Anamitra

We consider a typical version of the EPR paradox where an electron and a positron are produced collinearly in the +y and -y with the spins polarized in the positive and negative z directions. The pair is emitted with zero linear and zero spin-angular momentum.
If the positron is detected in the spin state -1/2 the electron has to be in the state +1/2 no matter what be the distance between them.
Now let the spin of the positron change to +1/2. This change is some sort of a random change. But what is a random change? Is it something that occurs without a stimulus or is it something in response to some SIGNAL.One may consider signals responsible for such changes.These signals may be of "internal" or "external" origin. In our case we have an isolated two-particle system. Of course we have our interacting detector[we assume a pair of detectors for the two particles]
I am assuming that the change in spin of the first particle is due to some signal from the second one coupled with the interaction from the detector which may be represented by a separate signal. So the second particle while releasing the signal had an estimate of the type of change it might induce on the first one[and the expected time for that change] even if it is a large distance from it! With such considerations we can always devise signal models of the EPR paradox consistent with Special Relativity .
An example:
We have two particles A and B and a pair of detectors to measure their spin values.
There is a change in the spin value of A,as observed by the detector.
Reason: Signal from B+Internal Signal of A+Detector Interaction[which may be represented by another signal]
As the particles separate after their creation, each is assumed to be continuously influenced by signals emitted by the other. So the presence of the signal from B[and acting on A] is mandatory for all instants.

Particle B: can emit external signals B1 and B2[but not simultaneously]
Particle A can emit internal signals of the type A1 and A2[But not simultaneously]
Detector Interaction [on A] may be represented by DS [detector signal]
The model assumes:
(B1+Ds-A1) acting on existing plus half spin of A changes it minus half spin.This signal combination leaves the minus half spin state of A unchanged
(B1+Ds-A2) acting on existing plus half spin of A changes it minus half spin.This signal combination leaves the minus half spin state of A unchanged
In so far as spin is concerned B1+Ds-A1 and B1+ds-A2 produce the same effect on A. They may produce different effects in other areas [ex: they may induce changes in the position and the momentum of A in a manner consistent with the uncertainty principle]

(B2+Ds-A2) acting on existing minus half spin of A converts it to the plus half spin state.It leaves the plus half spin state unaffected.
(B2+Ds-A1) acting on existing minus half spin of A converts it to the plus half spin state.It leaves the plus half spin state unaffected.
(B2+Ds-A2) and (B2+Ds-A1) produce the same effect on the spin state of A. They may produce different effects in other areas.
B knows that:
B1 always changes the spin of A from +1/2 to minus ½. It leaves the +1/2 state unaffected.
B2 always changes the spin of the spin of A from -1/2 to +1/2. It leaves the -1/2 state unaffected.
Therefore B is aware of the spin state of A[and the time of expected change]
[Important point to note : As the particles separate after their creation, each is continuously under the influence of signals emitted by the other. We don’t have signals like Ds-A1 or Ds-A2.Either B1 or B2 have to be present[Assumption]. Consideration of Ds-A1 or Ds-A2 may be important for a single particle system]]

One may formulate much better models to make the EPR paradox consistent with Special Relativity[in relation to the finite speed of signal transmission]. One may formulate interesting models to contradict Special Relativity. We simply exclude them as inadmissible options .Of course there may be a large number of successful options.

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Anamitra

One of the basic aspects of the EPR paradox is that it indicates at infinitely fast signals, violating Special Relativity.
In the initial example of the first posting,measurements of spin of the first particle may record change from+1/2 to -1/2. The second particle has to adjust its spin instantaneously, suggesting infinitely fast signals.If the second particle has some participation in producing the change in the spin state of the first particle,it has an estimate of the change that is going to occur along with an estimate of the time of such occurrence.So when the spin state of the first particle changes from +1/2 to second -1/2, the spin of the second one can change from -1/2 to +1/2. This does not require infinitely fast signals, since the second particle has a prior information of the change that is to take place with the second one.

The basic point is that any recorded change is due to the impact of signals.

I have tried to provide a "logical model" of how infinitely fast signals may be avoided in relation to the EPR paradox.

SpectraCat

One of the basic aspects of the EPR paradox is that it indicates at infinitely fast signals, violating Special Relativity.
In the initial example of the first posting,measurements of spin of the first particle may record change from+1/2 to -1/2. The second particle has to adjust its spin instantaneously, suggesting infinitely fast signals.If the second particle has some participation in producing the change in the spin state of the first particle,it has an estimate of the change that is going to occur along with an estimate of the time of such occurrence.So when the spin state of the first particle changes from +1/2 to second -1/2, the spin of the second one can change from -1/2 to +1/2. This does not require infinitely fast signals, since the second particle has a prior information of the change that is to take place with the second one.

The basic point is that any recorded change is due to the impact of signals.

I have tried to provide a "logical model" of how infinitely fast signals may be avoided in relation to the EPR paradox.
I suggest that you read up on what entanglement is, then you will see why your model doesn't make any sense. A couple of quick comments:

1) You cannot "change" the spin of one member of an entangled pair in the sense you describe, you can only measure it

2) Q.M. says that if you measure the state of one member of an entangled pair, you immediately know the state of the other member

3) any theory involving normal (i.e. lightspeed or slower) communication between the members of the entangled pair would fall under the realm of Local Hidden Variable Theories (at least I think it would), and these have been shown to be inconsistent with the predictions of Q.M. (c.f. Bell's theorem).

DrChinese

Gold Member
One of the basic aspects of the EPR paradox is that it indicates at infinitely fast signals, violating Special Relativity.
In the initial example of the first posting,measurements of spin of the first particle may record change from+1/2 to -1/2. The second particle has to adjust its spin instantaneously, suggesting infinitely fast signals.If the second particle has some participation in producing the change in the spin state of the first particle,it has an estimate of the change that is going to occur along with an estimate of the time of such occurrence.So when the spin state of the first particle changes from +1/2 to second -1/2, the spin of the second one can change from -1/2 to +1/2. This does not require infinitely fast signals, since the second particle has a prior information of the change that is to take place with the second one.

The basic point is that any recorded change is due to the impact of signals.

I have tried to provide a "logical model" of how infinitely fast signals may be avoided in relation to the EPR paradox.
Well, entangled particles do not "change" spins from one prior value to another. And I don't see anything which indicates that they are able to "anticipate" something at light speed.

Anamitra

I suggest that you read up on what entanglement is, then you will see why your model doesn't make any sense. A couple of quick comments:

1) You cannot "change" the spin of one member of an entangled pair in the sense you describe, you can only measure it

2) Q.M. says that if you measure the state of one member of an entangled pair, you immediately know the state of the other member

3) any theory involving normal (i.e. lightspeed or slower) communication between the members of the entangled pair would fall under the realm of Local Hidden Variable Theories (at least I think it would), and these have been shown to be inconsistent with the predictions of Q.M. (c.f. Bell's theorem).
With the entangled pair I can measure the spin of an individual member by using some device[and this automatically assigns a definite value to the other]. With two successive measurements I get the values +1/2 and -1/2 with a particular member of the entangled pair. This is definitely a change.

Now I would like to ask DrChinese and SpectraCat whether this change is something without a cause or reason[something miraculous] or could it be that some signal is responsible for it?

Anamitra

A change occurs without cause or reason----that itself is a spooky thing.
If you are to believe in that you are simply creating a license to consider any type of "spooky action" resulting from it.

Anamitra

A change occurs without cause or reason---that itself is a spooky thing.

If one is to believe in that he is simply creating a license to consider any "spooky action" resulting from it.

SpectraCat

With the entangled pair I can measure the spin of an individual member by using some device[and this automatically assigns a definite value to the other]. With two successive measurements I get the values +1/2 and -1/2 with a particular member of the entangled pair. This is definitely a change.
After the first measurement on *either* member of the pair, the pair ceases to be entangled. So any further measurements on *either* member of the pair do not have any "long-range" effect on the other.

Also, it is not clear why you think measuring one member of the pair a second time would change the observed value of the spin for that particle. This will not happen if you make the *same* measurement twice (for example, if you measure the z-component of the spin two times in succession). If you make a *different* measurement (for example if you measure the z-component, followed by the x-component), then you may get a different result the second time, but you will not be measuring the same property of the particle. This is all basic stuff specified in the normal postulates of QM.

Now I would like to ask DrChinese and SpectraCat whether this change is something without a cause or reason[something miraculous] or could it be that some signal is responsible for it?
Given my comment above, what specific change are you talking about?

Anamitra

Even if the entanglement breaks on making a measurement, certain important points can be made:
The particles since their creation have been under the influence of each other. Each is taking the full share of “signals” emitted by the other. There is no scope of dissipation if we consider a closed system.
A large distance of separation should not be an issue so long as the signals are not being dissipated.
Each particle knows fully well the type of signal the other is receiving at any particular moment of time [in a manner consistent with relativity]
The observer designing the device for the measurement of spin knows the nature of spin theoretically. The same holds for the particle/particles.They are also supposed to have the intelligence of what spin could be or what spin is.
Therefore each particle has the intelligence of predicting the result of the measurement when the other interacts with the device.
It knows fully well what type of signal the other is receiving at any moment of time and what would happen[as the result of a measurement] if it,ie the other particle, interacted with a gadget at that moment

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JesseM

Therefore each particle has the intelligence of predicting the result of the measurement when the other interacts with the device.
It knows fully well what type of signal the other is receiving at any moment of time and what would happen[as the result of a measurement] if it,ie the other particle, interacted with a gadget at that moment
But if you want to be consistent with relativity, then at the time particle #1 encounters a measurement device with a certain setting, particle #1 will not yet know the setting of the device that was encountered by particle #2 at a space-like separation. So the only way particle #1 can guarantee that it will give the same result as #2 if they both encounter the same setting is if particle #1 knows in advance what result #2 would give to any possible setting #2 might have encountered, so no matter what setting #1 encounters, it can give the same result that it knows #2 must have given if #2 encountered that same setting. But if that's true, then from that you can derive various Bell inequalities, yet these inequalities are violated in QM.

Anamitra

But if you want to be consistent with relativity, then at the time particle #1 encounters a measurement device with a certain setting, particle #1 will not yet know the setting of the device that was encountered by particle #2 at a space-like separation. So the only way particle #1 can guarantee that it will give the same result as #2 if they both encounter the same setting is if particle #1 knows in advance what result #2 would give to any possible setting #2 might have encountered, so no matter what setting #1 encounters, it can give the same result that it knows #2 must have given if #2 encountered that same setting.
The observer can design different types of gadgets [use different settings] to measure the spin of a particle. Whatever setting or design he uses , he has a theoretical idea of spin--what it could be--1/2 or -1/2. The observer has an intelligence regarding the concept of spin,before the observation is made.
The particles can have the same or similar intelligence.[Each particle knows that a gadget acting on the other at any moment can generate a result called spin--and what this thing means theoretically]

Particle B knows the type of signal it has been emitting so long and also the nature of the signal that has reached A at any moment. It can conclude that if a measurement is conducted by some gadget at this moment[and this moment could be any moment], what result it should obtain.[Selection of half or minus half would depend on the nature of the signal that has reached A at the time of measurement.]

JesseM

The observer can design different types of gadgets [use different settings] to measure the spin of a particle. Whatever setting or design he uses , he has a theoretical idea of spin--what it could be--1/2 or -1/2. The observer has an intelligence regarding the concept of spin,before the observation is made.
The particles can have the same or similar intelligence.[Each particle knows that a gadget acting on the other at any moment can generate a result called spin--and what this thing means theoretically]

Particle B knows the type of signal it has been emitting so long and also the nature of the signal that has reached A at any moment. It can conclude that if a measurement is conducted by some gadget at this moment[and this moment could be any moment], what result it should obtain.[Selection of half or minus half would depend on the nature of the signal that has reached A at the time of measurement.]
Your comment doesn't really seem to address what I said. Do you agree or disagree that if relativity is respected, the particles must have communicated in advance what result they would give for all three possible measurements in a Bell experiment, since neither one knows in advance which measurement the experimenter will choose to do? If you agree with that, do you understand that this automatically implies the particle statistics should respect various Bell inequalities, yet these inequalities are actually violated in QM?

Anamitra

The particles are continuously in communication with each other through signals,since their creation.[in a manner consistent with Special Relativity]
Each particle has theoretical knowledge of spin--that it could be plus half or minus half. At any moment it has to select one to know the result of measurement on the other by a device or detector. This selection is accomplished by the fact that it knows the type of signal hitting the other at any moment.[In fact the first particle had emitted this signal]

JesseM

You didn't answer my questions at all!
Do you agree or disagree that if relativity is respected, the particles must have communicated in advance what result they would give for all three possible measurements in a Bell experiment, since neither one knows in advance which measurement the experimenter will choose to do?
Please answer with a statement telling me whether you agree or disagree with this specific statement. Just so there's no confusion, please actually use the word "agree" or "disagree" in your response. And if the answer is "agree", then my follow-up question was:
If you agree with that, do you understand that this automatically implies the particle statistics should respect various Bell inequalities, yet these inequalities are actually violated in QM?
Again, if your previous answer was "agree", then please give a straightforward answer to whether you understand and agree with this statement or not.

DrChinese

Gold Member
The particles are continuously in communication with each other through signals,since their creation.[in a manner consistent with Special Relativity]
Each particle has theoretical knowledge of spin--that it could be plus half or minus half. At any moment it has to select one to know the result of measurement on the other by a device or detector. This selection is accomplished by the fact that it knows the type of signal hitting the other at any moment.[In fact the first particle had emitted this signal]
Your thinking fails because you are simply imagining that all this is possible. But when you put numbers to it - as Bell did - it falls apart. If you do NOT understand Bell, you should stop right now and learn the argument. I host a web page to help understand this:

http://drchinese.com/David/Bell_Theorem_Easy_Math.htm

eaglelake

Anamitra - You keep insisting that the particles are in communication by emitting signals. There is no such thing! This your own imagination at work. It has nothing to do with quantum mechanics. Experimentalists are very clever and, yet, not one of them has ever detected your mysterious "signals" which interact with particles.

You refuse to give up on the classical concept of separability - that the two particles are independent entities. This is not true at the quantum level. In quantum events, such entangled particles are a single entity. It is more correct to talk of the "pair" rather than of each particle separately. And a single measurement result consists of the two values + and -. But, you treat them as two different results, where the second result depends on a signal from the first particle. EPR treated the two particles as independent particles, as you do, each with their own spin, and got erroneous results. EPR-like experiments have been done and the results agree with the non-separability principle of quantum mechanics. If you seriously study quantum mechanics long enough, then eventually you must reject some of the most fundamental axioms of classical physics. Quantum events are not classical! If we accept this, then there is no EPR paradox.

I apologize for being so harsh. I know it is difficult to accept empirical evidence that goes against our "common sense" and intuition. We all struggle with this stuff.
Best wishes

Anamitra

Particles can always communicate by electromagnetic or gravitational disturbances.

But the real issue is the violation of Bell's inequality in quantum mechanical examples
Lets start of the familiar example given at the beginning of posting 1

$${P}_{QM}{(}{a}{,}{b}{)}{=}{-}{a}{.}{b}$$ --------------- (1)

[a and b are unit vectors along directions where spin is detected]

$${P}_{h}{(}{a}{,}{b}{)}{=}{\int}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$ ----------------------- (2)
[lambda is the hidden variable and rho is the distribution function; "h" represents hidden variable in the subscript of P]

[ I am considering an example from Liboff, Introductory Quantum Mechanics,Chapter 11,Section Bell's Theorem]
The above two probabilities seem to be un-normalized ones, if different angles between "a" and "b" are considered.

If one integrates modulus of Cos[theta] from zero to pi he gets 2.
The normalization coefficient for the first result seems to 1/2

If a constant coefficient is considered on the RHS of eqn (2) Bell's inequality is not changing.

But the value half seems to be resolving the conflict at least for this example.

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JesseM

Particles can always communicate by electromagnetic or gravitational disturbances.

But the real issue is the violation of Bell's inequality in quantum mechanical examples
Lets start of the familiar example given at the beginning of posting 1

$${P}_{QM}{(}{a}{,}{b}{)}{=}{-}{a}{.}{b}$$ --------------- (1)

[a and b are unit vectors along directions where spin is detected]

$${P}_{h}{(}{a}{,}{b}{)}{=}{\int}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$ ----------------------- (2)
[lambda is the hidden variable and rho is the distribution function; "h" represents hidden variable in the subscript of P]

[ I am considering an example from Liboff, Introductory Quantum Mechanics,Chapter 11,Section Bell's Theorem]
The above two probabilities seem to be un-normalized ones, if different angles between "a" and "b" are considered.
The left hand sides are not meant to be probabilities, rather they are meant to be expectation values for the product of the two results (+1 representing the result "spin-up", -1 representing the result "spin-down", so the product is positive if they give the same result and negative if they give different results). Likewise A(a,λ) is not a probability but just a deterministic function that is either equal to +1 or -1 depending on the value of the detector setting a and the hidden variables λ. I don't know if you agree that perfect correlations on measurements with the same detector setting imply the result should depend deterministically on the detector setting and the hidden variables, because you didn't answer the simple agree/disagree question in my previous post.

Anamitra

Let us consider the expectation value of -Cos theta[Equation (1)].

p1 probability of getting 1/2 on a
p2:probability of getting 1/2 on b
1-p1: prob of getting -1/2 on a
1-p2: prob of getting -1/2 on b
We have,

Expectation=(-)Cos[theta]=1/2p1*1/2 p2+ 1/2p1*(-1/2)(1-p)+(-1/2)(1-p1)1/2p2+(-1/2)(1-p1)(-1/2)(1-p2)

Or,
p1+p2=2Cos[theta]+2p1p2+1/2

For theta=0,
p1+p2>2

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JesseM

Let us consider the expectation value of -Cos theta[Equation (1)].

p1 probability of getting 1/2 on a
The results at any given detector angle are always represented as +1 or -1, you can't get 1/2 as a measurement result. Also of course in QM the results are not independent, so
P(+1 for particle #1 measured on setting a AND +1 for particle #2 measured on setting b)
is not equal to
P(+1 for particle #1 measured on setting a)*P(+1 for particle #2 measured on setting b).

Instead, in the experiment considered involving electron spins, the probability that both will give +1 would be 1/2*sin^2(theta/2), where theta is the angle between setting a and setting b (see [URL [Broken]).[/URL] That's also the probability that both will give -1, while the probability that particle #1 gives +1 and particle #2 gives -1 is 1/2*cos^2(theta/2), and likewise for the probability that particle #1 gives -1 and particle #2 gives +1. So, the expectation value is:

(+1)*(+1)*1/2*sin^2(theta/2) + (-1)*(-1)*1/2*sin^2(theta/2) + (+1)*(-1)*1/2*cos^2(theta/2) + (-1)*(+1)*1/2*cos^2(theta/2) =
sin^2(theta/2) - cos^2(theta/2)

And from the trigonometric identities here, sin^2(theta/2) = (1 - cos(theta))/2, while cos^2(theta/2) = (1 + cos(theta))/2, so:

sin^2(theta/2) - cos^2(theta/2) = (-cos(theta) - cos(theta))/2 = -cos(theta)

...which is the same as the negative dot product of vectors a and b, equation (1) from the textbook.

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Anamitra

The detector considers values as +1 and minus-1 instead of +1/2 and -1/2.

But I can always take the actual Sz eigenvalues +1/2 and -1/2 and get - 1/4 Cos theta instead of - Cos (theta)as the expectation value.

If one takes -1/4 Cos (theta) as the expectation value the conflict between Bell's inequality and Quantum Mechanics does not arise!.

JesseM

The detector considers values as +1 and minus-1 instead of +1/2 and -1/2.

But I can always take the actual Sz eigenvalues +1/2 and -1/2 and get - 1/4 Cos theta instead of - Cos (theta)as the expectation value.

If one takes -1/4 Cos (theta) as the expectation value the conflict between Bell's inequality and Quantum Mechanics does not arise!.
Of course it does, the inequality would just be modified if you use different numbers to label "spin-up" and "spin-down" measurements. Look at the step at the top of p. 406 (p. 4 of the pdf) of Bell's derivation, he invokes equation (1) which says that A(b,λ) = ±1 in order to justify equating the integrand [A(a,λ)*A(c,λ) - A(a,λ)*A(b,λ)] to the integrand A(a,λ)*A(b,λ)*[A(b,λ)*A(c,λ) - 1]...in other words, he's using the fact that A(b,λ)*A(b,λ)=+1. If you instead labeled the results with +(1/2) and -(1/2), then A(b,λ)*A(b,λ)=+(1/4), so that step wouldn't work. Instead the second integrand would have to be A(a,λ)*A(b,λ)*[4*A(b,λ)*A(c,λ) - 1].

Continuing on with steps analogous to the ones in Bell's paper, we can then apply the triangle inequality (which works as well for integrals as it does for discrete sums) to show that if $$P(a,b) - P(a,c) = \int d\lambda \rho(\lambda) A(a,\lambda)A(b,\lambda)[4A(b,\lambda)A(c,\lambda) - 1]$$, that implies $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \mid \rho(\lambda) A(a,\lambda)A(b,\lambda)[4A(b,\lambda)A(c,\lambda) - 1] \mid$$. Since $$\rho(\lambda)$$ is always positive and A(a,λ)*A(b,λ)=±(1/4) that becomes $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) (1/4) \mid [4A(b,\lambda)A(c,\lambda) - 1] \mid$$, then since 4*A(b,λ)*A(c,λ) - 1 is always equal to -2 or 0, its absolute value is always equal to 1 - 4*A(b,λ)*A(c,λ) so we now have $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) (1/4) [1 - 4A(b,\lambda)A(c,\lambda)]$$ or $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) [(1/4) - A(b,\lambda)A(c,\lambda)]$$. Bell then notes that $$\int d\lambda \rho(\lambda) [-A(b,\lambda)A(c,\lambda)] = P(b,c)$$ and since $$\int d\lambda \rho(\lambda) = 1$$ this gives an inequality equivalent to the one Bell derives:

(1/4) + P(b,c) ≥ |P(a,b) - P(a,c)|

With the measurement results labeled +(1/2) or -(1/2), there's no way this inequality can be violated in a realist theory that respects relativity.

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DrChinese

Gold Member
Particles can always communicate by electromagnetic or gravitational disturbances.

But the real issue is the violation of Bell's inequality in quantum mechanical examples
Lets start of the familiar example given at the beginning of posting 1

$${P}_{QM}{(}{a}{,}{b}{)}{=}{-}{a}{.}{b}$$ --------------- (1)

[a and b are unit vectors along directions where spin is detected]

$${P}_{h}{(}{a}{,}{b}{)}{=}{\int}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$ ----------------------- (2)
[lambda is the hidden variable and rho is the distribution function; "h" represents hidden variable in the subscript of P]

[ I am considering an example from Liboff, Introductory Quantum Mechanics,Chapter 11,Section Bell's Theorem]
The above two probabilities seem to be un-normalized ones, if different angles between "a" and "b" are considered.

If one integrates modulus of Cos[theta] from zero to pi he gets 2.
The normalization coefficient for the first result seems to 1/2

If a constant coefficient is considered on the RHS of eqn (2) Bell's inequality is not changing.

But the value half seems to be resolving the conflict at least for this example.
If you understand this, why won't you take the time to understand the Bell argument BEFORE you go off the deep end? It's not that hard. Really. I mean REEEEEEEEEALLY.

Anamitra

We may consider the function[integral]for the expectation value of spin:

$${P}_{h}{(}{a}{,}{b}{)}{=}{\int}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$

We may broaden the choice of the integral by including weights f(lambda,t) and g(lambda,t) for the spin values A and B. rho is normalized to unity as usual but we make,

Integral [rho*g*f]d lammda =n [which is not unity but some finite number suited to our testings]

[ By the above weights I have tried to take care of the time of measurement. Suppose for each instant we have the same distribution function rho for lambda[Or we may consider distribution functions].The weights are taking care of the cumulative effects of the distribution functions over the small interval of measurement.]

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