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Schrodinger Equation for Constrained Particle

  1. Sep 12, 2013 #1
    Hi, I am confused about how we obtain a part of the Schrodinger equation for a particle of mass m that is constrained to move freely along a line between 0 and a.

    Equation:
    [itex]\frac{d^{2}ψ}{dx^{2}}[/itex]+([itex]\frac{8∏^{2}mE}{h^{2}}[/itex])ψ(x)=0


    Where does the value in the parenthesis come from and what is this value for? How do we arrive at this part of this equation??

    I'd really appreciate some help with this I have a quiz in the morning for physical chemistry and I really need to do well.

    Thank you,

    Chris
     
  2. jcsd
  3. Sep 12, 2013 #2

    atyy

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    Does this help?

    http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_Theory/Trapped_Particles/Particle_in_a_1-dimensional_box [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Sep 12, 2013 #3
    Actually, I've looked at that and am still confused... I don't get where that comes from..
     
  5. Sep 12, 2013 #4

    atyy

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    Can you be more specific where you get confused in http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_Theory/Trapped_Particles/Particle_in_a_1-dimensional_box [Broken] ?
     
    Last edited by a moderator: May 6, 2017
  6. Sep 12, 2013 #5
    Yes, under Step 3 where it says "If we then solve for k by comparing with the Schrödinger equation above, we find: k=" what I said in parenthesis. The 8pi^2 value. Where does this come from? I'm not sure how this is derived.
     
  7. Sep 12, 2013 #6

    atyy

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    2 equations bfore Step 3 we have:

    ##-\dfrac{\hbar^2}{2m} \dfrac{d^2\psi(x)}{dx^2} = E\psi(x)##,

    which can be rearranged to

    ## \dfrac{d^2\psi(x)}{dx^2} = -\dfrac{2mE}{\hbar^2}\psi(x)##,

    which you can compare with the equation in step 3 to get the result (remembering that ##\hbar = \frac{h}{2\pi}##)
     
  8. Sep 12, 2013 #7
    Wow.. I can't believe I missed that. So is this being done by the reasoning of separation of variables? Correct me if I'm wrong, but we have to separate them because E varies differently than ψ(x)? Thank you very much for pointing out my silly mistake!

    Also, this doesn't really apply to this same question, but you seem very knowledgeable, so do you know how this trig occurs:
    knowing: A=(c[itex]^{2}_{1}[/itex]+c[itex]^{2}_{2}[/itex])[itex]^{1/2}[/itex] and c[itex]_{2}[/itex]= Acos([itex]\phi[/itex])
    solve for [itex]\phi[/itex]
    which yields: [itex]\phi[/itex]=sin[itex]^{-1}[/itex][itex]\frac{c_{2}}{(c^{2}_{1}+c^{2}_{2})^{1/2}}[/itex]=tan[itex]^{-1}[/itex][itex]\frac{c_{2}}{c_{1}}[/itex]
    I'm not sure how we use the inverse sin to find the phi in the cos function.
     
  9. Sep 12, 2013 #8

    atyy

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    In http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_Theory/Trapped_Particles/Particle_in_a_1-dimensional_box [Broken] the starting equation was the time-independent Schroedinger equation or TISE, in which E is a constant for each energy level.

    However, the TISE is usually derived en route to solving the time-dependent Schroedinger equation (TDSE), which is the more general equation. To solve the TDSE, we usually start by assuming separation of variables, which splits the TDSE into two equations, one of which is TISE.

    The TISE has many solutions called the energy eigenfunctions, each of which corresponds to a different E.

    The solutions of the TDSE can be constructed from the solutions of the TISE.

    I'm terrible at trig, maybe someone else can help, or ask it in the math forums below?
     
    Last edited by a moderator: May 6, 2017
  10. Sep 12, 2013 #9
    Thank you very much. I did make a post in the math forum.
     
  11. Sep 13, 2013 #10

    atyy

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    I hope my explanation wasn't too confusing. At any rate, there's usually plenty of people here who can help.
     
  12. Sep 13, 2013 #11
    Good deal, two buddies and I are studying for our physical chemistry, the quantum mechanics portion, quiz that is tomorrow. It's as if our textbook expects us to know a bunch of things without it telling us.
     
  13. Sep 13, 2013 #12

    atyy

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    Is it a typo? I think it could make sense if c[itex]_{1}[/itex]= Acos([itex]\phi[/itex]).

    By squaring A=(c[itex]^{2}_{1}[/itex]+c[itex]^{2}_{2}[/itex])[itex]^{1/2}[/itex] and recognizing Pythagoras's theorem, it could mean that c[itex]_{1}[/itex] and c[itex]_{2}[/itex] are the sides of a right angle triangle, and A is the hypotenuse.

    c[itex]_{1}[/itex]= Acos([itex]\phi[/itex]) means that c[itex]_{1}[/itex]/A=cos([itex]\phi[/itex]), so c[itex]_{1}[/itex] is the side adjacent to [itex]\phi[/itex] since A is the hypotenuse.

    So sin([itex]\phi[/itex]) would be opposite/hypotenuse which would be [itex]\frac{c_{2}}{A}[/itex], and tan([itex]\phi[/itex]) would be opposite/adjacent which would be [itex]\frac{c_{2}}{c_{1}}[/itex]
     
  14. Sep 13, 2013 #13
    Yeah, I'm assuming this is just a typo unless one of the math genius gets back to me and says otherwise. It's very disturbing though because I spent probably 30-45 minutes earlier today digging through trig stuff to figure out where I was going wrong, since it was printed like this in the solutions manual to our textbook and our teacher's handwritten solutions he posts online has the same, supposed, error.
     
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