1
$\begingroup$

A particle is moving in one dimension under a potential $V(x)$ such that, for large positive values of $x$, $V(x) \approx kx ^\beta$, where $k>0$ and $\beta$ $\geq$ 1. If the wave function in this region has the form $\psi(x)\sim \exp(-x^\lambda)$, Then show that, $$\lambda=\frac{\beta}{2}+1.$$


I am not sure how to approach this problem. I thought of using Variational Method, but ended up in a complicated Gaussian integral. Is there any other approach as to reach to this result.

$\endgroup$
2
  • 3
    $\begingroup$ What have you tried? Provide your working on the problem and it will greatly help people try and spot specifically where you went wrong. $\endgroup$
    – Triatticus
    Commented May 12, 2022 at 19:15
  • 2
    $\begingroup$ Have you tried plugging this wave function in the time-independent Schrödinger equation ? $\endgroup$ Commented May 12, 2022 at 20:20

1 Answer 1

1
$\begingroup$

We can use one dimensional time-independent Schrödinger equation, $$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\Psi(x)+V(x)\Psi(x) = E\Psi(x)$$

Given that $\Psi(x)\approx e^{-{x^\lambda}}$ and $V(x)\approx kx^\beta$, so we get

$$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}e^{-{x^\lambda}}+kx^\beta e^{-{x^\lambda}} = Ee^{-{x^\lambda}}$$ which gives $$-\frac{\hbar^2}{2m}\lambda[(\lambda-1)x^{\lambda-2}-\lambda x^{2(\lambda-1)}]+kx^{\beta}= E$$ Now we assume $\lambda>0$, then $O(x^{\lambda-2})<O(x^{2(\lambda-1)})$, so we may neglect the first term in the limit of large x. We can now write, $$\frac{\hbar^2}{2m}\lambda^2 x^{2(\lambda-1)}+kx^{\beta}\approx E$$ Differentiating both sides wrt x we get $$2\frac{\hbar^2}{2m}\lambda^2 (\lambda-1) x^{2(\lambda-1)-1}+k\beta x^{\beta-1}\approx 0$$ Now notice that the RHS of the equation is 0 while LHS is x dependent. So this equation can only be valid for all x if $$2(\lambda-1)-1=\beta - 1$$ because if the two are not equal then both the x terms will be linearly independent and their linear combination cannot yield zero. Finally we get, $$\lambda = \frac{\beta}{2}+1$$

$\endgroup$

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.