# Transmission coefficient

The transmission coefficient can be calculated as

$$T=\left ( 1+\frac{1}{4}\frac{\left ( p_{0}^{2}-p_{1}^{2} \right )^{2}}{p_{0}^{2}p_{1}^{2}} \sin^2 \left( \frac{p_1 a}{\hbar} \right )\right )^{-1}$$ where $p_0 = \sqrt{2mE}$, and $p_1 = \sqrt{2m ( E - V_0 )}$. What happens to the transmission coefficient when $p_1 a =n\pi\hbar$? What is the relationship between the de Broglie wavelength of the particle in region $0\leq x\leq a$ and the width of the potential barrier in this case.

Here's what I think, for very large energy $E$ the comparison reduces to $$T = \left( \sin^2 \left( \frac{p_1 a}{\hbar} \right) \right)^{-1} \quad \text{and} \quad T = \left( \sin^2 \left ( n \pi \right) \right)^{-1} \, .$$

I'm very much left clueless at this point. Some hints would be helpful.

• Please specify the kind of (quantum?) barrier you are referring to. – Gert Jan 1 '16 at 3:30
• @Gert The barrier in question is a step barrier (or Rectangular barrier) – Physkid Jan 1 '16 at 4:14
• @Gert I might have some insights. Begining with the original equations for T. When $p_{1}a=n\pi\bar{h}$, the equation reduces to one-evidently, Sin of an integral number of pi is always 0. Thus, T=1. This is the transmission resonance condition where transmission of the particle is at its maxima. – Physkid Jan 1 '16 at 5:00
• Ask yourself: if you read this post would you be able to understand what the person were asking? The post starts off with a gigantic formula for a transmission coefficient but it doesn't even say what system we're dealing with. Please describe the setup of the problem in the post. – DanielSank Sep 21 '18 at 4:53

Because

$\sin^2(\frac{n\pi h}{h}) = 0$

Then T is just $1^{-1} = 1$

You must be asking what is the wavelength inside and outside the barrier

when the particle is outside the barrier it must have more energy and thus a smaller wavelength

Outside the barrier

$p_0 = \sqrt{2mE} = \sqrt{2mK_0} = \sqrt{2m \frac{p^2}{2m}} = p_a$

and the De Broglie equation states that

$\lambda = \frac{h}{p}$

so $\lambda_1 = \frac{h}{p_a}$

We are taking $E$ as $K$ because it is actually $E-V_0$ where $V_0$ is 0

Inside the barrier

$p_1 = \sqrt{2m(E-V_0)}$

$p_1 = \sqrt{(2mK_1)}$

$p_1 = p_b$

$\lambda_2 = \frac{h}{p_b} = \frac{1}{n\pi}$

$p_a > p_b$

so $\lambda_1 < \lambda_2$

which makes sense because the kinetic energy is greater outside

• $\sin^2\left(n\pi h\right)$ should just be $\sin^2\left(n\pi\right)$. – Chris Apr 7 '18 at 8:38
• sorry i'lll fix it – Tim Crosby Apr 7 '18 at 9:19