# Bias in the results of superposition measurements on IBMQ Backends, qiskit

Dear people on this forum,

I was doing some research, and I created this circuit in qiskit

Please bear in mind that I am really new to this field, and I do not retain much knowledge yet. Therefore I am grateful to whoever will help me out here.

qc = qiskit.QuantumCircuit(1, 1)  # Make a circuit with 1 QuBit and 1 bit.
qc.u(np.pi / 2, 0, 0, 0)  # Rotate Q1 to |+⟩. Emulation of a Hadamard gate via rotation
qc.measure([0], [0])  # Measuring the qubit.


It should be correct to assume that such a circuit should have 50% chance of a 0 outcome, and 50% of a 1 outcome.

But when I ran in on three IBMQ backends, namely IBMQ-Santiago, IBMQ-Lima, and IBMQ-Armonk, there always was a higher percentage of 0 outcomes. Sometimes, depending on which backend, a higher difference, and sometimes a lesser.

Each run ran on the backend with 8192 shots.

See the tables and graphs down below:

As you can see, the bias from the Armonk backend could be a mere statistical error. But it is a clear bias when you check the data from Santiago or Lima. And this could not be merely explained by quantum noise, correct?

Could finally someone explain the nature of such a phenomenon?

Mattia

While its hard to make strong claims about the noise characteristics of any given quantum device, one explanation for what you're observing is readout error. For superconducting qubits readout error tends to be asymmetric: the probability $$p(0|1)$$ of observing a "0" after performing measurement on a computational basis state $$|1\rangle$$ tends to be higher than $$p(1|0)$$, the probability of observing "1" after measuring $$|0\rangle$$. You can verify this for yourself by looking at the Prob meas0 prep1 and Prob meas1 prep0 for the different systems offered by IBM.

Physically, this makes sense since usually $$|0\rangle$$ is defined as the ground state while $$|1\rangle$$ is an excited state. During the measurement process (for superconducting qubits this is usually on the order of $$\sim 1\, \mu s$$, which is quite long compared to gate durations) there is a tendency for the excited state $$|1\rangle$$ to relax to its ground state during the measurement process. However, as long as the the environment (i.e. dilution refrigerator) is cool enough the probability of exciting the ground state $$|0\rangle$$ tends to be much lower. Considering just thermalization effects, a back-of-the-envelope approach to this is to compute the Boltzmann ratio, which describes the equilibrium distribution for a two level system in terms of environmental temperature: $$$$\frac{p_0}{p_1} = \exp\left(hf_{01} / kT\right) \tag{1}$$$$

Plugging in some very rough values for superconducting qubits, $$T\approx 10 \, \text{mK}$$, $$f_{01} \gtrsim 1 \, \text{GHz}$$, Equation (1) gives something like $$p_0 \gtrsim 100 p_1$$, showing a strong tendency towards relaxation due to thermal effects (on a long enough timescale!). Of course this relaxation tendency will compete with other decoherence effects that may excite the qubit (e.g. $$T_1$$ decay will tend to depolarize your qubit state towards the maximally mixed state, which won't actually change the statistics you're trying to observe).

It is useful to consider a model for readout error to check whether this could be the source of your issue. Let $$\mathbf{p}' = [p_0, p_1]^T$$ be the vector of observed probabilities for the bits "0" and "1". A rudimentary model for readout error is to consider the measured bits for each qubit as resulting from a stochastic process described by the equation $$$$\mathbf{p}' = R \mathbf{p} \tag{2}$$$$
where $$\mathbf{p}$$ is the vector of "true" bit probabilities that we would have observed with perfect measurement, and $$R$$ is the matrix
$$$$R = \begin{pmatrix} p(0 |0) & p(0 |1) \\ p(1 |0) & p(1 |1) \end{pmatrix}\tag{3}$$$$
From this expression we can think about how readout error would affect the observed distribution when the true distribution for the state $$|+\rangle$$ given as $$\mathbf{p}_{|+\rangle} = [0.5, 0.5]^T$$: $$$$\mathbf{p}_{|+\rangle}' = R \begin{pmatrix} 0.5 \\ 0.5\end{pmatrix} =\frac{1}{2} \begin{pmatrix} 1-p(1|0) + p(0|1) \\ 1 - p(0|1) + p(1|0) \end{pmatrix}\tag{4}$$$$
where I've simplified using $$p(0|0) + p(1|0) = 1$$. From this we can see that $$p(0|1) \geq p(1|0)$$ will result in $$p_0' \geq p'_1$$ - the observed distributions will contain too many zeros. We can further check how well readout error explains the discrepancies in the measurements by computing $$\mathbf{p} = R^{-1} \mathbf{p}'\tag{5}$$ This depends a lot on the accuracy of device diagnostics, when you performed you experiments, etc. but computing Equation (4) using $$p(0|1)$$ and $$p(1|0)$$ diagnoistics taken from IBM's page just now (averaged over each device) and the probabilities $$\mathbf{p}'$$ you provided for Santiago and Lima I found $$p_0$$ for the corrected distributions to be 0.49 and 0.51 respectively, which suggests that readout error could explain much of the discrepancy in the observed distributions.
For better accuracy this calculation should be run using recent readout error diagnostics for $$p(0|1)$$ and $$p(1|0)$$ collected at the same time you run your experiment. I believe IBM also offers some readout error mitigation tools in their software package but you should make sure it uses fresh diagnostic data since these values tend to drift on very short timescales.