1
$\begingroup$

Consider an electron with total energy $E>V_2$ in a potential well with $$V(x)= \begin{cases} \infty & x< 0 \\ V_1 & 0< x< L \\ V_2 & x>L \end{cases} $$ where $V_2>V_1>0$.

We can determine that $$\phi_E(x)= \begin{cases} 0 & x< 0 \\ A\sin(kx)+B\cos(kx) & 0< x< L \\ Ce^{qx}+De^{-qx} & x>L \end{cases} $$ where $k^2=\frac{2m_e E}{\hbar}$ and $q=k\sqrt{V_1-E}$.

We can also apply the boundary condition at $x=0$ to determine that $$\phi_e(x)=A\sin(kx)$$ for $x\in[0,L]$.

We can also apply boundary conditions at $x=L$ to find that $$A\sin(kL)=De^{-qL}$$ $$Ak\cos(kL)=-Dqe^{-qL}$$ (since $C=0$ due to the corresponding positive exponent), and $$k\cot(kL)=-q$$

I need to calculate the de Broglie wavelength of the electron in the regions defined by the potential, if $V_1 = 10.0$ eV, $V_2 = 20.0$ eV, $E=30$ eV.

For the region where $0\le x\le L$, if I'm correct, the de Broglie $\lambda$ is determined by $\frac{h}{p}$, where $p=\hbar k=\sqrt{2m_e E}$, and so we can find $\lambda_{\text{de Broglie}}=8.962639\times 10^{-18}$ m.

For the region where $x<0$, we can't find this length because of the infinite potential.

For the region where $x>L$, if $p=\hbar q$, then it appears that $p$ is complex-valued, and then I'm in doubt whether I was correct to have dropped the coefficient $C$ above in my derivations of equations.

Would appreciate some help.

$\endgroup$

closed as off-topic by CuriousOne, user36790, ACuriousMind, honeste_vivere, John Rennie Jul 20 '16 at 16:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – CuriousOne, Community, ACuriousMind, honeste_vivere, John Rennie
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ It's not actually very clear to me what the question here is - you say you would "appreciate some help" but what is your question? Just how to calculate the deBroglie wavelength correctly? $\endgroup$ – ACuriousMind Jul 20 '16 at 13:37
3
$\begingroup$

As indicated in the answer to your previous question, since as $E(=30\:\mathrm{eV})$ is higher than $V_2(=20\:\mathrm{eV})$, that particle is not bound, it's not an eigenstate of the system's Schrödinger equation (it's a scattered state). Its wave function for $x \to +\infty$ would something like: $$\psi=c_1e^{-ikx}+c_2e^{+ikx}$$ ... where both complex parts represent the incoming and reflected waves respectively. That's essentially the wave function of a free particle.

Its de Broglie wave length where $x>L$, is calculated using:

$$p=\hbar k=\sqrt{2m_e (E-V_2)}$$

But for bound particles with $E<V_2$, the wave function $\psi$ (noted as $\phi$ by you) is Real over the entire domain $0\leq x \leq+\infty$.

For $0 \leq x \leq L$, then $p=\hbar k=\sqrt{2m_e (E-V_1)}$ (bound particle).

and then I'm in doubt whether I was correct to have dropped the coefficient $C$ above in my derivations of equations.

You're entirely correct to drop $C$ ($C=0$) for bound particles but note that not doing so wouldn't make $\psi(x)$ Complex in that area of the domain. Classically a particle with $E<V_2$ could not enter into that $x \leq L$ area but a quantum particle tunnels into the classically forbidden area.


1D SE solution for free particle of energy $E$ in constant potential field $V$: $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$$ $$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-V)\psi=0$$ $$k^2=\frac{2m}{\hbar^2}(E-V)$$ $$\frac{d^2\psi}{dx^2}+k^2\psi=0$$ For: $$k>0$$ $$\implies \psi(x)=c_1e^{-ikx}+c_2e^{+ikx}$$

$\endgroup$
  • $\begingroup$ Since the particle is unbound, did I need to drop the coefficient $C$? I think the function should decay in $x>L$. Also, for $x>L$, $q=k\sqrt{V_1-E}$, not $k\sqrt{E-V_1}$ do you think this is correct? $\endgroup$ – sequence Jul 20 '16 at 3:19
  • 1
    $\begingroup$ For the unbound particle: 1. $0\leq x \leq$, then $k=\sqrt{2m_e (E-V_1)}/\hbar$. 2. $x>L$, then $k=\sqrt{2m_e (E-V_2)}/\hbar$. The amplitude of the wave function drops a bit when going from $V_1$ to $V_2$. However, it does not decay further. Solve the SE for a free particle in a potential $V$ (with $V<E$). $C \neq 0$. $\endgroup$ – Gert Jul 20 '16 at 12:13
  • $\begingroup$ Ooopsie. That should have read the wave number drops a bit when [...]. Not the amplitude. $\endgroup$ – Gert Jul 20 '16 at 15:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.