Required math: calculus
Required physics: Schrödinger equation
Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.25.
We’ve seen that the bound state of a particle in a delta function potential wellhas the wave function
We can work out the mean values of position and momentum, and of their squares, in the usual way.
For , since the wave function is even and the function is odd, we have
since the integrand is odd.
For the mean square position, we get
where we used software to do the integral.
Now for the momentum. We get
We can split this integral into two parts joined at the origin.
The derivative is therefore an odd function. Since the original wave function is even, the product of the two is odd, so .
Calculating is a bit trickier, since the derivative above is discontinuous at the origin. If we define the step function
we can write the derivative above as
We’ve seen that the derivative of the step functioncan be taken as the delta function
so the second derivative of the wave function is
Observing that everywhere, we get
After all this we can now calculate :
Going from the second to the third line, we used the property of the delta function
The uncertainty principlefor the bound state of the delta function potential is therefore