The simplest derivation of position operator for momentum space

[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?

 

[stevendaryl]

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]

 

[stevendaryl]

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

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

 

[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.

 

[stevendaryl]

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.

I seem to remember that. I changed it.

 

[atyy]

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.

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.

 

[cryptist]

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]

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.