3
$\begingroup$

The step potential is of the form

$$V= \begin{cases} 0 & x< 0 \\ V_0 & x>0 \\ \end{cases}$$

For simplicity, let $V_0>0$. If we consider $0<E<V_0$ first: Define $k=\frac{\sqrt{2mE}}{\hbar}$ and $l=\frac{\sqrt{2m(V_0-E)}}{\hbar}$, whereform we can retrieve

$$\psi(x)=\begin{cases} Ae^{ikx}+Be^{-ikx} & x< 0 \\ Fe^{lx}+Ge^{-lx} & x>0 \\ \end{cases}$$

via the time independent Schrödinger equation.

Now, I am told to disregard $Fe^{lx}$ for various reasons in order to obtain

$$\psi(x)=\begin{cases} Ae^{ikx}+Be^{-ikx} & x< 0 \\ Ge^{-lx} & x>0 \\ \end{cases}$$

For example, in a similar question Tim Crosby's answer makes the argument that $\psi$ must remain normalizable as $x\to \infty$ for a bound state, but $Fe^{lx}$ blows up, so it must be that $F=0$. However, I am under the impression that no bound state can exist on the step potential, so we have a scattering state. Normalization is not important in scattering states anyway, so I don't see how this works.

SACHIN's answer agrees with my thoughts, stating that we forget about the bound state analysis and instead look at the physical significance of the constants $F,G$. Namely, $F$ represents a reflection coefficient to waves coming from positive infinity, yet there are not obstacles to reflect particles, hence $F=0$. This is not completely satisfactory for me either, as I am unsure whether $Fe^{lx}$ is a wave traveling to the right at all, as would be implied by $F$ being a reflection coefficient. In fact, if I tack on time dependence:

$$Fe^{lx}e^{-iEt/\hbar}=Fe^{lx-iEt/\hbar}=Fe^{l(x-iEt/l\hbar)}$$

I do get a wave traveling to the right, or do I? The speed $iE/l\hbar$ is imaginary! It was under my impression that speed has to be real, but fine, I will entertain this for now. Another worry of mine is that I do not infer $F$ to necessarily be a reflection coefficient at all. If we assume a particle is coming from the left of the step potential, can't $Fe^{lx}$ represent a transmitted wave? For instance, we treat $Fe^{ikx}$ in a delta potential barrier like a transmitted wave.

All in all, I have 2 possible answers to my question in the above mentioned thread, yet they seem to contradict each other. This has left me deeply confused. so I decide to ask myself:

  • Why can we disregard $Fe^{lx}$?

  • Can speed be imaginary in this context?

Also, it can be perhaps inferred that I am not that well versed in quantum mechanics, so I must graciously ask for any possible answers to be understandable to somebody who has only read the first 2 chapters of Griffiths's Introduction to Quantum Mechanics, if at all possible.


EDIT:

After some more thinking and Frotaur's suggestion in the comments, I have come up with another plan: We can try to form the general solution wavefunction as a linear combination of eigenfunctions of energy, and from that $F$ can be found to be $0$.

So, considering $l=i\frac{\sqrt{2m(E-V_{0})}}{\hbar}$ when $E\geq V_0$, we have the eigenfunctions

$$\Psi_{k}(x,t)=[Fe^{lx}+Ge^{-lx}]e^{-iEt/\hbar}, \qquad x>0, \quad k=\frac{\sqrt{2mE}}{\hbar}$$

from which arises the general solution via completeness postulate.

$$\Psi(x,t)=\int_{0}^{\infty} \phi(k)\Psi_{k}(x,t)dk=\int_{0}^{\infty} \phi(k)[Fe^{lx}+Ge^{-lx}]e^{-iEt/\hbar}dk$$

The plan is to assume $F\neq 0$ and take the limit $x\to \infty$ and see whether $\Psi\to 0$. If $\Psi\not\to 0$, this is not a physically realizable state, hence we can state that $F=0$ for physically manifestable wavefunctions and we're done. The first part is to eliminate $Ge^{-lx}$ from the equation as it disappears exponentially as $x\to \infty$. Thus, we have:

$$\lim_{x\to \infty}\Psi(x,t)=\lim_{x\to \infty}\int_{0}^{\infty} \phi(k)[Fe^{lx}]e^{-iEt/\hbar}dk$$

$$=F\lim_{x\to \infty}\int_{0}^{\infty}\phi(k)e^{lx-iEt/\hbar}dk$$

(be aware $l,E$ are not independent of $k$.)as long as this is not 0, we can state $F=0$, but that's much easier said than done... If the answer can solve this limit, I deem it as a satisfactory answer to my question.

$\endgroup$
4
  • $\begingroup$ Normalisation itself is not that important. But normalisability is. $\endgroup$ Aug 3, 2022 at 8:29
  • $\begingroup$ Not all waves have a defined velocity. An evanescent wave, like the one in $x>0$, doesn't propagate, so it doesn't have any velocity. $\endgroup$
    – Miyase
    Aug 3, 2022 at 8:32
  • $\begingroup$ Just a comment as it kinda is a cop-out. A differential equation problem is not well-defined so long as you don't have boundary conditions. In this particular problem, you could add as a boundary condition that $\psi(x)\rightarrow 0$ when $x\rightarrow \infty$, which will correctly discard the unwanted term. You would have to physically motivate that boundary condition though (to which I don't have a good answer to), which is the reason why I call this a cop out. $\endgroup$
    – Frotaur
    Aug 3, 2022 at 14:30
  • $\begingroup$ To answer the first comment, if normalisability where very important, we should also disregard the plane wave states, which are also non-normalisable. Perhaps a more careful treatment using wave-packets instead of plane waves would give a natural reason to discard the $e^x$ solution. $\endgroup$
    – Frotaur
    Aug 3, 2022 at 14:31

2 Answers 2

1
$\begingroup$

Your two questions are not in principle connected. Question 2 is not well defined, since the solution is not a travelling wave, so it doesn't have a speed.

The interesting Question is Question 1.The four constants $A$, $B$, $E$ and $F$ are connected by two boundary conditions at $x=0$, giving $ A+B=E+F $ and $(ik/l)(A-B)=(E-F)$. We need two more conditions to get a definite answer. One of these we can take to be that at $x<0$ there is an incoming wave of specified amplitude: that is, $A$ is given. The other condition must come from the behaviour for $x>0$. If we leave this open for the moment we can find two solutions, the usual one, $\psi_1$, with $A$ given and $F=0$, and another one with $A=0$: $\psi_2(x)=e^{(-ikx)}$ for $x<0$, and $\psi_2(x)=\cosh(lx)-(ik/l)\sinh(lx)$ for $x>0$. We can add any multiple of $\psi_2$ to the usual solution to obtain a solution with the prescribed value of $A$: $\psi_1+\alpha\psi_2$. $\psi_2$ describes a wavefunction with a constant negative probability flux (in other words, probability flux flowing in the negative direction) in both regions. $\psi_1$ has zero probability flux everywhere. $\psi_2$ could be part of a larger scattering problem with incoming flux from $x>0$, but if that is not possible in the actual situation described then the wavefunction cannot contain this solution, so $\alpha=0$.

In summary, I agree with Sachin.

$\endgroup$
8
  • $\begingroup$ I assume you mean $\sinh(lx)$, or did you actually mean $\sinh(x)$ as you wrote? I see, so $\psi_{2}$ would represent a wavefunction of a particle coming from positive infinity, correct? So $\textrm{exp}(-ikx)$ is the transmitted wave, right? Do $\cosh(lx)$ and $\sinh(lx)$ represent anything or is it just that $\cosh(lx)-(ik/l)\sinh(lx)$ in this combination represent BOTH the incoming wave from the right and the reflected wave? $\endgroup$
    – pjq42
    Aug 3, 2022 at 15:57
  • $\begingroup$ Yes of course thank you. $\endgroup$
    – CWPP
    Aug 3, 2022 at 16:52
  • $\begingroup$ But wait... The probability flux isn't linear necessarily, right? The probability flux of a sum of wavefunctions is not always the sum of the probability fluxes evaluated individually. So isn't it technically possible for $\psi_{1}+\alpha\psi_{2}$ to have a nonnegative probability flux, thus meaning $\alpha$ isn't necessarily 0? $\endgroup$
    – pjq42
    Aug 3, 2022 at 17:56
  • $\begingroup$ I mean, just peripherally speaking, aren't there cross terms to consider? $\endgroup$
    – pjq42
    Aug 3, 2022 at 18:18
  • $\begingroup$ Well spotted! I was thinking along the same lines. I didn't choose the smartest pair of solutions. I should have chosen the same $\psi_2$ and $\psi_1=\psi_2^\star$. These have the nice properties that the fluxes of the two are equal and opposite, and the cross terms in the flux between them if they are superposed vanish. I've never seen this solution before and I'm still working out its properties! $\endgroup$
    – CWPP
    Aug 3, 2022 at 21:00
1
$\begingroup$

Coefficients of a wave function are probability amplitude. So A and B are amplitude for free particle or electron. When a particle is subjected to constraints like step potential barrier then coefficient of wave function should be different, that are F and G.

Now, probability of a particle is becoming less as it pass through barrier, because potential barrier is against particle's energy. But coefficient F is not alone to determine probability, it is multiplied with positive exponential function which keeps increasing with distance in forward direction. After passing unit distance it is growing, which means probability increases more than one.

So term containing Fe$^{lx}$ is not physically possible, so it is disregarded. Also F is reflection or transmission coefficient, while step function is not bounded in forward direction. Question is why F is reflection coefficient and not G. Then answer is because G is feasible as decaying coefficient for propogating in barrier.

$\endgroup$

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.