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Homework Statement
Electron is in a 1-D potential well of depth 20eV width 0.2 nm in his ground state N=1. What is the energy of the ground state? Write the normalized wavefunction of the ground state. What is the probability, to find the particle outside the well?
Homework Equations
Stationary Schrodinger equation.
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.
http://shrani.si/?i/7n/3Ad2qT9a/screenshot-from-2013-08-.png
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 Let's 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)
Attachments
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