# Transformation Function - Position & Momentum Operators

1. Jun 9, 2009

### dengar768

1. The problem statement, all variables and given/known data
I am currently studying for my quantum physics exam and I am trying to derive the Transformation function:
⟨x'│p' ⟩=Nexp{(ip' x')/ℏ}
2. Relevant equations

⟨x'│p' ⟩=Nexp{(ip' x')/ℏ}

3. The attempt at a solution

Now I get how to get to
p'⟨x'│p' ⟩=-iℏ d/dx' ⟨x'│p' ⟩

but i cant work out how to get both the p' and the x' into the expontential and im not sure how to explain where the N comes from. Every example i find seems to jump from p'⟨x'│p' ⟩=-iℏ d/dx' ⟨x'│p' ⟩ straight to ⟨x'│p' ⟩=Nexp{(ip' x')/ℏ} without any explanation.
Would anybody be able to explain how to do this?
Thanks

Last edited: Jun 9, 2009
2. Jun 9, 2009

### diazona

Well, you have a differential equation,
$$p \langle x \vert p \rangle = -i\hbar \frac{\mathrm{d}}{\mathrm{d}x} \langle x \vert p \rangle$$
do you know how to solve it?

3. Jun 9, 2009

### dengar768

No. Is this problem easier than im making it?

4. Jun 9, 2009

### Cyosis

Do you know how to solve $a f(x)=f'(x)$? This is exactly the same.