1

I'm not an expert and I don't know if already exists something like the following model, but I wrote a simple (mental) model to understand the basic concepts of quantum mechanics. I want to know if this model is valid or not.

Red-or-Black (RoB) Card Model

Suppose to have two cards, one is red and the other is black. Both the cards have the other side with the same color (e.g. white). A special machine can shuffle the cards truly random and then puts the cards turned down on the table with only the backs visible (i.e. the white side). Nobody knows where is the red and the black card, so each card is red or black with the 50% of probability. In other words each card is red and black at the same time (superposition).

This model seems work because when you turn up one of the two cards (measurement), you're fixing its state to a single state (or red or black). At the same time you're fixing also the state of the other card (entanglement). If you repeat the measure turning down and then turning up one of the two cards the result is always the same (the state is fixed). Also the no-cloning theorem works because you can't copy the color of a card until you turn up the card. Is this correct?

Another strange thing is the observer definition. If I view the color of a card I'll fix its state, but this is valid also if my cat views the color of that card. So, who/what is the observer? Can be the card itself? The card have the color printed over its surface, so the card knows its state, then I can't understand if it's right to say that the card is in a superposition because the state should be already fixed by itself (i.e. there is only a single state).

RobotMan
  • 123

1 Answers1

1

The model is good as a probability model for quantum mechanics, up to the "superposition", where you say:

Nobody knows where is the red and the black card, so each card is red or black with the 50% of probability. In other words each card is red and black at the same time (superposition).

No , just the state of the card is unknown, it is not half black and half red, which is what superposition means.

So, who/what is the observer? Can be the card itself? .... so the card knows its state

The "knows" in QM needs an interaction/measurement. The cat is OK. The self interaction is not a measurement . This is also true in quantum mechanics.

anna v
  • 233,453
  • Today I thought again about this model. You say: "just the state of the card is unknown, it is not half black and half red" but how can you prove that? When the card is covered it can be both red and black at the same time, you can't prove the card has a single state. Any kind of measurement you'll try to do, will fix its state. – RobotMan Feb 09 '16 at 17:53
  • @RobotMan you are going into metaphysics. classical physics knows that the card is either black or red. – anna v Feb 09 '16 at 18:15
  • what do you mean with "knows"? is there a way to prove that? – RobotMan Feb 09 '16 at 20:20
  • @RobotMan classical mechanics , statististical, thermodymics, classical electromagnetism and all models derived from them have as an "axiom" that underlying levels are deterministic to derive their very successful models for measuring and predicting situations, including probability distributions. The validation is in the success of the models macroscopically. Indeterminacy found experimentally forced the introduction of quantum mechanics. . There are still physicists trying to eliminate indeterminacy in QM by underlying deterministic theories (cf 't Hooft) – anna v Feb 10 '16 at 04:28
  • Ok. Another consideration is that we can't prove the opposite idea, i.e. when the card is covered we can't prove that its both red and black at the same time. In a real system, using photons, we can mathematically prove that the diagonal base is a linear combination of the computational basis. Right? – RobotMan Feb 10 '16 at 06:04
  • @RobotMan I am not into a computational lingo, i am a simple experimental physicist. The simple fact about the cards being in a fixed state can be ascertained by a kibitzer, i.e. a third party to the game who does not disclose knowledge. – anna v Feb 10 '16 at 06:24