# Search results

1. ### Canonical momentum for Dirac adjoint field

should be 'acting to the left' for ψbar
2. ### Canonical momentum for Dirac adjoint field

I know that π = iψbarγ0. My confusion concerns the notion of the derivative in the Lagrangian 'acting to the right' for ψ or 'acting to the left' for ψbarγ0 . If I ignore that confusion then I can see that the canonical momentum would be 0.
3. ### Canonical momentum for Dirac adjoint field

I read that the canonical momentum for Dirac adjoint field is zero. Why is that?
4. ### Matrix element of momentum

Thanks to all. I believe Matterwave has put me on the right track. I was trying to evaluate the integral containing the partial derivative. If I evaluate the integral using the OP p and then substitute the partial derivative at the end everything falls into place.
5. ### Matrix element of momentum

OK. But the differential gives (ipn/h)eipnx/h ... what happens to the p?
6. ### Matrix element of momentum

Can somebody explain how you get: <pm|p|pn> = (1/2π)∫e-ipmx/h (-ih∂/∂x) eipnx/h dx ..... = ih∂δ(pm - pn)/∂x Conceptually, I am having a problem with how the inner product is formed when a derivative is involved (i.e. the ..... steps)
7. ### Slater Determinant for simple covalent bond

I have read that when 2 Hydrogen atoms come together their individual spacial wavefunctions overlap in the following way: ψsymmetric = ψa + ψb ... bonding case ψasymmetric = ψa - ψb ... antibonding case How do you express this in terms of the Slater Determinant?
8. ### Physical Meaning of Bra-kets

Thank you very much for your response. I had done a little research prior to my post and concluded that this was indeed a complicated issue. I think I will stick with the beginner view for my purposes, given that I am re-learning a lot of this stuff 25 years after leaving college! Thanks again.
9. ### Physical Meaning of Bra-kets

Re: Reply #3 <x|y> and <x|p> are interpreted as inner products correct? Thus, I assume these can expanded in terms of integrals involving delta functions. <x|p> is easy to figure out but I don't understand how to get δ(x-y) for <x|y>.