Can a Digital Computer Simulate Quantum Entanglement to Any Precision?

A digital computer can simulate the predictions of an entangled quantum system to any chosen numerical precision, under standard assumptions about the system and its inputs. That does not mean every simulation is practical: for a general system of many qubits, the amount of computation can grow exponentially with the number of qubits.

The word “simulate” can also mean two different things. A computer can calculate the probabilities and correlations predicted by quantum mechanics. That is not the same as physically producing an entangled state, or explaining entanglement using local classical variables.

What a digital simulation represents

For $n$ qubits, a pure quantum state can be written as

$$|\psi\rangle = \sum_{x=0}^{2^n-1} a_x |x\rangle,$$

where each $a_x$ is a complex amplitude and the probabilities sum to one: $\sum_x |a_x|^2=1$. A straightforward classical simulator stores the $2^n$ amplitudes, applies the specified quantum operations to them, and calculates measurement probabilities.

In principle, the amplitudes can be stored and calculated to as many decimal places as needed. Increasing numerical precision reduces rounding error. For a finite system whose initial state and operations can be specified to the required accuracy, this gives a systematic way to approximate its predictions as closely as desired.

There is a qualification: “any precision” does not mean that a finite computer can represent every possible real number exactly. The inputs must themselves have a usable specification—for example, finite formulas or computable numerical approximations. If an input contains an arbitrary, uncomputable real number, no ordinary algorithm can necessarily produce its digits.

Worked example: simulating an entangled pair

Consider the two-qubit Bell state

$$|\Phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt{2}}.$$

A state-vector simulator stores four amplitudes, in the order $|00\rangle, |01\rangle, |10\rangle, |11\rangle$:

$$\left(\frac{1}{\sqrt{2}},\,0,\,0,\,\frac{1}{\sqrt{2}}\right).$$

If both qubits are measured in the computational basis, the simulator assigns probability $1/2$ to outcome $00$, probability $1/2$ to $11$, and zero to the other outcomes. A sampled run therefore returns either $00$ or $11$; repeated runs reproduce the predicted frequencies, up to statistical fluctuations.

This state is entangled: it cannot be written as a product of one state for the first qubit and one state for the second. The simulator represents the joint state directly, so it retains the correlation between the outcomes. It does not need to assign each qubit a separate pre-existing value.

With finite-precision arithmetic, $1/\sqrt{2}$ is approximated by a decimal. Using more digits makes the calculated probabilities closer to their exact values. Sampling still has its own statistical uncertainty: even if probabilities are calculated with high numerical precision, a finite number of simulated trials may not reproduce them exactly.

Why Bell’s inequality does not forbid this calculation

Entanglement is often discussed through Bell tests. For a singlet pair, quantum mechanics predicts the correlation $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot\mathbf{b}$ for spin measurements along directions $\mathbf{a}$ and $\mathbf{b}$. With suitable choices of measurement directions, the Clauser–Horne–Shimony–Holt (CHSH) expression reaches $2\sqrt{2}$, greater than the local-hidden-variable bound of $2$.

A classical computer can calculate that quantum prediction from the joint state and the measurement settings. But that calculation is not a local-hidden-variable model. Bell’s result rules out reproducing all the relevant quantum correlations with a model that meets the theorem’s assumptions, including locality and suitable independence of the measurement settings. A centralized program can calculate joint probabilities, and a program with communication or access to both settings can generate matching samples; neither supplies the forbidden local explanation.

The cost of simulating many qubits

A general state vector for $n$ qubits has $2^n$ complex amplitudes. Doubling the number of qubits squares that count, so memory and computation quickly become limiting. Raising the numerical precision also costs resources, though the exponential growth with the number of qubits is usually the larger obstacle.

This is not a universal lower bound for every quantum system. Some states and circuits have exploitable structure. For example, tensor-network methods can simulate many one-dimensional systems efficiently when their entanglement remains limited. Other restricted circuit families also admit efficient classical algorithms. Conversely, highly entangled, generic circuits can be difficult to simulate exactly or sample from efficiently as their size grows.

Thus, “arbitrary precision in principle” and “efficient for a large system” are different claims. A classical computer can often approximate a finite system’s predictions as accurately as requested, while taking impractically long or requiring impractically large memory.

Common mistakes

• Confusing numerical precision with efficient simulation. More digits can improve an answer, but they do not remove exponential growth in the size of a general state-vector calculation.
• Treating a simulation as physical entanglement. The program stores numbers encoding a quantum state’s predictions; its memory cells are not thereby entangled qubits.
• Assuming Bell’s theorem prevents classical computation. It excludes a specific class of local explanations, not classical calculation of quantum-mechanical predictions.
• Ignoring sampling error. Precisely calculating a probability is different from estimating it through a finite number of simulated measurement trials.

For small systems, direct state-vector simulation is a clear way to study entanglement and test calculations. For large systems, first look for structure—such as limited entanglement—that permits a more efficient method.