Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.5.

The delta function well gives rise to a wave function that decays exponentially either side of the delta function:

We can normalize in the usual way:

By symmetry, and

Therefore

A plot of is shown, with vertical yellow lines indicating , for the case :

The probability that the particle lies outside is

In this case, the probability of being greater than one standard deviation is a constant, independent of .

chrisWhy does the e^(-iwt) term disappear?

Is it because we square it, or does it have to do with the nature of the exponential…?

gwrowePost authorWhen we calculate , we multiply by so gets multiplied by giving 1.

HalWhy do the powers of the exponentials at (2) have different powers for the different limits of integration?

gwrowePost authorBecause of the absolute value in equation 1. For , .

HalThanks! And at (3) How does the 2 arise?

gwrowePost authorBecause the two integrals in equation 2 are equal.

SigexWhen you do the integral, I get – 1/2L, but by you getting sqrt(L) you seem to have lost the minus sign some how. How did you lose the minus sign that comes from the expoential on e in the integral?

gwrowePost authorNote that since is always positive, the integral cannot be negative.

Cindyhello

i have some questions about eq6

i used gamma function here and thought i was stupid after looking your calculation

did u do integration by parts in eq6 ??

the equation is simple but i have no idea to the answer only by “look”

and how to know

“By symmetry, =0

i wonder how to just have a look and get the results

(after i do the calculation and thought i was stupid again……

gwrowePost authorYou could use the gamma function to do equation 6 after transforming the variable to :

Or you could integrate by parts twice to get rid of the .

As for , it’s the integral of an odd function () multiplied by an even function over an interval symmetric about the origin, so the answer is zero.

AnonymousI tried to verify = 0 by using eq 6 (replacing x^2 by x). Is this correct? (I am not getting zero, maybe I missed out something.)

gwrowePost authorYou can see that by noting that since is an even function, then is an odd function and the integral from to of any odd function is zero. If you want to modify equation 6, you need to replace by and also extend the lower limit on the integral to but it’s better to use the symmetry argument.