The Observer Effect

[studiot]

1 hour ago, Luc Turpin said:

What is LCAO bonding method?

In the double-slit experiment with photons being fired one at the time, is there superposition? if so? when or where?

Genady has answered youquestion about the slits, with   - yes you guessed it  - a linear combination.

 

 

OK so let's get a handle on superposition so we can see move on to see how is can be employed in QM.

Like before I will start with a couple of classical examples which are easier to countenance.

(I hope you have fun translating the English)

 

Superposition basically means 'in the same place'

By 'place' we mean that the effect of process B anywhere that process A exerts an influence.

Linear superposition requires that "The resultant effect of B superposed on A is the sum of A acting alone and B acting alone"

In other words it is a linear combination of A and B.

 

I am being deliberately a bit vague about place to leave it open ended.

Place can be a real region of space or it can be an abstract region of a theoretical mathematical space.

 

Right so real world example 1 from electronics.

So translated into electronic terms the superposition principle becomes

When there is more than one voltage source in a circuit the effect due each (current flows) may be considered as if each were the only source and the resultant current flows obtained by adding the individual flows obtained.

Here is a simple worked example.
I have not given details of the individual current calculations, but only note they themselves are part of linear mathematics via Ohm's Law.

 

 

Right so those are individual spot values, about as simple as you can get.

I hope you realise that some currents are negative because they are going in the opposite direction from the assumed.

 

The next example is from mechanics.

A simple beam carries two loads, its own self weight and a point load in the middle.

This time I have given a formula rather than specific numbers to show that linear combinations can be for more complicated things than simple point values.

Superposition says that the deflection at any point (I am just showing the centre) is the sum of the deflections due to each load acting alone.

 

 

As you can see neither individual deflection is anywhere near linear.

Yet we can place them in a linear combination (( i like your word) package.

 

 

So we come to quantum mechanics.

Our model, the Schrodinger Equation, can be solved explicitly only in a very few cases and these are for atoms.

Most of the material world is made of molecules, which are packages of atoms bonded together by chemical bonds.

Each electron in an atom has its own wavefunction.

These wave functions describe atomic orbitals, where the electrons reside.

 

When two atoms bond they approach each other untill an orbital of one overlaps the space of an orbital form the other atom.

Part of QM tells us that a single orbital can contain zero, one or two electrons.

If both of these overlapping atomic orbitals contain exactly one electron they atoms may bond.

That is a new 'molecular orbital' is formed, called the bonding orbital.

This new orbital has a different wavefunction from either of its constituents.

We are only just beginning to be able to solve one of these molecular orbitals directly with our best computers.

In the meantime we do what every scinetist and engineere does.

They ask themselves will a linear combination do the trick?

Yeha enter KJW's formula (with a small adjustment)  see  1  on my attachment.

Try the orbital of the combination as  the sum of the individual orbitals.

This works directly for a few molecules for example the hydrogen or oxygen molecules, where both atoms are of the same element.

But most molecules are made of a variety of elements.

The nuclei of different elements have different 'pulling power' on electrons so to reflect this I have taken hydrogen and oxygen and added constan coefficients p and q to the linear combination to compensate for this, just as we did with ax + by +  cz before.

So this is the final form of the LCAO method of describing what may be the most important natural process to all of us.

LCAO  means Linear Combination of Atomic Orbitals

and it works pretty well in most cases.

 

 

 

[swansont]

To add to Genady’s response: you don’t know the exact trajectory, so the photon passes through both slits and interferes with itself

[Luc Turpin]

 

29 minutes ago, swansont said:

To add to Genady’s response: you don’t know the exact trajectory, so the photon passes through both slits and interferes with itself

Got that, we do not know the trajectory and interferes with itself, but did I not read in the thread above that in a superposition of state, there is no energy, but a single quanta of energy has, well, energy.  Where did I go wrong. That's why I was talkiing about a single photon being emitted. 

58 minutes ago, studiot said:

Genady has answered youquestion about the slits, with   - yes you guessed it  - a linear combination.

OK so let's get a handle on superposition so we can see move on to see how is can be employed in QM.

 

Great! great! need to assimilate all of this info! I will get back to you sometime today.

Great also on "linear combination"; think I got that.

Again, appreciated! learning a lot! still confused on certain things, and misinterpreting things, but staying focused

[swansont]

1 hour ago, Luc Turpin said:

Got that, we do not know the trajectory and interferes with itself, but did I not read in the thread above that in a superposition of state, there is no energy, but a single quanta of energy has, well, energy

You did read that, but perhaps you saw that it was disputed and no example was given to support the assertion. In this example, half the energy is in each state.

[studiot]

3 hours ago, Genady said:

Yes, superposition of states resulting from a photon passing through different slits.

If the state after passing through slit 1 is ψ1 and the state after passing through slit 2 is ψ2 , then after passing the screen the state of the photon is the superposition, 12√ψ1+12√ψ2 .

12√ψ1+12√ψ2.

 

 

2 hours ago, Luc Turpin said:

Got that, we do not know the trajectory and interferes with itself, but did I not read in the thread above that in a superposition of state, there is no energy, but a single quanta of energy has, well, energy.  Where did I go wrong. That's why I was talkiing about a single photon being emitted.

So can you explain how Genady came by this formula ?

[Luc Turpin]

2 hours ago, swansont said:

You did read that, but perhaps you saw that it was disputed and no example was given to support the assertion. In this example, half the energy is in each state.

Understood! Meaning no energy is lost just shared; Hesitant, but does this have anything to do with the law of conservation of energy?

1 hour ago, studiot said:

 

So can you explain how Genady came by this formula ?

A linear combination (superposition) of both slit one and slit 2?????

 

4 hours ago, studiot said:

Genady has answered youquestion about the slits, with   - yes you guessed it  - a linear combination

 

OK so let's get a handle on superposition so we can see move on to see how is can be employed in QM.

Like before I will start with a couple of classical examples which are easier to countenance.

(I hope you have fun translating the English)

 

Superposition basically means 'in the same place'

By 'place' we mean that the effect of process B anywhere that process A exerts an influence.

Linear superposition requires that "The resultant effect of B superposed on A is the sum of A acting alone and B acting alone"

In other words it is a linear combination of A and B.

 

I am being deliberately a bit vague about place to leave it open ended.

Place can be a real region of space or it can be an abstract region of a theoretical mathematical space.

 

Right so real world example 1 from electronics.

So translated into electronic terms the superposition principle becomes

When there is more than one voltage source in a circuit the effect due each (current flows) may be considered as if each were the only source and the resultant current flows obtained by adding the individual flows obtained.

Here is a simple worked example.
I have not given details of the individual current calculations, but only note they themselves are part of linear mathematics via Ohm's Law.

 

 

Right so those are individual spot values, about as simple as you can get.

I hope you realise that some currents are negative because they are going in the opposite direction from the assumed.

 

The next example is from mechanics.

A simple beam carries two loads, its own self weight and a point load in the middle.

This time I have given a formula rather than specific numbers to show that linear combinations can be for more complicated things than simple point values.

Superposition says that the deflection at any point (I am just showing the centre) is the sum of the deflections due to each load acting alone.

 

 

As you can see neither individual deflection is anywhere near linear.

Yet we can place them in a linear combination (( i like your word) package.

 

 

So we come to quantum mechanics.

Our model, the Schrodinger Equation, can be solved explicitly only in a very few cases and these are for atoms.

Most of the material world is made of molecules, which are packages of atoms bonded together by chemical bonds.

Each electron in an atom has its own wavefunction.

These wave functions describe atomic orbitals, where the electrons reside.

 

When two atoms bond they approach each other untill an orbital of one overlaps the space of an orbital form the other atom.

Part of QM tells us that a single orbital can contain zero, one or two electrons.

If both of these overlapping atomic orbitals contain exactly one electron they atoms may bond.

That is a new 'molecular orbital' is formed, called the bonding orbital.

This new orbital has a different wavefunction from either of its constituents.

We are only just beginning to be able to solve one of these molecular orbitals directly with our best computers.

In the meantime we do what every scinetist and engineere does.

They ask themselves will a linear combination do the trick?

Yeha enter KJW's formula (with a small adjustment)  see  1  on my attachment.

Try the orbital of the combination as  the sum of the individual orbitals.

This works directly for a few molecules for example the hydrogen or oxygen molecules, where both atoms are of the same element.

But most molecules are made of a variety of elements.

The nuclei of different elements have different 'pulling power' on electrons so to reflect this I have taken hydrogen and oxygen and added constan coefficients p and q to the linear combination to compensate for this, just as we did with ax + by +  cz before.

So this is the final form of the LCAO method of describing what may be the most important natural process to all of us.

LCAO  means Linear Combination of Atomic Orbitals

and it works pretty well in most cases.

 

 

Here is my understanding of the above

• Superposition - everywhere A affect-influence B
• electronics - current superposition is linear
• Beam - Total deflection (concentrated, center deflection) is non linear
• QM - LCAO is linear
  • Schrodinger Equation applicable to atoms
  • Each electron has a wavefunction
  • Wave functions describe orbitals with 0, 1, or 2 atoms
  • two atoms with one atom orbital each bond, creating a molecule and new bonding orbital; new orbital different from either two
  • A linear combination is used to calculate new bonding orbital?
  • adding a constant coefficient when nucleis are of different elements

[swansont]

1 hour ago, Luc Turpin said:

Understood! Meaning no energy is lost just shared; Hesitant, but does this have anything to do with the law of conservation of energy?

Yes. There’s no violation of the law when you have a superposition.

[sethoflagos]

34 minutes ago, swansont said:

Yes. There’s no violation of the law when you have a superposition.

But what about the 2nd Law?

neutron => 1/2 neutron + 1/2 (proton + bits) => neutron 

... might be a bit problematic.

(though I'm sure I'm not the first to ask this question and somehow the problem must 'go away')

Edited December 15, 2023 by sethoflagos

[swansont]

51 minutes ago, sethoflagos said:

But what about the 2nd Law?

neutron => 1/2 neutron + 1/2 (proton + bits) => neutron 

... might be a bit problematic.

(though I'm sure I'm not the first to ask this question and somehow the problem must 'go away')

What, precisely, is the problem?

[sethoflagos]

35 minutes ago, swansont said:

What, precisely, is the problem?

I'm sure you'll be able to enlighten me why there isn't one. However, on the face of it, the superposition contains on average more particles; many more ways to distribute momentum and therefore, one might think, a higher total entropy.  

[studiot]

2 hours ago, Luc Turpin said:

Understood! Meaning no energy is lost just shared; Hesitant, but does this have anything to do with the law of conservation of energy?

A linear combination (superposition) of both slit one and slit 2?????

 

Here is my understanding of the above

• Superposition - everywhere A affect-influence B
• electronics - current superposition is linear
• Beam - Total deflection (concentrated, center deflection) is non linear
• QM - LCAO is linear
  • Schrodinger Equation applicable to atoms
  • Each electron has a wavefunction
  • Wave functions describe orbitals with 0, 1, or 2 atoms
  • two atoms with one atom orbital each bond, creating a molecule and new bonding orbital; new orbital different from either two
  • A linear combination is used to calculate new bonding orbital?
  • adding a constant coefficient when nucleis are of different elements

This is a really good time to review.

 

A linear combination (superposition) of both slit one and slit 2?????

 

Genady's formula is a linear combination of two fators.

The effect of slit A and the effect of slit B

That is the combination of the wave function of slit A divided by root2 and the wavefunction of slit B divided by root2

That is the new wavefunction after the barrier with the slits.

The root2 occurs because, if you look back several members have noted that the probability of that wavefunction is proportional to the square of the wavefunction, not the wavefunction itself.

It is observed that exactly 50% of the light passes through each slit which means that the probability of each is 1/2 (swasont has noted that probabilities must add up to 1)  so when we take the square root of psi squared/2 we get psi/√2

 

This was also shown on the slits experiment attachment I posted as the pattern built up particle by particle.

Superposition - everywhere A affect-influence B

Yes

electronics - current superposition is linear

Yes, but both the defining equation (ohms law) and the linear combination are linear.

Beam - Total deflection (concentrated, center deflection) is non linear

Not quite.
The formula the deflection due to neither load is linear, but these formulae can be used in a linear combination.

LCAO is linear

yes

Schrodinger Equation applicable to atom

Yes and to other particles.

Each electron has a wavefunction

yes

Wave functions describe orbitals with 0, 1, or 2 atoms

not atoms, electrons in orbitals.

two atoms with one atom orbital each bond, creating a molecule and new bonding orbital; new orbital different from either two

Again electrons in orbitals not atoms.

A linear combination is used to calculate new bonding orbital?

Yes

adding a constant coefficient when nucleis are of different elements

Yes but two constants, one for each nuclues and therefore each wavefunction  (look back at my attachment)

 

14 minutes ago, sethoflagos said:

I'm sure you'll be able to enlighten me why there isn't one. However, on the face of it, the superposition contains on average more particles; many more ways to distribute momentum and therefore, one might think, a higher total entropy.  

Surely entropy only applies to material things not 'potential things' ?

[Genady]

35 minutes ago, studiot said:

It is observed that exactly 50% of the light passes through each slit which means that the probability of each is 1/2 (swasont has noted that probabilities must add up to 1)  so when we take the square root of psi squared/2 we get psi/√2

However I don't know why it is \(\frac 1 {\sqrt 2}\psi_1 +\frac 1 {\sqrt 2}\psi_2\) rather than \(\frac 1 {\sqrt 2}\psi_1 -\frac 1 {\sqrt 2}\psi_2\). Any suggestions?

[swansont]

43 minutes ago, sethoflagos said:

I'm sure you'll be able to enlighten me why there isn't one. However, on the face of it, the superposition contains on average more particles; many more ways to distribute momentum and therefore, one might think, a higher total entropy.  

Sure. But the superposition also includes the low-entropy state, so the average of the two (weighted by their amplitudes) is less than the entropy of the final state, so there’s no conflict with the 2nd law.

[sethoflagos]

1 minute ago, swansont said:

Sure. But the superposition also includes the low-entropy state, so the average of the two (weighted by their amplitudes) is less than the entropy of the final state, so there’s no conflict with the 2nd law.

Which entropy state is lower than that of the incoming (or more pertinently, the outgoing) neutron?

There's clearly no issue with the decay path: it's the 'nearly decayed' back to 'as you were' path that's doing my head in.

[Luc Turpin]

1 hour ago, studiot said:

This is a really good time to review.

 

A linear combination (superposition) of both slit one and slit 2?????

 

Genady's formula is a linear combination of two fators.

The effect of slit A and the effect of slit B

That is the combination of the wave function of slit A divided by root2 and the wavefunction of slit B divided by root2

That is the new wavefunction after the barrier with the slits.

The root2 occurs because, if you look back several members have noted that the probability of that wavefunction is proportional to the square of the wavefunction, not the wavefunction itself.

It is observed that exactly 50% of the light passes through each slit which means that the probability of each is 1/2 (swasont has noted that probabilities must add up to 1)  so when we take the square root of psi squared/2 we get psi/√2

 

This was also shown on the slits experiment attachment I posted as the pattern built up particle by particle.

Superposition - everywhere A affect-influence B

Yes

electronics - current superposition is linear

Yes, but both the defining equation (ohms law) and the linear combination are linear.

Beam - Total deflection (concentrated, center deflection) is non linear

Not quite.
The formula the deflection due to neither load is linear, but these formulae can be used in a linear combination.

LCAO is linear

yes

Schrodinger Equation applicable to atom

Yes and to other particles.

Each electron has a wavefunction

yes

Wave functions describe orbitals with 0, 1, or 2 atoms

not atoms, electrons in orbitals.

two atoms with one atom orbital each bond, creating a molecule and new bonding orbital; new orbital different from either two

Again electrons in orbitals not atoms.

A linear combination is used to calculate new bonding orbital?

Yes

adding a constant coefficient when nucleis are of different elements

Yes but two constants, one for each nuclues and therefore each wavefunction  (look back at my attachment)

 

Surely entropy only applies to material things not 'potential things' ?

Meant electrons not atoms for orbital bond; was not paying attention

Understanding is much better; implications and what relates to what will take time.

Yes, two constants one for each nuclues and therefore each wavefunction; got it!

Need to put into practice what I learnt to get more acquainted with it.

[swansont]

2 hours ago, sethoflagos said:

Which entropy state is lower than that of the incoming (or more pertinently, the outgoing) neutron?

There's clearly no issue with the decay path: it's the 'nearly decayed' back to 'as you were' path that's doing my head in.

There is no outgoing neutron

[sethoflagos]

34 minutes ago, swansont said:

There is no outgoing neutron

We break the superposition by 'observing' and see a neutron?

[Genady]

13 hours ago, Genady said:

However I don't know why it is 12√ψ1+12√ψ2 rather than 12√ψ1−12√ψ2 . Any suggestions?

I've figured out the answer to my question above.

It is 12√ψ1+12√ψ2 rather than 12√ψ1−12√ψ2 because when the two wave functions are in phase they interfere constructively, and this requires the + rather than − in their superposition expression.

Edited December 16, 2023 by Genady

[Luc Turpin]

3 minutes ago, Genady said:

I've figured the answer to my question above.

It is 12√ψ1+12√ψ2 rather than 12√ψ1−12√ψ2 because when the two wave functions are in phase they interfere constructively, and this requires the + rather than − in their superposition expression.

I am getting better at following your reasoning, which is very sound.

[StringJunky]

50 minutes ago, Luc Turpin said:

I am getting better at following your reasoning, which is very sound.

You are in good hands with the people currently contributing.

[Luc Turpin]

3 minutes ago, StringJunky said:

You are in good hands with the people currently contributing.

And so gratefull for it

[swansont]

11 hours ago, sethoflagos said:

We break the superposition by 'observing' and see a neutron?

How does this happen?

How was the superposition created in the first place?

Does it even make sense to talk about the second law for such a system?

[sethoflagos]

46 minutes ago, swansont said:

How does this happen?

How was the superposition created in the first place?

Does it even make sense to talk about the second law for such a system?

Not the foggiest. Hence points 1) to 5) of my earlier post that you seem to have no interest in.

[studiot]

2 hours ago, Genady said:

I've figured out the answer to my question above.

It is 12√ψ1+12√ψ2 rather than 12√ψ1−12√ψ2 because when the two wave functions are in phase they interfere constructively, and this requires the + rather than − in their superposition expression.

Agreed, I'm glad you figured it.

2 hours ago, Luc Turpin said:

I am getting better at following your reasoning, which is very sound.

 

Basically you have what you need for this topic in all the back posts now.

Have a happy Christmas digesting it.

2 hours ago, Genady said:

I've figured out the answer to my question above.

It is 12√ψ1+12√ψ2 rather than 12√ψ1−12√ψ2 because when the two wave functions are in phase they interfere constructively, and this requires the + rather than − in their superposition expression.

Agreed. I'm glad you figured it out.

 

2 hours ago, Luc Turpin said:

I am getting better at following your reasoning, which is very sound.

Just to emphasis a couple of points.

 

The Schrodinger eqaution is about particles, which is why we need to work in terms of photons.

The solution, psi has some wavelike features.

 

In the case of the slits;

Between the source and the barrier with the slits psi looks the same and exhibits these simple wavelike features but there is no interference or diffraction so we cannot really distinguish between waves and particles, if we don't interact with the beam.

When the beam reaches the barrier there is an interaction.
As a result of that interaction there is a change of solution to the Schrodinger equation.
However in this short section we cannot offer a solution (wavefunction) It is indeterminate.

But the slit barrier is classed as an 'observer' because it interacts with the beam.

Beyond the barrier the new wavefunction settles down and now includes the effect of the interaction with the slit barrier.

If the geometry of the setup is wrong eg the photgraphic film or screen is too close we will not see an inteference pattern, but at a suitable distance, about 10 times the interslit distance the inteference/diffraction develops.

 

This is a bit like the classical situation of nice tidy laminar flow in a pipe that leads to a sudden large pipe expansion and turbulence, followed by a funnelling back into the normal pipe diameter and laminar flow again. But this second laminar flow will often contain pulses or waves.

The laminar flow regions are predictable, but the turbulent flow region caqn only be dealt with on a statistical basis.

 

You now have all you need in the back posts here so

 

Have a Happy Christmas digesting it all.

Edited December 16, 2023 by studiot

[Luc Turpin]

1 hour ago, studiot said:

Agreed, I'm glad you figured it.

 

Basically you have what you need for this topic in all the back posts now.

Have a happy Christmas digesting it.

Agreed. I'm glad you figured it out.

 

Just to emphasis a couple of points.

 

The Schrodinger eqaution is about particles, which is why we need to work in terms of photons.

The solution, psi has some wavelike features.

 

In the case of the slits;

Between the source and the barrier with the slits psi looks the same and exhibits these simple wavelike features but there is no interference or diffraction so we cannot really distinguish between waves and particles, if we don't interact with the beam.

When the beam reaches the barrier there is an interaction.
As a result of that interaction there is a change of solution to the Schrodinger equation.
However in this short section we cannot offer a solution (wavefunction) It is indeterminate.

But the slit barrier is classed as an 'observer' because it interacts with the beam.

Beyond the barrier the new wavefunction settles down and now includes the effect of the interaction with the slit barrier.

If the geometry of the setup is wrong eg the photgraphic film or screen is too close we will not see an inteference pattern, but at a suitable distance, about 10 times the interslit distance the inteference/diffraction develops.

 

This is a bit like the classical situation of nice tidy laminar flow in a pipe that leads to a sudden large pipe expansion and turbulence, followed by a funnelling back into the normal pipe diameter and laminar flow again. But this second laminar flow will often contain pulses or waves.

The laminar flow regions are predictable, but the turbulent flow region caqn only be dealt with on a statistical basis.

 

You now have all you need in the back posts here so

 

Have a Happy Christmas digesting it all.

Many, many thanks to you

yes, will need time to digest it all