[alancalverd]

f the sentence is meaningless, how did Milliken do his experiment? What did he assume?

There are no assumptions in Millikan's experiment. He "simply" discovered that charge is quantised - it's a clever and quite difficult experiment but yielded a remarkably accurate value of the quantum of charge.

[alancalverd]

Then Energy seems to be a Calculable entity.

It is a quantity, not an entity. See reply #7 above.

[Eternal Student]

Hi.

I am not sure I agreed with you regarding yard sticks.

    You've asked a lot of sensible questions about yard sticks and you are recognising the problems.   The main point is precisely that real physical sticks do tend to be made of something and they could change dimensions for all sorts of reasons.   When distance is defined as how far light travels in a unit of time then we avoid the need to worry about any of this.

As per what I understand you to be saying  those yard sticks remain unchanged in length

 
    I think that was from a post I made a while back.   In that post the sticks were "idealised sticks" in that we were trying to consider what happens if light slows down and for that we needed our sticks to be immune to whatever it was that might have caused light to slow down.  Don't spend too long worrying about that - the key point is that sticks are not good ways to measure a distance with, we do not know exactly how they would or should behave.   For example, we wouldn't know exactly which effects we wanted our "idealised sticks" to be immune to. 
    Let's re-phrase this:   Just for that particular situation you need to assume that we had access to a "god-given" stick of length 1 metre which was 1 metre and for evermore shall be 1 metre (and god knew what a "distance" was supposed to be better than we do, so he/she knew precisely what effects that stick had to be made immune to).

I was wondering if the force of electric attraction was proportional to the speed of em radiation -ie the speed of light.

    The answer is yes, assuming Maxwells equations continue to model the behaviour of light.   The speed of light, c, is given by   c =   1 / √(με)  where  ε and μ are the permittivity and permeability of space (things that will influence how strong the electric or magnetic attractions will be).  So if c changes over time then at least one of μ or ε  must be changing with time.   Assuming it's ε then we have:  c decreasing  => ε must be increasing  => the electric attraction between charges is proportional to 1/ε  =>  so that would have been decreasing.   That might end up with the atoms in the stick being less strongly pulled together so that the stick might grow longer.  (It might also be that the attraction is now so weak the molecules can't even hold together and the stick falls apart).  However, the reason or cause for light to be slowing down was deliberately left arbitrary and hypothetical - it may be that it wasn't following Maxwells equations in that future due to some as yet unknown physics.
       Anyway, this sort of lengthening of sticks would be precisely the opposite of what you want to happen if you're hoping to measure the same distances with sticks and with light being allowed to travel.   Half the speed of light and you should half the distance it travels in 1 second.   But since your sticks have grown you can't even fit half the number of sticks into that new length measurement,  you'd be lucky to get one-third as many sticks into that length measurement.  So it would now be even easier to conclude that measuring distances with sticks or with a travel time of light leads to disagreements (if the speed of light changes over time).
     Overall, the important point is that yard sticks are not the best way to measure a distance - there are far too many questions and complications about how sticks might behave.

Best Wishes.

[varsigma (OP)]

There are no assumptions in Millikan's experiment.

Sorry, I just can't see how that could be possible.
I still don't understand why it's meaningless to assume that the electron's properties occupy the same place. I just don't get what you mean.

Instead of "in the same place", perhaps I should say the mass, charge, and spin of an electron are all in phase.
p.s. I realised I should qualify that last sentence with "under normal circumstances".

[alancalverd]

Mass and charge are not wave functions and thus do not have a phase. You can align the spins of a group of electrons but in the absence of an external magnetic field, the concept of spin phase is meaningless.

What Millikan (and countless subsequent undergraduates) did was to measure the voltage required to prevent a charged droplet from falling under gravity between two parallel plates. It turned out that the required voltage has discrete values, from which he deduced that charge is quantised. Clever, difficult (I've never known an undergraduate  get it to work first time), but no assumptions.

[varsigma (OP)]

Mass and charge are not wave functions and thus do not have a phase.

The mass and charge of an electron aren't a part of its wavefunction? Or have I misinterpreted you?

What Millikan (and countless subsequent undergraduates) did was to measure the voltage required to prevent a charged droplet from falling under gravity between two parallel plates. It turned out that the required voltage has discrete values, from which he deduced that charge is quantised. Clever, difficult (I've never known an undergraduate  get it to work first time), but no assumptions.

Well, I think any experiment has to make assumptions. One assumption I think Milliken had to make was that oil droplets can be charged, that electrons stick to them and stay stuck in the experiment.

[alancalverd]

The mass and charge of an electron aren't a part of its wavefunction? Or have I misinterpreted you?

You have misinterpreted physics. A wavefunction is the mathematical  model we use to describe the probability of finding an object at a point in space. The electron doesn't "have" a wavefunction, but we assign one to it. You are not alone in your misconception, by any means!

If the droplets weren't charged, they wouldn't be prevented from falling by the electric field. If the charge dissipated, the stationary droplet wouldn't remain stationary. Nobody said anything about electrons. The experiment demonstrates that charge is quantised.

[varsigma (OP)]

You have misinterpreted physics. A wavefunction is the mathematical  model we use to describe the probability of finding an object at a point in space. The electron doesn't "have" a wavefunction, but we assign one to it. You are not alone in your misconception, by any means!

Ok. If there's a Schrodinger equation that describes the probability of finding an electron near a proton, say in a Hydrogen atom, in what sense does it not describe the position of the electron's mass or any other property of that electron?

If the droplets weren't charged, they wouldn't be prevented from falling by the electric field. If the charge dissipated, the stationary droplet wouldn't remain stationary. Nobody said anything about electrons.

Nobody said anything about electrons and having measured their charge? I thought that was the whole point of the experiment.
How then did Milliken know what he measured?

I still fail to see how anyone can do an experiment--any experiment--and not make assumptions.

[Eternal Student]

Hi.

If there's a Schrodinger equation that describes the probability of finding.......

    Minor detail:   It's not the Schrodinger equation that describes these things,  it's the wave function which appears in that equation.
   [I've edited this post to remove more details, the post was getting too long].

in what sense does it (the wave function) not describe the position of the electron's mass* or any other property of that electron?

   Various ways exist.   These are some of the easiest ones to explain:
1.   The wave function may be in a superposition of states.   Until a measurement is made a single unique value of that property cannot be assigned to the object.
2.    Measurement of one property will cause a wave function collapse and the wave function is then changed.   This can alter the value of other properties of the object.  To say this more clearly, you cannot measure all the properties one after the other and hope you'll know everything at the end.   As you continue measuring more things you will unavoidably mess up some of the previous things you measured.

I hope that makes some sense.

   * You actually mentioned "mass" as one example of a property.   Simple QM models will assume the mass of a particle is a fixed unchanging constant.  You want the mass as a parameter to generate the Schrodinger equation.  The above discussion applies to all the "other properties" of the electron you've ever mentioned (e.g. spin, momentum, position  etc.) and the general spirit of it should apply to mass.
- - - - - - - - - - - - - -
    I'm not sure that quantum mechanics needs to be too important for your discussion.   For large scale models we can assume objects have properties with definite values, these exist at all times and these properties are located with the object.
   No model in physics is perfect and no teacher would try to present the most complete or accurate model of physics to their class.   Instead you ( @varsigma ) can make a decision about how much detail or simplification is useful and just pick a model that is sufficient for the purpose.
     You didn't say much about the original discussion you were having .....

Hi. I recently had a discussion online about the subject of physics, in which I posted something about simplification, and how that seems to be where physics starts, at least.

    I assumed you might be wanting to get some new ideas for that other online discussion from this web forum.  That's why I have sometimes rambled on about a very small detail or niche area.
    You don't have to - but would you be able to tell me a bit more about that other online discussion  and/or  about what you're hoping to obtain from this discussion here on this forum?
 
Best Wishes.

[alancalverd]

Ok. If there's a Schrodinger equation that describes the probability of finding an electron near a proton, say in a Hydrogen atom, in what sense does it not describe the position of the electron's mass or any other property of that electron?

You are beginning to get the picture. The wave function does what I said - it maps the probability density of finding the electron. It is not a property of the electron, because it is a function of the environment (it's different for an electron in a hydrogen atom compared with a hydrogen molecule), not just the entity.
 

Nobody said anything about electrons and having measured their charge? I thought that was the whole point of the experiment.
How then did Milliken know what he measured?

I repeat: Millikan demonstrated that charge is quantised. How did he know what he measured? Whenever an oil drop was stationary he looked at the voltmeter. No assumptions (other than that the voltmeter measured volts). You (and almost everyone else) have made the assumption that the quantum is the charge of an electron. Why not just read the Wikipedia article?

I think the best experiments are true null investigations, like dropping a couple of rocks from a tower. No room for any assumptions: what you see is what you get.

[alancalverd]

Measurement of one property will cause a wave function collapse and the wave function is then changed. 

I seriously disparage this statement!

You can plot the outcome of dice throws as a wave function. You throw the dice and get a number. You haven't done anything to the hypothetical wave function, just chosen one value of it, which you could not predict. If you roll the dice again, you will get an equally unpredictable number. If you had "collapsed" the wave function, you would have restricted the range of future possibilities - the gambler's fallacy.

It is perfectly true that if you measure any property of a subatomic particle you will have altered its state in some way, e.g. by bouncing a photon off it, and thus biased its future state and wave function because you have changed its energy and momentum, but the notion of "collapse" rather militates against Heisenberg.

[geordief]

Measurement of one property will cause a wave function collapse and the wave function is then changed.

I seriously disparage this statement!

You can plot the outcome of dice throws as a wave function. You throw the dice and get a number. You haven't done anything to the hypothetical wave function, just chosen one value of it, which you could not predict. If you roll the dice again, you will get an equally unpredictable number. If you had "collapsed" the wave function, you would have restricted the range of future possibilities - the gambler's fallacy.

It is perfectly true that if you measure any property of a subatomic particle you will have altered its state in some way, e.g. by bouncing a photon off it, and thus biased its future state and wave function because you have changed its energy and momentum, but the notion of "collapse" rather militates against Heisenberg.

So collapsing the wave function is a bit like swatting a fly ?

You kill it but there is always another one.....

[alancalverd]

It's just a pointless and confusing expression, implying that a wave function has more significance than a mere model.

Do you collapse the wave function of a pair of dice? No, you make an observation of an event that you can't predict, though you have a very good idea of the likelihood  of scoring less than 2 (not possible), 7 (highly probable) or 12. The only difference is that Heisenberg's limit on the precision with which you can simultaneously know the position and momentum of an object is a bit more important for an electron than for macroscopic dice.

[Eternal Student]

Hi.

I seriously disparage this statement!

    OK.     
     It's still perfectly true unless the wave function was already in an eigenstate of what you were about to measure.   Even then you can still say that the measurement would "collapse" the wave function but in this situation that new function will just turn out to be the same wave function.   
     In most situations two operators representing different observables do not share a common set of eigenfunctions, so that measuring one forces a collapse to a state that cannot be an eigenstate of the other operator.
     Rather than write out the mathematics in something like Dirac notation for this, I suspect most of the readers would just prefer a concrete experiment to be discussed.
    So the experiment similar to the one in post #20 is one of the clearest demonstrations.
    Have some polarised light and start sending it through polarising filters.   Have a first filter that will collapse the wave function so that the lights polarisation is entirely in the x-axis direction.   Put that through a second filter which permits only light polarised along the y-axis to pass and no light will get through.   However, if you add a third filter turned at 45 degrees between the x- and y-axis and insert that in between the two other filters,   then you will now get some light to pass the last filter.      The measurement of the polarisation along the 45 degree axis has forced a wave function collapse which has now made sure that the wave function is back in a superposition of states for polarisation along the x-axis or y-axis.
      This has been discussed in several other threads and here's the YouTube video that tends to be chosen to go with it.   ( "Three polarising filters: a simple demo...." ,  available on YouTube,  duration ~ 1 min 30 seconds).
 

You can plot the outcome of dice throws as a wave function.

     It's an awful example but if you really want to use it we can.
     Consider an observable we can call "the last dice roll result".   If it was a 6 then that is now a fact (actually it is still a random variable but a very trivial one - it has only one possible value and a 100% probability of having that value).  All of the characteristics we want are still being exhibited.   We have a situation comparable to wave function collapse, roll the dice and the value of the last dice roll has to be updated immediately.   Throw the dice again.   If the dice roll you obtained was a 5 then that replaces the previous dice roll value and that is now the last dice roll result.
    About the gamblers fallacy  -->  They are just betting on the wrong things.   Bet on the last dice roll result because there has been a wave function collapse, the wave function is now one where the last dice roll will always be a 6 (or whatever it was).  They will win every time (until a new measurement is made).

Best Wishes.

[Eternal Student]

Hi.

Sorry, I spent so long writing my reply,  you ( @alancalerd ) had written another one in between.

Do you collapse the wave function of a pair of dice?

   Yes.   You haven't really clearly defined what you are considering as a "wave function" but I'll just assume it involves a description of the system that is sufficient to predict the outcome of dice rolling (and nothing much else).   Exactly how you're going to get that into the Schrodinger equation or what you are considering as the Hamiltonian is questionable  -  but hey, whatever, this is 2023 and we'll just go with it.   Let's just assume you are conceptualizing the dice as some system that is described by something which is loosely like a wave function.
   Prior to making any measurements your wavefunction presumably has these characteristics:   Each die can take one of six values  {1,2,3,4,5,6} with equal probability = 1/6 for each value.   The evolution of your wave function is not all that interesting by whatever version of the time dependant Schrodinger equation it follows.  As time progresses, the probability of getting a particular die roll result doesn't change.
    Now make a dice roll and measure the result (let's say it's a 1 and a 5 = 6 total).  That dice roll result is clearly not following anything like your old model any longer.   The wave function is not the same as it used to be, it has been changed, there was something analogous to a wave function collapse.   Your wavefunction is now one where the dice roll result would always be a 1 and a 5 making a total of 6.    Future results might still follow your random prediction model but the previous results certainly don't.

- - - - - - - - - -
     On a very minor and tangential topic.  The entire notion of Random Variables in pure mathematics is quite an interesting one.  With Quantum mechanics it is now reasonable that probability is something that exists in nature instead of being only a theoretical or abstract mathematical concept.   Since the number of people likely to be interested in pure mathematics is not high, I'll just leave off that discussion.
  - - - - - - - -

It's just a pointless and confusing expression, implying that a wave function has more significance than a mere model.

    Well,  the premise of QM is that every system can be described by a wave function and that wave function describes everything that is knowable about the system.   So, for example, if the mathematics says that you can't determine all the components (x, y and z- Axis components) of angular momentum simultaneously  (which they do), then many physicists will assume that  this would be true and you can not actually do that in reality.
   What I'm saying is that for many physicists, Quantum mechanics is of earth-shaking importance that changes our understanding of what reality might actually be.   You could choose to consider that a wave function is the best example of a property that a system can have and the only property you would need to assume it has.  All other observable properties that it might have can be determined from that wave function.  Some properties will never be knowable simultaneously or just are not properties that the system can have.  Change that wave function somehow and you can change the behaviour including various physical properties of the system.   If you know the wave function has not changed then you can assert that the system has not changed its properties or behaviour etc...    The wave function is the over-arching governor of, or register of properties for, the system.
    But it is just a model and not the law of our nation.   So you don't HAVE to assume the wave function has any underlying significance if you don't want to.

Best Wishes.
(LATE EDITING to fix some spelling / grammar - which still isn't perfect.).

[alancalverd]

You haven't really clearly defined what you are considering as a "wave function" but I'll just assume it involves a description of the system that is sufficient to predict the outcome of dice rolling

No! A Schrodinger wave function does not (cannot) predict the position of an object, but the probability density of its distribution in space. You can't predict the outcome of a dice roll but you can write down the probability density of each possible outcome. Same thing.

[alancalverd]

What I'm saying is that for many physicists, Quantum mechanics is of earth-shaking importance that changes our understanding of what reality might actually be.

I think you are confusing physicists with philosophers. There is nothing earthshaking about quantum mechanics - it is a good model of what happens: the everyday currency of physicists.

100 years ago it was a new model, but since it (a) explained  a lot of things that didn't make sense in a continuum/billiard ball model and (b) predicted a few things we hadn't yet observed, it was just a better way of doing business. Like relativity.   

[alancalverd]

Have a first filter that will collapse the wave function so that the lights polarisation is entirely in the x-axis direction.   Put that through a second filter which permits only light polarised along the y-axis to pass and no light will get through.   However, if you add a third filter turned at 45 degrees between the x- and y-axis and insert that in between the two other filters,   then you will now get some light to pass the last filter.      The measurement of the polarisation along the 45 degree axis has forced a wave function collapse which has now made sure that the wave function is back in a superposition of states for polarisation along the x-axis or y-axis.

Er, no. There being no polarisation along the 45 degree axis (because you said the first filter confined the beam to x polarisation only) the intermediate filter cannot have "measured" anything. What it did was to rotate the plane of polarisation between its input and output. If each filter had measured rather than reformatted the incoming beam, there would be very little loss and the superposition would result in close to 100% transmitted intensity overall.

The system determines the wave function, not the other way around. On proton, one electron, spherical distribution. Two protons, two electrons - a dumbell ¹H₂ molecule  Add a neutron, and it's ³He - spherical again! The wave function has a lot of significance - it helps us predict the behavior of a system. But wave function "collapse" doesn't.

[Eternal Student]

Hi.

We're at risk of hijacking @varsigma 's post entirely.    I'll keep this short.
  .....(Stuff removed to keep it really short....)...

   If you're not happy with light and polarising filters,  then you can do much the same experiments and get the same results using  atoms or electrons (instead of light)  and their spin (instead of polarisation).   For that you replace the filters with Stern-Gerlach apparatus.

Best Wishes.

[alancalverd]

Interesting that the Wikipedia article, when discussing sequential S-G  systems, talks about "measuring" when a particle passes through a nonhomogeneous magnetic field. When we use magnetic fields to select regions for analysis by spin resonance (e.g. MRI) we talk about polarising or forcing, not measuring*.  Wikipedia partially redeems itself with

Given that the input to the second S-G apparatus consisted only of z+, it can be inferred that a S-G apparatus must be altering the states of the particles that pass through it.

(my italics)

which is much more reasonable, and also makes sense if applied to the 45 degree optical polariser.

I may be a bit pedantic in distinguishing between segregating (black sheep to the left, white  to the right) and measuring (counting the sheep in each pen after segregation) but that's the residual chemist in me: qualitative and quantitative analysis are not the same thing. The "triple S-G" experiment simply selects white sheep then arbitrarily paints them red or blue.


*the measurement phase of MRI comes later: we listen to the radiofrequency emission as the selected spins relax and realign to the primary field.