# Calculate the Energy Levels of an Electron in a Finite Potential Well

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In summary, the conversation discusses finding the energy states of a particle in a finite well by defining the wave function over three domains and using the time independent Schrodinger equation. The even and odd parity wave functions are derived and it is shown that there will always be an even parity solution for any given value of V0. The computation of energy levels for a specific L and V0 is also discussed and compared to the energy levels for an infinite well. It is found that the ground state energy level for the finite well is lower than that of the infinite well.

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Homework Statement
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Relevant Equations
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Section 1
To find the energy states of the particle, we define the wave function over three discrete domains defined by the sets ##\left\{x<-L\right\}##, ##\left\{-L<x<L\right\}##, and ##\left\{L<x\right\}##. The time independent Schrodinder equation is
\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V_(x)-|E|]\psi

Outside of the well, setting ##V(x)=V_0## gives
\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V_0-|E|]\psi

The characteristic polynomial is
$$\lambda^2-\frac{2m[V_0-|E|]}{\hbar^2}=0$$ with the roots
\lambda=\pm \sqrt{\frac{2m[V_0-|E|]}{\hbar^2}}\Rightarrow \pm \alpha
and the general solution is

\psi=C_1e^{\alpha x}+C_2e^{-\alpha x}

Inside of the finite well, setting ##V(x)=0## gives

\frac{d^2\psi}{dx^2}=\frac{-2m}{\hbar^2}|E|\psi

which has the following characteristic polynomial

\lambda^2+\frac{2m}{\hbar^2}|E|=0

with the roots

\lambda=\pm i\sqrt{\frac{2m}{\hbar^2}|E|}\Rightarrow\pm i\sqrt{q}

and the general solution is

\psi=C_1e^{iqx}+C_2e^{-iqx}\Rightarrow \psi= C_3cos(qx)+iC_4sin(qx)

The real and imaginary parts of a complex valued solution form a fundamental set of solutions of the second order linear ODE, giving

\psi=A_1cos(qx)+A_2sin(qx)

The even parity wave functions are

\psi=Ce^{\alpha x}|_{\left\{x<-L\right\}}

\psi=Acos(qx)|_{\left\{|x|<L\right\}}

\psi=Ce^{-\alpha x}|_{\left\{L<x\right\}}

the odd parity wave functions are

\psi=-Ce^{\alpha x}|_{\left\{x<-L\right\}}

\psi=Asin(qx)|_{\left\{|x|<L\right\}}

\psi=Ce^{-\alpha x}|_{\left\{L<x\right\}}

Section 2
For the even solution, we require ##\psi \in C^1(\pm L)\Rightarrow \psi## is once continuously differentiable at ##x=\pm L##

\psi_{in}(\pm L)= \psi_{out}(\pm L)\Rightarrow Acos(qL)=Ce^{-\alpha L}

\psi'_{in}(\pm L)= \psi'_{out}(\pm L)\Rightarrow Aqsin(qL)=\alpha Ce^{-\alpha L}

we divide equation (18) by (17) and do a change of variables using $$x=qL$$ and $$r^2=\frac{2mV_0L^2}{\hbar^2}$$ The even solutions are given implicitly by

xtan(x)=\sqrt{r^2-x^2}

For the odd functions, we again require that ##\psi \in C^1(\pm L)##

\psi_{in}(\pm L)= \psi_{out}(\pm L)\Rightarrow Asin(qL)=Ce^{-\alpha L}

\psi'_{in}(\pm L)= \psi'_{out}(\pm L)\Rightarrow Aqcos(qL)=-\alpha Ce^{-\alpha L}

and divide equation (20) by (21). Following a change of variables for ##x## and ##r^2## we have

-x cot(x)=\sqrt{r^2-x^2}

Section 3
(19) and (22) are transcendental equations whose solutions are obtained using numerical approximations. The even solutions are given by

\{ \{y=xtan(x)\}\cap \{y=\sqrt{r^2-x^2}\}|x\in [0,\infty)\}

and the odd solutions are given by

\{\{y=-xcot(x)\}\cap\{y=\sqrt{r^2-x^2}\}|x\in [0,\infty)\}

It is worth mentioning the solutions exist in the domain
\{(x,y)\in {\rm I\!R} \times[0,\infty)\}

and the solution space is symmetric in the ##x## and the ##-x##. For wells with

V_0<\frac{(\hbar \pi)^2}{8mL^2}\Rightarrow L<\frac{\hbar \pi}{\sqrt{8mV_0}}\Rightarrow r<\frac{\pi}{2}

we use the fact ##r=\frac{\pi}{2}## is in the exceptional set defined by $$\varepsilon(f)=\Big\{\alpha\in\bar{\mathbb{Q}}:f(\alpha)\in\bar{\mathbb{Q}}\Big\}$$ where ##Q## denotes the field of rational numbers.
$$0=-\Big(\frac{\pi}{2}\Big)cot\Big(\frac{\pi}{2}\Big)=\sqrt{\Big(\frac{\pi}{2}\Big)^2-x^2}$$
It follows that

\{\{y=-xcot(x)\}\cap\{y=\sqrt{r^2-x^2}\}|(x,r)\in [0,\infty)\times{[0,\frac{\pi}{2}})\}=\{\emptyset\}

so there are no odd parity energy states. In the meanwhile, we find

\{ \{y=xtan(x)\}\cap \{y=\sqrt{r^2-x^2}\}|(x,r)\in [0,\infty)\times[0,\infty)\}\neq\{\emptyset\}

so there will always be an even parity solution for any ##V_o##. In particular, we find that ##\{y=xtan(x)\}\cap \{y=\sqrt{r^2-x^2}\}\}## where ##r\in[0,\frac{\pi}{2})## is the graph of ##y=xtan(x)## in the region ##\{|x|\leq.934\}##. So, there exists only one energy state of even parity for
$$V_0<\frac{(\hbar \pi)^2}{8mL^2}$$
Section 4
Setting ##L=1nm## and ##V_0=1eV##, we compute
$$r^2=\frac{2mV_0L^2}{\hbar^2}=\frac{2(9.1\times10^{-31}kg)(1.6\times10^{-19}J)(10^{-9})^2m}{(1.054\times10^{-34}Js)^2}\approx26.2729J$$
we plot

y=xtan(x)

y=-xcot(x)

y=\sqrt{26.2729-x^2}

and find
$$x_{even}=\{1.312, 3.86\}$$
and
$$x_{odd}=\{2.608, 4.964\}$$
We compute the energy using

E=\frac{x^2\hbar^2}{2mL^2}

giving
$$E_{1even}\approx1.05\times10^{-20}J$$
$$E_{2even}\approx9.09\times10^{-20}J$$
and
$$E_{1odd}\approx4.15\times10^{-20}J$$
$$E_{2odd}\approx1.50\times10^{-19}J$$
The expression for the energy levels of the infinite well is

E_n=\frac{n^2\pi^2\hbar^2}{8mL^2}

giving
$$E_1\approx1.50\times10^{-20}J$$
$$E_2\approx6.02\times10^{-20}J$$
so the ground state energy level for the finite well is lower.

Last edited by a moderator:
Looks good overall.

Eq 27 and 28: Normally ##\emptyset## is already the empty set. What you wrote is a non-empty set containing the empty set as element.

Section 4:
r is dimensionless, not in J

eV would be a more natural unit for the energy levels, especially as the potential is given in eV as well.

docnet
mfb said:
Looks good overall.

Eq 27 and 28: Normally ##\emptyset## is already the empty set. What you wrote is a non-empty set containing the empty set as element.

Section 4:
r is dimensionless, not in J

eV would be a more natural unit for the energy levels, especially as the potential is given in eV as well.
Thank you! I see what I did there.

Some useful combinations to memorize are
\begin{align*}
hc &= 1240~{\rm nm\cdot eV} \\
\hbar c &= 197.3~{\rm nm\cdot eV~(or~MeV\cdot fm)} \\
mc^2 &= 511000~{\rm eV}
\end{align*} where ##m## is the mass of the electron.

Using these and strategically introducing factors of ##c##, you can avoid a lot of unit conversions. For example, you had
$$r^2 = \frac{2mV_0L^2}{\hbar^2} = \frac{2(mc^2)V_0 L^2}{(\hbar c)^2} = \frac{2(511000~{\rm eV})(1.0~{\rm eV})(1.0~{\rm nm})^2}{(197.3~\rm{eV\cdot nm})^2} = 26$$

docnet
vela said:
Some useful combinations to memorize are
\begin{align*}
hc &= 1240~{\rm nm\cdot eV} \\
\hbar c &= 197.3~{\rm nm\cdot eV~(or~MeV\cdot fm)} \\
mc^2 &= 511000~{\rm eV}
\end{align*} where ##m## is the mass of the electron.

Using these and strategically introducing factors of ##c##, you can avoid a lot of unit conversions. For example, you had
$$r^2 = \frac{2mV_0L^2}{\hbar^2} = \frac{2(mc^2)V_0 L^2}{(\hbar c)^2} = \frac{2(511000~{\rm eV})(1.0~{\rm eV})(1.0~{\rm nm})^2}{(197.3~\rm{eV\cdot nm})^2} = 26$$
Thank you sir