In the article Quantum Algorithm Implementations for Beginners I found the following sentence

Entanglement makes it possible to create a complete $2^n$ dimensional complex vector space to do our computations in, using just $n$ physical qubits.

My taught process:

Let there be three qubits $(Q_1, Q_2, Q_3)$ with corresponding Hilbert spaces $\mathcal{H_1^{2}}$,$\mathcal{H_2^{2}}$,$\mathcal{H_3^{2}}$. Using tensor products I can write a $2^n$ dimensional Hilbert space as

\begin{equation} \mathcal{H^{2^3}} = \mathcal{H_1^{2}} \otimes\mathcal{H_2^{2}}\otimes \mathcal{H_3^{2}}. \end{equation}

So with my logic, I've created a $2^n$ dimensional Hilbert space without entanglement.

My question:

Why is entanglement making it possible and where I'm wrong?


2 Answers 2


The error lies in the assertion that $\mathcal{H^{2^3}} = \mathcal{H_1^{2}} \otimes\mathcal{H_2^{2}}\otimes \mathcal{H_3^{2}}$ is without entanglement. For example, $\frac{|000\rangle+|111\rangle}{\sqrt2}\in\mathcal{H^{2^3}}$ is the GHZ state famous for exemplifying one type of tripartite entanglement.

Nevertheless, the misunderstanding has an interesting lesson about the nature of entanglement. Note that we do not explicitly use or assume the existence of entanglement anywhere in the construction of $\mathcal{H^{2^3}}$. Instead, entanglement emerges as a result of using the tensor product in the definition. In physics terms, the tensor product expresses the fact that we allow$^1$ $\mathcal{H^{2^3}}$ to contain superpositions of product states such as $|000\rangle$ and $|111\rangle$. It turns out that most such superpositions are in fact entangled states. This is related to the fact that entanglement is not a primitive notion of quantum mechanics and does not explicitly appear in any of its postulates. Instead, it is a consequence of the postulate that defines the state space of a composite system to be the tensor product of the state spaces of the component systems.

$^1$The question whether a superposition of two states of a physical system is allowed or not is not a trivial matter. See superselection rules for more details on when superpositions are disallowed.


Adam's answer points out that your proposed space, the tensor product of $n$ qubits, does have entanglement.

Possibly one source of confusion (for me, anyways ^_^) may be conflating the tensor product (or really, Kronecker product) with a more familiar operation from set theory, the Cartesian product. The Kronecker product $\mathcal{H}^{2^3}=\mathcal{H}_1^{2}\otimes\mathcal{H}_2^{2}\otimes\mathcal{H}_3^{2}$ is certainly $2^n$ dimensional and includes entangled states.

Meanwhile, the Cartesian product $(\mathcal{H}^{2})^3=\mathcal{H}_1^{2}\times\mathcal{H}_2^{2}\times\mathcal{H}_3^{2}$ is the space of all factorizable $n$-qubit states and has only $2n$ dimensions. You can certainly write down elements of this space as though embedded in $2^n$ dimensions, as you would when multiplying out the full statevector, but you just don't have that many degrees of freedom.

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    $\begingroup$ Why only $2n$ dimensions? $n$ qubits can be in $2^n$ basis states, hence the dimension of the space is also $2^n$ . $\endgroup$ Feb 4, 2022 at 7:32
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    $\begingroup$ My pardons; after reading Adam's answer I've realized my answer conflates the tensor product of spaces with the tensor product of individual qubit states. If each state is parameterized by just two dimensions, their product has only $2n$ parameters. But this would describe the Cartesian product of the spaces, not the tensor product. It's possible the OP made this conflation at some point as well, hence prompting the question, so I'll edit my answer to address this. $\endgroup$
    – jecado
    Feb 4, 2022 at 18:41

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