# Normalize the wave function

1. Oct 16, 2014

### jimmycricket

1. The problem statement, all variables and given/known data
Normalize the wave function ,$\psi(x)$, where $$\psi(x)=\frac{1}{1+ix}$$.

2. Relevant equations

3. The attempt at a solution
$$\langle\psi\mid\psi\rangle= \int_{-\infty}^{\infty}\frac{1-ix}{1+x^2}\frac{1+ix}{1+x^2}dx=\int_{-\infty}^{\infty}\frac{1}{1+x^2}=\left. arctan(x)\right|_{-\infty}^{\infty}=0$$ since arctan(x) is an odd function.
Now I know a normalized wavefunction satisfies $$\langle\psi\mid\psi\rangle=1$$.
I dont know how to manipulate the given wavefunction in order to satisfy the condition.

2. Oct 16, 2014

### ShayanJ

The problem is, only when the integrand is odd, you can say the symmetric integral is zero. But in your case, the integrand is even!
Here's how it should be:
$\int_{-\infty}^\infty \frac{1}{1+x^2}dx=2 \int_{0}^\infty \frac{1}{1+x^2}dx=2 \tan^{-1}(x) |_0^\infty=(2k+1)\pi$
I should say I don't know what k to choose or even its important what k to choose or not!!!

3. Oct 17, 2014

### Orodruin

Staff Emeritus
The integrand is continuous and so must the primitive function be. Use this to fix k.

4. Oct 17, 2014

### ShayanJ

Well, this is what I think about k.
When people want to define $\tan^{-1}(x)$, they restrict the domain of $\tan(x)$ to $(-\frac \pi 2,\frac \pi 2)$ to make it one-to-one. So the range of $\tan^{-1}(x)$ is $(-\frac \pi 2,\frac \pi 2)$ and so $\lim_{x\rightarrow \infty} \tan^{-1}(x)=\frac \pi 2$.
Also it seems to me that the normalization constant doesn't have that much physical significant!

5. Oct 17, 2014

### Orodruin

Staff Emeritus
In this particular problem, it is fine to pick any branch of arctan as any branch is a primitive function to the integrand (I could add an arbitrary constant on top to boot!). The only important thing is that you evaluate both limits of the integral in the same branch.

The normalisation constant will be important when you start evaluating expectation values, unless you want to write
$$\langle x\rangle = \frac{\langle \psi| x |\psi\rangle}{\langle \psi |\psi\rangle}$$
everywhere ... Better to just have computed $\langle \psi |\psi\rangle$ once and for all.

6. Oct 17, 2014

### ShayanJ

Yeah, that's the right way of looking at it. I missed this in my first post. Considering this, $\tan^{-1}(x)|_0^\infty$ is equal to $\frac \pi 2$ anyway!

But if we choose $|\phi\rangle=\frac{|\psi\rangle}{\sqrt{\langle \psi|\psi\rangle}}$ to be our wave function, then the normalization constant is not important.

Last edited: Oct 17, 2014
7. Oct 17, 2014

### Orodruin

Staff Emeritus
Well, then you have just normalised it using the normalisation constant $1/\sqrt{\langle \psi|\psi\rangle}$ ... This is the entire point of computing the normalisation constant.

8. Oct 17, 2014

### ShayanJ

Yeah, I know what you mean. But I was saying that if it happens that the normalization integral can give multiple values, as I wrongly thought is the case here, then its not important which one to choose.