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I am unsure as to the nature of the potential

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  1. Dec 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle with mass m and electric charge e is confined to move in one dimension along the x -axis. It experiences the following potential:
    ##V(x) = {\infty}## for ## x{\lt0}##
    ##V(x) = -e^2/4\pi\epsilon_0x## for ## x \geq 0 ##
    (Note: the way the question is written down features no parentheses around ##4\pi\epsilon_0x##
    a) Describe the potential experienced by the particle.
    b) (This question just asks to write down the TISE which I can do).
    c)For the region ## x \geq 0 ##, by substituting in the Schrödinger equation, show that the wave function
    ## u(x)= Cxexp(\text{-}\alpha x) ##
    can be a satisfactory solution of the Schrodinger equation so long as the constant ##\alpha## is suitably chosen. Determine the unique expression for ##\alpha## in terms of m , e and other fundamental constants. Note that C is a normalisation constant.
    2. Relevant equations
    TISE

    3. The attempt at a solution
    For a) I am unsure to what the potential actually is and for c) I took the second derivative of u(x) and substituted it into the Schrodinger equation but it was more of a stab in the dark then anything else. I also thought about using boundary conditions to find an equation for ##\alpha## but given that my only boundary condition is 0, I've not managed to figure out a way of finding anything meaningful from doing this.
     
    Last edited: Dec 21, 2015
  2. jcsd
  3. Dec 21, 2015 #2
    Based on the offered eigenfunction in part (c), I think we can safely conclude that the the potential is [itex]V(x) = \frac{e^{2}}{4\pi\epsilon x}[/itex] for [itex]x\geq 0 [/itex]. What does this remind you of?

    Yes, "taking the second derivative of u(x) and substituting it into the Schrodinger equation" is the right approach. Can you show your work on this?
     
  4. Dec 21, 2015 #3

    blue_leaf77

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    There should be a negative sign in the potential, otherwise ##\alpha## will be negative and the wavefunction won't be normalizable.
     
  5. Dec 21, 2015 #4

    Ray Vickson

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    Note on proper LaTeX use: never write ##exp(-\alpha x)##: it looks ugly and can lead to ambiguities. Write, instead, ##\exp(-\alpha x)##, which looks much better and is unambiguous. You do that by writing "\exp" instead of "exp". The same holds for sin, cos, tan, ln, log, max. min, lim, etc.---all the standard functions. You get ##\sin##, ##\cos##, ##\tan##, ##\ln##, ##\log##, ##\max##, ##\min##, ##\lim## instead of ##sin##, ##cos##, ##tan##, ##ln##, ##log##, ##max##, ##min##, ##lim##, etc.

    Also, you do not need to say "exp(\text{-} \alpha x)"; LaTeX is smart enough and powerful enough to know how to parse the much simpler "\exp(-\alpha x)". The "\text{-}" plays no role here, so is a waste of your time and resources.
     
  6. Dec 21, 2015 #5
    Thank you for the advice, I've never used LaTex before and I used the text on the minus sign because I preferred the shorter length.
     
  7. Dec 21, 2015 #6
    Yes, sorry, I missed the minus sign
     
  8. Dec 23, 2015 #7

    Simon Bridge

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    (a) I concur with the others here - the potential is the coulomb potential - with a hard surface at x=0, or you can think of x as a radial distance.
    Later in your education you will be solving the SE with this sort of thing as a radiual component - right now your prof seems to want you to have some experience with the math so it's quite artificial.

    (c) You are on to a good start - after substituting, just use ordinary algebra to solve for alpha.
     
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