Is it possible to perform quantum computation between different Hilbert spaces?

Question

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?

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.