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Can you explain the working to this question? Probability - infinite potential well

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1. Question 3, I copied the way to do it from the board last year but don't understand it. Can you explain why there's a subtitle 2 beside the psi? And explain why the working is worked out like this?

Attachment 582508582510* *
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2. (Original post by Airess3)
Question 3, I copied the way to do it from the board last year but don't understand it. Can you explain why there's a subtitle 2 beside the psi? And explain why the working is worked out like this?

Attachment 582508582510* *
Why ?

It has to do with Born-interpretation of wavefunction.

is interpreted as the probability density that the quantum particle is at the position x.

Then the probability of the quantum particle is found in the interval of is

It seems that the wavefunction has something missing. See the link below.
http://hyperphysics.phy-astr.gsu.edu...ntum/pbox.html

You may want to ask your classmate for a copy of the solution. It seems that your integration is wrong.
3. (Original post by Eimmanuel)
Why ?

It has to do with Born-interpretation of wavefunction.

is interpreted as the probability density that the quantum particle is at the position x.

Then the probability of the quantum particle is found in the interval of is

It seems that the wavefunction has something missing. See the link below.
http://hyperphysics.phy-astr.gsu.edu...ntum/pbox.html

You may want to ask your classmate for a copy of the solution. It seems that your integration is wrong.
I asked the professor for some help but now I'm stuck on how he got from the 2nd last step to the last step (this is just part of the working):

*

How did he integrate that? Is there a specific formula that I can use? and how did he turn the 2/a into 1/a?*

And now I've tried to do the rest of the working but the a's don't cancel out?

*
4. It's just normal integration. Note that there's a 2/a outside one bracket and a 1/2 immediately to its right. So multiplying them (during the integration step) gives 1/a. That's not affected by integration because x is the integration variable (not a).

Regarding the a not cancelling: check your very last step.

General tip, either use decimals or (better) only fractions. Something like '2*pi*(0.25a)/a' is unnecessarily complicated notation. (2*pi*a)/(4*a) is so much easier to clean up in a single step without a mistake. It's also faster to write in an exam.
5. (Original post by mik1a)
It's just normal integration. Note that there's a 2/a outside one bracket and a 1/2 immediately to its right. So multiplying them (during the integration step) gives 1/a. That's not affected by integration because x is the integration variable (not a).

Regarding the a not cancelling: check your very last step.

General tip, either use decimals or (better) only fractions. Something like '2*pi*(0.25a)/a' is unnecessarily complicated notation. (2*pi*a)/(4*a) is so much easier to clean up in a single step without a mistake. It's also faster to write in an exam.
Thanks, I understand it now. I cleaned up the last line and got 0.25-1/2pisin(0.5pi) and got 0.24.... and not the 0.091... it's supposed to be?
6. (Original post by Airess3)
Thanks, I understand it now. I cleaned up the last line and got 0.25-1/2pisin(0.5pi) and got 0.24.... and not the 0.091... it's supposed to be?
7. (Original post by Eimmanuel)
Thanks! Got it now. *

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