I'm given the ground state wave function $\psi(x)=A\operatorname{sech}(bx)$. Potential is not given but told that it goes to 0 at $\infty$. How to find the eigen value of energy in this state?

My approach so far: Using $\psi(x)$ in TISE, $$\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right]\psi(x) = E\psi(x) $$

EDIT after suggestions: $$ \frac{\hbar^2}{2m}Ab^2\operatorname{sech}(bx)(2\operatorname{sech}^2(bx) -1) + V(x)A\operatorname{sech}(bx) = EA\operatorname{sech}(bx) $$ Evaluating at $\infty$ $E=-\frac{\hbar^2b^2}{2m}$

Oh, i have messed up by converting hyperbolic to exponential. Thanks. A little surprising that it has got the same ground state energy of a $\delta$ potential

  • $\begingroup$ The RHS doesn't depend on x $\endgroup$ – Phoenix87 Feb 16 '15 at 8:43
  • $\begingroup$ Your equation appears to be incorrect $\endgroup$ – hft Feb 16 '15 at 8:44

First, you need to fix your equation for E. You seem to have divided out by $\psi$ in the V and E terms, but not in the kinetic term... among other issues (the kinetic term should end up proportional to sech^2-tanh^2)... Just recheck the derivatives.

After you fix the equation you can just evaluate the LHS at infinity with the known value of V(x)=0.

| cite | improve this answer | |
  • $\begingroup$ Thanks, I got the answer. A little aside, how to know if this potential has other excited states? $\endgroup$ – levitt Feb 16 '15 at 9:17
  • $\begingroup$ Yes, it has other excited states. The Hamiltonian has eigenfunctions that form a complete set, so there must be more eigenfunctions than just the one you were given. Since the one you were given was specified to be the ground state, the other states are excited states. $\endgroup$ – hft Feb 16 '15 at 19:18
  • $\begingroup$ Or are you asking whether there are other bound excited states? $\endgroup$ – hft Feb 16 '15 at 19:19
  • $\begingroup$ yes, other bound states. I am guessing they are finite in number. $\endgroup$ – levitt Feb 17 '15 at 2:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.