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The infinite square well

  • #1
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A particle is in ground state of an infinite square well. Find the probabilirt of finding the particle in the interval [tex]\Delta x = 0.002L[/tex] at x=L. (since delta x is small, do not integrate)

here's what I have:

[tex]\Psi*\Psi = P(x) = \frac{2}{L} sin^2 \left(\frac{ \pi x}{L} \right) \Delta x[/tex]

[tex]P = \frac{2}{L} sin^2 \left(\frac{ \pi L}{L} \right) 0.002L [/tex]
[tex]P = 2sin^2 \left(\pi \right) 0.002 [/tex]
[tex]P=0.004[/tex]

is this the correct method?
 
Last edited:

Answers and Replies

  • #2
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you can't just ignore the sin(Pi) term because it is 0. You need to use a first order approximation of sin(Pi+delta) and square the first order approximation in order to get the result you want.

~Lyuokdea
 
  • #3
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what is a first order approximation? and why would I use sin(Pi+delta) ?
 
  • #4
Meir Achuz
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UrbanXrisis said:
A particle is in ground state of an infinite square well. Find the probabilirt of finding the particle in the interval [tex]\Delta x = 0.002L[/tex] at x=L. (since delta x is small, do not integrate)

here's what I have:

[tex]\Psi*\Psi = P(x) = \frac{2}{L} sin^2 \left(\frac{ \pi x}{L} \right) \Delta x[/tex]

[tex]P = \frac{2}{L} sin^2 \left(\frac{ \pi L}{L} \right) 0.002L [/tex]
[tex]P = 2sin^2 \left(\pi \right) 0.002 [/tex]
[tex]P=0.004[/tex]

is this the correct method?
You made a trig mistake with [tex]sin(\pi)[/tex].
It should = zero.
I disagree with the hint.
You can expand [tex]sin^2(\pi-z)[/tex] for small z, and then integrate from L-\delta x to L. You could also integrate without the expansion, and just be careful about the numbers.
 
  • #5
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sorry, i used bad terminology in the hint, use Meir Archuz's hint, try to do a taylor series expansion for sin(x) around x=0 and see how that applies to your problem.

~Lyuokdea
 
  • #6
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wouldnt the possibilty be just = 0? I dont think this question was designed for the use of the taylor series.
 
  • #7
Physics Monkey
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The probability is not zero. You only get zero because you've made too rough of an approximation to the integral. To get a better estimate of the propability, try evaluating the wavefunction somewhere else in the interval, say at the midpoint.
 
  • #8
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delta x=L/2
[tex]P = \frac{2}{L} sin^2 \left(\frac{ \pi }{2} \right) 0.002L [/tex]
P=0.004?
 
  • #9
Physics Monkey
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Hi UrbanXrisis,

You have misunderstood me, but let me ask you, do you think your answer makes sense? Should the probability depend on the value of the wavefunction in the middle of the box? To clarify, what I mean was that you should perhaps look at the midpoint of the interval [tex] [ L - .002 L, L] [/tex].
 
  • #10
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i'm not sure. there is an example in my text that is exacly the same problem. "what would be the probability of finding an electron while in ground state in a very narrow region delta x = 0.01L wide centered at x=5L/8"

The way that the book goes about solving this question was using:
[tex]P = \frac{2}{L} sin^2 \left(\frac{ \pi (5L/8)}{L} \right) 0.01L=0.017[/tex]

so I am following their procedure
 
  • #11
qtp
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why would the probability not be zero at x=L ??
it should be zero AT the boundary but little to the left say [itex] L-\Delta x [/itex] it will be almost zero but very small
 
  • #12
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UrbanXrisis said:
i'm not sure. there is an example in my text that is exacly the same problem. "what would be the probability of finding an electron while in ground state in a very narrow region delta x = 0.01L wide centered at x=5L/8"

The way that the book goes about solving this question was using:
[tex]P = \frac{2}{L} sin^2 \left(\frac{ \pi (5L/8)}{L} \right) 0.01L=0.017[/tex]

so I am following their procedure
but is this what I should follow when doing my problem?
 
  • #13
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no, because then you will get an answer that is zero.

For the spread centered around a point like x=5L/8, the above approximation is ok, because you can assume that (5+.01)L/8 is approximately similar to 5L/8.

However, you cannot do this for the point x=L because sin(Pi*x/L) is exactly 0 at that point. Thus sin(Pi*(1-.01)*L/L) is not approximately the same as sin(Pi).

Have you done work with Taylor Series? How does sin(x) change when x is close to zero? Find the first non-zero term in the taylor expansion, and use that in order to find a non-zero answer. For most classes, the first non-zero term is a good enough answer.

~Lyuokdea
 
  • #14
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but according to qpt:

qtp said:
why would the probability not be zero at x=L ??
it should be zero AT the boundary but little to the left say [itex] L-\Delta x [/itex] it will be almost zero but very small
 
  • #15
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that's exactly right, but your answer gives you a number that is exactly zero, because the sin (Pi) is 0. To find the small but non-zero answer, you need to expand sin(Pi) into it's taylor series.

Physics monkey gave another good suggestion which gives you a non-zero answer without depending on Taylor series, simply find the midpoint of the range [L-.002, L] and then use that value, as shown in the example which you quoted from the book, to find a first order approximation of the answer.

~Lyuokdea
 
  • #16
Meir Achuz
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Just do the simple integral of sin^2 from .998L to L.
If you can't get enough accuracy, then expand either before or after integrating.
 

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