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The RMS displacement of particle undergoing a random walk
Consider a 1D random walk: A particle moves left or right in unit jumps on
the x-axis; at each step it has an equal probability of moving left or right.
We want to know the probability that after N jumps the particle is exactly at a
distance d from where it started, where
d is measured in units of jump.
Let the total number of jumps to the right be n1 and that to the left be n2, where,
N = n1 + n2. Then the NET displacement or distance of the particle must be
(
)
d = n1 - n2 = n1 - N - n1 = 2n1 - N
Consider the sequence of jumps: n1n2n2n1n1. The probability of this sequence
or any other particular sequence of 5 jumps is just (1/2)5 . The number of such
sequences containing 3 right jumps and 2 left jumps is the number of ways we
can select 3 positions out of 5 for right jumps (or 2 for left jumps).
To understand how to think about this let’s consider a little simpler problem. First
since
N = n1 + n2
N n1 n2
= + = p+q
N N N
Now ask if we take three jumps what are the
possible outcomes?
qqq, qqp, qpq, pqq, qpp, pqp, ppq, ppp
There are a total of 8 possible outcomes and now let’s evaluate the probability of 0, 1,
2 and 3 jumps to the right.
P(0) = q 3
P(1) = qqp + qpq + pqq = 3q 2 p
P(2) = qpp + pqp + ppq = 3qp 2
P(3) = p 3
Note that ( p + q)3 = p 3 + 3p 2 q + 3pq 2 + q 3 and terms containing the same powers of p and q
have been added since the order of left and right jumps won't affect the net distance traveled.
You can see that the probability of 0, 1, 2, or 3 jumps to the right and 3, 2, 1, or 0 jumps
to the left is just one term in the expansion of (p +q)3.
Now generalizing we can say that the probability of getting k jumps to the right out of
a total of N jumps is just one term in the binomial expansion of (p +q)N.
Binomial Theorem
(
æ
ö
æ
ö
æ
ö
æ
ö
a + b = å ç n ÷ a n-k bk = a n + ç n ÷ a n-1b1 + ç n ÷ a n-2 b2 + ... + ç n ÷ a1bn-1 + bn
è 1 ø
è 2 ø
è n -1 ø
k =0 è k ø
)
n
n
æ n ö
n!
=
C
n,
k
=
ç k ÷
n - k !k!
è
ø
( ) (
)
Consider the sequence of jumps: n1n2n2n1n1. The probability of this sequence
or any other particular sequence of 5 jumps is just (1/2)5 . The number of such
sequences containing 3 right jumps and 2 left jumps is the number of ways we
can select 3 positions out of 5 for right jumps (or 2 for left jumps).
æ 5ö
5!
C 5,3 = ç ÷ =
è 3ø 5 - 3 !3!
( )
(
)
More generally, for N jumps the number of ways for which we can get n1 jumps to the
right and n2 jumps to the left is
æ Nö
N!
C N , n1 = ç ÷ =
N - n1 !n1 !
è n1 ø
(
)
(
)
If: p is the probability of a right jump and q = (1-p) is the probability of a left jump
Then the probability f(3) of exactly 3 right jumps out of 5 total jumps is just
one term in a binomial series:
()
( )
f 3 = C 5,3 p q
3
(5-3)
( )
= 10 × 1/ 2
5
More generally the probability of exactly n1 right jumps in N total jumps
Is just:
( )
(
)
n1
f n1 = C N , n1 p q
( N -n1 )
The RHS of this equation is just one term of a binomial expansion:
N
( p + q) = å C ( N ,r ) p
N
( p + q)
r=0
N
(
N -r
qr
)
( )
( )
(
)
(
)
= C N ,0 p N q0 + C N ,1 p N -1q1 + C N ,1 p N -2 q2 + ...C N , N -1 p1q N -1 + C N , N p 0 q N
Graph of
()
( )
f (x) = C ( 6, x ) (1/ 2) (1/ 2)(
f x = C 6, x p q
x
( N - x)
x
N=6
p = 1/2
6- x
)
f(x)
x
Some Definitions
Mean value of a variable, x:
x=
1
× å N i xi
N i
Mean value of a random variable, x whose probability function is f(x)
( )
x = å xi f xi
i
Standard deviation and variance of a variable, x:
( )
V (x) = å (xi - x )2 f xi
i
s x = V (x)
For the binomial probability function:
( )
(
)
n1
f n1 = C N ,n1 p q
x = Np
s x = Npq
( N -n1 )
Mean value of the number of steps to the right, n1: n1 = pN
Mean value of the number of steps to the left, n2:
n2 = qN
n1 + n2 = N ( p + q) = N
Since p +q = 1
Mean value of the number of displacement to the right: d = N( p - q)
Note that for p =q=1/2, d = 0
The variance of the displacement to the right can be evaluated as:
d = n1 - n2
Dd = d - d = (2n1 - N ) - (2n1 - N ) = 2(n1 - n1 ) = 2Dn1
(Dd )2 = 4 (Dn1 )2 = 4Npq
For , p = q = ½,
(Dd)2 = N; (Dd)2
1/2
= N 1/2
Normal or Gaussian distributions
For large N, the binomial distribution is tough to evaluate owing to the factorials
In C(N, n1) so we often use an approximation to the binomial distribution that
works well for large N. In order to get to this Gaussian distribution, we use
Stirling’s approximation for the factorials in addition to some algebraic manipulation:
Stirling’s
approximation
( )
ln n!= nln n - n + 1/ 2 ln(2p n)
ln n!= nn e- n 2p n
Gaussian distribution
f (x) ~
1
2p Npq
e
-( x- Np )2 /(2 Npq)
x = Np
s x = V (x) = Npq
(Dd )
2
1/2
= 2s
Calculation of the mean-square displacement from the time dependent diffusion
equation.
+¥
The number of particles, N is conserved at all times:
ò n(x,t) dx = N
-¥
Also, by definition:
1
x (t) =
N
2
+¥
ò
x 2 n(x,t) dx
-¥
To find how the mean-square displacement depends on time, multiply both sides
of the diffusion equation by x2 and integrate over x, i.e.,
+¥
+¥
2
2 ¶n
2 ¶ n
ò-¥ x ¶t dx = D-¥ò x ¶x 2 dx
+¥
+¥
¶
¶ 2
2 ¶n
2
x
dx
=
x
n
dx
=
N
x
ò-¥ ¶t
ò
¶t -¥
¶t
The LHS is easily integrated
The RHS is integrated by successive integration by parts:
+¥
+¥
+¥
+¥
2
+¥
é 2 ¶n ù
¶n
2 ¶ n
ò-¥ x ¶x 2 dx = êë x ¶x úû - 2-¥ò x ¶x dx = 0 - 2 éë xn ùû-¥ + 2-¥ò n dx = 2N
-¥
Applying the limits of integration, terms go to zero since abs(x) goes infinite.
Then setting the integrated L- and RHS equal we obtain the key result:
N
¶ 2
x = 2ND
¶t
x
2
1/2
= 2Dt