B Meaning of Wave Function Collapse

[member 648030]

That is not right.
The quantum mechanical state "superposition of A and B" is different from the quantum mechanical state "It is A or B and we don't know which yet", and there is no classical analogy for the former.

From the point of view of the measure (which is what interests us), it is the exact same thing. It is clear that the example of money is only a "way" to explain, but the probability that a head or a cross comes out by throwing the coin, or measuring the spin up or down, is the same probability in the two cases.
Rather, the coin, in its "state" of launch is a rotating object subject to gravity that can be described with the classical equations etc etc. while a quantum state is neither more nor less than the wave function, or the wave vector of its state

 

[Boing3000]

but the probability that a head or a cross comes out by throwing the coin, or measuring the spin up or down, is the same probability in the two cases.

No it is not.
You've got a probability for one event for the coin (that you cannot entangled with another coin)
And you've got two events for entangled spin, that cannot be described classically.

 

[member 648030]

You have a 50% chance of getting head, and 50% chance of getting cross. After all, you have a 50% chance of getting spin up and 50% spin down. I see no difference

 

[Boing3000]

You have a 50% chance of getting head, and 50% chance of getting cross. After all, you have a 50% chance of getting spin up and 50% spin down. I see no difference

The difference is obviously the correlation between probabilities when both particles are measure along some axis. This correlation varies with the angle$\frac{1}{2}cos\theta$ That's what make entanglement different.

 

[Nugatory]

You have a 50% chance of getting head, and 50% chance of getting cross. After all, you have a 50% chance of getting spin up and 50% spin down. I see no difference

If you try measuring the spin on the horizontal axis instead, you will see the difference.

The state "it is spin-up or spin-down" will, when measured on the horizontal axis, give you spin-left 50% of the time and spin-right 50% of the time.

The state "superposition of spin-up and spin-down" will give you spin-left 100% of the time.

 

[member 648030]

The difference is obviously the correlation between probabilities when both particles are measure along some axis. This correlation varies with the angle$\frac{1}{2}cos\theta$ That's what make entanglement different.

but who is talking about "both" the particles? I am considering only one particle. It is more than enough that I am considering any X object that is in some X state

 

[member 648030]

If you try measuring the spin on the horizontal axis instead, you will see the difference.

The state "it is spin-up or spin-down" will, when measured on the horizontal axis, give you spin-left 50% of the time and spin-right 50% of the time.

The state "superposition of spin-up and spin-down" will give you spin-left 100% of the time.

Obviously a coin is not exactly an electron ...

 

[Boing3000]

but who is talking about "both" the particles?

I am, because one quantum state describe all there is to know of many entangled particles. And there is no classical counterpart for "coins"

I am considering only one particle. It is more than enough that I am considering any X object that is in some X state

No, it is not enough, a least in the context of quantum "object".
You post #57 about "coin" is irrelevant to "wave function collapse" for many reasons:
1) A coin is never in a superposition of head and cross, so there is no "Wave function collapse" that apply to coin.
2) QM superposition is not a "either or" but an "and"
3) "Wave function collapse" is in itself is not relevant for understanding QM. Evolution of state is.

 

[member 648030]

I am, because one quantum state describe all there is to know of many entangled particles. And there is no classical counterpart for "coins"

I see no reason to use the concept of entaglement for the superposition principle, nor for the collapse of the wave function. You can take the simple example of the hydrogen atom and you will have state overlap and collapse without entagled states

No, it is not enough, a least in the context of quantum "object".
You post #57 about "coin" is irrelevant to "wave function collapse" for many reasons:
1) A coin is never in a superposition of head and cross, so there is no "Wave function collapse" that apply to coin.

Obviously, money is a classic and not a quantum object, but under the probabilistic aspects the behavior is exactly the same

2) QM superposition not a "either or" but an "and"

the possible states are logically linked by "and" before the measurement. In fact it is said that they are "overlapped" in the sense that they coexist. But from the point of view of the result you can get a result "or" another result "or" another one etc. " But you can not get a result "and" another result

3) "Wave function collapse" is in itself is not relevant for understanding QM. Evolution of state is.

It is a pity that the collapse of the wave function is one of the fundamental principles of quantum mechanics(Heisenberg, Dirac, Born etc).
After the measurement, as you know, the wave function ceases to be an overlap of "possible" states.
For any observable, the wave function is initially some linear combination of the eigenbasis of that observable. When an observer, experimenter, etc ..measures the observable associated with the eigenbasis , the wave function collapses from the full to just one of the basis eigenstates
(But I do not think I have to explain the theory you already know perfectly)
Unless you consider the interpretation "to many worlds", where the wave function continues to evolve in all the possible worlds in which it can give a certain result, etc

 

[stevendaryl]

I see no reason to use the concept of entaglement for the superposition principle, nor for the collapse of the wave function. You can take the simple example of the hydrogen atom and you will have state overlap and collapse without entagled states

The concept of entanglement is an essential part of QM. It's inevitable. If you have two systems interacting, then their states will become entangled. If a $\pi^0$ particle decays into an electron-positron pair (I think that's an allowable decay) then the spins of the two particles will be entangled.

All that it means for systems to be entangled is for the composite system to be in a superposition of states that fails to factor into a product state. For example, for spin, the state $\frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle)$ is an entangled state.

 

[Boing3000]

I see no reason to use the concept of entaglement for the superposition principle, nor for the collapse of the wave function. You can take the simple example of the hydrogen atom and you will have state overlap and collapse without entagled states

The reason is not for the superposition of state, but the number of "object" they describe. There is no classical analog.

Obviously, money is a classic and not a quantum object, but under the probabilistic aspects the behavior is exactly the same

No, they are not exactly the same. The "behavior" for classic object don't exist. You are kind of "undoing" a pure abstract "probabilization" of a totally not random object, coin, money or whatever. You may call it "Collapse" if you want to, but it only concern you abstract description.
In QM "collapse" is not even a thing. It is a word that some interpretation use to describe when the application of the Born rule "change" the state of the object "for real". It is part of the broader measurement problem, and is not about "potential ignorance" being "undone"

the possible states are logically linked by "and" before the measurement.

Not in classical mechanic. A coin is never in two state, and no genuine behavior can be associated with an AND state.

In fact it is said that they are "overlapped" in the sense that they coexist. But from the point of view of the result you can get a result "or" another result "or" another one etc. " But you can not get a result "and" another result

The result in QM often spans an infinities of observable (angle for spin). It is definitelly finite in classical mechanics.
But the point that in QM object behave as OR between observation, and the value only exist after (or is created by) after observation.

It is a pity that the collapse of the wave function is one of the fundamental principles of quantum mechanics(Heisenberg, Dirac, Born etc).

It isn't. The application of the Born rule is somewhat special, but not synonymous to collapse. It is interpretation dependent.

After the measurement, as you know, the wave function ceases to be an overlap of "possible" states.

But that wave function will only be on a eigenvalue for that observable basis only. It may still contains "overlap" on some other observable basis.
That is also a different with a "classic" collapse.

Unless you consider the interpretation "to many worlds", where the wave function continues to evolve in all the possible worlds in which it can give a certain result, etc

Or one of the many others that don't mention collapse at all.

 

[member 648030]

No, they are not exactly the same. The "behavior" for classic object don't exist. You are kind of "undoing" a pure abstract "probabilization" of a totally not random object, coin, money or whatever. You may call it "Collapse" if you want to, but it only concern you abstract description.
In QM "collapse" is not even a thing. It is a word that some interpretation use to describe when the application of the Born rule "change" the state of the object "for real". It is part of the broader measurement problem, and is not about "potential ignorance" being "undone"

The fact that a state has a value and another value means, from the point of view of the experimenter, one of two things:
1. the state is defined, but we do not know it
2. they are co-present multiple states simultaneously.

The first interpretation leads to a possible theory of hidden variables (Einstein), which however has been shown to be erroneous

The second remains, which is counterintuitive, and also "illogical", as if one thing is black and white simultaneously. But that's how things work. For example, in Feynman's theory of "virtual paths", a particle simultaneously follows infinite trajectories, which is completely meaningless (from a classical point of view).

Now, a coin is certainly a classic and not a quantum object, so, to say that it is "head" and "cross" simultaneously does not really make sense. But its behavior from the experimental point of view, is completely identical to the collapse of the wave function, or whatever you want to call it (change of state etc.)
After all, Einstein himself, when he criticized the QM, said: "God does not play dice!". All right, he did not say "God does not play heads or tails", but I think the meaning was just that.

 

[sshai45]

I want to comment here a bit about the "study the math" rebukes I see: the trouble here is that the formalism itself is, or *seems*, not complete enough, in a rather basic way, that may go underappreciated depending on how one is coming at it. And that is this: it provides for two basic operations on the quantum state vector, one of which is deterministic (Schrodinger equation), the other probabilistic ("collapse" law). The trouble is, suppose you wanted to take this description and now program a computer - we'll forget about the difficulty of having enough computing power, just imagine an ideal, infinitely powerful one with unlimited processing power and unlimited memory - to simulate a universe that might be described by this theory. (E.g. imagine a really good computer game, that is based on quantum-simulating everything in its world up from the atomic level.) For the theory's mathematics to "make sense", this is the criterion in which I imagine it. And when you do this, you run straight up into a big problem:

How do you choose which dynamic law to apply to the state of this universe at any given time step?

The collapse law says "when a 'measurement' happens", but it does not provide a mathematical equation to decide when that is happening,i i.e. something digestible by a computer. "Measurement" is not a mathematical term. Figuring out when/how requires interpretation, and that means different programmers will program effectively different dynamics.

That's the problem. "Study the maths" does NOT solve it.

 

[Boing3000]

The fact that a state has a value and another value means, from the point of view of the experimenter, one of two things:
1. the state is defined, but we do not know it
2. they are co-present multiple states simultaneously.

No, that's not the concerns of the experimenter. In both cases the probabilities are the same.
That's a concern for the model, and in the model everything is known, and everything is predictable (which both your point 1 and 2 are correct).

The first interpretation leads to a possible theory of hidden variables (Einstein), which however has been shown to be erroneous

The "hidden variable" must be non-local, that much is proved. However the universe does it, you can always call it "a hidden variable". But those "values" can spans huge swats of space (that's how entanglement is of critical important)

The second remains, which is counterintuitive, and also "illogical", as if one thing is black and white simultaneously.

I see grey things all the time, I don't find it illogical. Counter-intuitive is much more appropriate, but as long as conversation law are there, I personally find everything logical.

But that's how things work. For example, in Feynman's theory of "virtual paths", a particle simultaneously follows infinite trajectories, which is completely meaningless (from a classical point of view).

From a classical perspective, I found it totally logical. If I had to go somewhere "blinded" without any "guidance", I would try every-possible way, and kept the most efficient ones. That's totally meaningful for me that nature "kind of" does it all the time.

Now, a coin is certainly a classic and not a quantum object, so, to say that it is "head" and "cross" simultaneously does not really make sense.

I agree, but then it still is kind of useful/meaningfull. If the coin is not "fair", it is more "head" than "cross", but still both...

But its behavior from the experimental point of view, is completely identical to the collapse of the wave function, or whatever you want to call it (change of state etc.)

In QM superposition IS a thing of "reality". Negative probability and interference ALSO.
But classically there is nothing that change in the coin state on "collapse". It is never in superposition.

After all, Einstein himself, when he criticized the QM, said: "God does not play dice!". All right, he did not say "God does not play heads or tails", but I think the meaning was just that.

That's kind of unrelated. I think he just didn't like the stochastic only nature of the wavefunction.

 

[member 648030]

The concept of entanglement is an essential part of QM. It's inevitable. If you have two systems interacting, then their states will become entangled. If a $\pi^0$ particle decays into an electron-positron pair (I think that's an allowable decay) then the spins of the two particles will be entangled.

All that it means for systems to be entangled is for the composite system to be in a superposition of states that fails to factor into a product state. For example, for spin, the state $\frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle)$ is an entangled state.

It is not necessary to invoke entaglement
In fact, in general, I can represent any quantum state as a vector in a Hilbert space.
$| \psi\rangle=\sum_ {i = 1} ^ \infty c_i |\psi_i\rangle$
where $|\psi\rangle$ can be, for example, the wave function of a particle in a potential hole.
(or a hydrogen atom, to be more realistic)
As you can see, the $|\psi\rangle$ is not an entagled state at all, but rather the overlap of an infinite series of self-states by some operator (eg Energy)

 

[stevendaryl]

It is not necessary to invoke entaglement

In fact, in general, I can represent any quantum state as a vector in a Hilbert space.
$| \psi\rangle=\sum_ {i = 1} ^ \infty c_i |\psi_i\rangle$
where $|\psi\rangle$ can be, for example, the wave function of a particle in a potential hole.
(or a hydrogen atom, to be more realistic)
As you can see, the $|\psi\rangle$ is not an entagled state at all, but rather the overlap of an infinite series of self-states by some operator (eg Energy)

How you write it doesn't change whether it's entangled. The fact is that if you have a pair of particles that are in a superposition of

1. One state in which the first particle has spin-up and the other particle has spin-down
2. A second state in which the first particle has spin-down and the other particle has spin-up

then the spins of the two particles are entangled. Whether you write it as a product state or not doesn't change this.

Measuring the spin of one particle immediately tells you the spin state of the other particle, and vice-versa. But neither particle has a definite spin state prior to measurement.

 

[stevendaryl]

Now, a coin is certainly a classic and not a quantum object, so, to say that it is "head" and "cross" simultaneously does not really make sense. But its behavior from the experimental point of view, is completely identical to the collapse of the wave function, or whatever you want to call it (change of state etc.)

No, they aren't the same, from an experimental point of view. That's the whole point of Bell's inequality, to show that there are testable differences between quantum uncertainty and classical uncertainty.

 

[member 648030]

How you write it doesn't change whether it's entangled. The fact is that if you have a pair of particles that are in a superposition .

I do not see a pair of particles, let alone entagled. If you prefer, put the electron in a potential hole, or in any potential V. I still do not see two particles. Besides, I'm not interested in spin but only at energy levels. I still do not understand what entaglement has to do with it.

 

[member 648030]

No, they aren't the same, from an experimental point of view. That's the whole point of Bell's inequality, to show that there are testable differences between quantum uncertainty and classical uncertainty.

Suppose an apparatus similar to that of the Stern-Gerlach experiment. Suppose that you count from a beam of electrons those that have spin up (where the axis is chosen arbitrarily) and those with spin down. How many electrons will you count with spin up and how many with spin down? Or rather, the question is: how likely is (in your opinion) to find a spin up and to find a spin down. If you made a bet with a large sum where would you put it: on spin up or down?

 

[member 648030]

No, that's not the concerns of the experimenter. In both cases the probabilities are the same.
That's a concern for the model, and in the model everything is known, and everything is predictable (which both your point 1 and 2 are correct).

So in a quantum state measure, you are 100% sure to get a certain result, just like, knowing the initial conditions of a cannonball, can you predict exactly where it ends?

 

[stevendaryl]

I do not see a pair of particles, let alone entagled.

Yes, I know. Electrons and positrons are too tiny to see. But the point of saying that their spins are entangled is that you distant measurements that are correlated. You have a source of particle/antiparticle pairs. Out of each pair, Alice measures the spin of one particle, and Bob measures the spin of the other particle. Empirically, if you want to eliminate mentioning things that are not visible, the way things look is like this (simplified)

A spin measurement device has a dial that can be set to any number between 0 and 360. It has two lights, one on the left and one on right.

One "round" of the EPR experiment has the following steps:

1. Alice picks a number $\alpha$ and sets her device.
2. Bob picks a number $\beta$ and sets his device.
3. Charlie, halfway between them, presses a button (what it does can't be seen by you, so I won't mention it)
4. Either Alice's left light comes on, or her right light comes on.
5. Either Bob's left light comes on, or his right light comes on.
6. (Realistically, there are other possibilities, such as neither light coming on, but I'm oversimplifying)

The facts for the EPR experiment are these:

• If Alice and Bob choose the same number, then they always get opposite results.
• If they choose different numbers, then a fraction of the time $cos^2(\frac{\theta}{2})$, they get opposite results, and a fraction of the time $sin^2(\frac{\theta}{2})$ they get the same result (where $\theta = \beta - \alpha$).

So Alice's and Bob's results are strongly correlated. According to Bell's theorem, the correlation cannot be explained in terms of local hidden variables, but it can be explained in terms of entangled wave functions.

 

[stevendaryl]

So in a quantum state measure, you are 100% sure to get a certain result, just like, knowing the initial conditions of a cannonball, can you predict exactly where it ends?

In the famous EPR experiment, Alice and Bob are guaranteed 100% correlated results, but their individual results are completely unpredictable.

If Alice and Bob choose the same detector setting, then it is 100% certain that they will get opposite results: If Alice gets spin-up, Bob gets spin-down, and vice-versa. But it is completely unpredictable who gets which result.

 

[stevendaryl]

Suppose an apparatus similar to that of the Stern-Gerlach experiment. Suppose that you count from a beam of electrons those that have spin up (where the axis is chosen arbitrarily) and those with spin down. How many electrons will you count with spin up and how many with spin down? Or rather, the question is: how likely is (in your opinion) to find a spin up and to find a spin down. If you made a bet with a large sum where would you put it: on spin up or down?

To see the effects of entanglement, you have to have a source of entangled electron/positron pairs, and two different Stern-Gerlach devices. Then the statistics will be that:

• Each device will measure half of the particles to have spin-up and half to have spin-down.
• For any pair of particles, if one device measures spin-up for one of the particles, then the other device will measure spin-down for the other particle.

These facts by themselves don't imply that the particles are entangled. But the effects of entanglement are seen when the two Stern-Gerlach devices are not given the same orientation. Then the statistics are such that it is impossible to explain them using local hidden variables.

 

[stevendaryl]

Here's a game that summarizes the strangeness of EPR:

• Charlie, the dealer, deals out three cards to Alice, a left card, a middle card and a right card.
• He similarly deals out three cards to Bob.
• After the cards are dealt, Alice picks one card and Bob picks another. The remaining cards are left face-down.
• If Alice and Bob both pick the same position (left, middle or right), then their cards have opposite colors: If Alice's is red, Bob's is black, and vice-versa.
• If Alice and Bob pick different positions, then their cards have opposite colors 25% of the time and the same colors 75% of the time.

There is no way for Charlie to do this without either:

• Reading Alice's and Bob's minds to know which card they will pick, or
• Having trick cards that change color
• Charlie does some other trick (like switching Bob's cards around after Alice picks her card)

If Charlie tried to do it with regular cards and no tricks, then he would have to give one person two blacks and one red, and give the other person two reds and one black. But if he did that, then the probability that they would have opposite colors when they pick different positions is 1/3, not 1/4.

 

[Boing3000]

So in a quantum state measure, you are 100% sure to get a certain result,

Well, kind of. QM is verifyied 100% but only to the extend that you make many (many many) measures. It is a stochastic theory, meaning one outcome is very unpredictable. And that is not the case for classical mechanic.

just like, knowing the initial conditions of a cannonball, can you predict exactly where it ends?

No, very unlike the cannonball. Even if the is prediction is actually more difficult in reality that one may think (because of chaos). But with idealized cannonball (in vacuum etc... you can obtain very small margin of error).

 

[member 648030]

Well, kind of. QM is verifyied 100% but only to the extend that you make many (many many) measures.

Exactly like like throwing a coin or a nut ...

 

[member 648030]

Yes, I know. Electrons and positrons are too tiny to see. But the point of saying that their spins are entangled is that you distant measurements that are correlated. You have a source of particle/antiparticle pairs. Out of each pair, Alice measures the spin of one particle, and Bob measures the spin of the other particle. Empirically, if you want to eliminate mentioning things that are not visible, the way things look is like this (simplified)
...
So Alice's and Bob's results are strongly correlated. According to Bell's theorem, the correlation cannot be explained in terms of local hidden variables, but it can be explained in terms of entangled wave functions.

Ok everything's right what you say, but I do not understand why you talk about pairs of particles. I am considering the case of a single particle, if you want, the classic Schrodinger equation of the electron in the hydrogen atom.

 

[Boing3000]

Exactly like like throwing a coin or a nut ...

Exactly unlike the throwing of a coin which is at every single moment in a precise state. There is no stochastics involved.
The fact that you arrange the coin to be thrown in a random way (like using a hand) is where you get the illusion that the coin is in an unknown state and need a stochastic approach.

I hope you won't count on a "collapse" to win again future gambling robot. Because if they get to throw the coin, or even to bet after having watched you trough it ... you'll lose 99% of the time.
Meanwhile, you'll be as powerful as any robot to bet on spin of particles (because quanta behave very differently)

 

[member 648030]

Exactly unlike the throwing of a coin which is at every single moment in a precise state. There is no stochastics involved.
The fact that you arrange the coin to be thrown in a random way (like using a hand) is where you get the illusion that the coin is in an unknown state and need a stochastic approach.

I hope you won't count on a "collapse" to win again future gambling robot. Because if they get to throw the coin, or even to bet after having watched you trough it ... you'll lose 99% of the time.
Meanwhile, you'll be as powerful as any robot to bet on spin of particles (because quanta behave very differently)

So when Einstein (I repeat) saying the famous phrase "God does not play dice!" (evidently referring to the probabilistic character of the QM), Bohr should have answered: "Albert, but a dice is not a random object! But did you study classical physics? Did you give it the general physics exam?

 

[PeterDonis]

The fact that a state has a value and another value

This is not the case. The state has only one value.