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How looks the wave function for dice roll? How to calculate probability from it? Thank you very much for help.

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closed as unclear what you're asking by stafusa, By Symmetry, ZeroTheHero, sammy gerbil, M. Enns May 12 '18 at 16:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hi and welcome to the Physics SE! What are your thoughts? Please note that you are expected to have thoroughly searched for an answer before asking your question. And it's important to detail where you're stuck and why, in order to attract good answers. You can consider checking the advice on writing good questions. $\endgroup$ – stafusa May 10 '18 at 0:45
  • $\begingroup$ A particle flies as a wave (in an uncertain state), but is observed at a certain random place. A dice also flies in an uncertain state and upon stopping is observed in a certain random state. The main similarity here is randomness. Quantum mechanics does not predict a specific result, only its probability based on the value of the wave function. The dice equivalent here is the probability of 1/6. This is essentially the (squared) value of your wavefunction. It does not predict a specific outcome. – $\endgroup$ – safesphere May 10 '18 at 7:05
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Naïvely:

A six-sided die is a system with one measurable property - "roll". There are six equally likely "roll" states: 1, 2, 3, 4, 5, 6. Let's form an orthonormal basis from these states: $$ \def\braket#1#2{\langle#1|#2\rangle}\def\bra#1{\langle#1|}\def\ket#1{|#1\rangle} \ket{1}, \ket{2}, \ket{3}, \ket{4}, \ket{5}, \ket{6} \\ \braket{i}{j} = \delta_{ij} $$ From these we can create an ensemble "die" state, which is a linear superposition of these states: $$ \ket{\psi} = \frac{1}{\sqrt{6}} \big( \ket{1} + \ket{2} + \ket{3} + \ket{4} + \ket{5} + \ket{6} \big) $$ Clearly we have a 1/6 chance of measuring this state to be in any of our 1 - 6 "roll" states. An explicit example of this: the probability of rolling a 4 is: $$ \vert\braket{4}{\psi}\vert^2 = \frac 1 6 $$

None of this is quantum mechanics, I've merely borrowed its notation to define what is meant by a fair die. Yet, one might remark at how similar this setup is to the elementary example of a spin 1/2 particle. If one so wished they could treat this system in the same way, introducing something analogous to Pauli matrices. Perhaps in this regime time-evolution could model the rolling of the die (or multiple dice). However, this seems like a lot of work to create a toy model that in my opinion is not of much interest.

That said, I did find a paper in which dice are modelled quantum mechanically and used to demonstrate the violation of Bell's inequality. I can't vouch for its credibility but you might find it an interesting read. "Quantum dice", M.Bianchi: https://arxiv.org/pdf/1301.3038.pdf

I'm afraid I can't be of much more help. Without a more specific question it's hard to gauge what you were asking and what your academic background is.

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  • $\begingroup$ thank you for your detailed explanation. I have also one question. If I take fully charged capacitor and connect parallel six capacitors to it, how looks the superposition? $\endgroup$ – Marianna Kalwat May 10 '18 at 15:28
  • $\begingroup$ The quantum mechanical modelling of electronic circuits is far outside my scope of knowledge. I am unsure of how one would model capacitors and the flow of electrical charge simply. The only suggestion that springs to mind would be to analyse the system classically. Perhaps in doing this the question could be rephrased in a way that does have a simple quantum mechanical analogue. For instance, maybe the oscillating electric fields could be modelled by coupled quantum harmonic oscillators. $\endgroup$ – Vielbein May 10 '18 at 17:00
  • $\begingroup$ If on the main capacitor wave function is $U(t) = U_{0}e^{-tω}$ and on the one of six is $U(t) = U_{1}(1-e^{-tω})$. On the other five is the same wave function with the same amplitude $U_{1}$. $\endgroup$ – Marianna Kalwat May 10 '18 at 17:56

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