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

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?

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.
• 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? Apr 21 at 13:32