**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

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BobHi,

When you took the derivative of : [exp(-ma|x|/h^2) (2H(x)-1)]

how did you get a (2H(x)-1)^2 term? When I try to do it, only (2H(x)-1) term ever appears.

Thanks!

Bob

growescienceYou need to apply the product rule:

Since:

the first term gets a factor of .

GeorgeI believe that in your mean squared average you are missing an x^2, like so:

$\langle x^{2}\rangle=\int_{-\infty}^{\infty}x^{2}\psi^{2}\left(x\right)dx$

growescienceFixed now. Thanks.

PeroI avoided the integration for p^2 by calculating the expected value of V, hence the expected value of T (E is given). V and T are both E/2. Then getting p^2 from that.

growescienceI don’t think that argument works here, since the average potential is negative (it’s a delta-function

well) so you can’t set it equal to kinetic energy (which is always positive). You getEven if you try to set the absolute value of this to , there’s still that factor of 2 which messes things up and gives which is twice the correct answer. I think that is true only for certain potentials (like the harmonic oscillator, where everywhere).

PeroWe have and now , so and as required.

growescienceSorry,

I can’t

understand what you’re doing……. what is E- and how does this relate to V?

PeroSorry, wordpress is chewing up what I type. In words: E is the energy of the system, which came out when the equation was solved in the first place. The expected value of V can be easily derived, as you did above. The expected value of T must be the difference E-V. This gives the expected value of p^2 without any integration.

growescienceAh, I see. Yes, that will work. What confused me was your saying that . Actually, , but other than that, your argument gives a clever way of finding .