1 answer

2. (a) *D1* Let A be a fluctuating quantity like the position, velocity, or energy of...

Question:

2. (a) *D1* Let A be a fluctuating quantity like the position, velocity, or energy of a molecule. At any given time, we can quantify the fluctuation of A away from the average (A) by the quantity δ = A-(A). Show that the mean square fluctuation in A can be written in terms of its mean and mean square values. «6A)2-(A2>-(A)2. The following properties of averages might be useful (X and Y are fluctuating quantities, c is a constant) · 〈XY〉 = 〈X) 〈Y) only if X and Y are not correlated. .e., the outcome of X does not depend on Yand vice versa (2 pts) (b) *D1* Generalize this result to the case of two different fluctuating quantities A and B, i.e., show that 〈6A8B〉 = 〈AB) _ 〈A)(B). (2 pts) *D1* quantities A and B. To see this, consider tossing a pair of coins A and B. We will assign values of-1 and 1 for heads and tails, respectively. The coins are fair, so 〈A)-〈B) 0. Calculate (6ASB) for the following cases: (c) The quantity 46B) is often used to measure the correlation between two fluctuating i. A and B are statistically independent, i.e., they behave like normal coins ii. By some dark magic, B always shows the same result as A iii. By some even darker magic, B always shows a result opposite to A Explain in what sense the values of 〈5.45B) can be interpreted as a measure of correlation. (2 ts

2. (a) *D1* Let A be a fluctuating quantity like the position, velocity, or energy of a molecule. At any given time, we can quantify the fluctuation of A away from the average (A) by the quantity δ = A-(A). Show that the mean square fluctuation in A can be written in terms of its mean and mean square values. «6A)2-(A2>-(A)2. The following properties of averages might be useful (X and Y are fluctuating quantities, c is a constant) · 〈XY〉 = 〈X) 〈Y) only if X and Y are not correlated. .e., the outcome of X does not depend on Yand vice versa (2 pts) (b) *D1* Generalize this result to the case of two different fluctuating quantities A and B, i.e., show that 〈6A8B〉 = 〈AB) _ 〈A)(B). (2 pts) *D1* quantities A and B. To see this, consider tossing a pair of coins A and B. We will assign values of-1 and 1 for "heads" and "tails", respectively. The coins are fair, so 〈A)-〈B) 0. Calculate (6ASB) for the following cases: (c) The quantity 46B) is often used to measure the correlation between two fluctuating i. A and B are statistically independent, i.e., they behave like normal coins ii. By some dark magic, B always shows the same result as A iii. By some even darker magic, B always shows a result opposite to A Explain in what sense the values of 〈5.45B) can be interpreted as a measure of correlation. (2 ts

Answers

(a)

left langle left ( delta A ight )^2 ight angle = left langle left ( A - left langle A ight angle ight )^2 ight angle = left langle A^2 + left langle A ight angle^2 - 2Aleft langle A ight angle ight angle

= left langle A^2 ight angle + left langle left langle A ight angle^2 ight angle - left langle 2Aleft langle A ight angle ight angle

A2 (A) (Note that left langle A ight angle^2 is square of mean value and is a constant)

2 (A

= left langle A^2 ight angle - left langle A ight angle^2

(b)

left langle delta A delta B ight angle = left langle left ( A - left langle A ight angle ight )left ( B - left langle B ight angle ight ) ight angle

= left langle AB + left langle A ight angle left langle B ight angle - Aleft langle B ight angle - left langle A ight angle B ight angle

= left langle AB ight angle + left langle left langle A ight angleleft langle B ight angle ight angle - left langle A ight angleleft langle B ight angle - left langle A ight angleleft langle B ight angle

= left langle AB ight angle + left langle A ight angleleft langle B ight angle - left langle A ight angleleft langle B ight angle - left langle A ight angleleft langle B ight angle

= left langle AB ight angle - left langle A ight angleleft langle B ight angle

(c)

i. For independent events, left langle AB ight angle = left langle A ight angleleft langle B ight angle

left langle delta A delta B ight angle = left langle AB ight angle - left langle A ight angleleft langle B ight angle = 0

ii.

left langle delta A delta B ight angle = left langle AB ight angle - left langle A ight angleleft langle B ight angle = left langle AB ight angle (Given left langle A ight angle = left langle B ight angle = 0 )

If B always shows the same result as A. then A = B

left langle delta A delta B ight angle = left langle AB ight angle - left langle A ight angleleft langle B ight angle = left langle AB ight angle = left langle A^2 ight angle = left langle B^2 ight angle

iii.

If B always shows the opposite result as A. then A = -B or B = -A

left langle delta A delta B ight angle = left langle AB ight angle - left langle A ight angleleft langle B ight angle = left langle AB ight angle = left langle -A^2 ight angle

= - left langle A^2 ight angle = -left langle B^2 ight angle

If A and B are independent events, then correlation is 0.  left langle delta A delta B ight angle = 0.

If B always shows the same result as A, then correlation is positive.

If B always shows the opposite result as A, then correlation is negative.

.

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