I have been through the Young's double slit experiment. It's a direct proof or instance of showing that a wave is collapsed via observation or measurement, and shows no interference patterns.

I want to do something similar in a quantum circuit, not visual interference but something to distinguish between a collapsed qubit and an uncertain qubit. So that I can notice the difference while alternatively putting and omitting measurement on the pathway of a superposed qubit.


I followed @AdamZalcman 's answer and coded the whole thing up on a notebook

enter image description here


1 Answer 1


Consider a quantum circuit on one qubit initialized in the state $|0\rangle$ and consisting of two Hadamard gates where we can insert a measurement between the Hadamards.

The input state is $|0\rangle$, so the state after the first Hadamard gate is

$$ H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) $$

which is an equal superposition of $|0\rangle$ and $|1\rangle$ with zero relative phase.

Case 1: measurement absent

In the absence of measurement, the second Hadamard causes constructive interference of amplitudes on the $|0\rangle$ state (in green below) and destructive interference of the amplitudes on the $|1\rangle$ state (in red below)

$$ \begin{align} H\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) &= \frac{1}{2}(|0\rangle +|1\rangle)+ \frac{1}{2}(|0\rangle -|1\rangle)\\ &= \left(\color{green}{\frac{1}{2} + \frac{1}{2}}\right)|0\rangle + \left(\color{red}{\frac{1}{2} - \frac{1}{2}}\right)|1\rangle\\ &= \color{green}{1}|0\rangle + \color{red}{0}|1\rangle =\\ & = |0\rangle. \end{align} $$

Case 2: measurement present

Now suppose we insert a measurement in the computational basis between the Hadamards. Since we are measuring $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ in the computational basis, we have an equal chance of collapsing the qubit into the $|0\rangle$ state and the $|1\rangle$ state. After state collapse we end up in

$$ H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) $$

in the former case and

$$ H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) $$

in the letter case. Regardless of which case takes place, the interference observed previously does not occur.

Comparison to Young's experiment

In practice, in order to observe the difference in behavior, we need to add a terminal measurement after the second Hadamard to both circuits. As in the case of interference patterns in the double slit experiment, the difference manifests itself in the different output probability distribution.

In the case of Young's experiment it is the distribution of photon locations and in the case of the above circuits it is the distribution of $0$s and $1$s output from the terminal measurement. The first circuit yields measurement outcome $|0\rangle$ every time it is executed. The second circuit yields the outcome $|0\rangle$ half the time and the outcome $|1\rangle$ half the time.

In a sense, $|0\rangle$ corresponds to the bright band in the interference pattern and $|1\rangle$ corresponds to the dark band. When a measurement is added between the Hadamards, the pattern disappears and both bands register half of the "brightness".

  • $\begingroup$ Okay cool. Why didn't I think of this. I was hovering over all complex circuits and this was it. Actually I implemented the thing in python on a notebook as soon as I read you answer and voila it's working, Thaaanks a lot. $\endgroup$
    – Sayan Dey
    Jan 18, 2021 at 10:48
  • 1
    $\begingroup$ I'm glad it worked! Thanks for asking a cool question! $\endgroup$ Jan 18, 2021 at 16:44

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