# Noise in quantum computing

Consider a quantum computer, and one specific qubit $$q$$ in it. With the computer, we perform a series of operations which manipulate the qubits including $$q$$. I am interested in the time interval between two operations. Because of noise, the wavefunction of $$q$$ should change with time. The state at the beginning of the next operation, is different from the state at the end of the previous operation. How small should the amount of noise and the time between operations be to get a working quantum computer?

• Just because the Hamiltonian is time independent doesn't mean that the state of the system doesn't change in a well defined manner. A single spin in a constant magnetic field "calculates" a time dependent rotation or precession, if you will. It is not clear to me why one would exclude this process from the definition of "calculation". Maybe I don't understand your question correctly. Commented Jan 19 at 0:49
• Ok I will reformulate and clarify Commented Jan 19 at 0:50
• The details of your example are a little garbled, but you are getting at a genuine point that the time $\Delta$ over which the gate is applied does have to be carefully controlled for a quantum gate to work correctly. There will be some theoretically ideal value of $\Delta$, but in reality a physical device will not operate this exact duration, and that will introduce an error into the calculation. Obviously you need to control this error so it is small in some appropriate sense. In the paper I linked below, the laser pulses have to have very specific durations and amplitudes. Commented Jan 19 at 2:16
• Yes, timing matters, a lot. Commented Jan 19 at 2:27
• This effect only matters inasmuch as $E_1-E_0$ varies between qubits, that said, $E_1-E_0$ does vary between qubits in practice. Commented Jan 19 at 2:37

It's pretty rare to work with energy eigenstates in quantum computing, since:

(a) normally an explicit form of the Hamiltonian would be pretty convoluted and time dependent, since it would depend on what gates you apply to your system and when, and solving the corresponding Schrodinger equation would probably be hard.

(b) the energy eigenbasis isn't particularly useful since the input and output of the computation is done in the computational basis where you can easily measure the information stored in the individual qubits (for example, if you used electron spins as qubits, you'd tend to work in a basis where the basis states corresponded to spin up or spin down). Actually since the Hamiltonian is time dependent, it doesn't really make sense to talk about an energy eigenbasis in the sense you are assuming in your question, which is implicitly based on the time-independent Schrodinger equation.

However, superposition is certainly important in quantum computing. The trick is to perform a series of operations (represented as as a time-ordered sequence unitary operators acting on the quantum state) on the state so that you achieve a special superposition at the end, with a high probability associated with the computational basis state that represents the correct answer of the computation, and a low probability associated with other computational basis states. The time evolution does matter, in the same sense that time evolution matters on a classical computer -- the sequence of operations you perform transforms the original state representing the input into the final state representing the output of the computation.

The way this is achieved (ie, the specific way you put together unitary operations to get an output state that answers a computational problem) is through ingenuity and hard work. The field that studies this kind of problem is quantum algorithms, and of course the most famous is Shor's algorithm.

• Thank you for your answer! This highlights something that I forgot to make clear: I am only asking about the time interval in between of two operations. I updated my question to incorporate that. Commented Jan 19 at 0:42
• @Riemann Normally one considers the entire operation from beginning to end, which is represented as a unitary operator $U$. In a specific context, you can implement a given $U$ by operating with a given Hamiltonian over a period of time. For example, let's say you want $U=e^{i\alpha}\mathbf{1}$ for some $\alpha$ (this is boring but just an example). You could implement this unitary on a free particle with momentum $p$ by waiting a time $t=\alpha \hbar/p^2$, since then $U=e^{i Ht/\hbar} = e^{i \hat{p}^2 t/\hbar}$, which acts like $e^{i \alpha}\mathbf{1}$ on a state with momentum $p$. Commented Jan 19 at 0:52
• In other words... a quantum algorithm is defined in terms of unitaries, and an algorithms researcher normally wouldn't go down to the "physical layer" to specify the specific Hamiltonian that would lead to that unitary when you exponentiate it. However, if you are actually building quantum logic gates, then you need to know what Hamiltonian will act on the state to implement a given unitary, and the details of that depend on how you represent the qubits in reality (eg spins, photons, ...), and how you can act on those qubits (eg, magnetic fields, optical devices, ...) Commented Jan 19 at 0:54
• I am indeed interested in the "physical layer". How do current quantum computer implementations resolve this problem? Commented Jan 19 at 1:14
• @Riemann It depends on the details of how you implement the qubits. Just doing a little googling, here is a specific example of how a Welsh-Hadamard gate (a specific operation on 3 qubits) is implemented using superconducting qutrits: arxiv.org/abs/2003.04879. There's no simple and general answer. However if you want to attract people who can give more detailed answers, I'd ask a different question titled something like "What kind of Hamiltonian can you use to implement X quantum gate?" where X is a specific gate, or a class of gates. Commented Jan 19 at 2:07

If you have a quantum computer, you can control its Hamiltonian. That is, you can tune the Hamiltonian to be zero, in a suitable basis. Then, the system does not change in time. (Or rather, if it does, it is because you can't control your system precisely enough.)

• I think this would correspond to controlling the difference $E_1-E_0$. Then the question is, how can people control it so precisely? (Given the value of Planck’s constant) Commented Jan 19 at 13:13
• @Riemann Just define your eigenstates in the rotating frame. Another way to phrase my answer is: Work in the Heisenberg picture. It is always about controlling different Hamiltonians relative to each other, and not the "absolute" Hamiltonian (whatever that even is, without fixing a basis through some reference) at least as long as it is about the single-qubit terms. Commented Jan 19 at 15:32

Which state $$q$$ is in at the moment that we perform the next operation, depends on two variables: the amount of noise and the length of the time interval. The goal is to keep both of them small. The following is a back-of-the-envelop calculation to estimate how small these have to be.

Some formulas

Let $$q= \alpha |0\rangle+ \beta |1\rangle$$ be the qubit. During the operations, there might be a bit of noise: for example a tiny magnetic field. From a physics perspective, there exists a hamiltonian $$H$$ which models that noise. Now $$q$$ satisfies the Schrödinger equation for this hamiltonian.

I am now going to make a few unreasonable assumptions, but they make the argument a lot easier. The goal is to get a grasp of the scales involved.

1. Assume that the (noise) hamiltonian is time-independent.

2. Assume that $$|0\rangle$$ and $$|1\rangle$$ are eigenstates of the hamiltonian.

This means that they both have an eigenenergy: $$E_0$$ and $$E_1$$. We deduce that the complex phases of the components of $$q$$ change in time:

$$q(t)= e^{E_0t/(i\hbar)} \alpha |0\rangle+ e^{E_1t/(i\hbar)} \beta |1\rangle$$ for all $$t$$ in the time interval.

Therefore, the relative phase between the 0 and 1 states of q changes with $$e^{(E_1-E_0)t/(i\hbar)}$$

Now let $$\Delta$$ be the length of the time interval between the two operations. Then the relative phase has shifted in total by a factor $$e^{(E_1-E_0)\Delta/(i\hbar)}$$

The relative phase depends on $$\Delta$$. If you wait a fraction of a second to perform the next operation, then you are working with a different state of $$q$$. This could matter because the results of computations depend on relative phase. Therefore, we want to have

$$\frac{(E_1-E_0)\Delta}{\hbar}<<1,$$ i.e. $$(E_1-E_0)<<\frac{\hbar}{\Delta}$$

Let’s say that $$\Delta$$ is 10 nanoseconds. The value of Planck’s constant is of the order $$10^{-15} eVs$$, so the value of $$E_1-E_0$$ has to be smaller than $$10^{-15}/10^{-8}=10^{-7} eV$$. For a good precision of 99.9%, we need something like $$10^{-10}eV$$.

To see how large this is, choose for example the following implementation: the qubit is an electron that is spin up or spin down. Then the energy gap is given by $$2\mu_BB$$ where $$\mu_B$$ is the Bohr magneton and $$B$$ is the magnetic field parallel to the z-axis. The value of $$2\mu_B$$ is $$10^{-4}eVT^{-1}$$, so the noise-magnetic field has to be at most $$10^{-10}/10^{-4}=10^{-6}T=1\mu T.$$ According to Wikipedia, this is the magnetic field produced by a blender. So a quantum computer can probably create a setting with less noise than that.

• Earth's magnetic field is on the order of 50uT... so some shielding is in order, even in this scenario. Commented Jan 19 at 14:30