Let's assume we have some one-dimensional Delta-potential $V(x)=V_0 \delta(x)$. Then I have found numerous problems where the approach for a wave function is $$\varphi(x)=\begin{cases}e^{ikx}+re^{-ikx},\ & x<0\\te^{ik'x},\ &x>0\end{cases}$$ I have two questions about this:

  1. The Schrödinger equation for this wave function outside of $x=0$ yields $\frac{\hbar^2 k'^2}{2m}=E=\frac{\hbar^2k^2}{2m}$. This means $k'=k$. Is this correct? Can we say that in general the wave vector $k$ must be the same if the wave propagates in the same potential (which outside of $x=0$ is just $V=0$)? And if not, why do we then have a legit approach where the reflected part and the incoming part of the WV in the area $x<0$ have the same wave vector?

  2. By definition, for the transmission coefficient $T$ we have $T=\mid\frac{\varphi(\infty)}{\varphi(-\infty)}|^2=|t|^2$ which confuses me. Isn't $t$ already the coeffient of transmission? What else is $t$ if not the coeffiecient? And if it is the coefficient, what did I get wrong about the definition of $T$?


1 Answer 1

  1. You have understood this aspect correctly. The bottom line is: $\psi(x)$ is claimed to satisfy the time independent Schrodinger equation, so if in doubt, plug it in and check that it does!

  2. Transmission here is defined to be the ratio of two physically observable rates, namely $T = R$(transmit) $/ R$(incident) where $R$(incident) is the rate at which right-moving particles would be detected before the barrier if a detector were placed there, and $R$(transmit) is the rate at which right-moving particles would be detected after the barrier if a detector were placed there. These rates are proportional to the modulus-squared of the quantum amplitude associated with each plane wave, not the quantum amplitude itself, and they also involve a factor $k$ or $k'$ to account for the faster motion (higher flux) when the wavevector is high. To be precise, $$ T = \frac{ |t|^2}{ |1|^2 } \frac{k'}{k} $$ where I include the $|1|^2$ term to keep the logic clear (your incident wave has amplitude $1$) and the ratio of wavevectors obviously evaluates to $1$ when $k'=k$, but more generally this will not always happen. To understand this really fully you need to learn about the probability current or flux which is given by $$ {\bf j} = \frac{\hbar}{2 mi}(\Psi^* \nabla \Psi - \Psi \nabla \Psi^*) $$ (this expression can be related to the continuity equation which expresses the conservation of the number of particles, or if you prefer, the conservation of probability).

  • $\begingroup$ Thanks for your answer. The factor $k'/k$ confuses me however. For the coefficient of reflection we just have $R=r^2$, though? I would like to have a general "formula" for those given a specific WF, i.e. the one I gave above but with coefficients A and B in the section $x<0$ and C in the section $>0$, when $k'\neq k$. Can I say $T=|C/A|^2\cdot k'/k$ in that case? And $R=|B/A|^2? $\endgroup$
    – RedLantern
    Commented Dec 4, 2018 at 11:40
  • $\begingroup$ Yes. And there is also a good way to check your work: in any given case it should pan out that $T = 1 - R$, $\endgroup$ Commented Dec 4, 2018 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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