Why does an observer affect the electron?

[JesseM]

Then I don't understand what we are arguing about, or why you were having problems with what I said originally.

Well, I took it that you had a problem with what I said originally--you quoted me and said 'Again, there is a miconception here as implied by this post there the uncertainly principle is a "measurement" uncertainty, i.e. due to our technique'. Would you agree that if "measurement uncertainty" is defined in the way I am doing it, then the HUP is a form of measurement uncertainty, just one that has to do with fundamental physical laws rather than technique?

 

[ZapperZ]

Well, I took it that you had a problem with what I said originally--you quoted me and said 'Again, there is a miconception here as implied by this post there the uncertainly principle is a "measurement" uncertainty, i.e. due to our technique'. Would you agree that if "measurement uncertainty" is defined in the way I am doing it, then the HUP is a form of measurement uncertainty, just one that has to do with fundamental physical laws rather than technique?

Here's the problem I had with it from the way I understand it. You can have a "measurement uncertainty" even for the SINGLE measurement. But I have been asking how one gets the HUP out of such a thing, to prove that you can't, yet, you can still have a "measurement uncertainty". Even you have mentioned somewhere along the way that measurement uncertainty can contain the HUP plus other uncertainty, haven't you? So how can this and the HUP be the same thing?

The other problem that I have is that I don't think you have illustrated what you mean by such a thing clearly by using it for a particular example. You will notice that I keep trying to illustrate what I meant by using specific example, such as using a single slit case. I'd like to see how you extract such "measurement uncertainty", and how this is also the HUP.

Zz.

 

[JesseM]

Here's the problem I had with it from the way I understand it. You can have a "measurement uncertainty" even for the SINGLE measurement.

Only in the same sense that you have a "probability" for a single event--it's a notion that implicitly depends on the frequency of different outcomes in a large set of identical trials.

But I have been asking how one gets the HUP out of such a thing, to prove that you can't, yet, you can still have a "measurement uncertainty". Even you have mentioned somewhere along the way that measurement uncertainty can contain the HUP plus other uncertainty, haven't you? So how can this and the HUP be the same thing?

I think you're misunderstanding me, I didn't say "measurement uncertainty" is just the same thing as the HUP. Rather, the HUP is a type of measurement uncertainty, but there are also other forms of measurement uncertainty due to purely classical considerations like the resolving power of your measuring apparatus.

The other problem that I have is that I don't think you have illustrated what you mean by such a thing clearly by using it for a particular example. You will notice that I keep trying to illustrate what I meant by using specific example, such as using a single slit case. I'd like to see how you extract such "measurement uncertainty", and how this is also the HUP.

How I extract measurement uncertainty from what? Your example, or any example? If we are talking about the HUP as an example of measurement uncertainty, then you'd illustrate it in exactly the same way that you illustrate the HUP. For example, you could make an inexact momentum measurement followed by a precise position measurement immediately afterwards, and repeat over a near-infinite number of trials; then looking just at the subset of trials where the inexact momentum measurement found the momentum to be within a certain range, you could look at the range of different precise positions found immediately after in this same subset of trials, and see that the range of the momentum vs. the range of position obeys the relation given by the HUP.

If you want to illustrate an example of measurement uncertainty that is not due to the HUP but due to the particular measurement technique you're using, the basic idea would be similar, as I suggested in a previous post:

if you look at large number of trials where an imprecise precision-measuring device found the position to be at a certain pixel, and immediately afterwards a more precise position measurement was made, the uncertainty in the first measurement tells you the range of positions that will be found by the more precise measurement in the entire set of trials, given identical readings by the first device in each trial.

In every case, the idea is that you can determine the uncertainty in a given type of measurement by taking a large number of trials where you made that type of measurement and it gave you a certain answer, then immediately afterwards you made a much more precise measurement, whether of the same variable or a different variable; the "uncertainty" is in your prediction of the result of the second precise measurement, given only knowledge of the result of the first measurement. The wider the range of possible results of the second measurement in a large set of trials where the first measurement gave the same result on each trial, the greater the "measurement uncertainty" associated with the first measurement's ability to predict the value of the variable measured by the second measurement.

 

[ZapperZ]

Only in the same sense that you have a "probability" for a single event--it's a notion that implicitly depends on the frequency of different outcomes in a large set of identical trials.

But how are you to know this after you have performed just ONE set of measurement?

I think you're misunderstanding me, I didn't say "measurement uncertainty" is just the same thing as the HUP. Rather, the HUP is a type of measurement uncertainty, but there are also other forms of measurement uncertainty due to purely classical considerations like the resolving power of your measuring apparatus.

And I say that the HUP is not a type of measurement uncertainty. You can have an absolutely perfect measurement instrument giving you an ideal zero uncertainty, and you can still have the HUP, which isn't really an uncertainty, but rather than inherent spread of a value measured repeatedly. This is why I distinguised the HUP from the "measurement uncertainty". They are not of the same specie. You can improve the accuracy of your measurement of any observable independent of each other. I can make my measurement of the position as accurately as I want, independent of the accuracy of how I determine the momentum of that particle afterwards. They are not related and not coupled together as described by the HUP. But this doesn't mean that I have the ability to predict the outcome of the NEXT measurement, even when I have equipment to accurately determine both position AND momentum.

How I extract measurement uncertainty from what? Your example, or any example? If we are talking about the HUP as an example of measurement uncertainty, then you'd illustrate it in exactly the same way that you illustrate the HUP. For example, you could make an inexact momentum measurement followed by a precise position measurement immediately afterwards, and repeat over a near-infinite number of trials; then looking just at the subset of trials where the inexact momentum measurement found the momentum to be within a certain range, you could look at the range of different precise positions found immediately after in this same subset of trials, and see that the range of the momentum vs. the range of position obeys the relation given by the HUP.

No, in a single measurement, show me how you determine such "measurement uncertainty". Use ANY specific example as you wish. I happened to use the single slit. If you wish, you can use that. Show me exactly where, when single particle passes through the slit, you'd determine all the types of "uncertainty", and how such a thing would lead to the HUP.

If you want to illustrate an example of measurement uncertainty that is not due to the HUP but due to the particular measurement technique you're using, the basic idea would be similar, as I suggested in a previous post: In every case, the idea is that you can determine the uncertainty in a given type of measurement by taking a large number of trials where you made that type of measurement and it gave you a certain answer, then immediately afterwards you made a much more precise measurement, whether of the same variable or a different variable; the "uncertainty" is in your prediction of the result of the second precise measurement, given only knowledge of the result of the first measurement. The wider the range of possible results of the second measurement in a large set of trials where the first measurement gave the same result on each trial, the greater the "measurement uncertainty" associated with the first measurement's ability to predict the value of the variable measured by the second measurement.

But isn't this what I said way early on in this thread? I brought up the statistical nature of the outcome and clearly stated that the SPREAD in the value of the observable (and NOT the value of the uncertainty of a single outcome) is what goes into the HUP.

I make a position measurement $x_1$, and in that single measurement, I have a measurement uncertainty $\delta(x_1)$. I put it to you that this is not equal to, nor is what goes into the HUP, i.e. $\Delta(x_1)$.

I then make several more measurement, giving me $x_2, x_3, x_4,...$, each of them giving me a measurement uncertainty $\delta(x_1), \delta(x_2), \delta(x_3), ...$

Now what exactly is involved when I have to find $\Delta(x)$ that goes into the HUP? The $\delta 's$? Nope! It is all the values of $x_2, x_3, x_4,...$. This is where you get the average value and where the notion of an "average" value of anything is meaningful. That is why I said that even when you have an ideal measuring device that gives you ZERO measurement uncertainty, you will still have a spread in ALL of your measurement (not in a single measurement since we have zero measurement uncertainty in our ideal setup). It is this spread that is intrinsic to QM and consequently, intrinsic to our world.

Zz.

 

[JesseM]

But how are you to know this after you have performed just ONE set of measurement?

You can't, nor have I every claimed that you could. So why do you keep insisting on this condition, when it is not part of my definition of "measurement uncertainty"?

And I say that the HUP is not a type of measurement uncertainty.

Using your definition of "measurement uncertainty" or mine?

You can have an absolutely perfect measurement instrument giving you an ideal zero uncertainty

Not if you're talking about the uncertainty in two variables like position and momentum. I agree that you can measure either one individually with arbitrary precision, but I said before that my definition of "measurement uncertainty" can apply to pairs of variables rather than single variables.

you have and you can still have the HUP, which isn't really an uncertainty, but rather than inherent spread of a value measured repeatedly.

Of course it's an uncertainty, because you are uncertain about what the result of a precise measurement of variable #2 will be, given only knowledge of the result of a measurement of variable #1. Likewise, even if you measure variable #1 with great precision, if you immediately measure variable #2 and then immediately measure variable #1 again, you are uncertain about what the result of the second measurement of variable #1 will be, even if the time interval between the successive measurements is made arbitrarily small, which in classical mechanics would imply that the difference between the two measurements of the same variable should become arbitrarily small as well (on the other hand, if you measured variable #1 with great precision twice in succession without measuring variable #2 in between, then even in QM your uncertainty about the second measurement can be made as small as you wish).

This is why I distinguised the HUP from the "measurement uncertainty". They are not of the same specie.

If you use your definition, maybe not. If you use my definition, it's true that "measurement uncertainty due to equipment" is not of the same species as "measurement uncertainty due to HUP", but they are both members of the same genus, "measurement uncertainty". I don't see why you act as if there is something incoherent about my definition, or why your definition is the only one possible. Words and terms have no intrinsic meaning, if I wished I could use the term "measurement uncertainty" to mean the same thing as "tree", in which case a birch tree would be a type of "measurement uncertainty". As I said before, there is the sociological question of whether my use of "measurement uncertainty" is totally at odds with the way the term is used by scientists, but I did quote that abstract which suggests at least some scientists would describe the HUP as a form of "measurement imprecision", which is essentially the same term.

You can improve the accuracy of your measurement of any observable independent of each other. I can make my measurement of the position as accurately as I want, independent of the accuracy of how I determine the momentum of that particle afterwards. They are not related and not coupled together as described by the HUP. But this doesn't mean that I have the ability to predict the outcome of the NEXT measurement, even when I have equipment to accurately determine both position AND momentum.

But if we are to describe ordinary classical measurement uncertainty in operational terms (as opposed to taking a God's-eye-view where we always know the 'true' value of the position and can compare it to the result of a position measurement with a given apparatus), we must also talk about pairs of measurements, not single measurements. To determine the measurement uncertainty in the readings of a particular device, we need to have a more precise way of measuring the same variable, so that we can take two measurements in quick succession, the first using the device and the second using the more precise form of measurement. That way, we can look at the subset of trials where the device gave a certain reading, and then look at the spread of values of the more precise measurement among these trials--that will tell us the "measurement uncertainty" in the readings of the first device.

If you disagree, please explain how we can quantify plain old classical measurement uncertainty of an imprecise measuring-device, using only a SINGLE measurement with that device alone.

No, in a single measurement, show me how you determine such "measurement uncertainty".

Why do you think I need to show this, when it is not part of my definition of "measurement uncertainty" that it only applies to SINGLE measurements? That may be part of your definition, but it isn't part of mine.

But isn't this what I said way early on in this thread? I brought up the statistical nature of the outcome and clearly stated that the SPREAD in the value of the observable (and NOT the value of the uncertainty of a single outcome) is what goes into the HUP.

I don't understand what you mean by "NOT the value of the uncertainty of a single outcome" here. The value of the "uncertainty" of a single outcome also depends implicitly on the notion of a spread over multiple trials, just like the value of the "probability" of a single outcome depends implicitly on the frequency of that outcome in multiple trials. This is true of classical measurement uncertainty too, the "uncertainty" in a single reading of a given device corresponds to the spread of more precise values for that variable (whether measured by a more precise method immediately afterwards, or taking a God's-eye-view where the precise value of that variable is just known) in a large set of trials where the device returned that same reading. Of course, the HUP differs in that it's a type of uncertainty that only arises when you try to measure two different variables simultaneously (or in the limit as the time between measurements of different variables goes to zero), not a single variable; and, of course, it's also different in that it arises from fundamental laws of physics rather than the particular details of your measuring-device. But my definition of "measurement uncertainty" would subsume both types of uncertainty.

I make a position measurement $x_1$, and in that single measurement, I have a measurement uncertainty $\delta(x_1)$. I put it to you that this is not equal to, nor is what goes into the HUP, i.e. $\Delta(x_1)$.

I agree, but then nothing in what I have said implies they must be equal, they are two very different types of "measurement uncertainty" according to my definition.

Now what exactly is involved when I have to find $\Delta(x)$ that goes into the HUP? The $\delta 's$? Nope! It is all the values of $x_2, x_3, x_4,...$.

Likewise, when you talk about the classical uncertainty $\delta(x)$ due to the imprecision of your equipment, according to a frequentist you can only really undestand the meaning of this by looking at a large set of trials where your equipment returned that same reading, and then considering the more precise value of the position $x_1, x_2, x_3, ...$ on each of these trials (whether obtained by a more precise device which made a measurement immediately afterwards, or simply assuming we have omniscient knowledge of what the 'true' position was on each measurement). Without this implicit understanding the notion of classical "measurement uncertainty" would be meaningless, according to a frequentist.

 

[ZapperZ]

Then I give up. I have no clue what we are arguing about anymore.

Zz.

 

[quddusaliquddus]

lol ... seems my seemingly innocent question has created quiet a stir in the PF community! :D

 

[Dr.Brain]

Actually you can measure the position of an electron by bouncing light off it. As explained on this page (in the section 'Watching Electrons in the Double-Slit Experiment'), to resolve the electron's position with greater accuracy, you need to use light with a smaller wavelength, which means the photons will have more momentum (using DeBroglie's formula for the relationship between wavelength and momentum), and thus can impart more of their momentum to the electron. It turns out that the interference pattern is destroyed if the uncertainty in each electron's momentum is too large, so there's a minimum wavelength of light you can shine on the electrons and still get an interference pattern. When you actually calculate this minimum wavelength using the uncertainty principle, it turns out to be exactly equal to the distance between the slits...but to actually know which of the two slits it went through, you'd need a wavelength smaller than the distance between the two slits!

can pls osmeone tell me how can we mathematically prove that minimum wavelength of photon = distance between slits

 

[masudr]

Classical optics: given the Fraunhofer condition, the intensity distribution is the Fourier transform of the transmission function at the slit. What you will find is that if the wavelength of the light is comparable to the slit size then you will get a diffraction pattern.

Quantum mechanics: given that the slit determines the position of the particle, and the detection on the screen measures the momentum of the particle as it left the slit, we relate the two via a Fourier transform (see the Marcella paper that ZapperZ has quoted over and over). Again you will find is that if the wavelength of the light is comparable to the slit size then you will get a diffraction pattern.

Hey presto, it's the same answer.

 

[Farsight]

quddusaliquddus: I'll try to give you a mental picture:

In quantum physics (i.e. the double slit experiment with electron), why does the mere act of observing the electron affect the fact of whether its a wave or particle?

It doesn't.

An electron isn't something with a solid surface or a particular size. It's more like an electric standing wave. Imagine a pebble thrown into a pond. Plop! Now immediately freeze the image of the ripples, with a big peak in the centre. That's kind of what your electron usually looks like. Like a particle.

But fire it out of a gun and it spreads out like rubber onion rings. It now looks more like the ripples on the pond ten seconds after the stone went plop. Definitely like a wave. And remember there's no solid surface to it. It definitely isn't all in one place any more, and so it can be in two places at once.

But if any part of it snares on a detector or target screen and it stops, it snaps back into its original configuration and looks like a particle again.

 

[quantumcarl]

But if any part of it (an electron) snares on a detector or target screen and it stops, it snaps back into its original configuration and looks like a particle again.

Would the retina of the observer's eye be considered to be a target screen or detector?

If an electron is only a wave until it stops at your retina (and is then interpreted by the optic nerve and visual cortex as a solid) is this like saying that water isn't wet until you feel it?

 

[Farsight]

Would the retina of the observer's eye be considered to be a target screen or detector?

Yep.

If an electron is only a wave until it stops at your retina (and is then interpreted by the optic nerve and visual cortex as a solid) is this like saying that water isn't wet until you feel it?

Nope.

It's more like the situation where you close your eyes and feel a magnet's repulsion with another magnet. It feels kinda like there's something there, even though you ain't touching anything solid.

 

[masudr]

If an electron is only a wave until it stops at your retina (and is then interpreted by the optic nerve and visual cortex as a solid) is this like saying that water isn't wet until you feel it?

Electrons don't generally get to your retina. Photons do, though.

And, how much water does it take for it to feel wet?

 

[quantumcarl]

Yep.Nope.

It's more like the situation where you close your eyes and feel a magnet's repulsion with another magnet. It feels kinda like there's something there, even though you ain't touching anything solid.

Actually I don't know if I'm right to equate a retina with a detctor screen.

In the case of the detector... the observer observes the results of the electron becoming solid as it comes into contact with the screen. The observer's retina... as masudr points out, cannot even detect an electron on its own. Or, at least, we have not trained ourselves well enough to detect and electron hitting our retina.

So, with this being the case... it is the interpretation of the state of the electron given by the detector screen that the observer's eye is utilized to interpret.

So here we see that the eye is second party to the results of the wave collapsing (or whatever it does) and becoming, or behaving, like a solid.

In the case of a photon... again as masudr mentioned, I can see where the analogy of the eye being a detector plate or screen... because it is able to distinquish when the spectrum of em waves that is light is hitting it.

"Nature does nothing better than entertaining us." I must say.

 

[curious ron]

Ok, i have question. The double slit experiment is carried out in a vacuum so that foreign particles don't get in the way. So what about the so called quantum foam, particles and anti-particles appearing and disappearing, why do they not interference with the electron? or do they wipe each other out too quickly to make a difference.

 

[Farsight]

I for one don't know, ron. But this Lamb Shift on a bound electron is pretty interesting.

http://www.llnl.gov/str/May06/Beiersdorfer.html

"The Lamb shift is a tiny difference in the energy of an ion’s electron between two quantum states that are otherwise identical except for the shape of the electron’s orbit around the nucleus. Because of this dissimilarity in the electron’s orbit in the two states, the electron interacts slightly differently with virtual particles—photons, positrons, and electrons—that appear and disappear in quantum fluctuations. The resulting energy difference, called the Lamb shift, can be measured when the electron jumps from one energy level to another..."

 

[Norman Albers]

How close do the slits have to be to get notable interference? Is this momentum-dependent? . . . . Results of my wave packet study: when you Fourier analyze the Gaussian packet, you do get a delta function in k at the propagation value. what you also get is a nice smooth spread in transverse stuff a la $$e^-(k_T^2)$$

 

[RandallB]

question. The double slit experiment is carried out in a vacuum so ... particles don't get in the way.
So what about the so called quantum foam, ...

The double slit is not always done in a vacuum, Light experiments do not use a vacuum, and even if glass is put in front of one, or the other, or both slits the pattern is still displayed.
(Note that since the glass is transparent no information is gained from it to say which slit was used either.)
So in the case of an electron one could assume that whatever ‘quantum foam’ is; it must be ‘transparent’ to the electron as far as the slit experiment is concerned.

 

[spinfusion]

Returning to the central question

Most laymen are interested in this question because they want to know whether or not the universe is fundamentally deterministic. I am a laymen with background in philosophy and debate, and it is sometimes frustrating to read scientific types debating this question, because they often miss each others’ arguments, use idiosyncratic definitions, shift ground, etc.

Essentially, I agree with JesseM. HUP is a measurement uncertainty built into the laws of physics for simultaneously knowing momentum and location of quantum particles, because “observation” alters one if not both of the measured properties.

I see no philosophical reason to conclude from this impossibility of simultaneous measurement that matter is fundamentally probabilistic. On the other hand, I would not want scientists to complicate their formulas by positing a fundamentally unknowable exact location and momentum for quantum particles (waves, whatever).

I can’t pretend to answer the determinism question. But I have my theory, and I’m trying to reconcile it with the facts of this dual-slit electron experiment.

Here’s how I understand the indeterminacy crowd’s explanation of the slit anomaly: The electron alters its state from indeterminate to determinate upon coming into contact with matter, photons, etc. In the absence of these constraints, it spreads in a probability wave.

Since I don’t believe in indeterminacy, I must believe that there is a wave-to-particle, deterministic explanation for why no interference pattern is produced when one slit is observed.

I can visualize why one slit does not produce a spread, and why two slits produces an interference pattern. But I don’t understand why one slit observed and one unobserved doesn’t produce at least some interference. Shouldn’t the electron begins spreading into waves again immediately after contact with the photons? Even if their momentum or frequency is different than the unobserved slit’s electrons, shouldn’t there still be interference?

Thanks in advance for your answers.

 

[spinfusion]

Update:

I began reading up on QM in the almighty Wikipedia. The debate between Einstein and Bohr was fascinating.

All my reading has not convinced me that any scientific data compels acceptance of the indeterminacy school. However, I have learned to accept the impossibility of visualizing quantum phenomenon, which I now realize have nothing to do with my "visual" understanding of time, space, and locality.

Wow, what a universe.

 

[ZapperZ]

Most laymen are interested in this question because they want to know whether or not the universe is fundamentally deterministic. I am a laymen with background in philosophy and debate, and it is sometimes frustrating to read scientific types debating this question, because they often miss each others’ arguments, use idiosyncratic definitions, shift ground, etc.

You are forgetting that when scientists debates, we base our definition and principles on well-defined, unambiguous underlying mathematical description. There's nothing "idosyncratic" here. It is up to you to make the effort in understanding the mathematics to be able to figure out what is being talked about.

For example:

I can visualize why one slit does not produce a spread, and why two slits produces an interference pattern.

This is wrong. There IS a spread when electrons pass through ONE slit - it's called diffraction, something that I have used to illustrate the HUP. You have confused and mixed two different aspect of quantum mechanics - the HUP and the superposition principle. One can use the single slit diffraction (which I have many times) as a manifestation of the HUP, but the double slit is a manifestation of superposition, in this case, superposition of paths!

You need to be able to understand the physics involved here, first and foremost, before coming up with a "theory".

Zz.

 

[wavemaster]

Though I'm aware of HUP is talking about standard deviation, I'm curious about one thing. So I'd like to raise this question: in what precision can we measure position and momentum of one, single electron. Feel free to talk in terms of eigenstates and wavefunction collapse.

 

[selfAdjoint]

Though I'm aware of HUP is talking about standard deviation, I'm curious about one thing. So I'd like to raise this question: in what precision can we measure position and momentum of one, single electron. Feel free to talk in terms of eigenstates and wavefunction collapse.

See https://www.physicsforums.com/showthread.php?p=1044897#post1044897" discussion by ZapperZ and others about that issue

 

[Prince of Quarkness]

But this doesn't mean that I have the ability to predict the outcome of the NEXT measurement, even when I have equipment to accurately determine both position AND momentum.

But isn't there a point to be made here that position and momentum do not commute, ie: are not compatible, and so there's no meaning to saying that 'particle A has position (x,y,z) and momentum (Px, Py,Pz) at time t'.

Now that isn't _quite_ the HUP, since the (necessarily statistical) concept of Uncertainty isn't used.

But surely it does still represent a fundamental incompatibility between position and momentum, which isn't present in classical theory? So although it's a mislabelling to call it the HUP, the HUP is a result of that non-commuting property and so the terminological mistake isn't all that heinous.

-Dave

 

[ZapperZ]

But isn't there a point to be made here that position and momentum do not commute, ie: are not compatible, and so there's no meaning to saying that 'particle A has position (x,y,z) and momentum (Px, Py,Pz) at time t'.

But not commuting does mean you can't measure them with some degree of accuracy. Again, I will point out to the single-slit scenario. The only thing that affects the accuracy of my measurement of the position of the particle is the slit width, and the only that that affects the accuracy of my measurement of the momentum after it passes through the slit is the pixel size of the CCD.

However, since the position and momentum do not commute, my ability to tell you the momentum of the particle that passes through the slit depends intimately on the slit size. The smaller I make the slit size (I know the position even more), the less certain I can tell you its transverse momentum. Why is this not the HUP?

Zz.

 

[selfAdjoint]

Zapper, the single slit is clear in these discussions but explanation of it it not sufficient to pin down the phenomena around position/momentum non-commutativity and uncertainty. See the discussion and links on the "Delayed Choice Quantum Eraser" in the thread on Brian Greene's Beam Splitter Experiments.

 

[ZapperZ]

Zapper, the single slit is clear in these discussions but explanation of it it not sufficient to pin down the phenomena around position/momentum non-commutativity and uncertainty. See the discussion and links on the "Delayed Choice Quantum Eraser" in the thread on Brian Greene's Beam Splitter Experiments.

I have been following that, but I don't see the relevance here.

Maybe everyone is seeing it from the perpective of the THEORY and trying to make physical sense out of it. I don't. I approach it from the IDEAL measurement/experimental angle. How would I measure the x-position of a particle, and then measure the x-momentum of the particle? All I see here is that the accuracy of my position is dictated by the size of the slit, and the accuracy of the momentum is dictated by how well I can determine where the particle hit my detector. I know this can be done because I've done it! Now the accuracy of my prediction of the momentum is an entirely different beast.

Now, is there an "explanation" to such a thing beyond the QM description? How does one do that without delving into the various flavors of QM formulation and interpretation, which inevitably boils down, at least for now, to a matter of tastes?

Zz.

 

[Prince of Quarkness]

However, since the position and momentum do not commute, my ability to tell you the momentum of the particle that passes through the slit depends intimately on the slit size. The smaller I make the slit size (I know the position even more), the less certain I can tell you its transverse momentum. Why is this not the HUP?

I must be getting confused about the flow of the debate.

I thought that you had asserted that the HUP is meaningful only when considering groups of particles, when here you've just said that if you measure the position of a single particle more accurately, you are less certain of it's momentum.

I think what I've missed is that you're saying that the HUP doesn't reduce the accuracy of measurements, simply the accuracy of _predicting_ the results of, say, the momentum of a particle whose position has been measured.

i.e: Would I be right in saying that your issue here is that people are conflating the width of a gaussian wavefunction in momentum space with the width of the gaussian produced by experimental accuracies?

Would I be re-wording your own statements if I said that an exact measurement of position is entirely possible within the theory, but it is subsequently impossible to _predict_ the momentum, though it can be _measured_ with arbitrary precision?

-Dave

 

[ZapperZ]

I must be getting confused about the flow of the debate.

I thought that you had asserted that the HUP is meaningful only when considering groups of particles, when here you've just said that if you measure the position of a single particle more accurately, you are less certain of it's momentum.

I think what I've missed is that you're saying that the HUP doesn't reduce the accuracy of measurements, simply the accuracy of _predicting_ the results of, say, the momentum of a particle whose position has been measured.

i.e: Would I be right in saying that your issue here is that people are conflating the width of a gaussian wavefunction in momentum space with the width of the gaussian produced by experimental accuracies?

Would I be re-wording your own statements if I said that an exact measurement of position is entirely possible within the theory, but it is subsequently impossible to _predict_ the momentum, though it can be _measured_ with arbitrary precision?

-Dave

In the other thread, I stated that there is a difference between measuring "x" and "p", and measuring $\Delta(x)$ and $\Delta(p)$. The latter is what is contained in the HUP. There's no ambiguity or even confusion here, correct?

Next, each of the Delta's require a statistical averaging. It includes the square of the average value, and the average of a square value. This explicitly implies a statistical ensemble. OK so far?

If things so far have not caused anyone to have any constipation, then I don't see why what I have said earlier on the HUP would rouse any curiosity. Because if we buy what the mathematical description of the HUP has to say, then we know that

the HUP is a statistical experession of how well we know about the values of a pair of non-commuting observables when we know one of them to a particular certainty.

I make a measurement of x. The uncertainty is the width of the slit, let's say, which is $\Delta(x)$. Now, when I let it hit my detector, it will make a spot of a finite size. Since the location determines the momentum (such as that used in angle-resolved photoemission spectroscopy), then the CENTER of the spot is p, but the uncertainty in p is roughly the width of the spot. HOwever, and this is very important, this is NOT the uncertainty $\Delta(p)$ that is in the HUP. Why? Because if I make my CCD and detection better, I could get a cleaner signal (that's what people do sometime, by cooling the detector to LHe temperature to reduce thermal fluctuation). So already we know that the instrument uncertainty can be reduced INDEPENDENTLY of the $\Delta(x)$. This doesn't smell or look like the HUP, and it isn't!

So where is $\Delta(p)$? You make repeated measurement of the identical system. Shoot another, and another, and another, of the same particle prepared identically. Since your slit width doesn't change, your $\Delta(x)$ remains the same. However, the value of p that you measure may not be identical. In fact, if you make the slit small enough, the value of p will scatter all over the place! If you collect enough sampling of the values of all these p's, you will find not only the average value, but also a spread in the statistical variance of this value. This is the $\Delta(p)$ in the HUP!

If you apply this to what we know about statistics, the larger the value of $\Delta(p)$, then the less are we able to predict with a reasonable accuracy the value of p that we will get when we shoot the next identical particle. In fact, go to the extreme where the slit width is a delta function and you'll get a flat distribution of the value of p, which means your $\Delta(p)$ is infinite. The next value of p can attain any value imaginable.

This relationship is consistent with what we know of the HUP between two non-commuting observable.

Not sure if I've explained myself clearly enough, but I hope that is the last time I have to do that.

Zz.

 

[Prince of Quarkness]

Thanks for explaining that, and I'm now sorry to have put you to the trouble of saying it again. I quite agree with your stance.

-Dave

 

[ZapperZ]

Thanks for explaining that, and I'm now sorry to have put you to the trouble of saying it again. I quite agree with your stance.

-Dave

Well, if it makes just one person to finally understand what I was trying to get across, it was well worth it.

:)

Zz.

 

[gptejms]

Zz,tell me if
i) a single electron has a well defined x & p in your scheme of things?

ii)the wavefunction of a single particle has any meaning from your POV?--I mean do you believe that the wavefunction describes a multitude of particles or that it corresponds to a single particle?

 

[ZapperZ]

Zz,tell me if
i) a single electron has a well defined x & p in your scheme of things?

Before those are measured, no.

After they are measured, yes.

ii)the wavefunction of a single particle has any meaning from your POV?--I mean do you believe that the wavefunction describes a multitude of particles or that it corresponds to a single particle?

Note that in QM, you don't just write the wavefunction of a SINGLE particle - you have to know the potential geometry that it is in. I happen to know quite a bit about "single-particle" formalism, because in many-body systems, you reduce one intractable many-body problem into many one-body problem. This is where you get the single-particle spectral function.

So yes, you can write the "wavefunction" of one particle, or the wavefunction of a number of particles (assuming you have the ability to solve the Hamiltonian). I don't quite see how that relates to anything I have said.

Furthermore, I'm not so sure why this is my "POV". Can you point out to me which part of what I've said isn't consistent with the QM you get out of your textbooks?

Zz.

 

[Doc Al]

I think gptejms was asking if you thought the wavefunction describes the state of an individual particle or an ensemble of similarly prepared particles. (I don't think he was asking about many-particle systems.)

 

[ZapperZ]

I think gptejms was asking if you thought the wavefunction describes the state of an individual particle or an ensemble of similarly prepared particles. (I don't think he was asking about many-particle systems.)

Then I'm not sure if I understand that, because would that makes a difference?

For example, if I say that the probability of something to occur is 1/2, it means that if I perform it once, I would get a 50% chance of getting something. However, this could also mean that if I perform it on 100 identical system, I'd get half of them in the state that I want.

So is there any difference if it's just one or many, especially when they are non-interacting?

Zz.