An "infinite potential box"by iAlexN Tags: None 

#1
Dec1012, 12:00 PM

P: 3

This equation describes a particle in an "infinite potential box" with the width L, i.e.:
Note that I do not know if it would be called an infinite potential box in English, but basically the particle can only be found within this space; outside of this the potential is [itex]\infty[/itex] <xn> = [itex]\Psi_{n}[/itex](x)=[itex]\sqrt{\frac{2}{L}}[/itex][itex]\ast[/itex]sin([itex]\frac{n*pi*x}{L}[/itex]) 0<x<L I have the following energy states for the particle, E[itex]_{1}[/itex] and E[itex]_{3}[/itex], [itex]\varphi[/itex]=[itex]\frac{1}{\sqrt{2}}[/itex][itex]\ast[/itex](1>+3>) Which means that it must be found in either E[itex]_{1}[/itex] and E[itex]_{3}[/itex] The question is: How will the probability of finding the particle change depending on time? I am not entirely sure how to solve this problem. I start like this: [itex]\Psi_{1}[/itex](x,t)=e[itex]^{i*E_{1}*t/\hbar}[/itex]*[itex]\Psi_{1}[/itex](x,0) =[itex]\sqrt{\frac{2}{L}}[/itex][itex]\ast[/itex]sin([itex]\frac{1*pi*x}{L}[/itex])*e[itex]^{i*E_{1}*t/\hbar}[/itex] [itex]\Psi_{3}[/itex](x,t)=e[itex]^{i*E_{3}*t/\hbar}[/itex]*[itex]\Psi_{3}[/itex](x,0) =[itex]\sqrt{\frac{2}{L}}[/itex][itex]\ast[/itex]sin([itex]\frac{3*pi*x}{L}[/itex])*e[itex]^{i*E_{3}*t/\hbar}[/itex] Inserting this into: [itex]\varphi[/itex]=[itex]\frac{1}{\sqrt{2}}[/itex][itex]\ast[/itex](1>+3>) Gives (simplified): [itex]\varphi[/itex](x,t)=[itex]\frac{1}{\sqrt{L}}[/itex][itex]\ast[/itex](e[itex]^{i*E_{1}*t/\hbar}[/itex]*sin([itex]\frac{1*pi*x}{L}[/itex])+e[itex]^{i*E_{3}*t/\hbar}[/itex]*sin([itex]\frac{3*pi*x}{L}[/itex])) [itex]\varphi[/itex](x,t)[itex]^{2}[/itex]: I think this would give me how the probability changes over time, but I am not sure mathematically how this should be expressed or determined, because [itex]\varphi[/itex](x,t) is a big expression; assuming this is the right way to solve this problem. Thank you in advance! Oh, and since this is my first post on this forum I am not sure if this would qualify to be placed in "Advanced Physics", so feel free to move it to another category. 



#2
Dec1012, 12:27 PM

HW Helper
P: 1,391

One is to split ##\phi(x,t)## up into real and imaginary components, as ##\phi(x,t)^2## is then just ##(\mbox{Re}\phi(x,t))^2 + (\mbox{Im}\phi(x,t))^2##. You can also make things a bit easier by factor out the factor of ##\exp(iE_1t/\hbar)##, as you can shift ##\phi(x,t)^2## by any overall phase factor. You can then do what I suggested above. This may simplify your final expression slightly simpler. 



#3
Dec1012, 01:38 PM

P: 3

[itex]\frac{1}{\sqrt{L}}e^{iE_{1}t/\hbar}[/itex](sin([itex]\frac{pix}{L}[/itex])+e[itex]^{i(E_{3}E_{1}t)/\hbar}[/itex]*sin([itex]\frac{3pix}{L}[/itex])) However I am not sure how this will make it easier, because the e[itex]^{i(E_{3}E_{1})t/\hbar}[/itex] factor is still there. I also tried to separate them to Real and Imaginary parts but that resulted in long and cumbersome expression of cos and sin functions. Some thing like this: [itex]\frac{1}{\sqrt{L}}[/itex]*[(Cos(E[itex]_{1}[/itex]t/[itex]\hbar[/itex])*sin([itex]\frac{pix}{L}[/itex])+Cos(E[itex]_{3}[/itex]t/[itex]\hbar[/itex])*sin([itex]\frac{3pix}{L}[/itex])+i(sin(E[itex]_{1}[/itex]t/[itex]\hbar[/itex])*sin([itex]\frac{pix}{L}[/itex])sin((E[itex]_{3}[/itex]t/[itex]\hbar[/itex])*sin([itex]\frac{3pix}{L}[/itex])] The reason that I don't want a long expression once [itex]\phi(x,t)[/itex] has been squared is because I need to draw how [itex]\phi(x,t)^2[/itex] will change shape depending on time, or describe it with words. I think, however, that the separation to real and imaginary parts will make it easier, but it is difficult to analyze the expression for different values for t, other than 0. Any further advice how you could analyze how it will fluctuate depending on time? 



#4
Dec1012, 03:32 PM

HW Helper
P: 1,391

An "infinite potential box" 



#5
Dec1012, 04:47 PM

HW Helper
Thanks ∞
PF Gold
P: 4,510

When I have expressions that look messy, I often try using simpler notation. You can write the product that you need to expand as
##(Ae^{ia}+Be^{ib})(Ae^{ia}+Be^{ib})##. Try expanding using this notation. You will get 4 terms that are not complicated. Two of the terms can be combined using ##e^{i\theta}+e^{i\theta} = 2cos\theta## 


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