In our course we considered several examples of a particle in a potential wall. One case was that of a particle moving along the $x$ axis and the potential was a step function of the form:

$x<0 $: $V(x)=0$

$x>0 $: $ V(x)=V_0$

So we divide the region in two parts: Region 1: $x<0$ and Region 2: $x>0$

The wave function in the first region (x<0) was given as:


$A_1e^{ik_1x}$ represents the incident wave.

$B_1e^{-ik_1x}$ represents the reflecting wave (on the potential wall).

In the 2 region (x>0):


We consider that $B_2e^{-ik_2x}=0 \rightarrow B_2=0$, in other words there is not wave coming from the right side.

$A_2e^{ik_2x}$ is the transmitted wave.

While I understand, logically what is happening here for a matter wave, is similar to a electromagnetic wave/light wave when it passes throught mediums with different refracting index, two things I don't understand are:

  1. Where was this wave function expression $\psi_1=A_1e^{ik_1x}+B_1e^{-ik_1x}$ derived from?

  2. Is the state $\psi_1$ (or $\psi_2$ for that matter) of the system (particle) a superposition of eigenstates of a certain operator, or an eigenstate of an operator. Whether that Operator can or cannot be the Hamilton operator

  • $\begingroup$ tcm.phy.cam.ac.uk/~bds10/aqp/lec2_compressed.pdf Page 10 and further. $\endgroup$
    – Gert
    Mar 4, 2022 at 12:18
  • $\begingroup$ This is a nice pdf, but the information here is pretty condensed. Do you have a link that gives a better and broader explanation of this ? $\endgroup$
    – imbAF
    Mar 4, 2022 at 13:31
  • $\begingroup$ Not at hand. But webpages about the quantum potential barrier are really galore. Take your pick. $\endgroup$
    – Gert
    Mar 4, 2022 at 13:32
  • $\begingroup$ I have searched for a long time, for a derivation of the solution to the wave eq. in the form $u(x,t)=u_0e^{i(kx-\omega t)}$ but I haven't found one yet. Can you help me with one? $\endgroup$
    – imbAF
    Mar 4, 2022 at 14:56
  • $\begingroup$ I don't know that it exists in that form. The time-dependent SE is: $i\hbar \partial_t u(x,t)=\hat{H}u(x,t)$. The solution is assumed to be of the form $u(x,t)=\psi(x)\phi(t)$. Use this assumption to find $\phi(t)$ and $\psi(x)$ by separation of variables. $\psi(x)$ is the solution to the Time Independent SE: $\hat{H}\psi(x)=E\psi(x)$. The hatted quantity is the Hamiltonian operator. $\endgroup$
    – Gert
    Mar 4, 2022 at 15:33

2 Answers 2


Merely answering:

I have searched for a long time, for a derivation of the solution to the wave eq. in the form $u(x,t)=u_0e^{i(kx−\omega t)}$ but I haven't found one yet. Can you help me with one?

by OP in the in the comment section. You may have 'searched for a long time' because there's no solution of that form.

For a particle in $\text{1D}$ the Time Dependent Schrödinger Equation is (and where I'm using $\Psi$ instead of the OP's $u$):

$$i\hbar \partial_t \Psi(x,t)=\hat{H}\Psi(x,t)\tag{1}$$

where :


Assume ('Ansatz'):


Separation of variables: insert $(2)$ into $(1)$

$$i\hbar \psi \phi'=-\frac{\hbar^2}{2m}\psi'' \phi+\psi \phi V(x)$$ Divide by $\psi \phi$: $$i\hbar\frac{\phi'}{\phi}=-\frac{\hbar^2}{2m}\frac{\psi''}{\psi}+V(x)=E$$

where $E$ is a separation constant (total energy in this case). We get $2$ ODEs: $$i\hbar\frac{\phi'}{\phi}=E\tag{3}$$

and after minimal reworking:


$(4)$ is of course the Time Independent Schrödinger Equation, while $(3)$ solves to:


So $\Psi(x,t)$ takes the form:


The step potential of OP's question, $\psi(x)$ is derived here.

  • $\begingroup$ Is there a condition that must be met in order to do a separation of variables. In one of the books I read, only after it was explicitly said that the potential is time independent (and I assume the same goes for the Hamiltonian) we could do that separation. If the Potential, and consequently the Hamiltonian is time dependent, could you do that separation ? $\endgroup$
    – imbAF
    Mar 5, 2022 at 14:21
  • $\begingroup$ Separability is a property of linear PDEs like the SE, the Wave equation, the Heat equation etc. If $V=V(x,t)$ then the SE is no longer linear and separation is no longer possible (but other solution methodologies may exist) $\endgroup$
    – Gert
    Mar 5, 2022 at 16:48
  • $\begingroup$ why isn't SE no longer linear? How can you tell whether that's the case or no? $\endgroup$
    – imbAF
    Mar 5, 2022 at 23:12
  • $\begingroup$ Just look at the equation below $(2)$: it would become: $i\hbar\frac{\phi'}{\phi}=-\frac{\hbar^2}{2m}\frac{\psi''}{\psi}+V(x,t)=E$. Obviously separation has not been achieved because the $V(x,t)$ term can't be 'split'. $\endgroup$
    – Gert
    Mar 6, 2022 at 1:20
  • $\begingroup$ Thanks for accepting my answer. $\endgroup$
    – Gert
    Mar 6, 2022 at 1:23

To answer your first question, the equation you have to solve is the Schrödinger equation (nothing suprising) with the potential $V(x) = \delta(x)$. This makes sense as this means that there is no potential for $x < 0$ nor for $x > 0$ and leaves us with an infinite barrier at $ x = 0 $. This is where you get the solving of two different wave functions. Both equations have the following form (no time dependence here):

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

while keeping in mind that one is valid for negative positions and the other one for positive ones. It should be clear that the solution to this is a superposition of waves, one with positive and one with negative frequency. Then the next step is to impose continuity and derivability of the full solution, which implies that $\psi_1$ and $\psi_2$ must have the same values and the same derivative at $x = 0$. The last step would be to impose boundary conditions such as $B = 0$.

Now we are in position to answer your second question. $\psi_1$ and $\psi_2$ are themselves eigenstates, as should be clear from above. I understand that you are familiar with the free particle solution to the Schrödinger equation, and know that the energy eigenvalues are continious. The same applies here: there are two regions of free particle solutions so the full solution will be a continious superposition of all states (usually this is very messy and it is done by computer). This of course means that the operator giving rise to the eigenstates is the Hamiltonian operator, which also should be clear from above.

Please keep in mind this is only for the case you presented above, which implies that E > 0. For bound states (i.e. E < 0), there is only one posible state in the delta function potential.

  • $\begingroup$ I didn't downvote you $\endgroup$
    – imbAF
    Mar 6, 2022 at 10:41
  • $\begingroup$ Alright then, I was just surprised that I got downvoted twice, and maybe I could improve the answer if somebody gave me some reasons. People seem to downvote to discourage, instead of to improve the content of the forum :( $\endgroup$ Mar 6, 2022 at 10:43
  • $\begingroup$ People seem to downvote to discourage, instead of to improve the content of the forum To discourage what? Your answer doesn't answer the question, which is about a step potential, not a $\delta$ potential. $\endgroup$
    – Gert
    Mar 6, 2022 at 12:46
  • $\begingroup$ Alright, well you are completely right, I read over that part. But instead of just leaving a downvote they could also leave a comment. The difference between what I wrote and what I should have included is not that huge, just takes one edit you know? Not leaving comments when one downvotes discourages answering questions, because the feedback you get is to erase your answer, when you obviously didn't realise that you answered something wrong. $\endgroup$ Mar 6, 2022 at 13:41

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