Help Calculating probability between 2 limits for the ground state

  • #1
Hennessy
22
10
TL;DR Summary: Looking for help on a Intro to QM Problem

Hi All, THIS IS A GRADED PIECE OF WORK AT MY UNIVERSITY PLEASE DO NOT JUST GIVE ME THE ANSWER , I have made this post to see if what i've calculated seems reasonable, it sounds unlikely as 0.4 - 0.5L is in the middle of the well. The context is that we are in an infinite potential square well. The Question is : A quantum particle is trapped in a 1D infinite well with length L. Find the probability of finding the particle between 0.4L and 0.5L for the ground state. My answer is as follows:
$$ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi n x}{L}\right)$$ satisfies a free particle in a infinite square well. For the ground state we have $$ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi n x}{L}\right)$$ where n = 1 for the ground state and so becomes $$ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right)$$ Now we need to calculate $\lvert(\psi_x)\rvert^2$ which is $\psi^2(x) ={\frac{2}{L}} \sin^2\left(\frac{\pi x}{L}\right)$ Then to find the probability between 0.4 and 0.5 we need to evaluate this integral between those limits.
$$ \int_{0.4L}^{0.5L} (\psi_x)^2 dx = \frac{2}{L}\int_{0.4L}^{0.5L} \sin^2\frac{\pi x}{L} dx$$ Since $\sin^2$ cannot be integrated directly without producing Gaussian error functions we use the trigonometric identity that $\sin^2(Ax) = \frac{1 - \cos(2Ax)}{2}$ Where $A = \frac{\pi }{L}$ This means that our integral becomes $$ \frac{2}{L}\int_{0.4L}^{0.5L} \frac{1 - \cos(2\pi x)}{2} dx$$ Pulling out the $\frac{1}{2}$ leaves us with $$\frac{1}{L}\int_{0.4L}^{0.5L} 1-\cos(2 \pi x) dx$$ Computing this integral $$\frac{1}{L} \left[ x - \frac{L}{2\pi} \sin\left(\frac{2\pi x}{L}\right) \right]_{0.4L}^{0.5L} $$
Computing for the upper limit leads us to $$ \frac{1}{L} \left[ 0.5L - \frac{L}{2\pi} \sin(\frac{2\pi L}{2L})\right] = 0.5 -\frac{2\pi}{1} \times \sin(\pi) = 0.5$$ For the lower limit $$= 0.4 - \frac{L}{2 \pi} \sin(0.8 \pi) = 0.4$$ so the probability that you will find the particle in the ground state between our 2 limits is $$ 0.5-0.4 = P = 0.1$$

Am i on the right tracks or have i made a mistake somewhere ? i think i'm correct but just want to have others opinion on if i have done something wrong and if i have please tell me where and why im wrong so i can improve myself in physics, thank you <3.
 
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  • #2
You can't let ##L## pop in and out in the calculation! Look at the evaluation again..
 
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  • #3
BvU said:
You can't let ##L## pop in and out in the calculation! Look at the evaluation again..
which part are you refering to please ? oh is it the part where i wrote $$\frac{L}{2\pi} \sin(\frac{\pi x}{L})$$ ?
 
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  • #4
Seems a little small to me. Make a graph of ##sin^2 ## and see if the number is reasonable.......
 
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  • #5
hutchphd said:
Seems a little small to me. Make a graph of ##sin^2 ## and see if the number is reasonable.......
what i was thinking , i had two combative thought, one was it makes sense as its based in the middle of the function which to me is towards the peak of the wavefunction and so its a thin slice around the peak of the wavefunction which made it a low probability. My other thought was although P =1 to subtract 0.4+0.5 just seems way to cheap and easy and could of just been done without calculating any integrals or anything. So just confused where to go from here, i don't think my maths is incorrect but i might of made mistakes somewhere any advice?
 
  • #6
Yes. Make a graph and see if your result is reasonable. If you have screwed up, probably the result will look very wrong. If you do this and check the units carefully you will find almost every mistake on your own. As a careless person, I know how to check my work. Keeps me from blunders (usually). Check every step every time.
 
  • #7
hutchphd said:
Yes. Make a graph and see if your result is reasonable. If you have screwed up, probably the result will look very wrong. If you do this and check the units carefully you will find almost every mistake on your own. As a careless person, I know how to check my work. Keeps me from blunders (usually). Check every step every time.
yeah Sin^2(0.1x) looks very unreasonable, so now i need to figure out where i went wrong :(
 
  • #8
Use a little more care. Why ##sin^2(0.1x)##??
 
  • #9
Hennessy said:
THIS IS A GRADED PIECE OF WORK AT MY UNIVERSITY PLEASE DO NOT JUST GIVE ME THE ANSWER
Thread moved to advanced physics homework help.
 
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  • #10
PeterDonis said:
Thread moved to advanced physics homework help.
Apologies, i thought as it was introductory quantum mechanics i felt that was the most appropiate place to putt it, apologies.
 
  • #11
hutchphd said:
Use a little more care. Why ##sin^2(0.1x)##??
I have recalculated the integrals and now i have that the probability of finding the particle at the said point is 0.19 ?
 
  • #12
Hennessy said:
I have recalculated the integrals and now i have that the probability of finding the particle at the said point is 0.19 ?
apologies sin^2(x)=0.1 or sin^2(x)=0.19 , a very narrow slice, and 0.19 is a bigger width
 
  • #13
Hennessy said:
I have recalculated the integrals and now i have that the probability of finding the particle at the said point is 0.19 ?
This answer looks reasonable. Can you tell us what correction(s) you made to your original attempt?

Note that you are not calculating the probability that the particle is at some point. Rather, you are calculating the probability of finding the particle somewhere in the interval ##0.4 L < x < 0.5 L##. This is probably what you meant, but it is important to be precise with the language.
 
  • #14
Hennessy said:
I have recalculated the integrals and now i have that the probability of finding the particle at the said point is 0.19 ?
One trick is to calculate the integral from ##aL## to ##0.5L## and then you have an expression (in terms of ##a##) that you can sanity check. It should be 0.5 for ##a = 0## and 0 for ##a = 0.5##. Once you put the numbers in, it's harder to spot any mistakes.

I agree with that answer of 0.19.
 
  • #15
PS another trick is to set ##u = \frac x L## to see that the answer is independent of ##L##. That simplifies the integral and is something to remember for future calculations.
 
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  • #16
Hennessy said:
which part are you refering to please ? oh is it the part where i wrote $$\frac{L}{2\pi} \sin(\frac{\pi x}{L})$$ ?
No, where you wrote
$$
\frac{2}{L}\int_{0.4L}^{0.5L} \frac{1 - \cos(2\pi x)}{2} dx$$ when you meant $$\frac{2}{L}\int_{0.4L}^{0.5L} \frac{1 - \cos(2\pi x/L)}{2} dx$$

Anyway, let me show you a cheat:
https://www.wolframalpha.com/input?i=2/L+int+sin^2(pi+*x/L),+x=0.4L,0.5L

this is where you can also easily check that the normalisation ## \sqrt{2\over L}## is correct, too :wink:

##\ ##
 
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  • #17
TSny said:
This answer looks reasonable. Can you tell us what correction(s) you made to your original attempt?

Note that you are not calculating the probability that the particle is at some point. Rather, you are calculating the probability of finding the particle somewhere in the interval ##0.4 L < x < 0.5 L##. This is probably what you meant, but it is important to be precise with the language.
Hello so the correction i made was that i changed $$\frac{1}{L} \left[ x - \frac{L}{2\pi} \sin\left(\frac{2\pi x}{L}\right) \right]_{0.4L}^{0.5L}$$ and changed it to $$\frac{1}{L} \left[ x - \frac{1}{2\pi} \sin\left(\frac{2\pi x}{L}\right) \right]_{0.4L}^{0.5L}$$ this didn't change the value for my upper limit but did change the value for my lower limit which simplifies to

Computing for the upper limit leads us to $$ \frac{1}{L} \left[ 0.5L - \frac{1}{2\pi} \sin(\frac{2\pi L}{2L})\right] = 0.5 -\frac{2\pi}{1} \times \sin(\pi) = 0.5$$ For the lower limit $$= 0.4 - \frac{1}{2 \pi} (0.5878) \approx 0.31(2.s.f)$$ $\sin(0.8\pi) \approx 0.5878$ radians. So the probability that you will find the particle in the the interval between 0.4L and 0.5L is $$ 1 - 0.5+0.31 = P \approx 0.19$$

Thank you for making me more specific on my language, it is what i meant but thank you for literating it for me, i understand its important to understand these problems and use their vocabulary correctly so i can explain what im stuck on so thank you for putting what i couldn't say into words <3
 
  • #18
PeroK said:
PS another trick is to set ##u = \frac x L## to see that the answer is independent of ##L##. That simplifies the integral and is something to remember for future calculations.
so perhaps to simplify the integral after using a trig substitution is to use a change of variables to integrate it?
 
  • #19
BvU said:
No, where you wrote
$$
\frac{2}{L}\int_{0.4L}^{0.5L} \frac{1 - \cos(2\pi x)}{2} dx$$ when you meant $$\frac{2}{L}\int_{0.4L}^{0.5L} \frac{1 - \cos(2\pi x/L)}{2} dx$$

Anyway, let me show you a cheat:
https://www.wolframalpha.com/input?i=2/L+int+sin^2(pi+*x/L),+x=0.4L,0.5L

this is where you can also easily check that the normalisation ## \sqrt{2\over L}## is correct, too :wink:

##\ ##
DUDE WHAT THIS IS HACK ON LIFE BROOOOOOOOO, thank you so much
 
  • #20
Hennessy said:
so perhaps to simplify the integral after using a trig substitution is to use a change of variables to integrate it?
That's not what I meant. Use the substitution ##u = \frac x L## to eliminate ##L## from the calculation.

Note that physically the probability cannot depend on the width of the well. You could choose your units of length so the ##L = 1##. That's a physical (rather than mathematical) way to eliminate ##L##.
 
  • #21
This is now the updated Answer.

$$ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi n x}{L}\right)$$ satisfies a free particle in a infinite square well. For the ground state we have $$ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi n x}{L}\right)$$ where n = 1 for the ground state and so becomes $$ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right)$$ Now we need to calculate $\lvert(\psi_x)\rvert^2$ which is $\psi^2(x) ={\frac{2}{L}} \sin^2\left(\frac{\pi x}{L}\right)$ Then to find the probability between 0.4L and 0.5L we need to evaluate this integral between those limits.
$$ \int_{0.4L}^{0.5L} (\psi_x)^2 dx = \frac{2}{L}\int_{0.4L}^{0.5L} \sin^2\frac{\pi x}{L} dx$$ Since $\sin^2$ cannot be integrated directly without producing Gaussian error functions we use the trigonometric identity that $\sin^2(Ax) = \frac{1 - \cos(2Ax)}{2}$ Where $A = \frac{\pi }{L}$ This means that our integral becomes $$ \frac{2}{L}\int_{0.4L}^{0.5L} \frac{1 - \cos(2\pi x)}{2} dx$$ Pulling out the $\frac{1}{2}$ leaves us with $$\frac{1}{L}\int_{0.4L}^{0.5L} 1-\cos(2 \pi x) dx$$ Computing this integral $$\frac{1}{L} \left[ x - \frac{1}{2\pi} \sin\left(\frac{2\pi x}{L}\right) \right]_{0.4L}^{0.5L} $$
Computing for the upper limit leads us to $$ \frac{1}{L} \left[ 0.5L - \frac{1}{2\pi} \sin(\frac{2\pi L}{2L})\right] = 0.5 -\frac{2\pi}{1} \times \sin(\pi) = 0.5$$ For the lower limit $$= 0.4 - \frac{1}{2 \pi} (0.5878) \approx 0.31(2.s.f)$$ $\sin(0.8\pi) \approx 0.5878$ radians. So the probability of finding the particle on the interval
between 0.4L and 0.5L is $$ 1 - 0.5+0.31 = P \approx 0.19$$
 
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  • #22
Hennessy said:
DUDE WHAT THIS IS HACK ON LIFE BROOOOOOOOO, thank you so much
These days there are packages such as Maple or Mathematica or Wolfram Alpha that will do integrals and other calculations. And, the next generation of AI maths packages can't be far away. That said, you have to decide how much capability you need to carry out or follow complex algebraic calculations. Your course may demand that you can manage complex mathematics on your own. Perhaps the best advice for the time being is to use Wolfram to check things after you've done the integral yourself.
 
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  • #23
PeroK said:
These days there are packages such as Maple or Mathematica or Wolfram Alpha that will do integrals and other calculations. And, the next generation of AI maths packages can't be far away. That said, you have to decide how much capability you need to carry out or follow complex algebraic calculations. Your course may demand that you can manage complex mathematics on your own. Perhaps the best advice for the time being is to use Wolfram to check things after you've done the integral yourself.
Yeah thats what ive learned so far on my course. Its more of a attempt the question yourself and when your really completely stuck look at someone else who has solved something similar and try to map it on to my problem and justify the reasoning. I appreciate being made aware of tools like this though just so i can check if an answer is correct. I have only just started to use mathematica which im finding okay and only started using latex 2 weeks ago so i'm just getting introduced to this all. My anticipation is that i would like to work through these type of questions or most questions in general in exam conditions so i'm prepared for my exam in may. thank you and any further advice ?
 
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  • #24
Hennessy said:
Thank you and any further advice ?
There's too much. To emphasise one point. QM is sufficiently complicated mathematically that it is difficult not to make mistakes here and there. A full solution with no algebraic errors is hard to achieve. So, you need to develop good techniques to pick up on errors yourself. Keeping things general and algebraic for longer allows you to sanity check what you have.

As an example, let's take this integral:
$$2\int_a^b \sin^2(\pi x) \ dx = \int_a^b 1 - \cos(2\pi x) \ dx$$$$= (b-a) - \frac 1 {2\pi}\big [\sin(2\pi b) - \sin(2\pi a) \big ]$$Now, you can bang that into a spreadsheet in a few minutes for all the increments of 0.1 from 0 to 1 to get:

abP(a, b)
0​
0.1​
0.006​
0.1​
0.2​
0.042​
0.2​
0.3​
0.100​
0.3​
0.4​
0.158​
0.4​
0.5​
0.194​
0.5​
0.6​
0.194​
0.6​
0.7​
0.158​
0.7​
0.8​
0.100​
0.8​
0.9​
0.042​
0.9​
1​
0.006​
1.000​

That doesn't guarantee that we have the correct answer. But, after doing that I have a slightly warmer, fuzzier feeling!
 
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  • #25
Hennessy said:
Since $\sin^2$ cannot be integrated directly without producing Gaussian error functions
What Gaussian error functions?
Here is another hack for you, bro. Break down sines and cosines into complex exponentials, integrate, then reassemble into sines and cosines. If you start with a real integral, you should end up with a real result. Here is what I mean $$\begin{align} & \int \sin^2x~dx=\int\left(\frac{e^{ix}-e^{-ix}}{2i} \right)^2 dx=\int\left(-\frac{1}{4}\right)\left(e^{2ix}-2+e^{2ix} \right) dx\nonumber \\
& =-\frac{1}{4}\left(\frac{1}{2i}e^{2ix}-\frac{1}{2i}e^{-2ix}-2x \right)=-\frac{1}{4}\left(\frac{e^{2ix}-e^{-2ix}}{2i}\right)+\frac{x}{2} \nonumber \\& =-\frac{\sin(2x)}{4}+\frac{x}{2}. \nonumber \\
\end{align}$$You don't have to remember trig identities with this method other than how to express a sine and cosine in terms of complex exponentials which are easy to integrate.
 
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  • #26
While we are at it, using parts:
$$\int \sin^2 x \ dx = \sin x (-\cos x) - \int \cos x (-\cos x) \ dx$$$$= -\sin x \cos x + \int 1 - \sin^2 x \ dx = x -\sin x \cos x - \int \sin^2 x \ dx$$$$\implies \int \sin^2 x \ dx = \frac 1 2 \bigg [x -\sin x \cos x \bigg ] + C$$
 
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  • #27
kuruman said:
What Gaussian error functions?
Here is another hack for you, bro. Break down sines and cosines into complex exponentials, integrate, then reassemble into sines and cosines. If you start with a real integral, you should end up with a real result. Here is what I mean $$\begin{align} & \int \sin^2x~dx=\int\left(\frac{e^{ix}-e^{-ix}}{2i} \right)^2 dx=\int\left(-\frac{1}{4}\right)\left(e^{2ix}-2+e^{2ix} \right) dx\nonumber \\
& =-\frac{1}{4}\left(\frac{1}{2i}e^{2ix}-\frac{1}{2i}e^{-2ix}-2x \right)=-\frac{1}{4}\left(\frac{e^{2ix}-e^{-2ix}}{2i}\right)+\frac{x}{2} \nonumber \\& =-\frac{\sin(2x)}{4}+\frac{x}{2}. \nonumber \\
\end{align}$$You don't have to remember trig identities with this method other than how to express a sine and cosine in terms of complex exponentials which are easy to integrate.
tried putting it into an online integral calculator and it came back saying erg functions was all. So perhaps i should try to use complex forms to try and make some of the integrals easier as in general exponentials are easier to work with ?
 
  • #28
PeroK said:
There's too much. To emphasise one point. QM is sufficiently complicated mathematically that it is difficult not to make mistakes here and there. A full solution with no algebraic errors is hard to achieve. So, you need to develop good techniques to pick up on errors yourself. Keeping things general and algebraic for longer allows you to sanity check what you have.

As an example, let's take this integral:
$$2\int_a^b \sin^2(\pi x) \ dx = \int_a^b 1 - \cos(2\pi x) \ dx$$$$= (b-a) - \frac 1 {2\pi}\big [\sin(2\pi b) - \sin(2\pi a) \big ]$$Now, you can bang that into a spreadsheet in a few minutes for all the increments of 0.1 from 0 to 1 to get:

abP(a, b)
0​
0.1​
0.006​
0.1​
0.2​
0.042​
0.2​
0.3​
0.100​
0.3​
0.4​
0.158​
0.4​
0.5​
0.194​
0.5​
0.6​
0.194​
0.6​
0.7​
0.158​
0.7​
0.8​
0.100​
0.8​
0.9​
0.042​
0.9​
1​
0.006​
1.000​

That doesn't guarantee that we have the correct answer. But, after doing that I have a slightly warmer, fuzzier feeling!
saved this now , thank you!
 
  • #29
Hennessy said:
So perhaps i should try to use complex forms to try and make some of the integrals easier as in general exponentials are easier to work with ?
When dealing with the wavefunctions related to the particle in a box situation, complex exponentials can be very handy for casting them in terms of the known eigenstates. Consider the following problem.

A particle is in a box of width ##L## and is characterized by the wavefunction $$\psi(x)=N\sin^3\left(\frac{\pi~x}{L}\right).$$Find the value of the normalization constant ##N##.

To normalize the wavefunction, you might start with $$1=N^2\int_0^L\sin^6\left(\frac{\pi~x}{L}\right)dx$$and then try to do the integral by some method or other. I prefer the method of complex exponentials because they allow you to rewrite the sine cubed in terms of first powers of sines that are, conveniently, the eigenstates of the particle in a box. This in turn allows you to find the values of integrals without actually integrating by exploiting the orthonormality condition of the eigenstates. Here is what I mean.

$$\begin{align} & \sin^3u=\left(\frac{e^{iu}-e^{-iu}}{2i}\right)^3=-\frac{1}{4}\frac{\left(e^{3iu}-3e^{iu}+3e^{-iu}-e^{-3iu}\right)}{2i} \nonumber \\
& =-\frac{1}{4}\left[\sin(3u)-3\sin(u) \right]=\frac{3}{4}\sin(u)-\frac{1}{4}\sin(3u). \nonumber
\end{align}$$ Then with ##u=\dfrac{\pi~x}{L}## you can write $$\begin{align}
& \psi(x)=\frac{N}{4}\left[3\sin\left(\frac{\pi x}{L}\right)-\sin\left(\frac{3\pi x}{L}\right)\right]
=\frac{N}{4}\sqrt{\frac{L}{2}} \left[3\sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)-\sqrt{\frac{2}{L}}\sin\left(\frac{3\pi x}{L}\right)\right] \nonumber \\
& =\frac{N}{4}\sqrt{\frac{L}{2}} \left[3\psi_1(x)-\psi_3(x)\right].\nonumber
\end{align}$$where ##\psi_i(x)## are the normalized eigenstates of the particle in a box. Now you can write $$1=\frac{N^2}{16}\frac{L}{2}\int_0^L\left[3\psi_1(x)-\psi_3(x)\right]\left[3\psi_1(x)-\psi_3(x)\right]dx$$and take advantage of the orthonormality condition $$\int_0^L\psi_i(x)\psi_j(x)dx=\delta_{ij}$$ to reduce the above without integration to $$1=\frac{N^2}{16}\frac{L}{2}(9+1)\implies N=\frac{4}{\sqrt{5L}}.$$To complete the picture, the normalized given wavefunction is, in terms of the normalized eigenstates, $$\psi(x)=\frac{1}{\sqrt{10}}[3\psi_1(x)-\psi_3(x)].$$
 
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