Let us consider a protocol between Alice and Bob. Alice works in a $2^n$-dimensional Hilbert space $\mathcal{H}_A$, using $n$ qubits. Bob works in a $(1+2^n)$-dimensional Hilbert space using qdits. For instance, for $n=128$, Bob would work with two high-dimensional qdits, since $1+2^{128}$ is the product of two large primes.

Since $\mathcal{H}_B$ is isomorphic to $\mathbb{C}^{2^n+1}$, it is possible to write any state $|\psi\rangle_{\mathcal{H}_B}\in\mathcal{H}_B$ as: $$|\psi\rangle_{\mathcal{H}_B}=\sum_{i=0}^{2^n}\psi_i|i\rangle_{\mathcal{H}_B}\,.$$

Let us say that Alice prepares the following state: $$|\varphi\rangle_{\mathcal{H}_A} = \sum_{i=0}^{2^n-1}\varphi_i|i\rangle_{\mathcal{H}_A}$$ and sends it to Bob. Now, Bob wants to transform this state into: $$|\varphi\rangle_{\mathcal{H}_B} = \sum_{i=0}^{2^n-1}\varphi_i|i\rangle_{\mathcal{H}_B}+0\left|2^n\right\rangle_{\mathcal{H}_B}$$ and to apply an unitary matrix $\mathbf{U}_{\mathcal{H}_B}$ such that $\mathbf{U}_{\mathcal{H}_B}\left|2^n\right\rangle_{\mathcal{H}_B}=\left|2^n\right\rangle_{\mathcal{H}_B}$ and $\mathbf{U}_{\mathcal{H}_B}^\dagger\left|2^n\right\rangle_{\mathcal{H}_B}=\left|2^n\right\rangle_{\mathcal{H}_B}$ (that is, $\mathbf{U}_{\mathcal{H}_B}$ does not add the $\left|2^n\right\rangle_{\mathcal{H}_B}$ state in the superposition for any other state and is defined as the identity for this state). This would transform the state in : $$\mathbf{U}_{\mathcal{H}_B}|\varphi\rangle_{\mathcal{H}_B}=\sum_{i=0}^{2^n-1}\alpha_i|i\rangle_{\mathcal{H}_B}\,.$$ Finally, Bob transforms back this quantum state into a $n$-qubit quantum state and sends it back to Alice.

Does this make sense? Is it possible to consider such a protocol, where one converts a quantum state lying in an Hilbert space to one in another Hilbert space?


1 Answer 1


An $n$-level quantum system is an $n$-level quantum system, no matter if it's stored on $\lceil \log_2 n \rceil$ qubits or on $\lceil \log_3 n \rceil$ qutrits or other combinations of qudits.

There would be costs to interoperating computers that use qudits with different numbers of levels, because you have to do work to convert between them (e.g. having to run iterated long division under superposition to convert from binary to ternary, or wasting space on one to make conversion unnecessary by avoiding using the higher qudit levels that are not available on the other), but there's no fundamental barrier to interoperating them.

  • $\begingroup$ I get your point for basis states, but I fail to see why this is also true for a superposition of states. It means that Bob must have some way to "extract" the information from these $n$ qubits to store them on $2$ qudits (without violating the no-cloning theorem, so probably by somehow leaving "junk" in those $n$ qubits). If I'm not mistaken, your point is that there is no fundamental reason for which this interoperability wouldn't be possible, am I right? $\endgroup$
    – Tristan Nemoz
    Apr 21, 2021 at 13:32
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    $\begingroup$ @TristanNemoz You will need some minimal set of physical gates that make it possible, e.g. a qubit+qutrit gate that swaps the 3 states other than |11> of the two qubits for a the states of the qutrit. For qudits with many levels I think it would be sufficient to have a gate that toggles a qubit if the qudit is larger than some constant X, a controlled-by-qubit increment (mod n) gate for the qudit, a multiply by 2 (mod n) gate for the qudit, and a divide by 2 (mod n) gate for the qudit. There are a lot of different ways you could imagine decomposing the conversion into different physical gates. $\endgroup$ Apr 21, 2021 at 13:42
  • $\begingroup$ @TristanNemoz I recommend you try the following exercise 1) write a classical program that converts decimal numbers into hexadecimal numbers, 2) assuming a fixed number of digits, convert the program into a classical logic circuit, 3) replace all irreversible operations in the circuit/program with reversible ones. You can now run your circuit on a quantum computer and it will move a base-10 qudit state into a base-16 qudit state. $\endgroup$ Apr 21, 2021 at 13:46
  • $\begingroup$ I think what I wasn't thinking about in that you can use quantum gates acting on two registers that lie on different Hilbert spaces. I think I understand now, thanks! $\endgroup$
    – Tristan Nemoz
    Apr 21, 2021 at 13:54
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    $\begingroup$ @TristanNemoz They don't live in different Hilbert spaces. They live in a common Hilbert space that is the product of their individual Hilbert-space-when-isolated-s. You don't have to do anything special to make that happen, that's just the representation you have to use when trying to understand operations on collections of qudits. $\endgroup$ Apr 21, 2021 at 15:21

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