Evaluating complex integral problem

[Leechie]

Homework Statement

I'm having some trouble evaluating the integral
$$\int^\infty_{-\infty} \frac{\sqrt{2a}}{\sqrt{\pi}}e^{-2ax^2}e^{-ikx}dx$$
Where a and k are positive constants

Homework Equations

I've been given the following integral results which may be of help
$$\int^\infty_{-\infty} e^{-x^2} dx= \sqrt{\pi}$$$$\int^\infty_{-\infty} e^{-x^2}e^{-ikx} dx = \sqrt{\pi}e^{-\frac{k^2}{4}}$$

The Attempt at a Solution

The best I can come up with is
$$\int^\infty_{-\infty} \frac{\sqrt{2a}}{\sqrt{\pi}}e^{-2ax^2}e^{-ikx}dx=\sqrt{2a}e^{-\frac{k^2}{8a}}$$
But I'm not sure this is correct. Would someone be able to explain the steps to take to evaluate this integral so I can attempt to work it out myself. Thanks.

 

[PeroK]

Homework Statement

I'm having some trouble evaluating the integral
$$\int^\infty_{-\infty} \frac{\sqrt{2a}}{\sqrt{\pi}}e^{-2ax^2}e^{-ikx}dx$$
Where a and k are positive constants

Homework Equations

I've been given the following integral results which may be of help
$$\int^\infty_{-\infty} e^{-x^2} dx= \sqrt{\pi}$$$$\int^\infty_{-\infty} e^{-x^2}e^{-ikx} dx = \sqrt{\pi}e^{-\frac{k^2}{4}}$$

The Attempt at a Solution

The best I can come up with is
$$\int^\infty_{-\infty} \frac{\sqrt{2a}}{\sqrt{\pi}}e^{-2ax^2}e^{-ikx}dx=\sqrt{2a}e^{-\frac{k^2}{8a}}$$
But I'm not sure this is correct. Would someone be able to explain the steps to take to evaluate this integral so I can attempt to work it out myself. Thanks.

Are you talking about using those standard integrals you've posted? Or do you want to derive those standard integrals?

In the first case, it's just a simple substitution, isn't it?

In terms of the answer, I think that factor of $\sqrt{2a}$ should have canceled out.

 

[Leechie]

Sorry, I'll give you the full story. Its an assignment question in which I have to use the overlap rule to find the probability that a measurement of the particles energy will give the ground state. The given wave functions are:
$$Ψ(x,0)=\left(\frac{\sqrt{2a}}{\sqrt{\pi}}\right)^\frac{1}{4}e^{-ax^2}e^{-ikx}$$$$ψ(x)=\left(\frac{\sqrt{2a}}{\sqrt{\pi}}\right)^\frac{1}{4}e^{-ax^2}$$
The other integrals $\int^{\infty}_{-\infty}e^{-x^2}dx=\sqrt{\pi}$ and $\int^{\infty}_{-\infty}e^{-x^2}e^{-ikx}dx=\sqrt{\pi}e^{-\frac{k^2}{4}}$ where given in the question as integral values which may be useful.

The overlap integral I'm using is:
$$p=\left| \int^{\infty}_{-\infty}ψ^*(x)Ψ(x,0)dx \right|^2 $$
After substituting in the wave function values and simplifying a bit I've eventually ended up with:
$$p=\left| \frac{\sqrt{2a}}{\sqrt{\pi}} \int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx \right|^2$$
Which is where I'm getting stuck.

I was thinking I need to do something to $\int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx$ and get it into the form $\int^{\infty}_{-\infty}e^{-x^2}e^{-ikx}dx$. I've tried completing the square and using substitution which is how I got to the answer $\sqrt{2a}e^{-\frac{k^2}{8a}}$ but I'm not confident I've done it right. Really I just need some help evaluating $\int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx$. Thanks.

 

[PeroK]

Sorry, I'll give you the full story. Its an assignment question in which I have to use the overlap rule to find the probability that a measurement of the particles energy will give the ground state. The given wave functions are:
$$Ψ(x,0)=\left(\frac{\sqrt{2a}}{\sqrt{\pi}}\right)^\frac{1}{4}e^{-ax^2}e^{-ikx}$$$$ψ(x)=\left(\frac{\sqrt{2a}}{\sqrt{\pi}}\right)^\frac{1}{4}e^{-ax^2}$$
The other integrals $\int^{\infty}_{-\infty}e^{-x^2}dx=\sqrt{\pi}$ and $\int^{\infty}_{-\infty}e^{-x^2}e^{-ikx}dx=\sqrt{\pi}e^{-\frac{k^2}{4}}$ where given in the question as integral values which may be useful.

The overlap integral I'm using is:
$$p=\left| \int^{\infty}_{-\infty}ψ^*(x)Ψ(x,0)dx \right|^2 $$
After substituting in the wave function values and simplifying a bit I've eventually ended up with:
$$p=\left| \frac{\sqrt{2a}}{\sqrt{\pi}} \int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx \right|^2$$
Which is where I'm getting stuck.

I was thinking I need to do something to $\int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx$ and get it into the form $\int^{\infty}_{-\infty}e^{-x^2}e^{-ikx}dx$. I've tried completing the square and using substitution which is how I got to the answer $\sqrt{2a}e^{-\frac{k^2}{8a}}$ but I'm not confident I've done it right. Really I just need some help evaluating $\int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx$. Thanks.

Just do the substitution $u = \sqrt{2a}x$.

I think you just made an algebraic error the first time.

By the way, when you have a parameter in a standard integral, $k$ in this case, it's best to replace that with a parameter you are not going to use in your working. E.g. Use $\lambda$ here.

That makes it easier than using $k$ for two different things.

 

[Leechie]

Thanks for the tip!
I think this is where I'm getting confused. I thought the substitution $u=\sqrt{2a}x$ would mess the integral up and make it $\frac{1}{\sqrt{2a}}\int^{\infty}_{-\infty}e^{-u^2}e^{-ikx} dx$?

 

[PeroK]

Thanks for the tip!
I think this is where I'm getting confused. I thought the substitution $u=\sqrt{2a}x$ would mess the integral up and make it $\frac{1}{\sqrt{2a}}\int^{\infty}_{-\infty}e^{-u^2}e^{-ikx} dx$?

If you can't do integration then QM will be hard indeed. You must substitute $u$ all the way thru the integral. Just replacing $x$ in one place is no good.

 

[Leechie]

Ok, thanks for your help with this.