$\mathbb{C}^2 \otimes \mathbb{C}^2$ vs $\mathbb{C}^4$

Is there a difference between the following two Hilbert spaces: $$H_1 = \mathbb{C}^2 \otimes \mathbb{C}^2$$ and $$H_2 = \mathbb{C}^4$$?

Here's my confusion. For the following bases, $$H_1 = H_2$$ holds: $$\mathbb{C^4} =$$ span$$\{\lvert{000}\rangle, \lvert{100}\rangle , \lvert{010}\rangle , \lvert{110}\rangle \} =$$span$$\{ \lvert 00 \rangle \otimes \lvert 0 \rangle,\lvert 10 \rangle \otimes \lvert 0 \rangle, \lvert 01 \rangle \otimes \lvert 0 \rangle, \lvert 11 \rangle \otimes \lvert 0 \rangle\} = \mathbb{C}^2 \otimes \mathbb{C}^2$$. However, if I choose different bases, I couldn't manage to express $$\mathbb{C}^4 =$$span $$\{ \lvert{000}\rangle, \lvert{110}\rangle , \lvert{011}\rangle , \lvert{111}\rangle\}$$ in terms of $$\mathbb{C}^2 \otimes \mathbb{C}^2$$. So $$H_1 = H_2$$ holds depending on the basis set?

• Your question is a bit vauge. when you say $\mathbb{C^4} =$ span$\{\lvert{000}\rangle, \lvert{100}\rangle , \lvert{010}\rangle , \lvert{110}\rangle \}$, essentially you are working in a subspace of larger space $\mathbb{C^8}$. The basis $\{ |00 \rangle, |01 \rangle , |10 \rangle, |11 \rangle \}$ are sufficient for $\mathbb{C}^4$. Also $\mathbb{C}^2 = \text{Span}\{ |0 \rangle, |1\rangle \}$. Oct 19, 2023 at 21:43
• possible duplicate of quantumcomputing.stackexchange.com/q/1860/55
– glS
Oct 20, 2023 at 8:27

We say two Hilbert spaces $$A$$ and $$B$$ are isomorphic if there exists a linear isomorphism between the two spaces. That is, there exists a bijective map $$\phi: A \to B$$ such that
1. (Linear) For all scalars $$\alpha, \beta \in \mathbb{C}$$ and all vectors $$v,w\in A$$ we have $$\phi(\alpha v + \beta w) = \alpha \phi(v) + \beta \phi(w)$$
2. (Preserves inner products) For all $$v, w \in A$$ we have $$\langle \phi(v) , \phi(w) \rangle = \langle v, w\rangle\,.$$
It then follows that two Hilbert spaces of the same dimension are isomorphic (essentially the same). To see this suppose $$A$$ and $$B$$ have the same dimension d. Let $$\{v_1,\dots, v_d\}$$ be an orthonormal basis for system $$A$$ and let $$\{w_1,\dots,w_d\}$$ to be an orthonormal basis for system $$B$$. We can then define our isomorphism $$\phi : A \to B$$ to be the map that sends $$\phi(v_i) = w_i$$ and we just extend the map linearly. So any vector $$v \in A$$ can be written as $$v = \sum_i \alpha_i v_i$$ which through the map $$\phi$$ now corresponds to the vector $$w = \phi(v) = \phi(\sum_i \alpha_i v_i) = \sum_i \alpha_i \phi(v_i) = \sum_i \alpha_i w_i.$$ Through this identification it is clear that the map is bijective due to the uniqueness of writing any vector in the orthonormal bases. Moreover it preserves inner products as if we have $$v, v' \in A$$ then $$\langle v, v' \rangle = \langle \sum_i \alpha_i v_i, \sum_j \alpha_j' v_j \rangle = \sum_i \overline{\alpha}_i \alpha_i' = \langle \sum_i \alpha_i w_i, \sum_j \alpha_j' w_j \rangle = \langle \phi(v), \phi(v') \rangle\,.$$
For your case you can just make the identifications \begin{aligned} |00\rangle \leftrightarrow |000\rangle \\ |01\rangle \leftrightarrow |110\rangle \\ |10\rangle \leftrightarrow |011\rangle \\ |11\rangle \leftrightarrow |111\rangle \end{aligned} and you have your isomorphism.