Observer Effect- Photon detectors, how do they work? Split from: Double Slit Experiment

[Itoero]

1 hour ago, Mordred said:

When it comes to the wave-functions and superposition absolutely, the very act of measurement collapses the superposition state.

Those detectors  which are often used in a   Double Slit Experiment are measuring devices and collapse the superposition. This is the observer effect.

Edited February 4, 2019 by Itoero

[Eise]

50 minutes ago, Itoero said:

...and collapse the superposition. This is the observer effect

You say this half an hour after Swansont said that both often are confused?

[Itoero]

19 minutes ago, Eise said:

You say this half an hour after Swansont said that both often are confused?

why confused? My sincere apologies, I didn't read his reply.

 

Edited February 4, 2019 by Itoero

[Itoero]

9 minutes ago, swansont said:

 

No, it's not.

The observer effect can happen to systems that are not in superposition (it's not even required that the experiment be probing QM). Thus, they cannot be the same. The loss of  interference in a double slit can be from the observer effect (e.g. which-path information), but they are not synonymous. You can get which-path information without detection.

Yes but in here they do use detectors and it is about collapsing superposition. When you detect a photon you transform it's energy. Observing a photon changes the photon.

[swansont]

1 hour ago, Itoero said:

Yes but in here they do use detectors and it is about collapsing superposition. When you detect a photon you transform it's energy. Observing a photon changes the photon.

None of that contradicts what I have said. 

There are two things going on: you are collapsing the superposition, and you are destroying the photon. You can do either one, without doing the other. I can block one of the slits, so there is no longer a superposition, but not detect (and thus not destroy) the photon. I can detect (destroy) a photon that was not in a superposition.

 

[Itoero]

21 hours ago, swansont said:

None of that contradicts what I have said. 

There are two things going on: you are collapsing the superposition, and you are destroying the photon. You can do either one, without doing the other. I can block one of the slits, so there is no longer a superposition, but not detect (and thus not destroy) the photon. I can detect (destroy) a photon that was not in a superposition.

 

Yes but it's the observer effect like I said. Destroying a photon(also one that wasn't in superposition) implies transforming it's energy....

[Strange]

You can do the detection of which slit (using another, entangled, photon) after it has passed the slits. So you are not affecting the photon that went through the slits. 

And what does “transforming its energy” mean? The photon that goes through the slits is not changed by the fact it’s entangled partner is observed.  

[swansont]

37 minutes ago, Itoero said:

Yes but it's the observer effect like I said. Destroying a photon(also one that wasn't in superposition) implies transforming it's energy....

They aren't the same thing, even if you find an occurrence where both things are happening at the same time. That's all it is — two things happening at the same time. 

[Itoero]

3 hours ago, swansont said:

They aren't the same thing, even if you find an occurrence where both things are happening at the same time. That's all it is — two things happening at the same time. 

How do those photon detectors work?

When you destroy a photon, doesn't conservation of energy states the energy is transformed?

4 hours ago, Strange said:

You can do the detection of which slit (using another, entangled, photon) after it has passed the slits. So you are not affecting the photon that went through the slits. 

And what does “transforming its energy” mean? The photon that goes through the slits is not changed by the fact it’s entangled partner is observed.   

When you detect a photon you destroy it and conservation of energy states the kinetic/potential energy of the photon transforms. When a photon is detected then you have imo, the photoelectric effect.

[Strange]

10 hours ago, Itoero said:

How do those photon detectors work?

When you destroy a photon, doesn't conservation of energy states the energy is transformed?

When you detect a photon you destroy it and conservation of energy states the kinetic/potential energy of the photon transforms. When a photon is detected then you have imo, the photoelectric effect.

But that has nothing to do with entanglement. You just weren't very clear what you meant.

So the photon that is detected / destroyed is not the one that went the slits. So that second photon is not affected by the observer effect.

[Eise]

16 hours ago, Itoero said:

On 2/4/2019 at 6:34 PM, swansont said:

There are two things going on: you are collapsing the superposition, and you are destroying the photon. You can do either one, without doing the other. I can block one of the slits, so there is no longer a superposition, but not detect (and thus not destroy) the photon. I can detect (destroy) a photon that was not in a superposition.

Yes but it's the observer effect like I said.

It seems you want to use vaguer concepts again.

The 'observer effect' means very generally that measuring something in many cases means changing what you want to measure. A measurement is always a physical interaction with the system you are measuring, and unless you use an indirect way of measuring, using some phenomenon that happens anyway, if you are measuring or not, it implies a change of the system. A few examples:

- measuring the voltage between two points in an electrical circuit: the voltage will decrease slightly, because the voltmeter takes up a small part of the current

- measuring the pressure of a tyre: a bit of air must enter the pressure gauge, so you do not measure the pressure as it was before the measuring.

But essential to these is that one can try to make the effect as small as possible: by using a voltmeter with a high internal resistance (which presupposes that you can make the meter sensitive enough); by using a pressure gauge with as small as possible internal volume. Also, when one knows the physical parameters of the measurement device precisely, one can calculate the effect and so compensate for it.

But you do not have these options in QM: the uncertainty principle is intrinsic to what quantum particles are. It does not describe some principal problem with our measurement methods. It is true that even Heisenberg tried to explain his own uncertainty principle with a variant of the observer effect, but that is history.

The basic error arises from the fact that the wave function is not a physical object in itself:

Wikipedia:

Quote

When discussing the wave function ψ which describes the state of a system in quantum mechanics, one should be cautious of a common misconception that assumes that the wave function ψ amounts to the same thing as the physical object it describes. This flawed concept must then require existence of an external mechanism, such as the mind of a conscious observer, that lies outside the principles governing the time evolution of the wave function ψ, in order to account for the so-called "collapse of the wave function" after a measurement has been performed. But the wave function ψ is not a physical object like, for example, an atom, which has an observable mass, charge and spin, as well as internal degrees of freedom. Instead, ψ is an abstract mathematical function that contains all the statistical information that an observer can obtain from measurements of a given system. In this case, there is no real mystery in that this mathematical form of the wave function ψ must change abruptly after a measurement has been performed.

And last but not least:

Quote

The uncertainty principle has been frequently confused with the observer effect, evidently even by its originator, Werner Heisenberg.

 

[swansont]

14 hours ago, Itoero said:

How do those photon detectors work?

It depends on the detector

Quote

When you destroy a photon, doesn't conservation of energy states the energy is transformed?

Energy is conserved. This is irrelevant to the discussion.

Quote

When you detect a photon you destroy it and conservation of energy states the kinetic/potential energy of the photon transforms. When a photon is detected then you have imo, the photoelectric effect.

Your opinion is also irrelevant. The PEE is a well-defined interaction. It does not cover all photon absorptions. Excitation, for example, is not the PEE.

[Itoero]

On ‎2‎/‎6‎/‎2019 at 7:59 AM, Strange said:

But that has nothing to do with entanglement. You just weren't very clear what you meant.

So the photon that is detected / destroyed is not the one that went the slits. So that second photon is not affected by the observer effect.

Entanglement?

Superposition is not the same as entanglement.

Photons that hit the screen are also subject to the observer effect...

On ‎2‎/‎6‎/‎2019 at 11:46 AM, swansont said:

Energy is conserved. This is irrelevant to the discussion.

It is relevant. When photons are detected or hit the screen there kinetic/potential energy is transformed. A photon is destroyed but its energy is not.

 

On ‎2‎/‎6‎/‎2019 at 11:46 AM, swansont said:

Your opinion is also irrelevant. The PEE is a well-defined interaction. It does not cover all photon absorptions. Excitation, for example, is not the PEE.

Whether it's PEE or not depends on the material off the photon detectors or how  they work.

 

On ‎2‎/‎6‎/‎2019 at 10:47 AM, Eise said:

And last but not least:

Observer effect and HUP are not the same thing but they are very much related. When you measure the position or momentum then you apply a form of force which changes the position or momentum. The observer effect is not necessary only about physics...or about science...

Edited February 9, 2019 by Itoero

[Strange]

45 minutes ago, Itoero said:

It is relevant. When photons are detected or hit the screen there kinetic/potential energy is transformed. A photon is destroyed but its energy is not.

While trivially and obviously true, that has nothing to do with the double slit experiment. (And photons don't have kinetic energy because they are massless.)

47 minutes ago, Itoero said:

Observer effect and HUP are not the same thing but they are very much related.

Nope.

47 minutes ago, Itoero said:

When you measure the position or momentum then you apply a form of force which changes the position or momentum.

Nothing to do with the HUP.

48 minutes ago, Itoero said:

The observer effect is not necessary only about physics...or about science...

Just like your posts!

[swansont]

4 hours ago, Itoero said:

  It is relevant. When photons are detected or hit the screen there kinetic/potential energy is transformed. A photon is destroyed but its energy is not.

Energy is conserved regardless. It doesn’t matter if one worries about the observer effect, or if there is a superposition, or not. Neither of those depend on energy conservation. Energy conservation does not predict if those situations are in play.

4 hours ago, Itoero said:

Whether it's PEE or not depends on the material off the photon detectors or how  they work.

No, it depends on if you have ionized the target. 

4 hours ago, Itoero said:

Observer effect and HUP are not the same thing but they are very much related. When you measure the position or momentum then you apply a form of force which changes the position or momentum. The observer effect is not necessary only about physics...or about science...

They both can refer to posion and momentum, but they are distinct phenomena.

[Itoero]

23 hours ago, Strange said:

While trivially and obviously true, that has nothing to do with the double slit experiment. (And photons don't have kinetic energy because they are massless.)

You are wrong, it's odd that you believe that. Have you never heard about frequency or wavelength of photons? And you can deny it's validity but not its existence...photons have relativistic mass.

 

23 hours ago, Strange said:

Nope.

lol, you are faithing. You talk with your emotions.

 

19 hours ago, swansont said:

Energy is conserved regardless. It doesn’t matter if one worries about the observer effect, or if there is a superposition, or not. Neither of those depend on energy conservation. Energy conservation does not predict if those situations are in play.

Ok, but energy is conserved in the system. Photons in detectors or on the screen don't release their energy in the atmosphere.

 

19 hours ago, swansont said:

They both can refer to posion and momentum, but they are distinct phenomena

Observer effect is a lot more general. If measurement did not  alter  the phenomenon then you could measure momentum and position of one phenomenon as precise as possible and HUP would not exist.

Edited February 10, 2019 by Itoero

[Mordred]

HUP doesn't depend on measurement to be a cause, it is a fundamental constraint  that is present even without the object being measured. 

Edited February 10, 2019 by Mordred

[studiot]

1 hour ago, Itoero said:

Photons in detectors or on the screen don't release their energy in the atmosphere.

Of course they do.

If you shine a light on something is gets slightly warmer as it absorbs photons.
In turn that something warms the air in contact with its outer casing.

Edited February 10, 2019 by studiot

[swansont]

4 hours ago, Itoero said:

Observer effect is a lot more general. If measurement did not  alter  the phenomenon then you could measure momentum and position of one phenomenon as precise as possible and HUP would not exist.

No, this is not true. The measurement effect is NOT the source of the HUP. The latter is a consequence of the variables being Fourier transforms of each other 

 

[Strange]

5 hours ago, Itoero said:

You are wrong, it's odd that you believe that. Have you never heard about frequency or wavelength of photons? And you can deny it's validity but not its existence...photons have relativistic mass.

If by "relativistic mass" you mean energy, then obviously they do have energy. But, once again, you have moved the goalposts. No one denies that photons have energy. But they do not have kinetic energy and they do not have mass. 

5 hours ago, Itoero said:

You talk with your emotions.

Nope.

5 hours ago, Itoero said:

If measurement did not  alter  the phenomenon then you could measure momentum and position of one phenomenon as precise as possible and HUP would not exist.

This is just wrong, as you have been told multiple times.

[Itoero]

 

On ‎2‎/‎10‎/‎2019 at 6:17 PM, Mordred said:

HUP doesn't depend on measurement to be a cause, it is a fundamental constraint  that is present even without the object being measured. 

Observer effect is also a 'fundamental' constraint. Placebo controlled clinical trails are made to deal with the  observer effect. When for example wild animals are studied people make necessary arrangements to prevent animals from changing their behavior because they feel the are observed.https://en.wikipedia.org/wiki/Observer_effect

 

On ‎2‎/‎10‎/‎2019 at 11:06 PM, Strange said:

by "relativistic mass" you mean energy, then obviously they do have energy. But, once again, you have moved the goalposts. No one denies that photons have energy. But they do not have kinetic energy and they do not have mass. 

Why do you deny they have kinetic energy?

The energy of a photon, E (which can be considered as all kinetic energy since the proper energy = E0 = 0 and E = K + E0 = K), is related to the photon's frequency, f, by E = hf where h = Planck's constant = 6.626068 × 10-34m2kg/s.

Reference https://www.physicsforums.com/threads/photon-kinetic-energy.232106/
 

Edited February 12, 2019 by Itoero

[Eise]

1 hour ago, Itoero said:

Observer effect is also a 'fundamental' constraint.

I won't discuss that, but what the essential difference between the general observer effect is, and the HUP. The general observer effect is between physical objects: the object that is measured, and the object that measures. This is not so with the HUP. The wave function is not a physical object as other physical objects. And the HUP follows from the fact already mentioned by Swansont, that e.g. position and momentum are Fourier transformations of each other. It is not so that we have some physical effect on a physical system: their combination is not precise, independent from the fact if we measure it or not. And in not-measuring is definitely not an observer effect.

1 hour ago, Itoero said:

The energy of a photon, E (which can be considered as all kinetic energy since the proper energy = E0 = 0 and E = K + E0 = K), is related to the photon's frequency, f, by E = hf where h = Planck's constant = 6.626068 × 10-34m2kg/s.

You really think that Strange does not know that a photon has an energy according to  E = hf?

The question here is if you can call it kinetic energy. I am not sure, so I let it to him argue about this with you. But at least one difference is that you can make the kinetic energy of moving bodies 0 by slowing it down. You cannot do that with a photon. It always travels at c.

But knowing  that like to subsume as many concepts as possible under the same word, I understand your point... . 

Edited February 12, 2019 by Eise

[Itoero]

2 minutes ago, Eise said:

won't discuss that, but what the essential difference between the general observer effect is, and the HUP. The general observer effect is between physical objects: the object that is measured, and the object that measures. This is not so with the HUP. The wave function is not a physical object as other physical objects. And the HUP follows from the fact already mentioned by Swansont, that e.g. position and momentum are Fourier transformations of each other. It is not so that have some unknown physical effect on a physical system: their combination is not precise, independent from the fact if we measure it or not. And in not-measuring is definitely not an observer effect.

Ok but Hup is about the relationship between 2 measurements of one phenomenon and both measurements are subject to the observer effect. HUP and Observer effect are related but not causal.

 

21 minutes ago, Eise said:

You really think that Strange does not know that a photon has an energy according to  E = hf?

The question here is if you can call it kinetic energy. I am not sure, so I let it to him argue about this with you. But at least one difference is that you can make the kinetic energy of moving bodies 0 by slowing it down. You cannot do that with a photon. It always travels at c.

No, I just don't understand why she denies it's kinetic. Photons are always on the move (at c)and can scatter so it imo has kinetic/potential energy.

[swansont]

52 minutes ago, Itoero said:

Ok but Hup is about the relationship between 2 measurements of one phenomenon and both measurements are subject to the observer effect. HUP and Observer effect are related but not causal.

They are not related. The HUP is an inherent property of nature. The observer effect depends on how you do a measurement.

 

 

[Itoero]

On ‎2‎/‎12‎/‎2019 at 6:13 PM, swansont said:

They are not related. The HUP is an inherent property of nature. The observer effect depends on how you do a measurement. 

 

 

The observer effect and HUP are basically 2 constrains  on how we interpret/observe/study reality/nature. The measurement effect is only when you use measuring devices...the Observer effect in physics is mostly a measurement effect. You have several kinds of observer effects: https://en.wikipedia.org/wiki/Observer_effect   

The Hawthorne effect (also referred to as the observer effect) is a type of reactivity in which individuals (this can be any kind off lifeform) modify an aspect of their behavior in response to their awareness of being observed. This is for example why placebo controlled clinical trials are 'invented'. Or why people often have to hide themselves and use strong lenses when they film wildlife….it's to prevent wildlife from changing its  behavior because it's observed.

Another: In information technology, the observer effect is the impact on the behavior of a computer process caused by the act of observing the process while it is running.

In physics, the observer effect is the theory that simply observing a situation or phenomenon necessarily changes that phenomenon. This is often the result of instruments that, by necessity, alter the state of what they measure in some manner. In the double slit experiment,  the detectors and the screen are the instruments that alter the state of what they measure in some manner.  (like transforming energy and destroying wave behavior.