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Another Quantum Mechanics Infinite Sqr. Well Q?

  1. May 28, 2006 #1

    Mec

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    I hope someone here can help me a bit here and i'm just become a new member here, so just bear with me:

    A particle is trapped in an area form z=0 to z=s:
    a. find the expecation value for the 2nd hydrogen energy level.
    b. find the probability for an electron between .2s to .3s.
    Ok, i start with the wave function for a particle in a box,
    \Psi(x) = \sqrt{{\frac{2}{s}}\sin(n\Pi)\frac{x}{s}
    AND the 2nd energy level is when n = 3,
    Expecation vaue of x
    \int x\Psi^2(x) dx

    i take that expecation integral, came up with this [\frac{\theta^2}{4} - \frac{\theta\sin2\theta}{4} - \frac{\cos2\theta}{8}], and evulate 0 from 2\Pi
    am i doing this right?
    any help will be appreciated.:smile:
     
    Last edited: May 28, 2006
  2. jcsd
  3. May 28, 2006 #2

    Gokul43201

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    Where did the "hydrogen" atom come in here ?

    I'm going to enclose your Latex within [ tex ] [ /tex ] tags (and fix it a little bit) so I can read what you've done.

    Is this what you meant ? I had to correct the first equation some.
     
    Last edited: May 28, 2006
  4. May 28, 2006 #3

    Gokul43201

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    Edit : Wait. How did you go from n \pi x / s to x ? And why are you using n=3 ?
     
    Last edited: May 28, 2006
  5. May 28, 2006 #4
    The potential well is from [itex]x=0[/itex] to [itex]x=s[/itex]. Outside [itex]\psi(x,t)[/itex] is defined to be zero.

    I think [itex]\theta = n\pi x/s[/itex]. Remember that the integration to compute the expectation value of any dynamic variable will be carried out over all values of x. In this case, the integral reduces to an integral over [itex]0\leq x \leq s[/itex] as everywhere else the integrand is zero.

    EDIT: Didn't see Gokul's post while posting...
     
  6. May 28, 2006 #5

    Gokul43201

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    This makes no sense to me.

    1. Where did the hydrogen atom come in ?
    2. It doesn't say that you should find the expectation value "of the position of the electron". How do you know that is what it is asking for ?
     
  7. May 28, 2006 #6

    Mec

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    Isn't that the second energy level for hydrogen is 3, and 3 is the quantum number for 2nd energy level?
     
  8. May 28, 2006 #7
    The wavefunction you have obtained pertains to some particle in some energy state decided by the integer parameter [itex]n[/itex] (calculate the energy...). I don't understand how hydrogen comes into the picture....I agree with Gokul on this.

    PS--The hydrogen atom Schrodinger equation has an absolutely different form because for starters, the potential is different
     
    Last edited: May 28, 2006
  9. May 28, 2006 #8

    Mec

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    ok, this is my mistake here. i should have said this a bit clear. let me expalin it this way: find the expecation valude for the 2nd hydro. level for this sqr. well. And in fact the particle is traped between an area rang from z=0 to z=s.
    i hope this clear some of the confusion, and sorry about that...
     
  10. May 28, 2006 #9

    Gokul43201

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    No it's still not clear. There should be no mention of hydrogen in this problem. And you can find the expectation value of any function - including position (x), momentum, x^5, etc.

    Is this a translation from some other language ? If not, can you post the question exactly as it appears in your book/homework assignment/notes.

    PS : Henceforth, post other textbook/homework problems under the Homework Help subforum.
     
    Last edited: May 28, 2006
  11. May 28, 2006 #10

    Gokul43201

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    Assuming you wanted to calculate the expectation value of position for n=2, you wouuld have :

    [tex]\langle x \rangle = A \int _0 ^s x~sin ^2 \left( \frac{2 \pi x}{s} \right) ~dx = A \int _0 ^s \frac {x}{2} ~ \left[ 1 - cos\left( \frac{4 \pi x}{s} \right) \right] ~dx [/tex]

    Then if you wish, you can do a change of variables [itex] 4 \pi x/s \rightarrow \theta [/itex] and proceed.

    Alternatively, if you sketch the wavefunctions you might just notice that [itex]sin^2 \theta [/itex] is an even function with respect to the center of the well and x is an odd function. This should directly tell you the answer.
     
    Last edited: May 28, 2006
  12. May 29, 2006 #11

    Mec

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    Btw, Thanks for help though..
    Anyway, Here is the exact one problem if that help at all:
    Find the expectation val. for the 2nd hydrogen energy level for an infinite sqr. weell. Consider the particle trapped in the an area from z=0 to z=s.

    Anyway, what should the quantum number for the 2nd energy level for hydrogen, is 2 or 3, i kind of confused when reading the book here?
     
  13. May 29, 2006 #12

    Gokul43201

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    I hope that is not from a textbook. It is very badly written. For instance, there should be no mention of "hydrogen" in the question. The hydrogen atom is a completely different system with a very different potential from the square well potential. Secondly, it wants you to find the expectation value of some quantity...but it doesn't say which quantity. Honestly, you should complain to your teacher about this question. It is just wrong.

    The principal quantum number for the second energy level of the hydrogen atom is 2. The quantum number for the second exited state is 3.

    Is this question from a book ? What is the title of the book and who are the authors ?
     
  14. May 31, 2006 #13
    I think that problem is three dimensional
    the hydrogen is confined in z direction only, and moves in
    Coulombic central field in the other two dimensions!
    am i right???
     
  15. May 31, 2006 #14
    No; the electron is in a 3-D Coulomb potential.
     
  16. Jun 13, 2006 #15
    Gokul is right for sure, it just dosen't make sense, you only have to look at the boundary conditions to see that the question is wrong. Should the question be Find the expectation val. for the 2nd hydrogen energy level and for the infinite square well? This would make more sense as you are then looking at 2 seperate questions rather than one.
     
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