# Delta potential, negative energy, quantum particle

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.

• The negative energy implies that what you have found is a bound state. Meaning, the particle will be bound inside the delta potential. – Prasad Mani Oct 26 '16 at 18:12
• Can I think of this as taking a normal square potential well with width a and then letting a go to 0 ? – Luka8281 Oct 26 '16 at 18:24
• That is exactly how delta function potential is thought of asymptotically. – Prasad Mani Oct 26 '16 at 18:26

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.!

What is the physical interpretation of the negative energy, since the potential energy is equal to zero, except at origin?

First, we need to understand that Energy is relative, both in Quantum and in Newtonian mechanics. This means that different reference frames will determine different values for the energy of a given object at a given point in space and time.

In this case, we are allowing to have negative Energy values for delta potential energy by the following.

𝑉(𝑥)=−𝜆 𝛿(𝑥) (where 𝜆 is a positive number)

This means, in this reference frame we consider all Energy values with minimum 𝑉(𝑥) value that is negative infinity. Therefore, we are allowed to solve the Schrodinger Equation for all the negative and positive values of Energy.

In contrast, for the free particle problem's reference frame, we consider minimum 𝑉(𝑥) value is zero and all the Energy values should be positive for well-defined solutions.

However, the conservation of energy holds in any reference frame. This means that, in a single reference frame the total energy (KE + PE) will be constant over time.