# The simplest derivation of position operator for momentum space

1. Oct 14, 2014

### cryptist

Might be simple but I couldn't see. We can easily derive momentum operator for position space by differentiating the plane wave solution. Analogously I want to derive the position operator for momentum space, however I am getting additional minus sign.

By replacing $$k=\frac{p}{\hbar}$$ and $$w=\frac{E}{\hbar}$$ into the plane wave solution, we get
$$\Psi=e^{ipx/\hbar-iEt/\hbar}$$
Then taking the derivative with respect to momentum,
$$\frac{\partial\Psi}{\partial p}=\frac{ix}{\hbar}\Psi$$
Then I get,
$$\hat{x}=-i\hbar \frac{\partial}{\partial p}$$

It has additional minus sign. Where is my mistake and/or how do I derive the position operator for momentum space in the simplest way?

2. Oct 17, 2014

### stevendaryl

Staff Emeritus
Well, you need to think what is the relationship between a function in position-space and in momentum space. This is actually clearest if you use Dirac bra-ket notation, but if you haven't learned that, we can do without.

If you have a wave function $\psi(x)$ in position space, the corresponding function in momentum space is $\tilde{\psi}(p)$. The two are related as follows:

$\psi(x) = \frac{1}{\sqrt{2 \pi}} \int e^{i p x} \tilde{\psi}(p) dp$
$\tilde{\psi}(p) = \frac{1}{\sqrt{2 \pi}}\int e^{-i p x} \psi(x) dx$

(I'm leaving out the h-bars for simplicity; I hope you can figure out where they should go. I think every $p$ should be $p/\hbar$)

So notice that the inverse transform has a minus sign in the exponent. That makes it so that $\hat{x} = +i \frac{d}{dp}$ while $\hat{p} = -i \frac{d}{dx}$

[edit: changed the conventions to make forward and reverse transforms more symmetric]

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

### stevendaryl

Staff Emeritus
You can directly verify using these equations that:
$-i \frac{d}{dx} \psi(x) = \frac{1}{2 \pi} \int p e^{i p x} \tilde{\psi}(p) dp$
$+i \frac{d}{dp} \tilde{\psi}(p) = \frac{1}{2 \pi} \int x e^{- i p x} \psi(x) dx$

4. Oct 17, 2014

### dextercioby

The quantum mechanical convention is to put a sqrt(2pi) for each part of a Fourier transform, the conventionally direct one from coordinate space to momentum one and the reverse one which is from momentum space to coordinate space.

5. Oct 17, 2014

### stevendaryl

Staff Emeritus
I seem to remember that. I changed it.

6. Oct 17, 2014

### atyy

No one else uses this convention except physicists. That's because physicists love non-rigourous lazy ways of thinking, and are always trying to make their lives easy. :D By having the sqrt(2pi), it's the same forwards and back, and one doesn't have to remember where to put the 2pi.

7. Oct 20, 2014

### cryptist

Thank you for the answer. So, in order to derive position opeator for momentum space, I have to start with the wavefunction in momentum space. Not in position space. That was my mistake as far as I understand.