Finding potential given some conditions

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Homework Help Overview

The discussion revolves around finding the potential V(x) for a one-dimensional quantum system given the ground state and first excited state wave functions, along with specific boundary conditions. The context involves the time-independent Schrödinger equation (TISE) and the properties of quantum mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the potential V(x) by substituting the first excited state wave function into the TISE and manipulating the resulting equations. Some participants question the relationship between the potentials discussed in different threads, exploring how the context of each problem affects the potential forms.

Discussion Status

Participants are actively engaging with the mathematical derivations and exploring the implications of different potentials. There is acknowledgment of the harmonic oscillator potential in one response, and a clarification of differences between potentials in separate threads is being discussed.

Contextual Notes

Participants note that the potential in the original poster's problem is valid for all x, contrasting it with a previous thread where the potential was defined over a restricted interval. This highlights the importance of context in quantum mechanics problems.

CAF123
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Homework Statement


Denote the ground state and the first excited state of a 1D quantum system by ##\psi_{0}## and ##\psi_{1}##. If it is given that $$\psi_{1}(x) = x \psi_{0}(x)\,\,\,\text{and}\,\,V(0)=0$$ find the potential V(x).

Homework Equations


TISE

The Attempt at a Solution


If I sub ##\psi_1## into the TISE I get ##-\frac{\hbar^2}{2m} \psi_1'' + V(x)\psi_1 = E_1 \psi_1##(1). I can replace ##\psi_1## with ##x\psi_o## to give after differentiation, ##-\frac{\hbar^2}{2m} \left(2\psi_o' + x\psi_o''\right) + V(x)x\psi_o = E_1x\psi_o##. (2)

But I also know that ##-\frac{\hbar^2}{2m} \psi_0'' + V(x)\psi_0 = E_2\psi_0## Rearrange (2) yields $$-\frac{\hbar^2}{m}\psi_o' + x\left(-\frac{\hbar^2}{2m}\psi_o'' + V(x)\psi_o)\right) = E_1x\psi_o.$$The coefficient of x is precisely ##E_2\psi_o## so I end up with a diff eq for ##\psi_o##: $$\psi_o' = -\frac{m(E_2 - E_1)}{\hbar^2}x\psi_0 \Rightarrow \psi_o = K \exp\left(-\frac{m(E_1-E_2)}{\hbar^2} \frac{x^2}{2}\right)$$, K a constant.

So it then easy to get ##\psi_1## by multiplying ##\psi_o## by x. Subbing ##\psi_1## back into (1) and differentiating twice I get that $$V(x) = \frac{m}{2}\frac{(E_1-E_2)^2}{\hbar^2}x^2$$, neglecting the constant terms. Does this seem okay?
 
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Looks good. It's the harmonic oscillator with the frequency
\omega^2=\frac{(E_1-E_2)^2}{\hbar^2}
as it should be.
 
Hi vanhees71, I was wondering how the answer I got here differs from the potential in my other thread '1D Potential Step QM Problem'. Is it just the case that in that thread V was the harmonic oscillator to a restricted interval, whereas in this case it holds for all x? Hence the potentials in both threads are different.
 
eh? In the 1D potential step problem, I'm guessing the potential was simply a step potential. But here you have a completely different potential. different potential gives a different wavefunction.

edit: ah whoops, I didn't read your post carefully enough. OK so in your other thread, the potential was like a harmonic oscillator potential over some interval and some larger constant outside that interval? Well, still that potential is different to this potential, as you say, which is why you get different answers. In other words, different system, different answer.
 

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