Momentum Operator Construction

1. Nov 22, 2009

pzona

Quick question about how Schrodinger constructed the momentum operator. I understand that $$\widehat{H}$$ in the wave equation is the sum of the kinetic and potential operators, using the form H=T+V. The kinetic energy can be expressed by T=p^2/2m, and that's the last part I can follow. Most sources then give the momentum operator as a "substitution" which gives p $$\rightarrow$$ i$$\hbar$$$$\nabla$$. I have absolutely no idea where this comes from (I've yet to take a formal math class on anything beyond basic calculus, maybe this is the problem), could anyone provide some information on this?

2. Nov 22, 2009

physicsworks

$$\hat{\mathbf{p}} = -i \hbar \nabla$$.
Actually, it comes from the properties of Fourier transform, so
$$\overline{F(x,y,z)} = \int F(x,y,z) |\psi(x,y,z)|^2 dxdydz = \int \psi^{*}(x,y,z) F(x,y,z) \psi(x,y,z)dxdydz$$

$$\overline{F(p_x,p_y,p_z)} = \int F(p_x,p_y,p_z) |c(p_x,p_y,p_z)|^2 dp_xdp_ydp_z = \int c^{*}(p_x,p_y,p_z) F(p_x,p_y,p_z) c(p_x,p_y,p_z)dp_xdp_ydp_z$$
where $$c(p_x,p_y,p_z)$$ is the amplitude of de Broglie wave which has momentum $$\mathbf{p}(p_x, p_y, p_z)$$
can be generalized in case of entire rational function $$F(x,y,z)$$ (i.e., for example $$F(p_x) = \sum_n a_n p_x^n$$) and above equations becomes:

$$\overline{F(x,y,z)} = \int c^{*}(p_x,p_y,p_z) F \left(i \hbar \frac{\partial}{\partial p_x}, i \hbar \frac{\partial}{\partial p_y},i \hbar \frac{\partial}{\partial p_z} \right) c(p_x,p_y,p_z)dp_xdp_ydp_z$$.

$$\overline{F(p_x,p_y,p_z)} = \int \psi^{*}(x,y,z) F \left(-i \hbar \frac{\partial}{\partial x},-i \hbar \frac{\partial}{\partial y},-i \hbar \frac{\partial}{\partial z} \right) \psi(x,y,z)dxdydz$$
Consider $$F(p_x, p_y,p_z)=p_x$$. Then,
$$\overline{p_x}= \int \psi^{*}(x,y,z) \left( -i \hbar \frac{\partial}{\partial x} \right) \psi(x,y,z) dxdydz$$
In other words
$$\hat{p_x} = -ih \frac{\partial}{\partial x}$$

3. Nov 22, 2009

pzona

Alright, thanks a lot, that makes a little more sense now. I'm probably not at the level of mathematical skill I would need to be at to derive that on my own, but I can follow most of your steps. It's so interesting how hbar seems to keep coming up, even in the Fourier transformation. Thanks a lot, that was very helpful

4. Nov 23, 2009

physicsworks

Hbar comes from de Brogile wave :)