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1. The problem statement, all variables and given/known data

Electron of is in a 1-D potential well of depth $20eV$ width $d=0.2 nm$ 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?

2. 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.

enter image description here

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 finite potential well homework problem so take it easy on me.

I know, that in a standard finite potential well, which is symmetric 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 different 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 these 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}\frac{d}{2}\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 tangents 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 until this point are correct.

3. What I don't understand

I have to calculate energy of the ground state so I set $N=1$ in the last equation and calculated energy. I got result $37.64eV$ while the book says it is $4.4eV$... If I manage to calculate the energy I can afterwards calculate $L$ and $K$ which are needed by the wave function, so I need to start here I think.

Even if my obtained value for energy was correct I don't know on what criteria to decide which set of equations should I use (even or 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 which include $N$?

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


I will include the hyperlink to the derivation of the above equations (It is in Slovenian language but don't mind the language everything is there - I wrote it myself in latex just in case it might come handy)

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You have forgotten, in analysing your transcendental equations, that $E$ is in the left hand side too.

So, you cannot say that "tangent is repeated every $n\pi$ and derive the equation for energies". No, the left-hand side changes with $E$ too.

There is no problem to normalize $\psi_{odd}$ or $\psi_{even}$. The domains of definition of $\psi_1,\psi_2,\psi_3$ are different.

So $\int dx |\psi(x)|^2 = \int_{- \infty}^{-d/2} dx |\psi_1(x)|^2 +\int_{-d/2}^{+d/2} dx |\psi_{2}(x)|^2 +\int_{+d/2}^{+ \infty} dx |\psi_{3}(x)|^2 $

See Wikipedia for more details.

Note that you did not give the value of $W_p$ and $m$, so it is not possible to give numerical results.

[EDIT]

Looking at even solutions, and applying continuity of the first derivative, we get :

$$K = L ~ \tan(\frac{Ld}{2}) \tag{1}$$ This could be written :

$$u_0^2 - v^2 = v ~ \tan v \tag{2}$$ with : $$ u_0= \frac{d}{2} \sqrt{\frac{2mE_p}{\hbar^2}}, v = \frac{d}{2} \sqrt{\frac{2mE}{\hbar^2}} \tag{3}$$

Numerical calculus gives :

$$u_o^2= 5.245\tag{4}$$

Solving equation $(2)$ with numerical value $(4)$ (see plot1 and plot2) gives a numerical value for $v_1 = v(N=1)$:

$$v_1 = 1.08 \tag{5}$$

Finally, we have : $$E_1 = E_p \frac{v_1^2}{u_0^2}\tag{6}$$

So, numerically, it gives :

$$E_1 (eV) = 20 \frac{(1.08)^2}{5.245} = 4.45~ eV\tag{7}$$

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  • $\begingroup$ Thank zou for the info about normalization. Concerning the numerical value - the particle is an electron so just use its mass, but i really forgot to give the depth of a potential well which is $20eV$. Do you use mathematica? $\endgroup$ – 71GA Aug 6 '13 at 19:02
  • $\begingroup$ @71GA : I made an edit to the answer, with the solution $\endgroup$ – Trimok Aug 7 '13 at 8:11
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There is a good wikipedia article about this topic. But let me solve this my own way.

\begin{align} \large{\psi(x) = \left\{ \begin{aligned} Ce^{Kx}+De^{-Kx}\\ Ae^{\mathfrak{i}Lx}+Be^{-\mathfrak{i}Lx}\\ Ee^{Kx}+Fe^{-Kx} \end{aligned} \right. \begin{aligned} x &\le -d/{2}\\ -d/{2} &\le x\le d/{2}\\ d/{2} &\le x \end{aligned}} \end{align}

And our boundary conditions are:

\begin{align} \large{ \begin{aligned} \psi(x)&= \left\{ \begin{aligned} 0 & \ x \to -\infty \\ 0 & \ x \to\infty \end{aligned} \right. \\ &\Rightarrow \left\{\begin{aligned}D &= 0 \\ E &= 0 \end{aligned} \right. \end{aligned}} \end{align}

Also, we know that $\psi$ and its first derivative must be continuous. This gives us more equations:

\begin{align} \large{ \begin{aligned} Ce^{-\frac{Kd}{2}} &= Ae^{-\mathfrak i \frac{Ld}{2}}+Be^{\mathfrak i \frac{Ld}{2}} \\ CKe^{-\frac{Kd}{2}} &= A\mathfrak{i}Le^{-\mathfrak i \frac{Ld}{2}}-B\mathfrak{i}Le^{\mathfrak i \frac{Ld}{2}}\\ Fe^{-\frac{Kd}{2}} &= Ae^{\mathfrak i \frac{Ld}{2}}+Be^{-\mathfrak i \frac{Ld}{2}} \\ -FKe^{-\frac{Kd}{2}} &= A\mathfrak{i}Le^{\mathfrak i \frac{Ld}{2}}-B\mathfrak{i}Le^{-\mathfrak i \frac{Ld}{2}} \end{aligned} }\end{align}

These four equations are not independent(otherwise the answer would be unique, and the obvious answer is $0$). Writing it in the matrix form:

\begin{align} \begin{aligned} \left( \begin{matrix} e^{-\mathfrak i \frac{Ld}{2}} & e^{\mathfrak i \frac{Ld}{2}} & -e^{-\frac{Kd}{2}} & 0 \\ e^{\mathfrak i \frac{Ld}{2}} & e^{-\mathfrak i \frac{Ld}{2}} & 0 &-e^{-\frac{Kd}{2}}\\ \mathfrak i Le^{-\mathfrak i \frac{Ld}{2}} & -\mathfrak i Le^{\mathfrak i \frac{Ld}{2}} & -Ke^{- \frac{Kd}{2}} & 0 \\ \mathfrak i Le^{\mathfrak i \frac{Ld}{2}} & -\mathfrak i Le^{-\mathfrak i \frac{Ld}{2}} & 0 & Ke^{- \frac{Kd}{2}} \end{matrix}\right) \left(\begin{matrix}A \\ B \\ C \\ F \end{matrix} \right)=\left( \begin{matrix}0 \\ 0 \\ 0 \\ 0 \end{matrix}\right) \end{aligned} \end{align}

The determinant of this matrix has to be zero. It's not actually hard to calculate the determinant(it can be made lower-triangulated easily). The determinant is:

$$2 \mathfrak i e^{-d K} ((K^2-L^2) \sin (d L)+2 K L \cos (d L)) =0$$

This simplifies to:

$$(E_p- 2 E)\tan(d L)+ 2 \sqrt{E(E_p-E)}=0$$

This is a transcendental equation, and doesn't have algebraic solutions. But we can solve it numerically, to find the eigenvalues(or we can write $\tan\theta=\frac{2\tan\left( \frac{\theta}{2}\right)}{1-\tan^2\left( \frac{\theta}{2}\right)}$ and reproduce the equations given in the wikipedia article). This way we don't have to struggle with the symmetry of our eigenstates, and specially the ground state(which happens to be symmetric). Using Mathematica I was able to find two discrete eigenvalues for energies:

$$E=4.44364 \text{eV} \\ E=15.8826\text{eV}$$

This method has another benefit, and that's it clarifies the nature of the eigenvalues and the reason they appear.

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  • $\begingroup$ Wouldnt i get the energy of this state if i used hamiltonian on the wavefunction??? $\endgroup$ – 71GA Aug 6 '13 at 19:52
  • $\begingroup$ If you have the wavefunction, then sure. $\endgroup$ – Ali Aug 7 '13 at 1:04
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1) Energy of electron

$$E=\frac{n^2h^2}{8mL^2}$$ $E= 1.506 \times 10^{-18}\text{ Joules}$ which if you divide by electron charge ie $1.6 \times 10^{-19}$ will give $E$ in electron volts ($eV$)

i.e. $E = 9.4135\text{ eV}$, so thats the energy of electron at $n=1$.

2)Normalized Wave Function

Need to find $K$ for that first.

$K = \frac{4\pi^2}{h^2}\sqrt{2mE} = 1.489 \times 10^{44}$

for normalization then follow equation: 122 from this link

3) Probability can be found out by using equation: 121 from same link

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    $\begingroup$ This is only valid if the well is infinite, which is not the situation in this case. $\endgroup$ – PhysicalChemist Sep 21 '15 at 18:24

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