Can a Quantum Computer Run the von Neumann Architecture?

Abstract

At the core of nearly every modern computer is a central processing unit running the von Neumann architecture. This computer architecture gives computationally universal machines, and non-trivial control structures arise naturally, leading to high-level programming constructs.

At the core of the von Neumann architecture is the notion that program code may be stored and manipulated in the same way as data. A datum describing an operation may be stored and processed in the same way as any other form of data, but may also be ‘promoted’ to an operation, and applied.

Classically, this is well-studied—particularly from a categorical point of view. We consider such operations in the quantum setting, including Nielsen and Chuang’s orthonormal encoding, Abramsky and Coecke’s categorical foundations, the BBC protocol, and the Choi-Jamiołkowsky correspondence.

Obstacles to a quantum analogue of the von Neumann architecture are also considered, including the no-cloning and no-deleting theorems, the “no-programming principle”, and the Gottesman-Knill theorem.

Appendix

We consider the conditions required for a state vector to correspond to a unitary map, and show that this is intimately connected with questions of entanglement.

Definition 14

Let H denote a complex Hilbert space with orthonormal (computational) basis $\{\textbf{e}_i\}_{i=1 .. n}$. Then for each basis vector $\textbf{e}_i$ we define the left-span of $\textbf{e}_i$ to be the subspace of $H\otimes H$ generated by the basis vectors $\{ e_i\otimes e_j \}_{j=1}^n$. Dually, we define the right-span of $\textbf{e}_j$ to be the subspace of $H\otimes H$ generated by the basis vectors $\{ e_i\otimes e_j\}_{i=1}^n$. We denote these spaces by $lSpan(\textbf{e}_{\textbf{i}})$ and $rSpan(\textbf{e}_{\textbf{j}})$ respectively.

Our claim is that the vectors that are the names of unitary maps are exactly those that are equidistant to the left span and the right span of each basis vector e _i , and thus maximally entangled.

Theorem 9

Let H denote a complex Hilbert space with orthonormal basis $\{\textbf{e}_i \} _{i=1..N}$, and let $M:H\rightarrow H$ denote a linear map. Then the following two conditions are equivalent:

• (i) M is a unitary map.

• (ii) For each basis vector e _i , the norm of the projection of $\ulcorner M \urcorner$ onto either lSpan(e _i ) or rSpan (e _i ) is $\frac{1}{n}$.

Proof

($\textbf{(i)} \Rightarrow \textbf{(ii)}$ By definition of a unitary map, M satisfies

$$MM^\dagger = I \ \ \mathrm{and} \ \ I = M^\dagger M$$

(It is easier to state that $M^\dagger = M^{-1}$. However, interesting and computationally important C* algebras such as the Kuntz-Krieger algebras of [9] satisfy one-sided versions of these conditions, so we use them separately for future reference). Written in terms of matrix elements, we have that

$$\begin{aligned}( MM^\dagger )_{ik} = \sum_{j=1}^{n} m_{ij}\overline{m_{kj}} = \left\{ \begin{array}{lr} 1 & i=k \\ 0 & i\neq k \end{array} \right. \\ ( M^\dagger M )_{ik} = \sum_{j=1}^{n} m_{ji}\overline{m_{jk}} = \left\{ \begin{array}{lr} 1 & i=k \\ 0 & i\neq k \end{array} \right.\end{aligned}$$

These conditions can also be characterised as “the sum of the norms of the entries in each row is 1, as is the sum of the norms of the entries in each column”.

Moving to the name $\ulcorner M \urcorner\in H\otimes H$, we have

$$\langle \ulcorner M \urcorner | \ulcorner M \urcorner \rangle = \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}} = \left( \sum_{\alpha , \beta = 1}^n \left( \sum_{i , j=1}^n \langle \overline{m_{\alpha \beta}} (\textbf{e}_i\otimes \textbf{e}_j) | m_{ij} (\textbf{e}_i\otimes \textbf{e}_j) \rangle \right) \right)$$

Using the Kroneker delta notation, $\langle \textbf{e}_i\otimes \textbf{e}_j | \textbf{e}_\alpha \otimes \textbf{e}_\beta \rangle = \delta _{\alpha i} \delta _{\beta j}$, so

$$\langle \ulcorner M \urcorner | \ulcorner M \urcorner \rangle = \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}} = \left( \sum_{\alpha , \beta , i ,j = 1}^n \overline{m_{\alpha \beta}} m_{ij} \delta _{\alpha i} \delta _{\beta j} \right)$$

Hence, by the condition imposed on the matrix elements by the unitarity requirement,

$$\langle \ulcorner M \urcorner | \ulcorner M \urcorner \rangle = \frac{1}{N} \sum_{\alpha , \beta=1}^n \overline{m_{\alpha \beta}} m_{\alpha \beta} = 1$$

So the unitarity of M implies that $\ulcorner M \urcorner$ has norm 1.

For the next step, observe that we may isolate the individual $m_{ij}$ by

$$m_{ij} = \sqrt{n}.\langle \textbf{e}_i\otimes \textbf{e}_j | \ulcorner M \urcorner \rangle$$

and so the first unitarity condition gives that

$$1 = \sum_{j=1}^n \frac{1}{\sqrt{n}} \langle \ulcorner M \urcorner | \textbf{e}_i\otimes \textbf{e}_j \rangle \frac{1}{\sqrt{n}} \langle \textbf{e}_i \otimes \textbf{e}_j | \ulcorner M \urcorner \rangle$$

Hence

$$\frac{1}{n}= \sum_{j=1}^n \langle \ulcorner M \urcorner | \textbf{e}_i\otimes \textbf{e}_j \rangle \langle \textbf{e}_i \otimes \textbf{e}_j | \ulcorner M \urcorner \rangle$$

Using Dirac notation, for any orthnormal basis B the identity is given by $Id=\sum_{b\in B} |\textbf{b}\rangle \langle \textbf{b}|$, and so the inner product of vectors $\phi , \psi$ may be written as $\langle \phi | \psi \rangle = \sum_{\textbf{b}\in B} \langle \phi |\textbf{ b} \rangle \langle \textbf{b} | \psi \rangle$. From the definition of the space $lSpan(\textbf{e}_i)$ in terms of a basis set, we thus deduce that the projection of $\ulcorner M \urcorner$ onto the space $lSpan(\textbf{e}_i)$ has norm $\frac{1}{N}$.

The dual condition about the right spans $\{ rSpan(\textbf{e}_i) \}_{i=1}^n$ follows from the second unitarity condition.

(ii) ⇒ (i) Let $\psi = \ulcorner M \urcorner$ satisfy the left and right span conditions. We may write these fully as

$$\frac{1}{n}= \sum_{j} \langle \psi | \textbf{e}_i\otimes \textbf{e}_j \rangle \langle \textbf{e}_i \otimes \textbf{e}_j | \psi \rangle$$

and

$$\frac{1}{n}= \sum_{j} \langle \psi | \textbf{e}_i\otimes \textbf{e}_j \rangle \langle \textbf{e}_i \otimes \textbf{e}_j | \psi \rangle$$

respectively. The definition of the naming operation $\ulcorner \ \ \urcorner$ gives that

$$[M ]_{ij} = \sqrt{n} . \langle e_i\otimes e_j | \psi \rangle$$

and almost identical reasoning to above applied to the left span condition gives $\sum_{j=1}^n [M ]_{ij} . \overline{[M ]_{ij}} = 1$. Similarly, the right span condition gives that

$$\sum_{i=1}^n [M ]_{ij} . \overline{[M ]_{ij}} = 1.$$

From above, these are the two conditions required for unitarity, and hence our result follows.

Interpretation

Although the above is presented abstractly, a quantum computational interpretation is immediate: given a quantum register $r\in qByte$, and an observation on a single arbitrary qubit with respect to any orthonormal basis $\{ \textbf{b}_1,\textbf{b}_2\}$, then r is the name of a unitary map exactly when the observation of r gives either b ₁ or b ₂ with equal probability.