I Why Is the Probability of Finding a Particle in a Position Eigenstate Zero?

[bluecap]

Why is the probability of finding a particle in an eigenstate of position zero and not one?

When we say we have located a particle at a particular position - why is it always in a superposition of position eigenstates about that position. But still the probability should not be zero.

I need the math. Thank you.

 

[PeterDonis]

Why is the probability of finding a particle in an eigenstate of position zero and not one?

The underlying reason is that there is no such thing as a "position eigenstate". More precisely, a state with a precisely defined position is not normalizable; it is not contained in the set of square integrable functions, which is the applicable Hilbert space.

Mathematically, a position eigenstate is described by a Dirac delta function; a position eigenstate with eigenvalue $x$ is $\delta(x)$. But the Dirac delta "function" is not actually a function; heuristically, this is because it has zero value at all points but one, and an infinite value at that one point, whereas a true function can only have finite values at all points. To see why $\delta(x)$ has an infinite value at one point, consider the general rule that probabilities must add up to one; if we apply this to a position eigenstate, we find that we must have (heuristically)

$$
\int_{- \infty}^{\infty} \delta(x) dx = 1
$$

If you dig into how integrals are defined, you will find that, if $\delta(x)$ only has a nonzero value at one value of $x$, it cannot have any finite nonzero value at that $x$ and still satisfy the above.

More here:

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

When we say we have located a particle at a particular position

Strictly speaking, we can never say that. Physically, the reason is the uncertainty principle. Mathematically, that principle says, heuristically,

$$
\Delta x \Delta p \ge h
$$

where $h$ is Planck's constant, i.e., a nonzero real number. But if we could locate a particle at a single definite position, then $\Delta x$ would be zero, and the above uncertainty relation could not possibly be satisfied. So we can't actually locate a particle at a single definite position. The best we can do is to measure its position to be within some small but finite range.

 

[bluecap]

The underlying reason is that there is no such thing as a "position eigenstate". More precisely, a state with a precisely defined position is not normalizable; it is not contained in the set of square integrable functions, which is the applicable Hilbert space.

Mathematically, a position eigenstate is described by a Dirac delta function; a position eigenstate with eigenvalue $x$ is $\delta(x)$. But the Dirac delta "function" is not actually a function; heuristically, this is because it has zero value at all points but one, and an infinite value at that one point, whereas a true function can only have finite values at all points. To see why $\delta(x)$ has an infinite value at one point, consider the general rule that probabilities must add up to one; if we apply this to a position eigenstate, we find that we must have (heuristically)

$$
\int_{- \infty}^{\infty} \delta(x) dx = 1
$$

If you dig into how integrals are defined, you will find that, if $\delta(x)$ only has a nonzero value at one value of $x$, it cannot have any finite nonzero value at that $x$ and still satisfy the above.

More here:

https://en.wikipedia.org/wiki/Dirac_delta_function
Strictly speaking, we can never say that. Physically, the reason is the uncertainty principle. Mathematically, that principle says, heuristically,

$$
\Delta x \Delta p \ge h
$$

where $h$ is Planck's constant, i.e., a nonzero real number. But if we could locate a particle at a single definite position, then $\Delta x$ would be zero, and the above uncertainty relation could not possibly be satisfied. So we can't actually locate a particle at a single definite position. The best we can do is to measure its position to be within some small but finite range.

Thanks. Simon wrote it "When we say we have located a particle at a particular position - it is always in a superposition of position eigenstates about that position. That is how we describe it in maths right?" Well.. let's say we rewords it into "When we say we have located a particle within some small but finite range - it is always in a superposition of position eigenstates about that some small but finite range". Is this correct statement?

Why is it always in a superposition of position eigenstates about that small but finite range.. its related to your statement above "But the Dirac delta "function" is not actually a function; heuristically, this is because it has zero value at all points but one, and an infinite value at that one point, whereas a true function can only have finite values at all points"? Or not since we can't actually locate a particle at a single definite position and this "always in a superposition of position eigenstates??" no longer hold?

 

[PeterDonis]

"When we say we have located a particle within some small but finite range - it is always in a superposition of position eigenstates about that some small but finite range". Is this correct statement?

Since any state which is not an eigenstate is a superposition of eigenstates, then yes, this is true.

Why is it always in a superposition of position eigenstates about that small but finite range.

Because of the uncertainty principle.

 

[bluecap]

Since any state which is not an eigenstate is a superposition of eigenstates, then yes, this is true.
Because of the uncertainty principle.

Ok. In classical macroscopic objects, this is also true about positions of quantum particles being in some finite range and hence in superposition of eigenstates? So macroscopic objects are coarse graining of these microscopic degrees of freedom and we can say macroscopic objects are coarse graining of superposition of position eigenstates?

 

[PeterDonis]

its related to your statement above "But the Dirac delta "function" is not actually a function; heuristically, this is because it has zero value at all points but one, and an infinite value at that one point, whereas a true function can only have finite values at all points"?

This is related to the uncertainty principle in the sense that in any mathematical model that is consistent with the uncertainty principle, the "position eigenstates" must have properties like those of the Dirac delta function.

 

[PeterDonis]

In classical macroscopic objects, this is also true about positions of quantum particles being in some finite range and hence in superposition of eigenstates?

Since positions of quantum particles are always in superpositions of position eigenstates (because the eigenstates themselves are not normalizable), then yes, this is true.

So macroscopic objects are coarse graining of these microscopic degrees of freedom, and we can say macroscopic objects are coarse graining of superposition of position eigenstates?

What do you mean by "coarse graining"? Can you give a mathematical expression of this?

 

[dextercioby]

Why is the probability of finding a particle in an eigenstate of position zero and not one?

When we say we have located a particle at a particular position - why is it always in a superposition of position eigenstates about that position. But still the probability should not be zero.

I need the math. Thank you.

The coordinate observable has a purely continuous spectrum, hence the probabilistic interpretation we owe to M. Born needs adjustment.
For a point a in the continuous spectrum of A, the absolute probability of measuring it is ill-defined, as the theory of quantum mechanics only tells you the probability of measuring a in the open interval (b,c). If the interval (b,c) is the whole spectrum, then the probability is 1, of course. This could be rephrased to saying that the coordinate observable is "not sharp" (the b is strictly smaller than c, can never be equal). To obtain a true probability value, you need not only a spectral value, but also a spectral interval containing that value. In case of a discrete (countable) spectrum, the interval is not needed anymore, the spectral value is enough to obtain absolute theoretical probabilities.

To say "we have located the particle at a particular position" (call it "a" the value and A the coordinate observable having "a" in its spectrum) means that, if you measured A a 1.000.000 times on 1.000.000 identically prepared systems, you've got a statistical spread of measurement values around "a" with >99% probability of measuring "a" in a very small (yet non zero) interval around it. The larger the interval becomes (thus increasing even more the >99% probability towards 100%), then the less true it is that "we have located the particle at a particular position".

 

[bluecap]

The underlying reason is that there is no such thing as a "position eigenstate". More precisely, a state with a precisely defined position is not normalizable; it is not contained in the set of square integrable functions, which is the applicable Hilbert space.

Mathematically, a position eigenstate is described by a Dirac delta function; a position eigenstate with eigenvalue $x$ is $\delta(x)$. But the Dirac delta "function" is not actually a function; heuristically, this is because it has zero value at all points but one, and an infinite value at that one point, whereas a true function can only have finite values at all points. To see why $\delta(x)$ has an infinite value at one point, consider the general rule that probabilities must add up to one; if we apply this to a position eigenstate, we find that we must have (heuristically)

$$
\int_{- \infty}^{\infty} \delta(x) dx = 1
$$

If you dig into how integrals are defined, you will find that, if $\delta(x)$ only has a nonzero value at one value of $x$, it cannot have any finite nonzero value at that $x$ and still satisfy the above.

More here:

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

By the way. Is this valid also for all observable like spin or energy.. that if the particle state vector is described by a superposition of eigenstates for *any* observable, then the particle does not have a determined value for that observable? No exception?

Since positions of quantum particles are always in superpositions of position eigenstates (because the eigenstates themselves are not normalizable), then yes, this is true.
What do you mean by "coarse graining"? Can you give a mathematical expression of this?

When you look at a wall, you won't see the individual atoms but the whole macroscopic wall with less degrees of freedom.. is this not called "coarse graining"? What is your definition of coarse graining then?

 

[PeterDonis]

Is this valid also for all observable like spin or energy.. that if the particle state vector is described by a superposition of eigenstates for *any* observable, then the particle does not have a determined value for that observable?

Yes.

When you look at a wall, you won't see the individual atoms but the whole macroscopic wall with less degrees of freedom.. is this not called "coarse graining"? What is your definition of coarse graining then?

All of the degrees of freedom are still there, but you are only measuring a very small number of them; and the degrees of freedom you are measuring are not degrees of freedom of any individual atom. For example, if I say the wall has a certain "position", I really mean something like: the center of mass of the wall is at some point in space, and the dimensions of the wall, centered on that center of mass, have some particular set of values--and all values having some small but nonzero range associated with them instead of being single definite numbers, since I can only measure things with a finite accuracy.

Another way of looking at this is: if I make a measurement on a single quantum object (say an electron), I can often pin down its state exactly, or at least to within a very small range. For example, if I measure an electron's spin along a certain axis, I get a definite result--spin up or spin down--corresponding to a definite spin state for that electron.

But if I make a measurement on a wall, I get something like what I described above, which does not pin down the full quantum state of the wall, including all of its atoms, very well at all; there will be a huge number of possible quantum states for the wall that are consistent with the measurement result. And I label this whole huge collection of possible quantum states of the wall with a single "classical" label, something like: the wall is "here" (at a particular position with a particular size). In this sense, yes, by assigning a classical "state" to the wall I am coarse-graining over the space of all possible quantum states, partitioning that space into subspaces, each containing a huge number of individual quantum states, and each subspace labeled with a "classical" state label.

 

[bluecap]

Yes.

Why can't you pinpoint energy or spin or other observable exactly.. what HUP would be violated?

 

[PeterDonis]

Why can't you pinpoint energy or spin or other observable exactly..

That isn't what I said, and it isn't the question you asked. You asked, if the particle is in a superposition of eigenstates for any observable, is it true that the particle does not have a determined value for that observable. That is true, but note carefully that it's a conditional statement: if a particle is in a superposition.

The question it looks like you actually wanted the answer to is this: I said that a particle can't ever be in a single position eigenstate (i.e., not in a superposition of position eigenstates) because that would violate the uncertainty principle. Is this also true of the eigenstates of every observable? The answer to that question is no: there are plenty of observables for which particles can be in eigenstates without violating the uncertainty principle. Spin is one of them.

Energy is more problematic, because the energy observable--the Hamiltonian operator--also governs the time evolution of the system. So strictly speaking, if a system is in an exact eigenstate of the Hamiltonian, it can never change and nothing can ever happen to it. But that means we can never measure it, because measurement would be a happening--something would have to change in order to record the measurement result, and we would have to include that something in our analysis, and that would mean the system as a whole, including both measured quantum object and measuring apparatus, would not be in an eigenstate of the full Hamiltonian, including the interaction involved in the measurement.

Discussions of, for example, the energy levels of electrons in atoms being "stationary states" gloss over this; what is actually meant is that an electron which is in a definite energy level is in a eigenstate of a Hamiltonian $H_0$ which only describes the atom and the electron--not any other interactions involved. But in any real atom, there will be other terms in the full Hamiltionian $H$ besides $H_0$, terms that describe, for example, interactions that allow the electron to emit or absorb photons and jump to a lower or higher energy level. The electron is never in an eigenstate of $H$, because if it were, as above, it could never change. Similar remarks apply to any quantum system.

 

[bluecap]

That isn't what I said, and it isn't the question you asked. You asked, if the particle is in a superposition of eigenstates for any observable, is it true that the particle does not have a determined value for that observable. That is true, but note carefully that it's a conditional statement: if a particle is in a superposition.

The question it looks like you actually wanted the answer to is this: I said that a particle can't ever be in a single position eigenstate (i.e., not in a superposition of position eigenstates) because that would violate the uncertainty principle. Is this also true of the eigenstates of every observable? The answer to that question is no: there are plenty of observables for which particles can be in eigenstates without violating the uncertainty principle. Spin is one of them.

Energy is more problematic, because the energy observable--the Hamiltonian operator--also governs the time evolution of the system. So strictly speaking, if a system is in an exact eigenstate of the Hamiltonian, it can never change and nothing can ever happen to it. But that means we can never measure it, because measurement would be a happening--something would have to change in order to record the measurement result, and we would have to include that something in our analysis, and that would mean the system as a whole, including both measured quantum object and measuring apparatus, would not be in an eigenstate of the full Hamiltonian, including the interaction involved in the measurement.

Discussions of, for example, the energy levels of electrons in atoms being "stationary states" gloss over this; what is actually meant is that an electron which is in a definite energy level is in a eigenstate of a Hamiltonian $H_0$ which only describes the atom and the electron--not any other interactions involved. But in any real atom, there will be other terms in the full Hamiltionian $H$ besides $H_0$, terms that describe, for example, interactions that allow the electron to emit or absorb photons and jump to a lower or higher energy level. The electron is never in an eigenstate of $H$, because if it were, as above, it could never change. Similar remarks apply to any quantum system.

Thank you. For multi particle systems (such as multiple entangled particles or larger macroscopic object). Which is the case that you can apply one eigenstate to all of them at same time by considering the center of mass (etc.) in some finite range (energy, position or other observable)? I found this statement in Zurek article https://arxiv.org/pdf/0707.2832.pdf
"Therefore, instantaneous eigenstates of the reduced density matrix of the open systems will in general not coincide with the preferred pointer states singled out by einselection".

Note reduced density matrix of the open system is not just one particle but multi particles via decoherence so how can it mentioned "instantaneous eigenstates of the reduced density matrix" (I understand the concepts of density matrix).. does it mean center mass of them or each particle?

 

[atyy]

Strictly speaking, we can never say that. Physically, the reason is the uncertainty principle. Mathematically, that principle says, heuristically,

$$
\Delta x \Delta p \ge h
$$

where $h$ is Planck's constant, i.e., a nonzero real number. But if we could locate a particle at a single definite position, then $\Delta x$ would be zero, and the above uncertainty relation could not possibly be satisfied. So we can't actually locate a particle at a single definite position. The best we can do is to measure its position to be within some small but finite range.

No. A particle can be at a definite position upon measurement.

 

[jtbell]

A particle can be at a definite position upon measurement.

What measuring device gives us a definite position, without uncertainty?

 

[atyy]

What measuring device gives us a definite position, without uncertainty?

That's a different practical consideration. The use of the HUP by Peter Donis is a wrong application of a principle. The HUP does not prevent a particle from having a particular position.

It is true that the wave function of a particle cannot be a position eigenstate, because position eigenstates are not admissible wave functions. It is also true that the probability of a particle being at any particular position is zero, because position is a continuous variable. However, when the position of a particle is measured, the particle does have a particular position.

The HUP talks about measurements on two different ensembles of particles, and the variance of position in one ensemble, with the variance of momentum in a second ensemble.

 

[PeterDonis]

That's a different practical consideration.

No, it isn't. A "position eigenstate" is not normalizable and hence is not actually a valid state at all.

when the position of a particle is measured, the particle does have a particular position.

What state is the particle in after the measurement? A heuristic treatment might say it is in a "position eigenstate" (for an instant, anyway, since such a state quickly evolves into a superposition of position eigenstates), but such a treatment would not hold up under scrutiny since, as above, a position eigenstate is not actually a valid state, and the particle has to be in a valid state after measurement.

 

[bluecap]

Thank you. For multi particle systems (such as multiple entangled particles or larger macroscopic object). Which is the case that you can apply one eigenstate to all of them at same time by considering the center of mass (etc.) in some finite range (energy, position or other observable)? I found this statement in Zurek article https://arxiv.org/pdf/0707.2832.pdf
"Therefore, instantaneous eigenstates of the reduced density matrix of the open systems will in general not coincide with the preferred pointer states singled out by einselection".

Note reduced density matrix of the open system is not just one particle but multi particles via decoherence so how can it mentioned "instantaneous eigenstates of the reduced density matrix" (I understand the concepts of density matrix).. does it mean center mass of them or each particle?

I think you missed the message above Peter or it's so hard for this thread.

My question is simply if the system is composed of many components.. do you get the eigenstate of the system by the superposition of all the components or particles? For example for eigenstate of position in some finite range. Do you get superposition of all positions of all particles then get the eigenstate of the superposition?

 

[vanhees71]

That's a different practical consideration. The use of the HUP by Peter Donis is a wrong application of a principle. The HUP does not prevent a particle from having a particular position.

It is true that the wave function of a particle cannot be a position eigenstate, because position eigenstates are not admissible wave functions. It is also true that the probability of a particle being at any particular position is zero, because position is a continuous variable. However, when the position of a particle is measured, the particle does have a particular position.

The HUP talks about measurements on two different ensembles of particles, and the variance of position in one ensemble, with the variance of momentum in a second ensemble.

You are contradicting youself! Correct is that there is no position eigenstate, and the conclusion is that the position of a particle can never be accurately determined at point. That's indeed reflected in the uncertainty principle, $\Delta x \Delta p_x \geq \hbar/2$, telling you that the standard deviation of the position probability distribution is never 0.

Measuring the position is also always of finite resolution, no matter whether you use oldfashioned devices like a cloud chamber (the droplets condensing due to ionization due to a particle running through) or a modern pixel detector (where the pixels may be small and densely packed the better technology gets, but it's always of finite resolution). True is that you can measure the position of a particle in principle with arbitrary small uncertainty, but it's always a finite non-zero accuracy. The uncertainty relation tells you that you cannot prepare a particle with definite position in principle, but of course you can measure its position with much higher precision than the implied $\Delta x$, and this you must do to resolve the uncertainty $\Delta x$ with high significance.

 

[PeterDonis]

if the system is composed of many components.. do you get the eigenstate of the system by the superposition of all the components or particles? ]For example for eigenstate of position in some finite range. Do you get superposition of all positions of all particles then get the eigenstate of the superposition?

This is muddled and I'm not sure I can disentangle it. "Superposition of all positions of all particles" is meaningless; that's not what a superposition is. "Eigenstate of the superposition" is also meaningless.

Let me try to describe how QM models a system with just two particles and see if that helps. We assume both particles have no spin so the Hilbert space only contains information about position. The single-particle Hilbert space has a basis of wave functions $\psi(x)$, i.e., functions of a single position $x$. The two-particle Hilbert space has a basis of wave functions $\Psi(x_1, x_2)$, i.e., functions of two positions $x_1$ and $x_2$. These wave functions give the probability amplitude for one particle to be at position $x_1$ and the other particle to be at position $x_2$.

Now, before we can even talk about eigenstates of this two-particle system, we have to know what observable we are looking for eigenstates of. Suppose we are talking about the "joint position" observable: i.e., we measure, simultaneously, the positions of both particles. Then the eigenstates of this observable will be states like $\delta(x_1) \delta(x_2)$; I'm being somewhat sloppy with notation here, but hopefully the meaning is clear that this is a state with one particle exactly at $x_1$ and the other exactly at $x_2$. And, just as with the single particle case, this state is actually not a valid state at all; it's not normalizable.

But there are also other observables we could talk about: for example, a "center of mass position" observable. Without trying to specify here exactly how we would measure such an observable, we can see that the center of mass position, assuming that both particles have the same mass, is just $(x_1 + x_2)/2$, i.e., the average of the two particles' positions. For a given wave function $\Psi(x_1, x_2)$, we can calculate the probability $P(X)$ of observing a center of mass position $X$. But note that a given center of mass position $X$ does not require any particular positions $x_1$ or $x_2$ for the two particles themselves; all it tells us is their average. So eigenstates of the center of mass position observable will not be eigenstates of position for either individual particle; if we write a center of mass eigenstate down in the position basis we have been using, it will be a superposition of an infinite number of possible wave functions $\Psi(x_1, x_2)$ that satisfy the constraint that $(x_1 + x_2)/2 = X$.

 

[atyy]

No, it isn't. A "position eigenstate" is not normalizable and hence is not actually a valid state at all.

But a position eigenstate is not related to whether upon whether a sharp position measurement can be made.

In a sharp position measurement, the wave function of the ensemble can be any admissible wave function. When a measurement is carried out on a single particle, we get a particular position value. When this is repeatedly done on many members of the ensemble, we get a distribution of positions given by the square of the wave function.

What state is the particle in after the measurement? A heuristic treatment might say it is in a "position eigenstate" (for an instant, anyway, since such a state quickly evolves into a superposition of position eigenstates), but such a treatment would not hold up under scrutiny since, as above, a position eigenstate is not actually a valid state, and the particle has to be in a valid state after measurement.

For continuous variables, the collapse rule as usually stated (collapse to the eigenvector corresponding to the eigenvalue) is not correct. One has to use a more general collapse rule, eg. https://arxiv.org/abs/0706.3526, Eq 3 and 4, and section 2.3.2 discusses sharp position measurements.

 

[PeroK]

It is also true that the probability of a particle being at any particular position is zero, because position is a continuous variable. However, when the position of a particle is measured, the particle does have a particular position.

For a continuous distribution, you have probability density, not a finite probability for every number. If you take what you say literally then each time you take a measurement of position, the impossible has happened.

For example, the rational numbers have measure 0. The total probability, therefore, of getting any rational number in a continuous distribution is 0. For, let's say, a digital measuring device, every measurement of position would have to be a rational number. The probabiity of getting any measurement at all would then, by your argument, be zero.

 

[bluecap]

This is muddled and I'm not sure I can disentangle it. "Superposition of all positions of all particles" is meaningless; that's not what a superposition is. "Eigenstate of the superposition" is also meaningless.

What I meant by Eigenstate of the superposition is in the case of the double slit experiment, the Eigenstate of the superposition of the two slits is the detector positions.

Let me try to describe how QM models a system with just two particles and see if that helps. We assume both particles have no spin so the Hilbert space only contains information about position. The single-particle Hilbert space has a basis of wave functions $\psi(x)$, i.e., functions of a single position $x$. The two-particle Hilbert space has a basis of wave functions $\Psi(x_1, x_2)$, i.e., functions of two positions $x_1$ and $x_2$. These wave functions give the probability amplitude for one particle to be at position $x_1$ and the other particle to be at position $x_2$.

Now, before we can even talk about eigenstates of this two-particle system, we have to know what observable we are looking for eigenstates of. Suppose we are talking about the "joint position" observable: i.e., we measure, simultaneously, the positions of both particles. Then the eigenstates of this observable will be states like $\delta(x_1) \delta(x_2)$; I'm being somewhat sloppy with notation here, but hopefully the meaning is clear that this is a state with one particle exactly at $x_1$ and the other exactly at $x_2$. And, just as with the single particle case, this state is actually not a valid state at all; it's not normalizable.

But there are also other observables we could talk about: for example, a "center of mass position" observable. Without trying to specify here exactly how we would measure such an observable, we can see that the center of mass position, assuming that both particles have the same mass, is just $(x_1 + x_2)/2$, i.e., the average of the two particles' positions. For a given wave function $\Psi(x_1, x_2)$, we can calculate the probability $P(X)$ of observing a center of mass position $X$. But note that a given center of mass position $X$ does not require any particular positions $x_1$ or $x_2$ for the two particles themselves; all it tells us is their average. So eigenstates of the center of mass position observable will not be eigenstates of position for either individual particle; if we write a center of mass eigenstate down in the position basis we have been using, it will be a superposition of an infinite number of possible wave functions $\Psi(x_1, x_2)$ that satisfy the constraint that $(x_1 + x_2)/2 = X$.

In a macroscopic object like an apple. When you want to solve a single Eigenstate for the entire apple in position observable (again taking into account the not normalizable thing again just for sake of discussion). Do you use center of mass or joint position" observable of the entire apple or what kind of position observable is usually used?

 

[PeroK]

In a macroscopic object like an apple. When you want to solve a single Eigenstate for the entire apple in position observable (again taking into account the not normalizable thing again just for sake of discussion). Do you use center of mass or joint position" observable of the entire apple or what kind of position observable is usually used?

I can't imagine trying to use QM to study the motion of an apple!

 

[houlahound]

I wouldn't be surprised if some academic has already done it.

 

[PeterDonis]

a position eigenstate is not related to whether upon whether a sharp position measurement can be made.

What do you mean by "a sharp position measurement"? If all you mean is that the measurement gives some number, of course that's possible, because we can just define the measurement procedure to ignore the "error bars" and give us the best-guess position as "the measurement result". I agree that the status of "position eigenstates" is irrelevant to what you said if that's what you meant; but in that case, what you said is irrelevant to what we've been discussing.

But if you mean that the state of the measured system after the measurement has a "sharp position", then the status of "position eigenstates" is certainly relevant. See below.

the collapse rule as usually stated (collapse to the eigenvector corresponding to the eigenvalue)

Is actually irrelevant to what I'm saying. I'm not insisting that the system has to be in an eigenstate following a measurement; I'm just saying that it has to be in some valid state, and "position eigenstates" are not valid states. And only "position eigenstates" have a "sharp position". So it cannot be possible that a system has a "sharp position" after measurement if what we mean by that is that the state it is in after measurement has a sharp position.

 

[PeterDonis]

What I meant by Eigenstate of the superposition is in the case of the double slit experiment, the Eigenstate of the superposition of the two slits is the detector positions.

This is still meaningless. Try using math instead of ordinary language. What actual math are you referring to here?

When you want to solve a single Eigenstate for the entire apple in position observable

There is no such thing. Read my previous post again. Even for the two-particle quantum system I described there (let alone for an apple with something like $10^{25}$ particles), the center of mass observable does not pick out a single state for the system. It only picks out a superposition of all the possible states that have the measured value for the center of mass position, i.e., that have position values for each of the individual particles that average to the measured center of mass position. So there is no such thing as an "eigenstate" of the center of mass observable.

Do you use center of mass or joint position" observable of the entire apple or what kind of position observable is usually used?

It depends on what experiment you're running. If you're talking about trying to describe how we ordinarily observe objects like apples in ordinary life, nobody has ever written down an explicit observable for "the position we see the apple to be in". An apple is far too complex a system, and since, experimentally, apples follow the laws of classical physics to a very, very good approximation, nobody bothers trying to write down a quantum state, or space of quantum states, for them. We just use the classical laws, which are much, much easier to work with.

 

[atyy]

What do you mean by "a sharp position measurement"? If all you mean is that the measurement gives some number, of course that's possible, because we can just define the measurement procedure to ignore the "error bars" and give us the best-guess position as "the measurement result". I agree that the status of "position eigenstates" is irrelevant to what you said if that's what you meant; but in that case, what you said is irrelevant to what we've been discussing.

But if you mean that the state of the measured system after the measurement has a "sharp position", then the status of "position eigenstates" is certainly relevant. See below.

A sharp position measurement is one that carried out on an ensemble in a pure state yields measurement outcomes distributed as |ψ(x)|². Yes, I am talking about the measurement outcome - that can be distributed according to |ψ(x)|². The state of the particle certainly cannot be a position eigenstate after a measurement.

Is actually irrelevant to what I'm saying. I'm not insisting that the system has to be in an eigenstate following a measurement; I'm just saying that it has to be in some valid state, and "position eigenstates" are not valid states. And only "position eigenstates" have a "sharp position". So it cannot be possible that a system has a "sharp position" after measurement if what we mean by that is that the state it is in after measurement has a sharp position.

OK, so it looks like you are using traditional language, then I agree.

However, I'm not sure that's what the OP meant when he wrote that a particle has a particular position.

Did he mean that (1) a single instance of the particle, when measured in a sharp position measurement, yields a single particular position? This is possible.

Or did he mean (as you, and tradition mean) that (2) an ensemble of particles prepared in the same pure state all yield the same particular position when a sharp position measurement is made? This is not possible.

 

[bluecap]

This is still meaningless. Try using math instead of ordinary language. What actual math are you referring to here?
There is no such thing. Read my previous post again. Even for the two-particle quantum system I described there (let alone for an apple with something like $10^{25}$ particles), the center of mass observable does not pick out a single state for the system. It only picks out a superposition of all the possible states that have the measured value for the center of mass position, i.e., that have position values for each of the individual particles that average to the measured center of mass position. So there is no such thing as an "eigenstate" of the center of mass observable.
It depends on what experiment you're running. If you're talking about trying to describe how we ordinarily observe objects like apples in ordinary life, nobody has ever written down an explicit observable for "the position we see the apple to be in". An apple is far too complex a system, and since, experimentally, apples follow the laws of classical physics to a very, very good approximation, nobody bothers trying to write down a quantum state, or space of quantum states, for them. We just use the classical laws, which are much, much easier to work with.

In the popular Zurek Quantum Darwinism paper discussed in many threads, there is a passage inside

"When a quantum system gives up information, its own
state becomes consistent with the information that was
disseminated. Collapse" in measurements is an extreme
example, but any interaction that leads to a correlation
can contribute to such re-preparation: Interactions that
depend on a certain observable correlate it with the environment,
so its eigenstates are singled out, and phase
relations between such pointer states are lost."

Let's take the case of apple. It says that
"Interactions of the apple that depend on a certain observable correlate it with the environment, so the apple eigenstates are singled out, and phase
relations between such pointer states are lost".

So when Zurek mentioned "eigenstates". Does he meant for individual particle of the apple? And he said phase relations between such pointer states are lost. So Pointer States (I'm asking in a thread that has no replies) is the Eigenstates. But could the pointer states of the apple be each individual particle? I'm assuming it's the entire apple macroscopic thing hence asking if the apple can be modeled as one eigenstate or if many eigenstates, for what observable and what particle(s)? Let me continue the quotes:

"Negative selection due to decoherence is the essence of
environment-induced superselection, or einselection [7]:
Under scrutiny of the environment, only pointer states
remain unchanged. Other states decohere into mixtures
of stable pointer states that can persist, and, in this sense,
exist: They are einselected." (https://arxiv.org/pdf/0903.5082v1.pdf)

So what really are the pointer states of the apples? For each particle or for the entire apple (but you said in last message it's not possible for entire apple at once).

 

[bluecap]

A sharp position measurement is one that carried out on an ensemble in a pure state yields measurement outcomes distributed as |ψ(x)|². Yes, I am talking about the measurement outcome - that can be distributed according to |ψ(x)|². The state of the particle certainly cannot be a position eigenstate after a measurement.
OK, so it looks like you are using traditional language, then I agree.

However, I'm not sure that's what the OP meant when he wrote that a particle has a particular position.

Did he mean that (1) a single instance of the particle, when measured in a sharp position measurement, yields a single particular position? This is possible.

Or did he mean (as you, and tradition mean) that (2) an ensemble of particles prepared in the same pure state all yield the same particular position when a sharp position measurement is made? This is not possible.

atyy you are familiar with Zurek. When he talks of :

When a quantum system gives up information, its own
state becomes consistent with the information that was
disseminated. Collapse" in measurements is an extreme
example, but any interaction that leads to a correlation
can contribute to such re-preparation: Interactions that
depend on a certain observable correlate it with the environment,
so its eigenstates are singled out, and phase
relations between such pointer states are lost."

"Negative selection due to decoherence is the essence of
environment-induced superselection, or einselection [7]:
Under scrutiny of the environment, only pointer states
remain unchanged. Other states decohere into mixtures
of stable pointer states that can persist, and, in this sense,
exist: They are einselected."

What are the pointer states of the apple say in position? Please comment how you understand it because my question of position is related to Zurek's.

 

[PeterDonis]

So what really are the pointer states of the apples?

I don't think Zurek ever writes one down. And Zurek's paper is an attempt to tackle a very complex issue; I don't think you should even be trying to understand what Zurek is saying until you are crystal clear about much simpler cases. See below.

For each particle or for the entire apple (but you said in last message it's not possible for entire apple at once).

I still don't think you understand what I've been saying. Once again, before even trying to understand the case of an apple, you should try to understand the case of a two-particle quantum system, which I've posted about several times now.

First: for the two particle system, there is no such thing as "the state of particle 1" in isolation. There is only the state of the system. (Here we are using "state" to mean "wave function" or "state vector in Hilbert space".) If the system is such that we can specify a state for each individual particle, then it is not a quantum two-particle system; it is just two individual quantum particles that do not interact with each other at all, and no quantum interference is possible between them. (More precisely: the state of the two-particle system is separable, and therefore no quantum interference effects can be observed between the two particles.)

Second: if we construct an observable on the two-particle system like the "center of mass" observable, this observable does not pick out a single state of the two-particle system. It only picks out a subspace of the full Hilbert space of all possible states--the subspace that satisfies the constraint that the center of mass position (the average position of the two particles) is equal to the measured value of the center of mass observable.

Do you understand what the above statements mean? And do you understand how they make the things you have been saying meaningless?

 

[bluecap]

I don't think Zurek ever writes one down. And Zurek's paper is an attempt to tackle a very complex issue; I don't think you should even be trying to understand what Zurek is saying until you are crystal clear about much simpler cases. See below.
I still don't think you understand what I've been saying. Once again, before even trying to understand the case of an apple, you should try to understand the case of a two-particle quantum system, which I've posted about several times now.

First: for the two particle system, there is no such thing as "the state of particle 1" in isolation. There is only the state of the system. (Here we are using "state" to mean "wave function" or "state vector in Hilbert space".) If the system is such that we can specify a state for each individual particle, then it is not a quantum two-particle system; it is just two individual quantum particles that do not interact with each other at all, and no quantum interference is possible between them. (More precisely: the state of the two-particle system is separable, and therefore no quantum interference effects can be observed between the two particles.)

Second: if we construct an observable on the two-particle system like the "center of mass" observable, this observable does not pick out a single state of the two-particle system. It only picks out a subspace of the full Hilbert space of all possible states--the subspace that satisfies the constraint that the center of mass position (the average position of the two particles) is equal to the measured value of the center of mass observable.

Do you understand what the above statements mean? And do you understand how they make the things you have been saying meaningless?

So for 2-particle system, it is entangled and in pure state.. but it's component is not in pure state. But the entangled state is a superposition. So I understand your first case where you stated "for the two particle system, there is no such thing as "the state of particle 1" in isolation".. because it is entangled. In the second case. Are you saying that we can only get observable for each component and not the entire pure state? The component being same meaning as "subspace" of the full Hilbert space of all possible states? So we can only get observable by tracing the rest in the density matrix (only focusing on a subspace or component)?

How do you apply this concept to one particle that is in superposition of taking left or right slit. Going to the double slit experiments. The single particle is in superposition taking account left and right slits. And the detections in the screens are its eigenstates (?) Here what you mean the observable is only defined for each component or subspace.. you are like saying the screen detection is only for each slit.. but isn't it for both slits (hence pick out a single state of the two-slit paths)?

Thanks a while lot! Appreciated so much for all the help.

 

[PeterDonis]

So for 2-particle system, it is entangled and in pure state..

If the state is not separable, which is what we are discussing, yes.

but it's component is not in pure state.

Remember how I defined "state": a wave function/state vector. The components (the individual particles--note that this is one of several possible usages of the term "component", see below) do not have such a state at all; only the two-particle system does. If one uses the density matrix formalism, one can assign a "state" (a density matrix) to one of the particles individually, but I am not talking about that definition of "state", because we first need to get clear about the simpler case.

But the entangled state is a superposition.

That depends on the observable. For any pure state, it is always possible in principle to find an observable for which that state is an eigenstate, not a superposition. Since this is true of any pure state for any quantum system, it is true for any pure state of the two-particle system under discussion.

You need to take a step back and rethink your understanding in the light of what I have just said; I think your confusion between an entangled state and a superposition is underlying much of your confusion in this thread.

Are you saying that we can only get observable for each component and not the entire pure state? The component being same meaning as "subspace" of the full Hilbert space of all possible states?

It appears that you don't understand what a "subspace" is. Let me illustrate with a particular example. Suppose we have measured the center of mass position of a two-particle system to be $X = 0$; i.e., we have measured the center of mass of the two-particle system to be at the spatial origin. The full Hilbert space of the two-particle system is the set of all wave functions $\Psi(x_1, x_2)$ of two positions. Our measurement of the center of mass position restricts the state of the two-particle system to the subset $\Psi_0(x_1, x_2)$ for which $x_2 = - x_1$ (because the average position must be $0$), i.e., to the set of functions $\Psi(x_1, x_2)$ which only have nonzero values when $x_2 = - x_1$. This set of functions is a subspace of the full Hilbert space. But a "pure state" is a single function $\Psi(x_1, x_2)$, not a whole set of them. (Strictly speaking, this is only true if we have defined the wave functions appropriately, so that each one corresponds to a distinct ray in the Hilbert space; but we'll ignore that complication here.)

For another usage of the term "component", different from how I used it above, see below.

Going to the double slit experiments. The single particle is in superposition taking account left and right slits.

More precisely: the wave function at the detector can be expressed as a superposition of two "components", one for each slit; each component can be thought of, heuristically, as expressing the probability amplitude for the particle to arrive at a given position on the detector after passing through one of the slits. Here the "components" are parts of a single particle wave function, whereas earlier in this post, the "components" were individual particles in a multi-particle system. These are two different concepts and you need to be very careful not to confuse them.

And the detections in the screens are its eigenstates (?)

No. As I have said several times now, eigenstates depend on the observable. You continue to be very sloppy about how you are expressing these things, and I think this is contributing to your confusion. Also, you are confusing the double slit case with the two-particle case; they are not the same. See below.

Here what you mean the observable is only defined for each component or subspace.. you are like saying the screen detection is only for each slit..

No, that's not what I am saying at all. In the double slit experiment, we do not have a two-particle system. We have a single-particle system. So none of the things I've been saying about two-particle systems even apply to this case.

In the double slit experiment, if we are sloppy and talk about "position eigenstates" as if they were actual states, then after the particle hits the detector and is detected, it is in a position eigenstate. If we are not sloppy, then we have to say that after the particle hits the detector and is detected, it is in a state with a very narrow spread of amplitude as a function of position--heuristically, a state in which the particle's position is somewhere within a very small "box" (the size of the "box" corresponds to the spatial resolution of the detector). But these are states of a single-particle system, not a multi-particle system.

Also, you are still confused about what I'm saying about the two-particle system. I am not saying an observable like the "center of mass position" is "only defined for each component or subspace". Components and subspaces are different things (see above). The observable is defined on the entire Hilbert space; but for each possible measured value of the observable, there is a corresponding subspace of the Hilbert space that contains all the possible states that are consistent with that measured value, as I described earlier in this post.

 

[bluecap]

If the state is not separable, which is what we are discussing, yes.
Remember how I defined "state": a wave function/state vector. The components (the individual particles--note that this is one of several possible usages of the term "component", see below) do not have such a state at all; only the two-particle system does. If one uses the density matrix formalism, one can assign a "state" (a density matrix) to one of the particles individually, but I am not talking about that definition of "state", because we first need to get clear about the simpler case.
That depends on the observable. For any pure state, it is always possible in principle to find an observable for which that state is an eigenstate, not a superposition. Since this is true of any pure state for any quantum system, it is true for any pure state of the two-particle system under discussion.

You need to take a step back and rethink your understanding in the light of what I have just said; I think your confusion between an entangled state and a superposition is underlying much of your confusion in this thread.
It appears that you don't understand what a "subspace" is. Let me illustrate with a particular example. Suppose we have measured the center of mass position of a two-particle system to be $X = 0$; i.e., we have measured the center of mass of the two-particle system to be at the spatial origin. The full Hilbert space of the two-particle system is the set of all wave functions $\Psi(x_1, x_2)$ of two positions. Our measurement of the center of mass position restricts the state of the two-particle system to the subset $\Psi_0(x_1, x_2)$ for which $x_2 = - x_1$ (because the average position must be $0$), i.e., to the set of functions $\Psi(x_1, x_2)$ which only have nonzero values when $x_2 = - x_1$. This set of functions is a subspace of the full Hilbert space. But a "pure state" is a single function $\Psi(x_1, x_2)$, not a whole set of them. (Strictly speaking, this is only true if we have defined the wave functions appropriately, so that each one corresponds to a distinct ray in the Hilbert space; but we'll ignore that complication here.)

For another usage of the term "component", different from how I used it above, see below.
More precisely: the wave function at the detector can be expressed as a superposition of two "components", one for each slit; each component can be thought of, heuristically, as expressing the probability amplitude for the particle to arrive at a given position on the detector after passing through one of the slits. Here the "components" are parts of a single particle wave function, whereas earlier in this post, the "components" were individual particles in a multi-particle system. These are two different concepts and you need to be very careful not to confuse them.
No. As I have said several times now, eigenstates depend on the observable. You continue to be very sloppy about how you are expressing these things, and I think this is contributing to your confusion. Also, you are confusing the double slit case with the two-particle case; they are not the same. See below.
No, that's not what I am saying at all. In the double slit experiment, we do not have a two-particle system. We have a single-particle system. So none of the things I've been saying about two-particle systems even apply to this case.

In the double slit experiment, if we are sloppy and talk about "position eigenstates" as if they were actual states, then after the particle hits the detector and is detected, it is in a position eigenstate. If we are not sloppy, then we have to say that after the particle hits the detector and is detected, it is in a state with a very narrow spread of amplitude as a function of position--heuristically, a state in which the particle's position is somewhere within a very small "box" (the size of the "box" corresponds to the spatial resolution of the detector). But these are states of a single-particle system, not a multi-particle system.

Also, you are still confused about what I'm saying about the two-particle system. I am not saying an observable like the "center of mass position" is "only defined for each component or subspace". Components and subspaces are different things (see above). The observable is defined on the entire Hilbert space; but for each possible measured value of the observable, there is a corresponding subspace of the Hilbert space that contains all the possible states that are consistent with that measured value, as I described earlier in this post.

I tried to understand everything and I'm getting it thanks.
But if eigenstates are good only for one particle.. how is it useful for practical applications where you have more than one particle? (such as quantum chemistry)
What is the common applications of eigenstates when the system has more than one particle? Can you give an actual example of more than one particle system and they happily apply the concept of eigenstates? Thank you.

 

[PeterDonis]

I'm getting it

Unfortunately I'm still not sure you are. See below.

if eigenstates are good only for one particle..

I did not say they were. In principle, one can define observables for multi-particle systems such that the multi-particle system will be in one particular eigenstate of that observable when the observable is measured. However, in practice, it is often very difficult to find such observables that can actually be measured, and it gets more and more difficult the more particles a multi-particle system has. For a macroscopic object it is practically impossible. Which means that most of the time, when we talk about observables for multi-particle systems, we are talking about observables like the center of mass position observable, that don't put the system into a single particular eigenstate when they are measured. There might not even be any definable eigenstates for such an observable at all.

This leads to a question: why are you so interested in eigenstates? It seems to me that you would be better served by taking a step back and looking at the way QM actually models multi-particle systems mathematically. Your current understanding seems to be based on misconceptions.

 

[bluecap]

Unfortunately I'm still not sure you are. See below.
I did not say they were. In principle, one can define observables for multi-particle systems such that the multi-particle system will be in one particular eigenstate of that observable when the observable is measured. However, in practice, it is often very difficult to find such observables that can actually be measured, and it gets more and more difficult the more particles a multi-particle system has. For a macroscopic object it is practically impossible. Which means that most of the time, when we talk about observables for multi-particle systems, we are talking about observables like the center of mass position observable, that don't put the system into a single particular eigenstate when they are measured. There might not even be any definable eigenstates for such an observable at all.

This leads to a question: why are you so interested in eigenstates? It seems to me that you would be better served by taking a step back and looking at the way QM actually models multi-particle systems mathematically. Your current understanding seems to be based on misconceptions.

I'm interested in eigenstates because I want to know what observables were chosen in Decoherence Einselection (Environment Induced Superselection). Zurek wrote:

"Interactions that
depend on a certain observable correlate it with the environment,
so its eigenstates are singled out, and phase
relations between such pointer states are lost".

In an apple, what eigenstates were singled out by the environment in Decoherence, Einselection which cause "phase relations between such pointer states are lost"??

 

[PeterDonis]

I'm interested in eigenstates because I want to know what observables were chosen in Decoherence Einselection

Then I think you need much, much more background than you currently appear to have. As I said before, Zurek is trying to deal with a very complex and advanced subject in QM. Also, the way he is using the word "eigenstate" looks nonstandard to me. It might be appropriate for the particular advanced subject he is dealing with, but it certainly doesn't lead to a simple answer to this question of yours:

In an apple, what eigenstates were singled out by the environment in Decoherence, Einselection which cause "phase relations between such pointer states are lost"??

As far as I can tell, the short answer to this question is that there are no such eigenstates if we use "eigenstate" in the way that term is usually used. He is basically using "eigenstate" to mean "pointer state", but his "pointer states" are really what I was calling "subspaces" earlier; they aren't single states, they're huge sets of states that are all consistent with a particular measured value for some macroscopic observable. All he is saying about decoherence and einselection is that there is no way to measure quantum interference effects between the different possible "pointer states". None of this has anything to do with the questions you have been asking about eigenstates as they appear in simpler measurements, like measuring the position of a single particle or the double slit experiment.

 

[bluecap]

Then I think you need much, much more background than you currently appear to have. As I said before, Zurek is trying to deal with a very complex and advanced subject in QM. Also, the way he is using the word "eigenstate" looks nonstandard to me. It might be appropriate for the particular advanced subject he is dealing with, but it certainly doesn't lead to a simple answer to this question of yours:
As far as I can tell, the short answer to this question is that there are no such eigenstates if we use "eigenstate" in the way that term is usually used. He is basically using "eigenstate" to mean "pointer state", but his "pointer states" are really what I was calling "subspaces" earlier; they aren't single states, they're huge sets of states that are all consistent with a particular measured value for some macroscopic observable. All he is saying about decoherence and einselection is that there is no way to measure quantum interference effects between the different possible "pointer states". None of this has anything to do with the questions you have been asking about eigenstates as they appear in simpler measurements, like measuring the position of a single particle or the double slit experiment.

Peter. I have a last question that I appeal you to answer before you tell me to spend 4 years undergraduate coarse in QM and Ballentine. I need to know the answer to the following that is so basic. Here's my puzzle. You said in old archive that:

However, once again, the size of the disturbance relative to the size of the object matters. If you are measuring an object that has only one quantum building block, like an electron, any measurement you make is going to disturb it significantly--heuristically, because the measurement itself has a minimum size which is basically one quantum building block. (For example, if we try to measure the electron by bouncing photons off of it, the minimum measurement we can make is to use one photon.) But if you are measuring an object with 1025'>10251025 10^{25} building blocks, like a piece of wood, there are lots of ways to measure it without significantly affecting its state, simply because of the huge number of building blocks. In fact, measuring an object of that size is really no different from what its environment is continually doing to it anyway--which is part of Zurek's point.

But Zurek own Pointer States being Subspace can be re-prepared. See https://arxiv.org/pdf/0707.2832v1.pdf "This (iii) quantum Darwinism allows observers to use environment as a witness to acquire information about pointer states indirectly, leaving system of interest
untouched and its state unperturbed." Zurek has to create the complicated concepts of fragments just so we can't directly perturb the pointer states. And the pointer states are macroscopic. Can you give example of what it means to directly perturb the pointer states if there is no fragments.. like can you make apples change shape or something (just one example). And lastly. In conventional QM.. how come people don't talk about perturbing the subspaces and effecting macroscopic quantum effect? Please answer these last questions before you lock this thread again and censor any discussions about this. Thank you!

 

[PeterDonis]

Zurek own Pointer States being Subspace can be re-prepared. See https://arxiv.org/pdf/0707.2832v1.pdf "This (iii) quantum Darwinism allows observers to use environment as a witness to acquire information about pointer states indirectly, leaving system of interest
untouched and its state unperturbed."

What you quote from Zurek's paper does not say that pointer states can be re-prepared. It says that observers can acquire information about pointer states (which is not the same as "re-preparing" them) from the environment.

Also, where does Zurek say that pointer states are subspaces? That word does not appear at all in the paper linked to in the quote above. If you are basing this on what I've said previously about subspaces, don't; you need much, much more background than you currently have before you will be able to use that concept correctly.

Zurek has to create the complicated concepts of fragments just so we can't directly perturb the pointer states.

He uses the concept of fragments to explain how we can repeatedly make observations that give us information about the same pointer state, without directly perturbing the pointer state. It's not that we can't perturb the pointer state; in principle it should be physically possible to do that (see below). But that can't be what we are actually doing when we make ordinary observations; if we did, if that's what "observing" the pointer state was, then we would not be able to make repeated observations of the same pointer state that gave the same results. The only way that can be possible is if our observations don't actually interact with the pointer state at all, but instead interact with something else (the fragments in the environment) that has information about the pointer state.

Can you give example of what it means to directly perturb the pointer states if there is no fragments.

Not based on anything in Zurek's paper; he doesn't include any theoretical model of how to do that. Nor does anyone else that I'm aware of. In principle it should be physically possible (though not necessarily practical) to perturb any quantum state; but saying that's possible in principle is not at all the same as having a specific model for how to do it in a particular case.

In conventional QM.. how come people don't talk about perturbing the subspaces and effecting macroscopic quantum effect?

First, please see my comment above about the term "subspaces". You are not using it correctly here.

Second, nobody knows how to make a "macroscopic quantum effect" in practice (see above), so nobody talks about doing it. But experimentalists have been gradually increasing the size of systems that can be shown to have quantum interference effects under the right conditions. The double slit experiment, for example, has now been done with molecules having 810 atoms each and massing more than 10,000 amu. See here:

https://arxiv.org/abs/1310.8343

Please answer these last questions before you lock this thread again

I have done so; and now I am locking the thread after answering them.