Double slit probability question

[BvU]

In order to understand why an electron goes in one direction or the other you need to know all the details regarding the incoming electron and the microscopic structure of the slits.

You wouldn't get a single iota further. Especially not in the context of this thread.

 

[ueit]

You wouldn't get a single iota further. Especially not in the context of this thread.

So you are claiming that understanding the interaction that is responsible with the momentum transfer is irrelevant in understanding the momentum transfer, right?

 

[entropy1]

In quantum mechanics, wave function collapse is said to occur when a wave function—initially in a superposition of several eigenstates—appears to reduce to a single eigenstate (by "observation")

Thank you. So if I understand correctly, the probability (amplitude) of a collapse to occur is given by the wave function?

So it could tick off the detector, but it could also tick the wall of the box the detector is in, and so forth?

Or can a detector 'attract' a detection?

 

[Karolus]

Thank you. So if I understand correctly, the probability (amplitude) of a collapse to occur is given by the wave function?

So it could tick off the detector, but it could also tick the wall of the box the detector is in, and so forth?

Or can a detector 'attract' a detection?

I try to be simple. We have our own wave function, we say that is a function of the x coordinate of the type $f (x) = \cos(x)$. (Not normalizzabile etc, it does not matter) In this state, the electron is "everywhere" in the region$-\infty \leq x \leq \infty$ Where he is exactly, it is a question that has no sense.
I know that the probability that the electron is detected at point $x_0$ is proportional to $|f (x_0)|^2$. Suppose that this value is 0.3 (ie 30%).
Suppose we put the detector at the point $x_0$. The detector can detect the electron (with a click) or not. At the time that the electron has detected electron, f (x) does not exist anymore ... because our electron is so to speak collapsed at the point $x_0$, there where we put our detector.
So there is no attraction or anything, just a probability of detecting the particle at point x.
Things are a little more complex, because in this case, the detector, could be a "shot" photon at point $x_0$. if I have so many electrons equal, with the same wave function, 3/10 the photon will detect the electron at point $x_0$

 

[entropy1]

So, the electron 'is' not anywhere when not detected, but the probability it is detected at some position can be calculated from te wavefunction.

 

[ueit]

So, the electron 'is' not anywhere when not detected, but the probability it is detected at some position can be calculated from te wavefunction.

QM does not say that the electron is not anywhere and it doesn't say there is somewhere either. It just allows you to calculate the probability of detection.

 

[zonde]

You have tied yourself into logical knots and confusion by thinking classically - don't do that and your confusion will disappear.

Getting rid of confusion by all costs is not sensible. Scientific thinking is subset of classical thinking and it is not very sensible idea to get rid of that part of classical thinking. So if you can't be more specific I would say your suggestion is rather useless.

 

[BvU]

Scientific thinking is subset of classical thinking

That wouldn't be good ! Perhaps you meant it the other way around ?

 

[mike1000]

I try to be simple. We have our own wave function, we say that is a function of the x coordinate of the type $f (x) = \cos(x)$. (Not normalizzabile etc, it does not matter) In this state, the electron is "everywhere" in the region$-\infty \leq x \leq \infty$ Where he is exactly, it is a question that has no sense.

I like your example. Let's pretend that the wave function is given by Ψ(x) = cos(x). That is just the probability distribution for the location of the electron. It is a statement of our lack of knowledge of where the electron is located. (I know that cos(x) cannot be a true wave function because it is not square integrable, which is a requirement to have a finite dot product and finite probabilities.) It is not saying the electron is everywhere or no where, it is just the probability distribution associated with the electron at that time.

Position (and momentum) of the electron are random variables in quantum mechanics. As such their possible values are randomly distributed by some, yet to be understood, process.

In probability and statistics, a random variable, random quantity, aleatory variable, or stochastic variable is a variable quantity whose possible values depend, in random manner, on a set of random outcomes events.[1] It is common that the outcome depends on some physical variables that are not well understood.

https://en.wikipedia.org/wiki/Random_variable

 

[zonde]

That wouldn't be good ! Perhaps you meant it the other way around ?

No, I meant it the way I wrote it.
There is some basis from which you do any further thinking. You can't start any reasoning from nothing. And in order to have meaningful discussion we have to have common basis. In science this common basis is scientific method and any prerequisites that are required for application of scientific method.

 

[PeterDonis]

the probability (amplitude) of a collapse to occur is given by the wave function?

If you are using a collapse interpretation. But there are interpretations that don't have collapse (such as the MWI).

 

[entropy1]

If you are using a collapse interpretation. But there are interpretations that don't have collapse (such as the MWI).

I ment 'collapse' in the sense of 'detection'.

 

[entropy1]

So, the electron 'is' not anywhere when not detected, but the probability it is detected at some position can be calculated from te wavefunction.

QM does not say that the electron is not anywhere and it doesn't say there is somewhere either. It just allows you to calculate the probability of detection.

I thought it was a fairly straighforward question; but I think we are on the same track here.

 

[PeterDonis]

I ment 'collapse' in the sense of 'detection'.

That clarifies what you meant, but you should realize that this is very confusing terminology, since "collapse" has a precise technical meaning in QM, which is not the same as "detection". It's much better to say "detection" if that's what you mean.

 

[entropy1]

That clarifies what you meant, but you should realize that this is very confusing terminology, since "collapse" has a precise technical meaning in QM, which is not the same as "detection". It's much better to say "detection" if that's what you mean.

I was thinking that a detection and a collapse were similar in the sense that in case of position/momentum uncertainty a detection yields position-information and for that to be possible the wavefuntion has to collapse in a way that the position is known. (sorry for the layman way of expressing this)

I am aware that two different terms usually refer to different concepts.

 

[PeterDonis]

I was thinking that a detection and a collapse were similar

No, they're not, because "detection" is interpretation-independent: it's something that's directly observed. "Collapse" is interpretation-dependent: some interpretations of QM have collapse, some don't, and we don't directly observe collapse, we only directly observe detection.

in case of position/momentum uncertainty a detection yields position-information and for that to be possible the wavefuntion has to collapse

Only in collapse interpretations. In no collapse interpretations (the MWI, for example), it doesn't.

 

[Nugatory]

n case of position/momentum uncertainty a detection yields position-information and for that to be possible the wavefunction has to collapse in a way that the position is known.

That cannot be right... It sounds as if you are saying that we cannot make a position measurement if there is no collapse. But we already know that quantum mechanics works just fine, for measurements of position as well as everything else, without introducing the concept of collapse.

 

[bhobba]

So you are claiming that understanding the interaction that is responsible with the momentum transfer is irrelevant in understanding the momentum transfer, right?

No - he is claiming what the formalism of QM says.

Its simple - so simple many don't get it. It took me a while to get it - but that's just because its in plain sight and you gloss over the obvious.

The formalism is a theory about observations that occur in a common-sense classical world. Whats going on when not observed - momentum transfer yada yada yada the theory is silent on.

This raises the legit issue of QM - the issue ignored by Einstein and Bohr which is why they both have issues (I won't say wrong - its simply a blemish that's best fixed - Weinberg is a bit more prosaic). How does a theory that assumes a classical world from the start explain that world. Great progress has been made in fixing that up, but some problems remain. What those issues are (the factorization problem, key theorems elucidating the problem, and other key theorems associated with decoherence - there are others as well) are (at least in part) detailed here:
https://www.amazon.com/Understanding-Quantum-Mechanics-Roland-Omnès/dp/0691004358

Thanks
Bill

 

[bhobba]

No, they're not, because "detection" is interpretation-independent: it's something that's directly observed. "Collapse" is interpretation-dependent: some interpretations of QM have collapse, some don't, and we don't directly observe collapse, we only directly observe detection.

Go to a library, or in some other way, get a copy of Ballentine.

QM is based 2 axioms - none of which includes collapse.

I know this can be confusing because some texts have it as an actual axiom - it isn't. I remember having a long 'conversation' with the author of such a book. But he remained unconvinced despite pointing him to Ballentine.

But you can take it as a given there is no collapse in the formalism of QM - I know of no expert here (and many are professors who teach it) that says so. The author of the textbook simply has not thought it through carefully enough eg there is obvious no collapse in MW - if collapse was an axiom you wouldn't have that interpretation.

Thanks
Bill

 

[PeterDonis]

QM is based 2 axioms - none of which includes collapse.

Yes, I know, but there are interpretations of QM which have "collapse" in them. I was pointing out the same thing you are: "collapse" is not part of the actual theory of QM, it's only part of certain interpretations.

 

[bhobba]

Yes, I know, but there are interpretations of QM which have "collapse" in them. I was pointing out the same thing you are: "collapse" is not part of the actual theory of QM, it's only part of certain interpretations.

Sorry - I should have made it clear - I was simply elaborating on what you said to help others reading the thread.

I don't know why, but some simply will not let go off the idea - its insidious. It was actually an axiom in a textbook

Von-Neumann gave them many moons ago:
https://en.wikipedia.org/wiki/Dirac–von_Neumann_axioms

But then you get this:
http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html

So I suppose there is a reason - sigh.

Thanks
Bill

 

[ueit]

So you are claiming that understanding the interaction that is responsible with the momentum transfer is irrelevant in understanding the momentum transfer, right?

No - he is claiming what the formalism of QM says.

Its simple - so simple many don't get it. It took me a while to get it - but that's just because its in plain sight and you gloss over the obvious.

The formalism is a theory about observations that occur in a common-sense classical world. Whats going on when not observed - momentum transfer yada yada yada the theory is silent on.

1. QM certainly takes care about momentum transfer, otherwise it would fail to correctly predict the experimental results. We know that the electron acquires momentum when passing through the slits (otherwise instead of an interference pattern you would get a point, just like in the case no barrier is present between the source and screen). We also know that momentum conservation is obeyed in QM. Nothing stops you from using a microscopic description of the barrier, write down the Hamiltonian (which would include in this case explicitly the interaction between the electron and the charged particles in the barrier) and calculate the probabilities of finding the electron in different points on the screen. It might take you some time and processing power, but in principle it could be done.

2. Even if QM does not explicitly describe what the electron does at the slits you can still use logical inference. No barrier - you get a point. A barrier with a slit - diffraction pattern. A barrier with two slits - interference pattern. A barrier with no slit - nothing. What explanation do you have for this if the electron does not interact with the material of the barrier?

This raises the legit issue of QM - the issue ignored by Einstein and Bohr which is why they both have issues (I won't say wrong - its simply a blemish that's best fixed - Weinberg is a bit more prosaic). How does a theory that assumes a classical world from the start explain that world. Great progress has been made in fixing that up, but some problems remain. What those issues are (the factorization problem, key theorems elucidating the problem, and other key theorems associated with decoherence - there are others as well) are (at least in part) detailed here:
https://www.amazon.com/Understanding-Quantum-Mechanics-Roland-Omnès/dp/0691004358

[URL='https://www.amazon.com/dp/0691004358/?tag=pfamazon01-20[/URL]

I have red this book. It is good, but I disagree with the authors on some key points (entanglement). I don't find the consistent histories approach particularly appealing.

 

[Karolus]

Only in collapse interpretations. In no collapse interpretations (the MWI, for example), it doesn't.

In 'many-worlds interpretation of Hugh Everett, (MWI) it is assumed that as, for example, when the electron is detected, its wave function continues its evolution in another universe. In other words, many universes exist (infinite ??) where the wave function evolves in every universe according to one of its possible eigenstates ... As interpretation (as high-quality scientific) seems rather philosophical ...

 

[entropy1]

No, they're not, because "detection" is interpretation-independent:

Only in collapse interpretations. In no collapse interpretations (the MWI, for example), it doesn't.

That cannot be right... It sounds as if you are saying that we cannot make a position measurement if there is no collapse. But we already know that quantum mechanics works just fine, for measurements of position as well as everything else, without introducing the concept of collapse.

If an observable has eigenstates A and B, then, by measuring, we observe (detect) either A or B. You can omit the concept of 'collapse', but each 'world' in MWI yields a single eigenvalue. Each world of MWI is compatible with collapse, though all the worlds together (in MWI) don't need it. The real world must match QM (the formalism), and therefore in every 'world' the collapse interpretation is valid, just as well as the other valid interpretations. A detection selects an eigenvalue and thus collapse is valid. That is what I ment.

 

[PeterDonis]

In 'many-worlds interpretation of Hugh Everett, (MWI) it is assumed that as, for example, when the electron is detected, its wave function continues its evolution in another universe.

No, that's not what MWI actually says. It says that when an electron is detected, the wave function of the electron becomes entangled with the wave function of the measuring device, so that both of them split into "pieces", one for each possible measurement result. For example, if we measure the electron's spin, there are two possible results, which we'll call "up" and "down", so the wave function evolution looks like

$$
\vert e \rangle \vert M \rangle \rightarrow a \vert \uparrow \rangle \vert U \rangle + b \vert \downarrow \rangle \vert D \rangle
$$

where $e$ and $M$ are the starting states of the electron and the measuring device, $\uparrow$ is the "up" state of the electron, $U$ is the "measured electron up" state of the measuring device, $\downarrow$ is the "down" state of the electron, $D$ is the "measured electron down" state of the measuring device, and $a$ and $b$ are complex coefficients whose specific values will depend on the details of the state $e$.

Notice that there is nothing here about "other universes" or "other worlds"; there is just one wave function, which happens to have two terms after the measurement (and even that is dependent on the choice of basis that we made). Calling each term a different "world" is a sort of interpretation on top of an interpretation, so to speak; you can use the MWI without ever having to think of the different terms that way.

Each world of MWI is compatible with collapse

If by "collapse" you mean "picking out one term of the wave function and ignoring all the others, even though they're still there", yes, this is true. But the usual meaning of "collapse" is "all terms but one in the wave function actually disappear". That is why it's confusing to use the word "collapse" in a context where no particular interpretation is required.

 

[entropy1]

That is why it's confusing to use the word "collapse" in a context where no particular interpretation is required.

Then, how should I call it according to you? To me a measurement is still the selection of an eigenstate.

 

[PeterDonis]

how should I call it according to you?

"Detection". That's the word you used earlier.

To me a measurement is still the selection of an eigenstate.

Not in the MWI. In the MWI all possible measurement results occur.

 

[bhobba]

1. QM certainly takes care about momentum transfer, otherwise it would fail to correctly predict the experimental results..

You are falling into the trap.

All QM predicts is the outcome of observations. Whats going on when not observing its silent on.

Yes an electron going through an electric field will gain or loose momentum WHEN OBSERVED. Whats going on while in the field - who knows.

But gaining momentum when going through slits - can't follow that one. It changes direction when observed - but KE is unchanged hence the absolute value of momentum doesn't change - only direction when observed. When there are two slits you apply the principle of superposition which leads to the interference pattern - but again KE is not changed - momentum direction - yes but in a more complicated way due to the superposition.

As I said its so easy and in plain sight you ignore it. You must force yourself to NOT do that. That's all there is to it really. At all times remember - without an observation QM says nothing other than the probability of an observation if you were to do it. Now exactly how does an observation 'work'. Even defining in purely QM terms what an observation is. Start a new thread if you want to discuss that - but decoherence has shed a lot of light on it.

Thanks
Bill

 

[ueit]

You are falling into the trap.

All QM predicts is the outcome of observations. Whats going on when not observing its silent on.

Yes, but the electron is observed at the screen.

Yes an electron going through an electric field will gain or loose momentum WHEN OBSERVED. Whats going on while in the field - who knows.

Again, we are speaking about electrons that are observed.

But gaining momentum when going through slits - can't follow that one. It changes direction when observed - but KE is unchanged hence the absolute value of momentum doesn't change - only direction when observed.

Let's say the electrons coming from the source travel along Z axis, which is perpendicular on the barrier. Let's say that X is the axis connecting the slits in the plane of the barrier.

The initial momentum on X is 0, right?

When the electrons are observed they are found at some distance from the original direction, therefore they acquired some momentum on X, say mx.

So, the momentum on X has changed and momentum conservation requires that some other particle/particles acquired a momentum -mx.

When there are two slits you apply the principle of superposition which leads to the interference pattern - but again KE is not changed - momentum direction - yes but in a more complicated way due to the superposition.

It seems to me that you have a wrong idea about the momentum conservation law. Momentum is conserved as a vector. If the direction is changed, the momentum is changed. Momentum conservation requires that the total momentum on each of the three axis (X, Y and Z) remains constant. If the particle started with 0 momentum on X and ended with a non-0 momentum on X it means that some other particles acquired an opposite momentum on X.

The superposition principle cannot be a cause for the change in momentum. The cause can only be some interaction with other particles.

Andrei

 

[DrChinese]

The initial momentum on X is 0, right?

That's not correct in any particular case, as the momentum is undefined (poorly defined) in X. If it were well-defined, there would be no interference pattern building up.

 

[ueit]

That's not correct in any particular case, as the momentum is undefined (poorly defined) in X. If it were well-defined, there would be no interference pattern building up.

Interference or not, one can always determine the mean speed of the particles on any axis by dividing the distance traveled over the time taken from the emission to detection. If there is nothing between the source and screen the distance traveled on X is 0 (almost). If there is a barrier with slits you can get 0 for the central maximum but for the other fringes you will get a non-0 value.

Andrei

 

[DrChinese]

Interference or not, one can always determine the mean speed of the particles on any axis by dividing the distance traveled over the time taken from the emission to detection. If there is nothing between the source and screen the distance traveled on X is 0 (almost).

One can imagine there is something that can be calculated and labeled mean speed. But it is meaningless (no pun intended) because you still don't know where it went or what it was doing in the meantime (no pun intended here either).

Obviously uncertainty is a substantial issue as well. Making it difficult to comment about momentum in a particular direction.

 

[ueit]

One can imagine there is something that can be calculated and labeled mean speed. But it is meaningless (no pun intended) because you still don't know where it went or what it was doing in the meantime (no pun intended here either).

Obviously uncertainty is a substantial issue as well. Making it difficult to comment about momentum in a particular direction.

It does not matter what the particles are doing "in meantime". You just cannot measure a violation of momentum conservation. Think in this way:

1. The initial momentum of the particles on X (before the arrival at the slits) can be made arbitrarily small by placing the source very far away. Do you agree with this?
2. The final momentum of the particles on X has some value, proportional with the distance between the center and the last visible fringes. It cannot be arbitrarily small because in that case you would see one dot, not an interference pattern.

You can measure very accurately the time of emission, the time of detection, the position of the slits and the position of the dots on the screen. The only unknown is the time when the particle arrives at the slits, but that could also be determined by comparing the time of flight with and without the barrier in place. This gives you everything you need to determine both the initial and final momentum with any accuracy you want.

So, if you claim that momentum does not change at the slits it means that you have observed a violation of momentum conservation.

 

[DrChinese]

It does not matter what the particles are doing "in meantime". You just cannot measure a violation of momentum conservation. Think in this way:

1. The initial momentum of the particles on X (before the arrival at the slits) can be made arbitrarily small by placing the source very far away. Do you agree with this?
2. The final momentum of the particles on X has some value, proportional with the distance between the center and the last visible fringes. It cannot be arbitrarily small because in that case you would see one dot, not an interference pattern.

You can measure very accurately the time of emission, the time of detection, the position of the slits and the position of the dots on the screen. The only unknown is the time when the particle arrives at the slits, but that could also be determined by comparing the time of flight with and without the barrier in place. This gives you everything you need to determine both the initial and final momentum with any accuracy you want.

So, if you claim that momentum does not change at the slits it means that you have observed a violation of momentum conservation.

I never said you could measure a violation of momentum conservation. But I certainly reject the idea the particle is flying in a straight line like a little billiard ball, because we would see no interference if that were the case. How momentum might vary when we are not watching is unknown.

On the other hand, you cannot measure the initial position accurately AND expect to know much about initial momentum. Ditto with ending q and p.

 

[Ostrados]

It does not matter what the particles are doing "in meantime". You just cannot measure a violation of momentum conservation. Think in this way:

1. The initial momentum of the particles on X (before the arrival at the slits) can be made arbitrarily small by placing the source very far away. Do you agree with this?
2. The final momentum of the particles on X has some value, proportional with the distance between the center and the last visible fringes. It cannot be arbitrarily small because in that case you would see one dot, not an interference pattern.

You can measure very accurately the time of emission, the time of detection, the position of the slits and the position of the dots on the screen. The only unknown is the time when the particle arrives at the slits, but that could also be determined by comparing the time of flight with and without the barrier in place. This gives you everything you need to determine both the initial and final momentum with any accuracy you want.

So, if you claim that momentum does not change at the slits it means that you have observed a violation of momentum conservation.

Initial momentum is unknown and it is not 0 at x=0 how did you assume that?! And you can't know if it is small either.

You are thinking about particles as projectiles, as if they are billiard balls. Photon for example can never be standing still by have 0 momentum at x=0.