# What interesting properties can I measure with a 1 qubit state?

I am playing with the IBM Q and I would like to know what interesting properties I can measure with only a 1 qubit state. This is because things like Bell's inequalities, concurrence, PPT criterion, etc. are all for 2 qubit states (which I plan on doing, eventually).

• With a (single) qubit you could build a (simple) random number generator. More precisely, a random bit genearator. You can put a qubit in a superposition (with H-Gate) and then take a measurement. The result of the measurement is a random bit.
– user4961
Commented Apr 30, 2019 at 6:28
• What do you call "interesting"? You can do Deutsch's algorithm with a phase oracle, or two-state Grover. BTW, how do you measure PPT or concurrence in a QComp? Commented Apr 30, 2019 at 7:30
• I'm sorry. I was going to model the state in the simulator and calculate those properties analytically. Commented Apr 30, 2019 at 9:05
• You can measure the relative phase of one qubit relative to any other. But it's also interesting to ask what you can't measure. You cannot measure the global phase. My very limited understanding is that this is more profound than at first blush, being the stepping stone to quantum field theories. Commented May 8, 2019 at 2:06

One interesting property that you can measure with just 1 qubit is interference, and you can demonstrate that no classical (even probabilistic) computing can reproduce that effect. Specifically, you can demonstrate how interference occurs in a system with 1 qubit but not in a system of 1 probabilistic classical bit (or p-bit). This is because matrices that operate on qubits have to be unitary (i.e. $$U^{\dagger}U = I$$), whereas all matrices that operate on p-bits have to be right stochastic (positive square matrices whose rows add up to 1).
As an example of such a demonstration, we start with a $$0$$ state, which for both the qubit and p-bit can be represented as $$\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$. We then put them into a $$50$$-$$50$$ state. For the qubit, that means acting on it with an $$H$$ matrix. For the p-bit, that means acting on it with the matrix $$A = \begin{bmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{bmatrix}$$. You will see that in both cases, you will have a $$50$$-$$50$$ state (i.e. $$\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}$$ for the qubit and $$\begin{pmatrix} \frac{1}{2} \\ \frac{1}{2} \end{pmatrix}$$ for the p-bit). But now try acting on these new states again with their respective operators, and you'll notice that the final states will now give different probabilities. This is because the qubit will interfere with itself whereas the p-bit won't (which you can see by explicitly performing the matrix multiplication on those states).