# What will be the sign of the ground state energy?

A particle of mass $m$ moves in a 1 dimensional potential $V(x)$, which vanishes at infinity. The exact ground state eigenfunction is $\phi(x)=A\operatorname{sech}(\lambda x)$, where A and λ are constants.

Using Schrodinger equation in one dimension, I found my answer to be negative, can the ground state energy be negative? Or I am making some mistake while calculating it?

• depending on the potential, the energy can be negative which may be a manifestation of a bound state. – physicopath Jan 8 '18 at 8:33
• But here type of potential is unknown, then how can we say that? And if the ground state energy is negative then that state is bounded which means what? – Prashant Kumar Jan 8 '18 at 8:43
• I suggest you to have a look at, for example, quantum mechanical derivation of eigenfunctions and eigenenenergies for hydrogen atom (see feynmanlectures.caltech.edu/III_19.html). Keep in mind that to solve the SE you need to know the potential. If in your question the wave function is a given then it suggests that someone "knew" the potential and solved the SE and got the wave function. Bound state, again referring to hydrogen atom, means that a state in which the electron will be bound to the nucleus, that is, proton. – physicopath Jan 8 '18 at 8:58
• I tried to calculate the ground state energy. First I took the second order derivative, substituted it into the SE and imposed the boundary condition , V(x) goes to zero as x approaches infinity then i got the solution to be negative. – Prashant Kumar Jan 8 '18 at 9:03
• All right. I do not think I understand the details of your question exactly nevertheless having the energy negative itself can not be sign of mistake. – physicopath Jan 8 '18 at 9:18

If $\lim_{|{\bf r}|\to\infty}V({\bf r})=0$ asymptotically, then the bound states have negative energy $E_n<0$. In particular, the ground state energy $E_0$ would be non-positive. This is e.g. the case for hydrogen-like atoms.