Question

With what amplitude and frequency does the expectation value of the momentum of a proton (m=1.67x10-27 kg) in a ID box of length 0.02 nm oscillate with time? The particle has the initial wave function Voya- Vioys.

Answer 1

Given initial wave function:

$\psi=\frac{3}{\sqrt{10}}\psi_2-\frac{1}{\sqrt{10}}\psi_3$

m=

1.67 x 10-27 kg

length=a=0.02nm=

2 x 10-11 m

Let us find expectation value of momentum first then we will check its amplitude and frequency

We know

U2 = sin(272/L)

V3 = sin(372/L)

Hence $\psi=\frac{3}{\sqrt{10}}\psi_2-\frac{1}{\sqrt{10}}\psi_3$

3 U =- Tosin (27/L) - - Tosin(37.0/L)

As

p=-ih

Hence

4?- = ned

Þy = ih osin (268/L) - hrasin(370/L)

1 37 pu=-ibi. 3 27 10 I Cos(27.0/L) - 10 i cos(37.2/L))

PL 3 3 27 <upu >= -ih 37 Tosin (20/L)- V10 ./L) - =sin(31x/L) 10 T Cos(27x/L)- 1 J-L 10 I Cos(37.2/L))d.

< Apoy >= 70 L (3sin(2.x / L)–sin (37.x /L) cos(262/L)- 37 Cos(37 x /L))da

_ г. 18 9л < ур>= - sin(2пх/L)cos(2п.x/L) - sin(2пх/L)cos(3пх/L) — 3л . е sin(Злх/L)cos(2пх/L) + c sin(3п.x/L)cos(37x/L))dx -LL бл

-ih L 97 < ÝÞŲ >= 70 L sin(472/L) – (sin(571/L) + sin(–1a/L)) – 31 (sin(572/L) + sin(Tx/L) + 3+ sin(67x /L))dx J-L

As all functions are SINE which are odd functions whose integration is zero .i.e=$\int_{-L}^L f(x)=0$ if f(x) is odd

Hence

0=< non >

Hence amplitude of expectation value of momentum is zero and frequency if we take as COSINE function is $n\pi/2$ where n=1,2,.... for SINE function it is $n\pi$