# Quantum Joke (not a real joke, not a riddle)

Supposing I want to make a quantum joke, like writing this on a coffee machine:

$$| \text{Status}\rangle = \frac{1}{\sqrt{2}}\ \big( | \text{Working}\rangle \color{red}{\pm} | \text{Down}\rangle \big) \, ,$$

should I choose the $\color{red}{+}$ or the $\color{red}{-}$ sign, or is it the same? Why?

• Well, if all you're going to do is make a direct measurement of whether you're working or down, the sign doesn't matter for the outcome. Nov 29, 2015 at 15:46
• States don't have to be normalised. So you don't need the root two. But it adds a nice touch, a note of misplaced precision.... By the same token, you could multiply |working> by the imaginary unit $i$ to suggest that the status is a complicated affair, and the idea that it would be working is purely imaginary.... Nov 29, 2015 at 19:12
• As someone who is not familiar with quantum physics, this joke gives me a headache, along with the answers. Could someone give me a <500 chars explanation of this joke? Nov 29, 2015 at 22:16
• @NateKerkhofs Understanding quantum physics doesn't make it funny either. Nov 29, 2015 at 23:32

The signs are important for fixing an out of order machine. Define the states $|\pm\rangle$ as:

$$|\pm\rangle = \frac{1}{\sqrt{2}}\left[\left |\text{Working}\right\rangle\pm \left |\text{Down}\right\rangle\right]$$

And we define the observable $O$ as:

$$O = |+\rangle\langle + |\ - \ |-\rangle\, \langle -|$$

Suppose then that coffee machine is out of order. To fix it, you measure $O$ and then you measure if it is working, if not you repeat the procedure of measuring $O$ and then checking if it is working. At each step you have 50% probability that it will be found to be working.

• quantum machines seem to be really easy to fix via brute force! Nov 29, 2015 at 18:14
• Awesome answer! Ahaha nice! Nov 29, 2015 at 19:00
• How does the protocol for fixing the machine depend on whether the machine starts in state $|+\rangle$ or $|-\rangle$? Nov 30, 2015 at 4:38

Perhaps you don't want a quantum superposition, but just a statistical mixture:

$$\rho = \begin{pmatrix}1/2 & 0 \\ 0 & 1/2\end{pmatrix}$$

Although I'm not 100% sure that this will describe your situation any better...

If you want to declare indeterminacy and a probability of being either working or down you should use the vector notation:

       (working )


Status> =

       ( down)


Status being the column vector analogous to the column state vector of the wavefunction in a matrix representation

The user would be the operator :)