with the tensor product of the underlying Hilbert spaces - it is the image of the Segre embedding, being a smooth embedding

$$ \mathcal{P}(\mathcal{H}_1) \times \mathcal{P}(\mathcal{H}_2) \to \mathcal{P}(\mathcal{H}_1 \otimes \mathcal{H}_2), (\psi,\phi) \mapsto \psi \otimes \phi$$

where $\mathcal{P}(\mathcal{H}_1) \times \mathcal{P}(\mathcal{H}_2)$ are the separable states.1 In the language of projective varieties, this image is a $(m-1)+(n-1)$ dimensions projective subvariety of $\mathcal{P}(\mathcal{H}_1 \otimes \mathcal{H}_2)$, but since we should more properly see $m' = m - 1$ and $n' = n - 1$ - the dimensions of the projective spaces - as the dimensional of the actual spaces of states, we obtain indeed that the subvariety corresponding to the separable states has the sum of the dimensions of the individual states as its dimension.

1Note that on ordinary Hilbert spaces, this is not even injective, let alone an embedding in any proper sense, since $\psi \otimes \phi = k\psi \otimes \frac{1}{k}\phi$ means that $(\psi,\phi)$ and $(k\psi,\frac{1}{k}\phi)$ map to the same element of the tensor product space.

  • $\begingroup$ ah! algebraic geometry is everywhere! but I am such a poor fellow who doesn't understand, Thanks but sorry with my mathematics background I don't understand your answer $\endgroup$
    – Marso
    Sep 1, 2014 at 14:43
  • $\begingroup$ Very nice! Was really struggling to see the geometry here; this helps a lot. One very small nitpick: in the original notation, $\dim\mathcal{H}_1=n$, so $\dim\mathcal{P}(\mathcal{H}_1)=n-1$, so the separable states would be an $(n+m-2)$-dimensional variety here if I understand correctly. $\endgroup$ Sep 1, 2014 at 15:22
  • $\begingroup$ @Holographer: You're right, of course. $\endgroup$
    – ACuriousMind
    Sep 1, 2014 at 15:38

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