# Quantum Entanglement and corresponding probability of 2 Dices System

I am new to quantum entanglement and I want to have a simple example that can relate entangled states with probability. Therefore, I try to use the 2 dices system to examine whether am I really understand quantum entanglement. Suppose I have 2 dices, call dice A and dice B. They range from 1 to 6. Classically, they are independent since whatever I roll dice A, the result of dice A would not affect the result on dice B. However, if I impose the condition that the sum of result of dice A and dice B must obey ($$i + j = 3$$), where $$i$$ and $$j$$ are the result of the dice A and B respectively. I can construct an entangled state as follow: $$$$|\Psi \rangle = |i+j = 6\rangle = \sum^{6}_{i} \sum^{6}_{j} C_{ij} |i\rangle_{A} \otimes |j\rangle_{B} ~~,~~ |i \rangle_{A} \in \text{Dice A} , ~~,~~ |j \rangle_{B} \in \text{Dice B}$$$$ Where the state $$| \Psi \rangle = |i+ j =3\rangle$$ describes the possible combinations of $$|i\rangle \otimes |j\rangle$$ such that it satisfies the condition $$i+j =3$$. In such case, we can easily guess the possible outcomes which satisfies $$i+j = 3$$ without knowing the form of $$C_{ij}$$: $$$$|\Psi \rangle \propto |1\rangle_{A} \otimes |2 \rangle_{B} + |2\rangle_{A} \otimes |1 \rangle_{B}$$$$ Instead of thinking 1 possible outcome where $$i + j =3$$, I can also take $$i+j =2$$ and $$i+j =4$$ into account. Therefore, the state that describes all the possible outcomes above will be like this: $$$$\begin{split} | \Psi \rangle = |i+j=m; m=2,3,4 \rangle &\propto \ |1\rangle_{A} \otimes |1 \rangle_{B} \\ &+ |1\rangle_{A} \otimes |2 \rangle_{B} + |2\rangle_{A} \otimes |2 \rangle_{B} \\ &+ |1\rangle_{A} \otimes |3 \rangle_{B} + |2\rangle_{A} \otimes |2 \rangle_{B} + |3\rangle_{A} \otimes |1 \rangle_{B} \end{split}$$$$ Where the first line on R.H.S corresponds to the $$|i+j = 2 \rangle$$ , and the second line on R.H.S corresponds to $$|i+j = 3 \rangle$$, and so forth. Then, I want to ask a question regarding the normalization factor of the above states. Currently I have not fixed the normalisation factor of these states. The reason is that I am not sure which normalisation factor that I should use. For instance, let's us focus the case of $$|i+j = 3 \rangle$$: $$$$|i+j =3 \rangle = \alpha |1\rangle_{A} \otimes |2 \rangle_{B} + \beta |2\rangle_{A} \otimes |1 \rangle_{B} ~~,~~ |\alpha|^{2} + |\beta|^{2} = 1$$$$ There are infinitely many choices of $$\alpha$$ and $$\beta$$ to satisfy the normalisation. The most convenient choice is $$\alpha = \beta = \frac{1}{\sqrt{2}}$$, but I am not sure why this will work if I use this normalisation to calculate the classical probability of such system. Suppose I have the following normalisation for $$|i+j =3 \rangle$$ and $$|i+j=m; m=2,3,4 \rangle$$: \begin{align} |i+j =3 \rangle &= \frac{1}{\sqrt{2}} \Big( |1\rangle_{A} \otimes |2 \rangle_{B} + \beta |2\rangle_{A} \otimes |1 \rangle_{B} \Big) \\ |i+j=m; m=2,3,4 \rangle &= \frac{1}{\sqrt{6}} \Big(\ |1\rangle_{A} \otimes |1 \rangle_{B} + |1\rangle_{A} \otimes |2 \rangle_{B} + |2\rangle_{A} \otimes |2 \rangle_{B} + |1\rangle_{A} \otimes |3 \rangle_{B} \\ & \qquad \qquad + |2\rangle_{A} \otimes |2 \rangle_{B} + |3\rangle_{A}\otimes |1 \rangle_{B} \Big) \end{align} Classically, we know that there are total 6 states contribute to $$|i+j=m; m=2,3,4 \rangle$$. If we want to find the probability of the system sitting in $$|i+j=3\rangle$$, we can directly get the probability $$P(m=3) = \frac{2}{6}$$ since there are to micorstates in $$|i+j=m; m=2,3,4 \rangle$$ contribute to $$|i+j=3\rangle$$. Besides, we can also use quantum mechanics to solve the problem: $$$$P(m=3) = | \langle i +j =3 | i+j=m, m=2,3,4 \rangle|^{2} = ( \frac{1}{\sqrt{2}} \frac{1}{\sqrt{6}} \times 2 )^{2} = \frac{2}{6}$$$$

To summarise above , I want to know whether my thought is correct, whether I can use this two dices system to illustrate quantum entanglement? Secondly, I want to ask why the normalisation of $$|i+j =2\rangle$$ and $$|i+j=m, m=2,3,4 \rangle$$ are $$\frac{1}{\sqrt{2}}$$ and $$\frac{1}{\sqrt{6}}$$ respectively? My thought of the such normalisation factors comes from the assumption of equally probable of microstates in statistical mechanics. If there is any mistake that I made, could anyone help me to point out the mistake and correct me. Thank you.

• You cannot use dice to properly illustrate entanglement. Apr 3, 2021 at 15:39
• Thank you for your comment @NorbertSchuch. May I ask what is the reason why I cannot use dices as an example to illustrate quantum entanglement? Apr 3, 2021 at 15:42
• Because the very feature of quantum entanglement is that it exhibits correlations which can not be modeled using classical objects. So whatever your dice do, your analogy will be flawed, unless they can talk to each other instantly and can tell each other when they are being thrown. Apr 3, 2021 at 15:43
• Nope. Classical objects cannot be entangled. Apr 3, 2021 at 15:49
• My answer here gives a relatively easy example of an obstacle to using a classical system to emulate entanglement. The answer doesn't use the word "entanglement" (except in sources that it cites), but hopefully the connection is clear. Apr 3, 2021 at 16:16

To summarise above, I want to know whether my thought is correct, whether I can use this two dices system to illustrate quantum entanglement?

Not really, no.

The core roadblock is a problem that you waved aside very casually:

There are infinitely many choices of $$\alpha$$ and $$\beta$$ to satisfy the normalisation. The most convenient choice is $$\alpha = \beta = \frac{1}{\sqrt{2}}$$, but I am not sure why this will work if I use this normalisation to calculate the classical probability of such system.

The problem is not the normalization $$-$$ it is the relative phase of the two components of the superposition (i.e. the complex argument of $$\alpha$$ vs that of $$\beta$$).

In quantum mechanics, the superpositions $$|1⟩+|2⟩$$ and $$|1⟩-|2⟩$$ are completely different (i.e. completely distinguishable), and you cannot just ignore the information in the sign (or potentially complex phase) between the two states.

If you have a measuring device that can distinguish between those two superposition states ($$|1⟩+|2⟩$$ and $$|1⟩-|2⟩$$) on each of your individual dice, then you have a chance of having genuine, benchmarkable entanglement between them, as represented by e.g. a Bell inequality violation.

But in practice, for macroscopic objects, you don't have such control, and your experimental toolkit is unable to distinguish between, for example the entangled states \begin{align} |\Psi_1⟩ & = |1⟩\otimes|1⟩ + |2⟩\otimes|2⟩ \\ |\Psi_2⟩ & = |1⟩\otimes|1⟩ - |2⟩\otimes|2⟩ \end{align} (which, as far as QM is concerned, are completely different). If you can't distinguish between these two opposite alternatives, then your experiment is unable to tell genuine quantum entanglement apart from mere probabilistic superpositions of the states $$|1⟩\otimes|1⟩$$ and $$|2⟩\otimes|2⟩$$, which are completely classical.