Question

\(\Psi = \sqrt{\dfrac 2 L} sin{(\dfrac {2\pi x}{L})}\) for 0\(\le\)x\(\le\) L.
Determine the expectation value.


The expectation value is calculated as \[ = \int_{-\infty}^{+\infty} \Psi^* x \Psi dx\] where \(\Psi^*\) is the complex conjugate of \(\Psi\). Anyone know how to do the calculation? (a) Determine the expectation value of x. (b) Determine the probability of finding the particle near \(\dfrac 1 2\)L by calculating the probability that the particle lies in the range 0.490L\(\le\) x\(\le\) 0.510L. (c) What If? Determine the probability of finding the particle near \(\dfrac 1 4\)L by calculating the probability that the particle lies in the range 0.240L \(\le\) x \(\le\) 0.260L. (d) Argue that the result of part (a) does not contradict the results of parts (b) and (c).

Answer 1

@ybarrap
@agent0smith

Answer 2

@niksva

Answer 3

complex conjugate of wave function would be same
so we will get
\[\frac{ 2 }{ L } \int\limits_{0}^{L} x*\sin^2(\frac{ 2 pix }{ L })\]
is that clear?

Answer 4

But why is it the same? Aren't I working with complex numbers?

Answer 5

yeah we are working with the complex number
in this case we have the real wave function
not an imaginary wave function

Answer 6

How do you determine that? (Sorry, new to Quantum Physics)

Answer 7

u must have done the derivation of wavefunction for 1D potential well??????

Answer 8

sorry, computer crashed.

Okay, because it's oscillating, it has to be a Sin/Cos wave. Due to boundary conditions, It has to be the sin wave.

Answer 9

Is that what you're referring to?

Answer 10

sorry owl was not allowing me to come here :-P

Answer 11

Me too

Answer 12

Anyways ^^^^^^^
was that what you were talking about?

Answer 13

yeah because of the boundary condition we have sine wave

Answer 14

now can u determine expectation value?

Answer 15

Isn't it just integration?

Answer 16

Is there more to it? (I just haven't done calculus in a while...=( )

Answer 17

right now its just integration
use double cosine angle formulae
u will surely get the expectation value of x

Answer 18

I'm going to take a little while.

Answer 19

cool take your time

Answer 20

\[sin^2(\theta /2)={\dfrac 1 2} (1-cos\theta)\]
Is this the one you're talking about?

Answer 21

yeah

Answer 22

I'm not sure how to manipulate that since it's \(\theta /2\) on the inside.

Answer 23

can we write sin^2(theta) = 1/2 (1-cos(theta) ) ???????

Answer 24

I think I recognize that one.

Answer 25

i m now writing all the steps in one go

Answer 26

\(\huge \dfrac {1}{L} \int_{-\infty}^{+\infty} x[ 1-cos(2{\dfrac{2\pi x}{L}}) ] dx\)

Answer 27

That's what I get so far

Answer 28

\[\frac{ 2 }{ L }\int\limits_{0}^{L}x*\sin^2(n pix/L)\]
\[\frac{ 2 }{ L }\int\limits_{0}^{L} x*\frac{ 1 }{ 2 }(1-\cos(\frac{ 4 pix }{ L })) dx\]

Answer 29

Why 0 to L?

Answer 30

Limits should be in between 0 to L as it is a potential box of dimension L

Answer 31

Is this integration by parts? due to the x on the outside?

Answer 32

\[\frac{ 1 }{ L }\int\limits_{0}^{L} \ (x- xcos(\frac{ 4 pix }{ L })) dx\]
for xcos(4pix/L) go for integration by part while other one is simpler

Answer 33

ok so I get

\[\dfrac L 2-\dfrac 1 L\int_0^L {x cos({\dfrac {4\pi x}{L}})}dx\]

Answer 34

yeah now solve the integral
i feel the integral will come out to be zero as this is an odd function

Answer 35

How do you figure it's odd?? The fact that it's a sine function?

Answer 36

we have a function f(x) = xcos(4pix/L)
function is odd when f(-x) = - f(x)
in the above case condition f(-x) = -f(x) satisfies
hence it is an odd function

Answer 37

But its limits are from 0 to L, not -L to L

Answer 38

yeah it is
it was just my feeling :-)
we will solve this out dont worry

Answer 39

one sec; almost done with the integration by parts.

Answer 40

Ok yeah, I think you were right

Answer 41

This is my final answer

Answer 42

\(\Large \dfrac L 2-\dfrac {x}{4\pi}sin({\dfrac {4\pi x}{L}})\)

Answer 43

\[x*\frac{ \sin(\frac{ 4 \pi x }{ L } }{ \frac{ 4 \pi }{ L } }-\int\limits_{}^{}\frac{ d }{ dx }(x) \frac{ \sin \frac{ 4 \pi x }{ L } }{ \frac{ 4\pi }{ L } }\]
\[x*\frac{ \sin \frac{ 4 \pi x }{ L } }{ \frac{ 4 \pi }{ L } } - \int\limits_{}^{}\frac{ L }{ 4 \pi }* \sin (\frac{ 4 \pi x }{ L})dx\]
can u do it now?

Answer 44

\[\frac{ Lx }{ 4\pi } \sin \frac{ 4 \pi x }{ L } +\frac{ L^2 }{ 16 \pi^2 } \cos(\frac{ 4 \pi x }{ L })\]
now apply limit

Answer 45

I already told you my final answer, but it had x in it...

Answer 46

u did not apply limit right
u have to put upper limit L and lower limit 0 in place of x

Answer 47

my final answer is
\[\frac{ L }{ 2 } +\frac{ L }{ 16 \pi^2 }\]

Answer 48

\[\Large \dfrac L 2-\dfrac {x}{4\pi}sin({\dfrac {4\pi x}{L}})\]
so if I apply limits 0 and L to this?

Answer 49

I get L/2.... I'm so confused

Answer 50

u have done integration by parts wrong
see the steps done by me first

Answer 51

\[\frac{ Lx }{ 4\pi } \sin \frac{ 4 \pi x }{ L } +\frac{ L^2 }{ 16 \pi^2 } \cos(\frac{ 4 \pi x }{ L })\]
In the first term, why is there an L in the numerator? didn't the 1/L at the beginning cancel it out?

Answer 52

\[\frac{ \color {red }{\Large L} x }{ 4\pi } \sin \frac{ 4 \pi x }{ L } +\frac{ L^2 }{ 16 \pi^2 } \cos(\frac{ 4 \pi x }{ L })\]

Answer 53

@niksva I put the integral into wolfram alpha, it's L/2.

Answer 54

I need help with parts b-d.

Answer 55

@theEric

Answer 56

@roadjester in part b and c integral will will be same
only the limit has been changed
in the first part we are expecting the particle to be find in one dimensional well of length L
now in other parts , range of well has been reduced

Answer 57

this is your solution of first part
Last day I was just doing here on this forum
today i took the paper and pen

Answer 58

Thanks @niksva
I managed to solve the rest of the problem but thanks for the help.