Recent content by einai

I think I figured it out. I used the commutation relation p = - i m hbar*[H,x]. Thanks.

Thanks, Haelfix and lethe :redface: . I thought about using the operator form of p, but I wasn't sure how it acts on the energy eigenstate |E>. Can I just say that after it takes the x derivative of |E>, the state becomes orthorgonal to the original |E>, ie, <E|-i\hbar \frac{d}{dx}|E>...

Griffiths for undergrad and Sakurai for graduate.

Hi, I came across a problem which seems to be pretty simple, but I'm stuck :confused: . Given a Hamiltonian: H=\frac{\vec{p}^2}{2m}+V(\vec{x}) If |E> is a bound state of the Hamiltonian with energy eigenvalue E, show that: <E| \vec{p} |E>=0 ----------------------------------- So...

Hi, I came across a problem which seems to be pretty simple, but I'm stuck :confused: . Given a Hamiltonian: H=\frac{\vec{p}^2}{2m}+V(\vec{x}) If |E> is a bound state of the Hamiltonian with energy eigenvalue E, show that: <E| \vec{p} |E>=0 ----------------------------------- So...

Hi, In our class, we were asked whether it's practical to use a magnifying glass to focus sunlight to burn wood. One question really bothers me. Could somebody please explain to me how the focal ratio (focal length of converging lens divided by its diameter) affects the temperature of the...

Hi, In our class, we were asked whether it's practical to use a magnifying glass to focus sunlight to burn wood. One question really bothers me. Could somebody please explain to me how the focal ratio (focal length of converging lens divided by its diameter) affects the temperature of the...

Looks much better :)! I didn't use substitution. I just treated &int;&delta; (x-x0) &delta; (x-x1) dx as &int;f(x) &delta;(x-x1) dx so when I integrate it, it gives f(x) -> f(x1) = &delta; (x0-x1). I'm not sure whether this is a correct method, although it does give me the answer :D.

You need to put ; after &delta :D. And thanks for answering my question. I did get the same thing, but I wasn't sure whether that implies orthonormality. Now I know, since I got the solution from the prof. It does imply orthonormality, and I got it right! :)

Yeah, that should be the right formula. A positron basically is just an electron with a positive charge, so they have the same mass.

I think it means if they're the same function, the product should be integrated to one, otherwise it's zero? Hm...I multiplied 2 wavefunction and integrate them. It gave me another delta function.

I think you can find v from the energy of the positron by using the kinetic energy formula, since E = 22.5 eV << the positron rest mass (0.5 MeV). The mass is equivalent to the electron mass.

Thank you! That makes a lot of sense. :smile: However, I'm not sure whether I understand this part of the question - Now, consider the totality of these wave functions for different values of x0. Do they form an orthonormal set? Does it mean whether all the delta functions at different...

What's the wave function in coordinate space &Psi;x0(x') of a particle (in 1-D) located at a certain position x0? What about the wave function &Phi;x0(p') in momentum space? Now, consider the totality of these wave functions for different values of x0. Do they form an orthonormal set? The...

Quantum question again... What's the wave function in coordinate space &Psi;x0(x') of a particle (in 1-D) located at a certain position x0? What about the wave function &Phi;x0(p') in momentum space? Now, consider the totality of these wave functions for different values of x0. Do they...