Finding potential given some conditions

[CAF123]

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?

 

[vanhees71]

Looks good. It's the harmonic oscillator with the frequency
$$\omega^2=\frac{(E_1-E_2)^2}{\hbar^2}$$
as it should be.

 

[CAF123]

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.

 

[BruceW]

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.

 

[paranormal]

I appreciate your approach to solving this problem. Your use of the TISE and the given conditions to derive the potential V(x) is logical and well-explained. However, I would suggest adding more detail to your explanation of how you arrived at your final result, particularly in the step where you obtain V(x) from the differential equation. This will help others understand your thought process and make your solution more clear. Overall, your solution seems sound and your approach demonstrates a good understanding of quantum mechanics. Well done.