An Elementary Proof that Quantum Mechanics is a Retrocausal Theory

Computing is inherently logical. Bits behave predictably, doing exactly what they’re instructed to do. But what happens when they don’t? What if bits begin to behave unpredictably, changing their responses based on context? And what if they change those responses based on future conditions? In such a system, the output of a program might no longer be predictable. Instead, it might only be predictable in terms of probabilities.

The Setup

Qubit#0 is placed into a superposition of states, then it is passed in as the control to a Controlled-NOT gate and a Controlled-Hadamard gate. Given it is in a superposition of states, it can either be 0 or 1 and which one is fundamentally random.

If Qubit#0 has a value of 0, it will not activate either gates, so Qubit#0 will remain unaffected. Qubit#1 begins in the 0 state, and so it would leave in the 0 state. Hence, the outcome in that case would be 00.

If, on the other hand, Qubit#0 is instead a 1, then it will flip Qubit#1 to a 1, and then place it into a superposition of states, and so Qubit#1 has some random probability of being 0 or 1. Hence, the outcome in that case is either 01 or 11.

Therefore, the only possible outputs to this circuit are 00, 01, or 11. And we can just run the circuit to prove that.

Now, let’s imagine conduct three slight variations of this experiment.

Variation #1

At the end of the experiment, rather than immediately measuring Qubit#0, we instead send it away to our contestant, Alice, and ask her to try and make a prediction about Qubit#1 with it. Then, she can return to verify her prediction.

When Alice receives it, she decides to measure it on the X basis.

This gives her a somewhat different probability distribution.

Recall that Alice is measuring the least significant qubit. If she measures a 0, then she would be unlucky because the two possibilities would be 00 and 10, so she would have to guess. However, let’s assume she is lucky and measures 1. The only possibility there is 11.

She is then allowed to return and verify her prediction is accurate.

This new qubit represents her recording what she sees when she goes to verify it.

As you can see, the cases where she measures 1 and then predicts Qubit#1 is 1, what she records when she goes to verify it is also 1. That proves such a logical deduction is sound.

Variation #2

At the end of the experiment, rather than immediately measuring Qubit#1, we instead send it away to our contestant, Bob, and ask him to try and make a prediction about Qubit#0 with it. Then, he can return to verify his prediction.

When Bob receives it, he decides to measure it on the X basis.

This gives him a somewhat different probability distribution.

Recall that Bob is measuring the most significant qubit. If he measures a 1, then he would be unlucky because the two possibilities would be 10 and 11, so he would have to guess. However, let’s assume he is lucky and measures 0. The only possibility there is 00.

He is then allowed to return and verify his prediction is accurate.

This new qubit represents him recording what he sees when he goes to verify it.

As you can see, the cases where he measures 0 and then predicts Qubit#0 is 0, what he records when she goes to verify it is also 0. That proves such a logical deduction is sound.

Variation #3

Finally, we decide for the last round to send the two qubits to Alice and Bob at the same time, who live very far away in opposite directions. They both choose to again measure on the X basis because they both got so lucky the last time.

We can then compute the probability distribution for both of their measurements at the same time.

There isn’t a very high probability for 01, but that is the lucky state for both Alice and Bob as we mentioned before, because that allows them to make predictions regarding the other qubit with certainty.

Let’s assume that these two are just the luckiest people alive and they just so happen to measure 01. This is indeed a possibility that can occur with an 8.3% probability. That would mean Alice would be able to infer with certainty that Qubit#0 is a 1, and Bob would be able to infer with certainty that Qubit#1 is a 0.

…wait a minute…Qubit#0 is a 1 and Qubit#1 is a 0? Isn’t that the combination we said is logically impossible? If Qubit#0 is 0, then it would never activate the Controlled-NOT gate nor the Controlled-Hadamard gate, and so Qubit#1 could not possibly be anything other than 0.

What on earth is going on here?

Conclusions

An impossible scenario of 01 was reached, and this seems to only be possible to be reached if Alice and Bob both make the measurements on the X basis. If only one of them does, it’s not possible to be reached. This scenario is impossible from the logic gates alone, and whether or not it is possible to occur depends solely on what they choose to measure or do not measure later on.

This is strange because the particles can be spatially separated from one another and the decision as to what to measure occurs after the initial state of the system should have already been fixed. Even if we stick solely to analyzing the system with the wave function, when Alice and Bob make their separate measurements, they will reduce the wave function and make conclusions about the system based on that, and those conclusions lead to the seemingly impossible outcome of 10.

This seemingly impossible outcome really does only have a possibility to occur in the mathematics if both Alice and Bob decide to make the measurements, even though the measurements are made independently of one another, and even though the particles are spatially separated from one another.

You seem to run into the inevitable conclusion that the outcome of experiments in quantum mechanics depends upon what you are going to measure.