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I was solving a task involving bound states of an electron : We have a delta potential $V(x) = -\lambda \space \delta(x)$, find the energy of the only bound state. So I get $\sqrt{k}e^{i k x} , \space x<0 $ and $\sqrt{k} e^{-i k x},\space x>0$, where I used normalisation and the continuity condition. And then I integrate the stationary Schrödinger equation $H\psi=E\psi$ to get the only eigenvalue - bound state energy $E=- \frac{\lambda ^2 m}{2 \hbar^2}$. What is the physical interpretation of the negative energy, since the potential energy is equal to zero, except at origin, does this imply that the kinetic energy is negative ($H=T + V$)? There are some similar examples on this, but they are all done with tunneling, where there is a potential present.

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  • $\begingroup$ The negative energy implies that what you have found is a bound state. Meaning, the particle will be bound inside the delta potential. $\endgroup$ – Prasad Mani Oct 26 '16 at 18:12
  • $\begingroup$ Can I think of this as taking a normal square potential well with width a and then letting a go to 0 ? $\endgroup$ – Luka8281 Oct 26 '16 at 18:24
  • $\begingroup$ That is exactly how delta function potential is thought of asymptotically. $\endgroup$ – Prasad Mani Oct 26 '16 at 18:26
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For your confusion about whether kinetic energy is negative, just do this. You have the total energy $E$ = $-\frac{\lambda^2 m}{2\hbar^2}$. Find $\langle V \rangle$, the expectation value of the potential. Subtract that value from the energy $E$. What you get is the expectation value of kinetic energy $\langle T \rangle$, which is positive (try it out). Finding $\langle T \rangle$ directly for the wavefunction of delta function potential is nasty.!

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