Expectation Values-Particle in a Box - Colby College

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Colby College. Expectation Values – Particle in a Box. Most Probable Position: Maximum in Ψ2. Ψ(x) = . . . . . 2 a. ½ sin. . . . . .  nπx a. Ψ2.
Expectation Values – Particle in a Box Most Probable Position: Maximum in 2 2½ nπx (x) =   sin  a  a 

0.2

2 0.1

2 nπx 2(x) =   sin2   a  a  for n = 1



0

xmp = 0.5 a

0

2

4

6 8 x (cm)

10



=

- * o^  dx  -



if  is real and normalized: = -  o^  dx

   dx *

Average Position: =  * x  dx

integral “over all space”

½  2 nπx 2½ nπx =  sin    a  x  a sin a  dx -  a       



average x

a 2 2nπx =  0 x asin  a  dx

 average x 

2 dx = probability distribution

nπx dy nπ a = x= y a dx a nπ 2 2  a  n =      y sin2(y) dx a n 0 y=

2 2  a 2 y y sin 2y cos 2y n =      – – 4 8  0 a n  4

dx =

a dy nπ x2 x sin 2x cos 2x  x sin (x) dx = 4 – 4 – 8 2

sin(0) = sin(2n) = 0, cos(0) = cos(2n) = 1

2 2  2a  n   a = =  2 2  n    4  2

Momentum: p^  =

ħ d i dx

^

Ek =

p^ 2 ħ2 d2 =– 2m 2m dx2

  ħ d 

= - * p^  dx = - *    dx  i dx

p^  =

ħ d ħ 2½ d sin(nπx/a) ħ 2½ n nπx =   =     cos  i dx i  a dx i a  a   a 

=

ħ 2 n a nπx nπx    sin cos  dx i a  a  0  a   a 

=

ħ 2 n a  n  sin y cos y dy = 0   i a  a n 0

Colby College

n

n

 0 sin y cos y dy = [ ½ sin2y | 0 = 0