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I have just started learning about quantum computers, and my understanding of quantum mechanics is very limited. However, I can not find an answer to my question. So my question can be divided into two main sub-questions:

  • If a qubit can just be represented as two complex numbers (to represent the wave function of the superposition, why can't a computer just simulate a quantum computer with, say 32-bits per qubits? So, instead of the number of bits growing as $2^n$, it would be something like $k\times n$, where $k$ is some number of bits, e.g. 32. What is the fundamental difference between the qubit and a 32-bit set of two complex numbers

  • If a classical computer cannot simulate a quantum computer, how can a laser? I saw a video which uses lasers. Since lasers can simulate a quantum computer, why can't a classical computer?

    • Directly related to this video, why do you need to entangle the photons of the lasers to add more qubits? What role does entanglement play?
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Superposition isn't what causes problems for simulation, but rather the way multiple qubits are combined (which is what gives rise to the notion of entanglement). Yes, the state of a single qubit is represented by two numbers, but the state of, $n$ qubits is not represented by $n$ independent qubit states (i.e. $2n$ complex numbers), but by the tensor product of them, which is $2^n$ numbers.

More concretely, let's say we have three qubits. The state of the qubits is not

$$ | \psi \rangle = a_1 | 0 \rangle_1 + b_1 | 1 \rangle_1 + a_2 | 0 \rangle_2 + b_2 | 1 \rangle_2 + a_3 | 0 \rangle_3 + b_3 | 1 \rangle_3 $$

but rather

$$ | \psi \rangle = c_1 | 000 \rangle + c_2 | 001 \rangle + \cdots + c_7 | 110 \rangle + c_8 | 111 \rangle $$

Notice how we have a coefficient for every combination of all three basis states.

Quantum computation, at a basic level, is done via quantum gates; these are matrices that act on subsets of qubits, so a $1$-qubit gate is a $2 \times 2$ matrix, a $2$-qubit gate is $4 \times 4$, $3$-qubit gate $8 \times 8$ and so on. A quantum circuit consists of a sequence of these gates being multiplied by the state, one-by-one, and each multiplication entails modifying all $2^n$ numbers in the state; this is why simulation is so hard.

So how does a quantum computer sidestep this problem?

Now, different quantum computer hardware implementations work very differently, so let's look at your laser example. First off, assuming you're talking about trapped-ion devices, the laser doesn't simulate the quantum computer, but rather performs the quantum computing operations on the qubits (the ions) by coupling ions together; the photons aren't entangled, but the ions are.

A sequence of laser pulses will in effect alter the state of the qubits in a way mathematically described by performing a matrix multiplication. However, the laser isn't actually multiplying any numbers; that's just the way we describe the state. In other words, we're letting the physics of the laser and trapped ions do the computation for us.

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    $\begingroup$ Your first claim is a bit misleading. Entanglement does not necessitate the impossibility of classical simulation. For instance, you can prepare many highly entangled states using only Clifford circuits, which can be simulated efficiently by a classical computer, see Gottesman-Knill. Perhaps stick to just the claim that it's essentially due to "the way multiple qubits are combined". $\endgroup$
    – Condo
    Jan 11 at 22:44
  • $\begingroup$ Fair enough, I was using entanglement as a basic catch-all for combining qubits, since it's one of the two "pillars" that people talk about in QC, alongside superposition. I'll edit accordingly. $\endgroup$
    – Cody Wang
    Jan 11 at 23:07
  • $\begingroup$ the edit looks good! $\endgroup$
    – Condo
    Jan 12 at 14:17

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