# Normalising schrodinger pdf

1. Nov 22, 2009

### Identity

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

Normalise
$$\Psi(x,t) = Ae^{-\lambda |x|}e^{-i \omega t}$$

2. Relevant equations

(i.e. it must satisfy $$\int_{-\infty}^{\infty} |\Psi (x,t)|^2 dx=1$$)

3. The attempt at a solution

This is my first attempt at normalisation and I don't know how valid my reasoning is, can someone please check to see if I'm doing it right:

$$\int_{-\infty}^{\infty}|\Psi (x,t)|^2 dx = \frac{A^2}{\lambda} e^{-2 i \omega t}$$

$$\Rightarrow \frac{\lambda}{A^2}e^{2i \omega t} \int_{-\infty}^{\infty}|\Psi (x,t)|^2 dx =1$$

$$\int_{-\infty}^\infty \left(\frac{\sqrt{\lambda}}{A}e^{i\omega t} |\Psi|\right)^2dx =1$$

$$\therefore \Psi (x,t)_{normalised} = \frac{\sqrt{\lambda}}{A}e^{i\omega t} \times \left(Ae^{-\lambda |x|}e^{-i \omega t}\right) = \sqrt{\lambda}e^{-\lambda |x|}$$

thanks

2. Nov 22, 2009

### Staff: Mentor

First, recall that $|\Psi (x,t)|^2$ means $\Psi^*(x,t) \Psi(x,t)$. That is, you multiply $\Psi(x,t)$ by its complex conjugate $\Psi^*(x,t)$.

What is the complex conjugate of the $\Psi(x,t)$ that you were given?

3. Nov 22, 2009

### Identity

Oh, right! So I can't just square it?

The conjugate would be $$Ae^{-\lambda |x|}e^{i \omega t}$$

So I just times those together to find $$|\Psi|^2$$?

Are my other steps sound though (assuming that I fix the start up)?

Thanks again

4. Nov 22, 2009

### ideasrule

Your other steps will be completely different. For one thing, the e^ part becomes 1 and you won't have to worry about it. You should get a value for A, which makes sense because normalization means finding the amplitude A that goes in front of the wavefunction.

5. Nov 23, 2009

### Identity

Thanks for your help! Is this correct now?

$$\begin{array}{cclc}\int_{-\infty}^\infty |\Psi|^2dx &=& 2\int_{0}^\infty (Ae^{-\lambda x}e^{-i \omega t} Ae^{-\lambda x}e^{i \omega t})dx & \ \ \ \ \ \ \ \ \ \int_{-\infty}^\infty f(|x|)dx = 2\int_0^\infty f(x)dx \\\\ &=& 2\int_0^\infty A^2e^{-2\lambda x}dx & \\\\ &=& \dfrac{A^2}{\lambda}& \end{array}$$

so the answer is $$A = \sqrt{\lambda}$$?

Last edited: Nov 23, 2009
6. Nov 23, 2009

### ideasrule

Almost, but e^-λx * e^-λx = e^-2λx.

7. Nov 23, 2009

thanks!