# Why isn't spin-statistics applied in quantum computing?

Why isn't spin-statistics taken into account in quantum computing?

By spin statistics I mean the fact that fermion and boson states must by respectively totally anti-symmetric and symmetric. I say it's not taken into account because states like $$\vert 01\rangle$$ (and many others) are neither symmetric nor anti-symmetric but still appear to be valid states in quantum computing. I speculate there may be two possible answers:

1. These states appear in the math but are never created in practice.

2. For some reason qubits are considered distinguishable.

• The answer is 2: Qubits are distinguishable. Commented Aug 3 at 18:20
• In fact they need to be distinguishable else you will have a difficult time to address individual qubits and transform them. Commented Aug 7 at 17:18

Why isn't spin-statistics taken into account in quantum computing

... like $$|0 1\rangle$$...

The state $$|01\rangle = |0\rangle\otimes|1\rangle\;,$$ is a direct product state and is neither totally symmetric nor totally anti-symmetric.

1. For some reason qubits are considered distinguishable.

Yes. The two-qubit state to which you are referring is a direct product state and the individual spin states are considered distinguishable. This is why you can call one of them the "first" and write it to the right in the direct product notation, and you can call one of them the "second" and write it to the left in the direct product notation. (Or vice versa depending on your notational preference for big-endian or little-endian bit strings.)

In practice, if the two particles are identical (e.g., electrons), the distinguishability must arise from some physical feature of the quantum computing apparatus, like a physical separation in space of the two qubits that are considered distinguishable.

Those practical details are swept under the rug when we, for example, only consider the qubit to have spin degrees of freedom, since clearly any physical realization will have other degrees of freedom. E.g., we can discuss a silver atom as having two different $$S_z$$ spin eigenstates that are split by a magnetic field in a Stern-Gerlach apparatus. But clearly a silver atom can also move around in the usual three spatial dimensions too.

One difficulty of implementing a multi-qubit quantum computer is having to deal with these practical details.

• Don't the atoms (not ions!) in Stern-Gerlach only have two spin states? (At least, the figures in the original paper only show two lines.) Commented Aug 3 at 19:20
• I just read the paper (only 4 pages, and a nice read) -- definitely atoms. (Much easier to prepare, you just heat silver; and certainly also much easier to prepare as a beam, slow ions will probably be deflected much more easily.) Commented Aug 3 at 19:38

The symmetrization/antisymmetrization requirements for bosons or fermions are always present in principle, but in practice we can ignore them without getting the wrong answer for observables if the particles are well-separated compared to their individual wavefunction spreads. If you look around you will find many previous questions that delve into this in more details, such as this and this.

As an example, one possible quantum computing platform uses a string of ions held in a single trap. These are all identical particles. Their wavefunction spreads are set by the mass and harmonic trapping frequency as $$x=\sqrt{\hbar/m\omega}$$, which for usual parameters (m = 171 amu, $$\omega=2\pi\cdot$$100 kHz) is about 24 nm. The typical spacing between the ions, on the other hand, is set by the balance between the Couloumb repulsion and the overall trapping potential: $$1/2m\omega^2x^2=q^2/(4\pi\epsilon_0 x)$$ , which is typically around 1000 times larger (for the parameters above, and q=e, it is about 16 um). As a result, the distinguishability approximation is extremely good. Similar results apply to many other platforms such as neutral atoms in optical tweezers.

Then, there are also some popular quantum computing platforms, like superconducting circuits, in which the qubits are not in any way identical in the first place. Each superconducting qubit has its own detailed internal structure, set by the precise fabrication, and exchange statistics are not relevant even in principle.

Qubits are distinguishable because each one has, in addition to its value, an address in the quantum RAM. Writing A and B for the addresses (to avoid confusion with the values 0 and 1), the wave function might be more verbosely written as any of

$$\frac{1}{\sqrt2}\big(|0_A\rangle|1_B\rangle + |1_B\rangle|0_A\rangle \big)$$ $$\frac{1}{\sqrt2}\big(|0_A\rangle|1_B\rangle - |1_B\rangle|0_A\rangle \big)$$ $$|0_A\rangle|1_B\rangle$$

depending on whether the implementation uses identical bosons, identical fermions, or nonidentical subsystems for the two qubits. A and B represent some physical property that can be used to distinguish the subsystems, such as position. They are not unphysical labels any more than 0 and 1 are, so you don't need to (anti)symmetrize over them.

The extra terms don't affect the computation. You can work with only the term in which the addresses are ordered A, B, C, ..., and recover the other terms (if any) from it at any time. If you write down only that term, you can omit the address labels. That's what $$|01\rangle$$ ultimately means. It's a better notation because it's more concise and because it's less dependent on implementation details.

• More specifically, A and B are another component of the wavefunction, i.e. sth which is tensored with the kets. Commented Aug 4 at 12:43