An "infinite potential box"


by iAlexN
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iAlexN
iAlexN is offline
#1
Dec10-12, 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]

<x|n> = [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.
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Mute
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#2
Dec10-12, 12:27 PM
HW Helper
P: 1,391
Quote Quote by iAlexN View Post
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]

<x|n> = [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.
It looks like you have the right approach. ##|\phi(x,t)|^2## is indeed the probability; I'm afraid there's no way around calculating that. You can do some things to make the calculation a bit easier, though.

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.
iAlexN
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#3
Dec10-12, 01:38 PM
P: 3
Quote Quote by Mute View Post
It looks like you have the right approach. ##|\phi(x,t)|^2## is indeed the probability; I'm afraid there's no way around calculating that. You can do some things to make the calculation a bit easier, though.

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.
When I factor out e[itex]^{E_{1}t/\hbar}[/itex]:

[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?

Mute
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#4
Dec10-12, 03:32 PM
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P: 1,391

An "infinite potential box"


Quote Quote by iAlexN View Post
When I factor out e[itex]^{E_{1}t/\hbar}[/itex]:

[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.
The problem is fundamentally time dependent; you cannot entirely get rid of the exponentials, but this change does mean that when you calculate ##|\phi(x,t)|^2## you will get factors of ##\cos((E_3-E_1)t/\hbar)## and ##\sin((E_3-E_1)t/\hbar)## in your expression rather than sums or differences of ##\cos(E_3t/\hbar)## and ##\cos(E_1t/\hbar)##, etc.

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])]
Yes, as you can see, you have gotten something ugly because you didn't remove the exponential phase factor that I suggested. Also, if you're using LaTeX to type the hbar's, you should use it for the entire equation - it's easier to read that way.

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.
Once you've calculated the cleaned up version it will be easier to draw, because there will be several relatively nice values of t that you can plot.
TSny
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#5
Dec10-12, 04:47 PM
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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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