# Evaluating complex integral problem

1. Nov 18, 2016

### Leechie

1. The problem statement, all variables and given/known data
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

2. Relevant equations
I've been given the following integral results which may be of help
$$\int^\infty_{-\infty} e^{-x^2} dx= \sqrt{\pi}$$$$\int^\infty_{-\infty} e^{-x^2}e^{-ikx} dx = \sqrt{\pi}e^{-\frac{k^2}{4}}$$
3. 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.

2. Nov 18, 2016

### PeroK

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 cancelled out.

3. Nov 18, 2016

### 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.

4. Nov 18, 2016

### PeroK

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.

5. Nov 18, 2016

### 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 $\frac{1}{\sqrt{2a}}\int^{\infty}_{-\infty}e^{-u^2}e^{-ikx} dx$?

6. Nov 18, 2016

### PeroK

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.

7. Nov 18, 2016

### Leechie

Ok, thanks for your help with this.