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Particle in a potential well - not infinite :/

  1. Aug 6, 2013 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    Stationary Schrodinger equation.

    3. The attempt at a solution
    1st I draw the image of the well, so we can talk better - otherwise this makes no sense as it looks like a complex homework. In the image ##W_p## marks the potential energy but never mind i ll use ##E_p## notation in this problem.

    [Broken]

    Ok now that i have an image i can tell you what I already know and what is still unclear to me. And this is my 1st potential well (not infinite) problem so take it easy on me.

    I know, that in a standard potential well like this which is symetric we have two possible wavefunctions - one is odd ##\psi_{odd}## and one is even ##\psi_{even}##. They are both split into three separate functions which are diffrent for each interval - i will name those ##\psi_1,\psi_2,\psi_3##. Now let me write all those:

    For an odd solutions we have wavefunction:

    \begin{align}
    \psi_{odd} =
    \left\{
    \begin{aligned}
    \psi_1&=Ae^{Kx}\\
    \psi_2&=- \frac{ Ae^{-K\frac{d}{2}} }{ \sin \left( L \frac{d}{2} \right) } \sin \left( Lx \right)\\
    \psi_3&=-Ae^{-Kx}
    \end{aligned}
    \right.
    \end{align}

    where ##L=\sqrt{2mE / \hbar^2}## and ##K=\sqrt{-2m(E-E_p)/\hbar^2}##. These are the same even for "even solutions".

    For an even solutions we have wavefunction:

    \begin{align}
    \psi_{even} =
    \left\{
    \begin{aligned}
    \psi_1&=Ae^{Kx} \longleftarrow\substack{\text{same as for the odd solutions}}\\
    \psi_2&=\frac{ Ae^{-K\frac{d}{2}} }{ \cos \left( L \frac{d}{2} \right) } \cos \left( Lx \right)\\
    \psi_3&=Ae^{-Kx}
    \end{aligned}
    \right.
    \end{align}

    By applying boundary condition for matching derivatives on theese wavefunctions (even and odd) we always get "transcendental equation" - its LHS is different in case of odd and even wavefunctions while its RHS is the same in both cases:

    For an odd solutions we have transcendental equation:

    \begin{align}
    -\sqrt{\frac{1}{E_p / E -1} } = \tan\left(\frac{\sqrt{2mE}}{\hbar}\right)
    \end{align}

    For an even solutions we have transcendental equation:

    \begin{align}
    \sqrt{\frac{E_p}{E} - 1 } = \tan\left(\frac{\sqrt{2mE}}{\hbar}\frac{d}{2}\right)
    \end{align}

    Because the RHS is the same we can use the constraint that tangens is repeated every ##N\pi## and derive the equation for energies which we derived and it looks like this (solved for N):

    $$
    N = \frac{\sqrt{2mW}}{\pi\hbar} \frac{d}{2}
    $$

    1st i would like to know if my equations util this point are correct. Now Lets say i set N=1 in the last equation and calculate energy of the ground state. I get result 37.64eV while the book says it is 4.4eV... From this i presume i can calculate L and K and then continue mz calculation but...

    Now even if i calculated this energy correctly i don't know on what criteria to decide which set of equations should i use (even od odd). I am guessing that for ##N=1## i should take odd ones. I am guessing that for ##N=2## i should then take even ones, but what about for ##N=3##? Notice that in the wavefunctions there are no N... How do i plug N in my equations? Where should i put it and how do i derive equations with N?

    Oh and there is one more thing. I don't know how to normalise ##\psi_{odd}## or ##\psi_{even}##. Do I have to take a superposition of their subfunctions ##\psi_1,\psi_2,\psi_3##?

    I will include the derivation of the above equations in the attachment (It is in Slovenian language - i wrote it myself in latex just in case it might come handy)
     

    Attached Files:

    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 6, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    That does not work (if it would, you could get analytic solutions). You are not interested in the period of the tangent function, you are interested in equality of both sides.
    To get numerical solutions, you can graph both LHS and RHS and see where they intersect. That is possible both for odd and even solutions.
     
  4. Aug 6, 2013 #3
    I know that there is only a graphical method to solve this (everyone knows it - Griffith, Zetilli, hyperphysics...), but we had this in our test and we were supposed to calculate it somehow... Whithouth any kind of graph... What on earth!? Our proferor is a moron.

    He also stated that if we insert the E_p into that energz equation we can calculate the number of states from that equation if we round the result down to get the whole number. The equation was:

    $$ N_{max} \approx \frac{\sqrt{2mE_p}}{\pi\hbar} \frac{d}{2}$$

    I guess this is more of a bedtime story than a proven equation. Welcome to Slovenia. Everyone plays smart but in fact they re bunch of morons and our university sucks. I only wish i was from England...

    Can you at least tell me how to find the normalisation factor for the ground state? Should i do it like this (##\psi_1,\psi_2,\psi_3## are for the EVEN solutions)?

    $$\int\limits_{-\infty}^{-d/2} \psi_1 dx + \int\limits_{-d/2}^{d/2} \psi_2 dx + \int\limits_{d/2}^{\infty} \psi_3 dx =1$$
     
    Last edited: Aug 6, 2013
  5. Aug 6, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That is true, as each tan branch has at most one intersection with the other function. It also gives you an upper limit for the energy of the first state.

    You need the squared magnitude of the wave functions. Apart from that: right.
     
  6. Aug 6, 2013 #5
    How would i find the upper limit for the energy of the first state (did you mean the "ground state")? How do i calculate this? Is it maybe like this (i am only guessing)?
    $$W_{1~max} = \frac{E_p}{N_{max}}$$
    where formula for ##N_{max}## is the one you quoted.
    I am sloppy when pissed and get my heart to pump too much adrenaline in my head :)
     
  7. Aug 6, 2013 #6

    mfb

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    Staff: Mentor

    The state with the lowest energy (the first in your graph) is the ground state, indeed.
    The first state happens "before" (at lower energy) the tan has its first discontinuity.
     
  8. Aug 6, 2013 #7
    You mean the 1st graph in the pdf that i wrote? I am just checking.

    Nice, so now i know that i can at least calculate the upper limit of the ground state like this ##E_{1~max}=E_{p}/N_{max}##.

    Ok i understand that the first state is lower, but to get its real value there is no other way than using the graphical methods i guess.

    Please correct me if any of the statements i wrote are wrong.
     
  9. Aug 6, 2013 #8
    If i calculate ##N_{max}## i get the number which is below 1 - does this mean this potential well is impossible? Here is the result:

    $$N_{max} = \frac{\sqrt{2mE_p}d}{\pi\hbar2}=\frac{\sqrt{2\cdot9.109\times10^{-31}kg \cdot 20 \cdot 1.602\times10^{-19}J}~~0.2\times 10^{-9}m}{\pi \cdot 1.055\times10^{-34}Js\cdot 2} = 0.728$$
     
  10. Aug 6, 2013 #9

    mfb

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    2016 Award

    Staff: Mentor

    It is Nmax ≈, not =. I guess it can be wrong by one. Or your E_d is too low.
     
  11. Aug 6, 2013 #10
    IT just occured to me. We know that here we have the ground state so i know i have to use ##\psi_1,\psi_2,\psi_3## for an even solution. If i manage to calculate the normalisation factors then i can use hamiltonian and get the expectation value for energy of an electron... Wouldn't this work?
     
  12. Aug 6, 2013 #11

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You need the wave function to get the normalisation factors, and you need the energy levels to get the wave function.
    No, that does not work.
     
  13. Aug 6, 2013 #12
    So there is no way to solve this except to put the transcendental equation into a program like Mathematica to return the numerical value or draw the graphs and then find the intersections.

    Thank you I will tell this to the professor so he can fix it.
     
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