Recent content by peperone

This surely looks slick! :) But why is \partial^\nu\partial^\mu a symmetric second rank tensor?

To be honest, I don't understand what you're doing here (step 2 -> step 3)

Hm, that's true. Do you know a better way? I'm highly interested in alternative solutions.

Sorry, it was just a typo. I edited my post. Thanks for all the help. I think I got it now... Start with the Dirac equation: (i\gamma_\mu\partial^\mu - m)\Psi(x) = 0 Multiply the operator -(i\gamma_\mu\partial^\mu + m) (from left): (\gamma_\mu\partial^\mu\gamma_\mu\partial^\mu +...

Can anybody verify this? \gamma^\nu\partial_\nu\gamma_\mu\partial^\mu = \partial_\mu\partial^\mu Just as clarification: I am trying to show that every solution of the Dirac equation also solves the Klein-Gordon equation. So I started with the Dirac equation: (i\gamma_\mu\partial^\mu -...

I kinda felt it couldn't be that simple. :p Thanks anyway.

Hey there. In an exercise I was trying to show that every solution of the Dirac equation also solves the Klein-Gordon equation. Now I have two very simple questions: Is \gamma_\nu \gamma^\mu = \gamma^\nu \gamma_\mu ? And is \partial_\nu \partial^\mu = \partial^\nu \partial_\mu ...

Heh, that nearly was a simultaneous post. :wink: The problem formulation says: H = 0 for t<0 and H = the Hamiltonian posted in the beginning for t>0. This made me think of H as a perturbation which is "switched on" at t=0. Your solution looks nice, too. Edit: Infact, it leads to...

Hm... The following would make sense (mathematically). Let's say: |c_1(t)|^2 \Leftrightarrow p(|++\rangle) |c_2(t)|^2 \Leftrightarrow p(|+-\rangle) |c_3(t)|^2 \Leftrightarrow p(|-+\rangle) |c_4(t)|^2 \Leftrightarrow p(|--\rangle) Then, the initial conditions would be: |c_1(0)|^2 =...

Hi Concerning your questions: - What do you mean by labeling the matrix? For me, |++> = (1,0,0,0), |+-> = (0,1,0,0) and so on. But I guess that was not your question. - E is the identity matrix OK, my quantum physics course somehow follows Sakurai's books "Modern Quantum Mechanics" and...

My tutor emailed me: The \Delta is just a variable for the intensity of the spin coupling so the above assumption of leaving out the \Delta doesn't matter at all. Feel free to post your thoughts about the above calculations. :smile:

Hello. I had the same feeling when I looked at the Hamiltonian in this problem. For clarification I've emailed my tutor but I'm still waiting for a reply. For the moment, let's just think of the same Hamiltonian without the \Delta operator: H = (\frac{4}{\hbar^2})\;\vec{S_1}\cdot\vec{S_2}...

Nobody? :) PS: Wow, just discovered the tex code function. Here's the Hamiltonian in a more readable format: H=(\frac{4\Delta}{\hbar^2})\vec{S_1}\cdot\vec{S_2}

Hello. Consider a two object spin system. For t>0 the Hamiltonian is given as: H= 4*LAPLACE/h_^2 * S_1*S_2 The initial state is given as |+->. I need to calculate the probabilities of finding the system in one of the four possible states (|++>,|+->,|-->,|-+>) a) by solving the problem...