Why does quantum computing generally use matrix formulation of quantum mechanics rather than a continuous variable formulation?

Question

The book Elements of Quantum Computing draws a distinction between "Wave Mechanics" and "Matrix Mechanics" (p. 38). I believe the reason we use a matrix formulation is that we are not dealing with continuous quantities like position, but finite quantities like spin. But I would like to validate this assumption. The paper Quantum Information: A Brief Overview and Some Mathematical Aspects (arXiv) also notes this distinction on p. 3.

I believe my earlier version of this question mixed "Wave Mechanics" and "Matrix Mechanics" with Schrödinger and Heisenberg respectively. While Heisenberg was the first to introduce matrices to QM, his picture of QM is different than Schrödinger's. I believe that Von Neumann was able to unite these two pictures into the postulates we know and love today as described in the paper A Brief History of the Mathematical Equivalence Between the Two Quantum Mechanics (source).

Answer 1

Quantum computing, in general, does not use Heisenberg's matrix mechanics or Dirac's interaction picture.

I suggest that you carefully read the Wikipedia page on dynamical pictures. The differences between the Heisenberg picture, the Schrödinger picture and Dirac (interaction) picture are well summarized in the following chart. The development of matrix mechanics, as a mathematical formulation of quantum mechanics, is attributed to Werner Heisenberg, Max Born, and Pascual Jordan.)

In the widely used circuit model of quantum computing, the qubit states evolve in time as per the Schrödinger equation $|\Psi_s(t)\rangle = e^{-iH_st}|\Psi_s(0)\rangle$ (by application of quantum gates) while the observables are constant (note that observables in quantum mechanics are defined in terms of positive operator-valued measures). For details, read this section. So your claim that quantum computing (or quantum information theory for that matter) uses Heisenberg's matrix mechanics, is incorrect.

I presume your confusion arose because you saw matrices all around in quantum computing textbooks and thought they use some kind of "matrix mechanics", but in reality, Heisenberg's matrix mechanics is something completely different i.e., it's an algebraic approach to quantum mechanics using mathematical devices like ladder operators. Moreover, in Heisenberg's matrix mechanics quantum states remain constant while the operators evolve (see kludg's answer for an example). Nevertheless, DaftWullie's answer explains why matrices are preferable to using continuous-wave functions for low-dimensional finite systems like qubits in quantum computing if that's what you were confused about.

Nearly 100 years ago, von Neumann had already fairly rigorously shown the equivalence (under certain reasonable assumptions cf. this) among all the three pictures with his separable Hilbert space formulation of quantum mechanics (which really underpins all of modern quantum mechanics and quantum information theory); the three pictures differ only by a basis change with respect to time-dependency. It's just that one picture might be more useful than another in certain contexts.

Rotation of state vectors in the Hilbert space is easier to visualize than evolving observables and that's one of the reasons why the Schrödinger picture preferred in introductory approaches to quantum mechanics. And, in general, quantum computing just deals with "toy" systems from quantum mechanics i.e., finite-dimensional qubit or qudit systems, whereas very high-dimensional or infinite-dimensional systems are much more prevalent in the real-world (though, there are exceptions like continous variable quantum information that makes use of physical observables like the strength of electromagnetic fields whose values belong to continuous intervals).

Answer 2

Semantics aside, I’m assuming that your question is essentially “why do we use a matrix formulation of quantum mechanics rather than a continuous variable/differential equation/integral formulation” (I may be wrong and would welcome clarification) and nothing to do with the interaction picture and the like, which some other answers seem to be touching upon.

Largely, we use the bra-ket notation of Dirac, so we might write quantities such as $\langle\psi|\phi\rangle$. These are very abstract entities. You can choose to represent these abstract entities using either a vector/matrix formalism of an integral formalism. It doesn't matter.

However, we typically choose to use vectors just because it tends to make the maths easier. Vectors have, somewhat more obviously, properties such as orthogonality built into them, so I don't end up evaluating integrals all the time. Particularly when I want to talk about a two-level system, I could describe it using two continuous functions $\psi_0(x)$ and $\psi_1(x)$, but it's just overkill.

The sort of problem that I studied as an undergraduate would talk, for example, about the infinite square well. You'd be given some initial state $f(x)$, and asked what it looks like at some later time. Perhaps you have to evaluate the probability that the particle is on the right-hand side of the well. By far the quickest way of doing this is to decompose $$ f(x)=\sum_n\alpha_n\psi_n(x), $$ where $\psi_n(x)$ are the stationary states of energy $E_n$. The only thing I really care about is the discrete set of values $\alpha_n$, which are most conveniently stored as a vector. Then the solution would be $$ \sum_n\alpha_ne^{-iE_nt}\psi_n(x). $$ Next, I decompose the state of being on the right-hand side of the well as $$ \sum_n\beta_n\psi_n(x). $$ The probability that the final state is on that side is $$ \left|\sum_n\beta_n^\star\alpha_ne^{-iE_nt}\right|^2. $$ The only integrals that I need to do are to evaluate the $\alpha_n$ and $\beta_n$. The rest of the calculation is more easily done using orthogonality/linearity properties which are all present in the integral form, but are harder to see. So you very quickly start abstracting away from the integral form and are naturally led towards linear algebra.

Answer 3

The quantum circuit model does not use the Heisenberg picture of quantum mechanics. Also, what is "Dirac's Matrix Mechanics"? Quantum computing widely uses Dirac's bra-ket notation.

Quantum circuit model considers how state vectors of qubits are transformed by quantum gates which follows the Schrödinger picture of quantum mechanics.

---

PS: Maybe I need to say more. The difference between Schrödinger and Heisenberg pictures of QM is totally unrelated to the difference between differential equation and matrices.

Suppose I have a qubit $$|0\rangle=\begin{pmatrix} 1\\ 0 \end{pmatrix}$$ and apply Hadamard gate to it; as a result, the state of the qubit changes: $$ \begin{pmatrix} 1\\ 0 \end{pmatrix}\rightarrow\frac{1}{\sqrt{2}} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix} =\frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1 \end{pmatrix} $$ This is Schrödinger way of thinking, and it is used in Quantum Computing without saying it explicitly.

Heisenberg picture of QM is very different; It goes as follows: suppose I have $S_z$ spin operator, $$S_z=\frac{\hbar}{2}\sigma_z=\frac{\hbar}{2}\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}$$ and apply Hadamard gate to it: $$S_z\rightarrow\frac{1}{\sqrt{2}} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}S_z \frac{1}{\sqrt{2}} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}=\frac{\hbar}{2} \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}=S_x $$ That is, Hadamard gate transforms $S_z$ operator to $S_x$ operator (or transforms $\sigma_z$ Pauli matrix to $\sigma_x$ Pauli matrix).

This is Heisenberg way of thinking; I'm not sure it suits well to quantum circuit model.