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Topic: Particle in one-dimension box  (Read 5937 times)

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Offline Dolphinsiu

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Particle in one-dimension box
« on: December 03, 2008, 01:47:22 PM »
Consider a particle in a 1D box of length L. Let the particle be in an energy eignestate Yn(x)

(i) Calculate the probability P(1/4) of finding the particle in the left quarter (0 (< and equal to) x (< and equal to) L/4)

(ii) Give an explicit expression for P(1/4) for even and odd values of the quantum number n

(iii) Which value of n gives the maximum probability? What is the maximum value?

I have done part (i) and the answer is 1/4, but I get stuck with part (ii) and hence part (iii). Please help me! Thanks.

Offline Yggdrasil

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Re: Particle in one-dimension box
« Reply #1 on: December 03, 2008, 11:10:53 PM »
How did you do part (i)?  Do they specify which eigenstate you're looking at?

For (ii), the best place to start would be writing down the formula for the general equation for the wavefunction of the eigenstates then squaring it to find the probability density function.

Offline Dolphinsiu

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Re: Particle in one-dimension box
« Reply #2 on: December 08, 2008, 08:50:38 AM »
But if you say so, what is the difference between part (i) and (ii). I do part (i) by this method.

Yn(x) = sqrt.(2/L) sin (npix/L)

P(1/4)
= S(0 to L/4) [Yn(x)]2 dx
= 2/L S (0 to L/4) sin2(npix/L) dx
= 1/L S (0 to L/4) 1 - cos2(2npix/L) dx
= 1/4

Then I think your method used in part (ii) is not correct as the question needs explicit expression rather than a constant.

but how?...This question is found in my exam paper. Is this question itself full of mistakes?

Offline tamim83

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Re: Particle in one-dimension box
« Reply #3 on: December 15, 2008, 03:22:49 PM »
Your integration in part i is incorrect.  Try using

S(sin ax)dx = (x/2) - [(sin 2ax)/2a] + C

This gives you an expression in terms of n.  Then substitute n=2n (for even n) and n=2n+1 (for odd n) to get the respective expressions. 

Hope this helps.   ;D

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