Quantum computing refers (occasionally implicitly) to a "computational basis". Some texts posit that such a basis may arise from a physically "natural" choice. Both mathematics and physics require meaningful notions to be invariant under a change of basis.

So I wonder whether the computational complexity of a problem (say, the k-local Hamiltonian) depends on the basis of the n-qubit tensor product space $\mathbb{C}^{2^n}$ be what is (wrongly) called the "canonical" one?

Specifically, it seems that the "definition" for a Hamiltonian to act nontrivially only on k of the n qubits fails the sanity check of invariance under a change of basis.

Edit for clarification:

Let $U \in SU(2^n,\mathbb{C})$ be an arbitrary change of basis of the full Hilbert space and $H = A_{2^k} \otimes I_{2^{n-k}}$ be a Hamiltonian that acts nontrivially only the first $k$ qubits. Then $U^\dagger H U$ might not be of the same form as $H$ only acting on $k$ qubits.

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    $\begingroup$ Can you clarify your last sentence? As far as i’m concerned it doesn’t fail that sanity check. $\endgroup$
    – DaftWullie
    Oct 22, 2019 at 15:42
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    $\begingroup$ Welcome to QCSE, and this is a great question. However, please note that Jalex Stark was not wrong in calling the computational basis "canonical." He was using the term canon in plain English, not the mathematical term of art for a canonical morphism. I can see how this could be confused if English is not your first language. At worst, you could say his statement was potentially confusing. $\endgroup$ Oct 22, 2019 at 19:31

3 Answers 3


The computational basis is "natural" in the sense that it provides a practical representation of measurement outcomes. Other bases are also "natural" for other tasks, and bases cannot be interchanged aribtrarily if such a change impacts the underlying tensor structure of the system.

For example the Bell states form a basis in four dimensional Hilbert space representing two maximally entangled qubits. Bell states are not bilinear combinations of computational basis states, so we cannot simply choose to represent separable states over the Bell basis, or entangled states over the computational basis.

What seems to be missing from your assumptions is a recognition that the tensor product structure over the computational basis is of fundamental importance because it comports with measurement outcomes - our only connection to the quantum realm. In other words, the computational basis for an $n$-qubit system, $\vert \psi_0 \rangle$, is a tensor product space of the form $$\vert \psi_0 \rangle = \vert b_0 \rangle \otimes \vert b_1 \rangle \otimes ... \otimes \vert b_{n-1} \rangle \in (\mathbb{C}^2)^{\otimes n},$$ where $\vert b_i \rangle \in \mathbb{C}^2$. A transformation that preserves the tensor structure of $\vert \psi_0 \rangle$, i.e. does not induce entanglement, takes the form $$U = U_0 \otimes U_1 \otimes , ... , \otimes U_{n-1} \in U(2)^{\otimes n} \subset U(2^n).$$ Hamiltonians of this form are regularly employed in both theory and practice, where any given $U_i$ can act trivially or non-trivially.

However, generally in practice only a discrete set of single qubit transformations are available. By the Solovay-Kitaev theorem $U_i$ can be approximated to precision $\epsilon$ using $\Theta(\log^c(\tfrac{1}{\epsilon}))$ gates from a fixed discrete set, where $c$ is a constant between 1 and 2 (as far as I know its exact value is still an open problem).

When multilevel gates are applied to the system to induce entanglement, we still need to be able to track the relationship back to the tensor structure of $\vert \psi_0 \rangle$. For sufficiently large $n$, applying an arbitrary $SU(2^n)$ transformation to $\vert \psi_0 \rangle$ amounts to complete decoherence of the system because any meaningful description of the wavefunction is lost (i.e., breaking $SU(2^n)$ down into a Tensor Network representation, discussed below, becomes intractable.)

Such transformations need to be applied in very rigorous and non-trivial ways. As Carlton Caves said "Hilbert space is a big place," and it's easy to get lost. Notice, for example, that the space of linear transformations spanned by $U(2^n)$ is exponentially larger than that of $U(2)^{\otimes n}$, e.g. $U(2^n)$ has real dimension $2^{2n}$, whereas $U(2)^{\otimes n}$ has real dimension $4n$.

In practice, low energy systems can only access a relatively small portion of the full Hilbert space because physically realistic states are heavily constrained by locality. Tensor networks are a good example of a class of approaches to managing this challenge by limiting the description of a system to physically realistic regions of Hilbert space (see, e.g. section 3.3 here).

  • $\begingroup$ THANK YOU, particularly for the crucial sentence: "the computational basis is of fundamental importance because it comports with measurement outcomes". This however seems to contractict the principle that all physics (including quantum mechanics) should be independent of the basis under consideration: https://physics.stackexchange.com/questions/233949/basis-independence-in-quantum-mechanics $\endgroup$ Oct 23, 2019 at 17:08
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    $\begingroup$ An arbitrary 2-qubit system in the computational basis is $\vert \psi_0 \rangle = (\alpha_i \vert 0 \rangle + \beta_i \vert 1 \rangle \otimes (\alpha_j \vert 0 \rangle + \beta_j \vert 0 \rangle)$. This tensor transforms under $U(2) \otimes U(2)$ - an 8 real dimensional space of linear transformations. No transformation of that form can create a Bell state from a separable state. The transformation from a separable state to a Bell state requires a $U(4)$ transformation - a 16 real dimensional space of linear transformations. $\endgroup$ Oct 23, 2019 at 17:16
  • $\begingroup$ How is a Bell state, say $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ not a linear combination of computational basis vectors as you write? $\endgroup$
    – Marsl
    Oct 23, 2019 at 20:31
  • $\begingroup$ @Marsl It's not a bilinear (I should have said bilinear) combination of computational basis states because there does not exist two one-qubit states, $\vert \psi_1 \rangle$, $\vert \psi_2 \rangle$, such that $\vert \psi_1 \rangle \otimes \vert \psi_2 \rangle = \tfrac{1}{\sqrt{2}} (\vert 00 \rangle + \vert 11 \rangle)$. You can find a more detailed discussion/proof here. $\endgroup$ Oct 23, 2019 at 21:08

This is a partial answer addressing only what I know: Stoquastic Hamiltonians in the Monte Carlo sign problem. The TLDR is, yes, complexity may depend on the choice of basis.

A stoquastic Hamiltonian is one that has non-positive off-diagonal matrix elements in the standard basis. This class of Hamiltonians was quite famously studied first here.

In the world-line Monte Carlo method, one tries to estimate the quantum partition function $Z=\text{Tr}(e^{-\beta H})$ for inverse temperature $\beta$ by cutting it into many pieces and sampling from it. This is done by writing $$e^{-\beta H}=(e^{-\beta H/m})^m$$ The trace can now be written as $$\text{Tr}(e^{-\beta H})=\sum_i \langle i |e^{-\beta H})|i\rangle =\sum_{i_1,i_2,\dots,i_m}\langle i_1 |e^{-\beta H/m})|i_2\rangle\langle i_2 |\dots \langle i_m |e^{-\beta H/m})|i\rangle=\sum_J a_J $$ where $|i\rangle$ are basis vectors of canonical/computational/standard basis and $J$ is a shorthand for the index set $i_1,i_2,\dots,i_m$.

How does this relate to stoquasticity you might ask. Well, on the right hand side, I have written a big sum of terms I called $a_J$ that one can easily sample from using standard Monte Carlo methods just like one would estimate an integral via Monte Carlo sampling. However, in order for this to work, the summands should be positive! otherwise one samples from a quasi-probability distribution (i.e. a non-positive distribution). Positivity of the summands is ensured for stoquastic Hamiltonians since to first order $e^{-\beta H}=1-\beta H$ and the sums involve only products of certain entries of this matrix. However, non-stoquastic Hamiltonians exhibit what is famously known as the Monte Carlo sign problem.

It should be rather self-evident now, that this is a problem of choosing the right basis: Prominently, in the eigenbasis of $H$ there is no problem since $H$ is diagonal. However, finding the eigenbasis or other sign-problem free bases was shown to be NP-complete already for 3-local Hamiltonians.

If you want to learn more, I suggest you take a look at these two papers addressing the Monte Carlo sign problem and how to alleviate it.




I give this second answer to address a misconception that might be lurking in the question and to look at the local Hamiltonian problem: I am not entirely convinced by my answer and happy to hear what others have to say about it.

A Hamiltonian $H$, is first and foremost a Hermitian operator on the Hilbert space $\mathcal{H}=(\mathbb{C}^2)^{\otimes n}$, i.e. $$ H: \mathcal{H}\to \mathcal{H}$$

Consider for example the Hamiltonian $H=Z\otimes I$ acting on two qubit. This is an operator not a matrix: For example the $Z$ operator acting on the first qubit can be represented by the Pauli matrix everyone is familiar with but in order to do so, you have to (implicitly) pick the computational basis. Otherwise, this operator $Z$ just maps certain vectors to certain vectors. The notion of $k$-locality here (in this case $1$-locality), I believe, exists completely separately from the concept of a basis. That also means, that for any unitary $U$ that you want to conjugate $H$ with, you simply get a different Hamiltonian! (Of course, once you write down $H$ as a matrix in some basis, the conjugation by unitary matrices corresponds to a basis change)

Now, the local Hamiltonian problem is essentially about finding the smallest eigenvalue but formulated as a decision problem. It is a computational problem and thus we need to resort to picking a basis for our Hilbert space. Note that the decision problem would be trivial if you knew the eigenbasis of the Hamiltonian! Thus, it is entire clear again that the complexity of this computational problem is related to the choice of basis, but this is independent of the $k$-locality.


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