Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.3.

Probably the most common continuous probability density is the gaussian distribution, specified by

First, we need to normalize the distribution by finding . That is, we must have

The gaussian integral is very common, and the result is that

Although there is no closed form indefinite integral, the definite integral can be found by a cute trick.

We can now transform to polar coordinates using

Therefore

Using Maple (or integration by parts) we can work out the average and variance.

The distribution has the standard bell shape. Here’s a plot for and :

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Cindyis there any mistakes between equation14~16???

since prob. density in this problem is defined as A*e^[-lamda*(x-a)^2]

gwrowePost authorFixed now. Thanks.

Bùi Việt AnhI can’t understand how we use integration by parts to find the result. Could you explain me just a bit details, please?

gwrowePost authorThe integration by parts applies to . For , we can use the substitution to get

which can be done using (13). The integral because the integrand is an odd function.

For (15), use the same substitution to get

The integral can be done by parts

which again can be done using (13).

MooseI don’t know why I keep getting 1/lamda instead of 1/(2*lamda) for the constant in front of the last integral above. I get a factor of 2 that cancels, since I let u’=u^2, so du’=2u du, everything else matches perfect though. Hmm. I don’t know what I must be missing. Please excuse my oversight.

MooseWhoops, found it.

CoryI am having the same error as Moose. I used that same substitution, but still don’t see where my error is. Could someone please make a suggestion as to what to look at to resolve the issue?

CoryNever mind, I figured out that the substitution I chose provides a different answer. I am not sure why that is the case though. I get the author’s solution if I choose u’=u, but I get Moose’s extra factor of 2 with Moose’s substitution of u’=u^2.

gwrowePost authorIf you’re referring to the integral I don’t see why you need a substitution at all. As I say, you can do it by parts

which again can be done using (13).

CoryThanks for the quick response. I was incorrect before when I said substituting. I was using integration by parts both times, but I think I understand now why one of the arrangements that I tried does not work mathematically. Thank you again for posting these. It has helped immensely with my learning of the material.

ramonaI’m curious about (5), i accept it on a gut level, and i could explain a guess at why it makes sense–or why it seems to make sense to me… but i don’t feel like my “feeling” of it is really good enough to go on,in saying i actually “understand” it…

might you give some explanation why it works?

Joey ContrerasI was working on this very problem earlier today, but I noticed the problem says “look up any integrals you need.”

One of my buddies was telling me that there are specific integrals in the back of the Griffith’s book that are can simply be looked up, but that for anything beyond those integrals it’s customary to simplify one’s integral down to a form where that table in the back of the book can be applied.

Do you know if he’s right or if there are certain integrals which are necessary to solve problems like these? Part of my confusion with this problem is that I’m not sure when I have to simplify and when I can just look up what the integral evaluates out to.

Thanks.

gwrowePost authorThere aren’t any integral tables in the two books by Griffiths that I have (Electrodynamics and Quantum Mechanics) so you’d just have to look up the integrals in other books or on the web. Integrals can be tricky, which is why I use software (Maple) to do a lot of them. If you have to do them by hand, I’d try a few basic techniques like integration by parts, trig substitutions or other algebraic substitutions. If none of those works, you’ll just have to look them up.

Fahimif the integration of u*exp(-lu ^2) is

equal to zero,then why not

integration of x*exp{-l(x-a^2)} isn’t

zero?

gwrowePost authorSee the comment from 12 March 2016. is not an odd function () so it’s not anti-symmetric about so we wouldn’t expect its integral to be zero.