How Do I Normalize a Quantum Mechanics Equation With an Integral?

[Marthius]

I am starting to teach myself quantum mech. in preperation for this coming semester, however i have hit a mathamatical road block. I need to normalize the folowing equation.

$$P(x)=Ae^{-\lambda(x-a)^{2}}$$

Unfortunatly I do not take DiffEQ until this coming semester and I don't know how to take the following integral.

$$\int e^{-\lambda(x-a)^{2}}$$

Any help would be greatly appreciated (yes, I know that i should have taken DiffEQ before this...).

 

[HallsofIvy]

I am starting to teach myself quantum mech. in preperation for this coming semester, however i have hit a mathamatical road block. I need to normalize the folowing equation.

$$P(x)=Ae^{-\lambda(x-a)^{2}}$$

Unfortunatly I do not take DiffEQ until this coming semester and I don't know how to take the following integral.

$$\int e^{-\lambda(x-a)^{2}}$$

Any help would be greatly appreciated (yes, I know that i should have taken DiffEQ before this...).

You may not like my answer: That integral is
$$\frac{A\sqrt{\pi}}{2}erf(x-a)+ C$$
where "erf(x)" is the error function. It is defined by
$$erf(x)= \frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}dx$$

You have to understand that most integrable functions (in a technical sense "almost all integrable functions") cannot be integrated in terms of "elementary" functions- the kind that you learn about in basic algebra or calculus courses. That integral, which is used extensively in probability and statistics calculations is one such.

 

[Marthius]

Thanks for the answer, unfortunatly now i am more confused then ever. Given that as the integral of the function, how do i go about normalizing the function with A. Should i be ignoring the error function since no numerical values were given in the problem?

 

[scorpion990]

That indefinite integral is impossible to calculate in terms of elementary functions. However, since you're normalizing a wavefunction, you don't need the indefinite integral, but only the definite. I assume that at least one of your limits is infinity...

Like somebody else said... The mathematically rigorous way of going about it is to look into the error function. If you don't have too much experience with special functions, then there is an easy fix: Look at a table of integrals in the back of your quantum book! Virtually all quantum mechanics books have a short table of definite/indefinite integrals. Your integral should be in there in one form or another... You will more than likely need to use an elementary substitution.

 

[HallsofIvy]

Well, what do YOU mean by "normalizing" a function? While that anti-derivative cannot be done in "closed form" it turns out to be relatively easy to show that
$$\int_a^\infty e^{-\lambda(x-a)^2}dz= \frac{\sqrt{\pi}}{2\lambda}$$
That was the reason for the "$2/\sqrt{\pi}$" in the definition of erf(x). If I understand what you mean by "normalization" correctly, $erf(x- a)/\lambda$ is the normalization of your function.

 

[Marthius]

Well, what do YOU mean by "normalizing" a function? While that anti-derivative cannot be done in "closed form" it turns out to be relatively easy to show that
$$\int_a^\infty e^{-\lambda(x-a)^2}dz= \frac{\sqrt{\pi}}{2\lambda}$$
That was the reason for the "$2/\sqrt{\pi}$" in the definition of erf(x). If I understand what you mean by "normalization" correctly, $erf(x- a)/\lambda$ is the normalization of your function.

As it turns out, since I last posted this problem i have been able to solve my question. What you said here is exactly what I mean (withought realizing it). Thanks for the help.

 

[Marthius]

would I be correct in saying

$$\int_{-\infty}^\infty e^{-\lambda(x-a)^2}dz= \frac{\sqrt{\pi}}{\sqrt{\lambda}}$$

 

[slider142]

Yep, that's what you should get. Do you mean dz or dx?

 

[Marthius]

Yep, that's what you should get. Do you mean dz or dx?

i meant dx, z was just the substitution var I used, was an accident.