Question

@empty

Answer 1

\[\int\limits_{-\infty}^{\infty} x^2 e^{-a(x-b)^2}dx\] like I found the one for just x on wikipedia but not sure about thsi

Answer 2

symmetry can be used but the integrand itself is gross

Answer 3

I thought to use

Answer 4

Yeah, actually if you don't mind I can show you at trick for the x integral too.

Answer 5

Yeah go ahead im still getting used to this

Answer 6

Yeah you can totally use that, you need to use a u-substitution to get that and you're done

Answer 7

yeah thats what i thought but i thought its weird because then we have to do x=u+a

Answer 8

Substitute to get it into the form you want like you describe:

\[x-b = u\]\[dx = du\]

\[ \int\limits_{-\infty}^{\infty} x^2 e^{-a(x-b)^2}dx = \int\limits_{-\infty}^{\infty} (u+b)^2 e^{-au^2}du\]

Now just square that binomial:

\[ \int\limits_{-\infty}^{\infty} u^2 e^{-au^2}du\ + 2b \int\limits_{-\infty}^{\infty} u e^{-au^2}du\ +b^2 \int\limits_{-\infty}^{\infty} e^{-au^2}du\]

At this point, you might have already worked out those last two integrals, since that last one could be 1 if you already normalized it for example. You can also rearrange that leading one to be of the form that fits your thing cause of symmetry, also like you said.

Answer 9

Oh damn haha, I think I was getting close to that exact method I was just squaring it to

Answer 10

That totally makes sense

Answer 11

Yeah, in a sense, not simplifying the integrals can actually end up working out to your advantage sometimes because you'll end up with integrals you know the answer too which is a common trick. In fact, later you'll just assume you're dealing with normalized wave functions so stuff like \(\langle \psi |\psi \rangle = 1\) will just be something you don't even think about.

Answer 12

I don't know if you're familiar with the Bra-Ket notation yet, if not, it's really not worth worrying about, pretend I wrote \(\int \psi^* \psi dx = 1\) there, it's the same thing.

Answer 13

Ah yeah, good stuff haha.

Lol dude when our prof mentioned Bra-Ket and it was just < this being bra, and > this ket it gave me a good laugh, and yeah I already did some of this stuff, just sometimes these integrals can get tricky

Answer 14

That's why I'm practicing xD

Answer 15

Like I didn't know what you meant with the <|> and stuff but from chapter 1 you can see we use that notation for like expectation value

Answer 16

Yeah, the purpose of the bra-ket notation is that it actually is much much much more flexible. It essentially representing everything as matrix and vectors in a generalized way without reference to the representation. Right now, although you are probably not aware, you are representing everything with x, y, z coordinates. But you can have a wave function that is a function of its momentum coordinates as well for instance which contains the exact same information and the two are related to each other by a Fourier transform... Don't let me scare you though, just forget it if it bothers you haha.

Answer 17

Haha, no it's cool, I like seeing this stuff, the more we're exposed to this I think the better off we'll both be xD

Answer 18

True that haha. So I guess I'll give two little tricks to evaluating

\[\langle x \rangle = \int_{-\infty}^\infty x e^{-a(x-b)^2} dx\]

It's really the same trick I just did, but I like it so here goes:
\[x-b=u\]\[dx=du\]
\[\int_{-\infty}^\infty u e^{-au^2} du + b \int_{-\infty}^\infty e^{-au^2} du\]
By symmetry,
\[0 = \int_{-\infty}^\infty u e^{-au^2} du \]
By normalization (remember, shifting x to u+b doesn't change the area under the curve if it's integrated infintely everywhere),
\[1 = \int_{-\infty}^\infty e^{-au^2} du\]
Therefore,
\[\langle x \rangle = b\]

Of course, I am assuming that for this entire argument that your wave function is normalized.