Well, just looking at your expression for probability, we know you made a mistake somewhere because it depends on a, which is a unit-full quantity. Since probability is unitless, it shouldn't depend on the value of a. And in your quantity in the parentheses, the units don't match up, since one...
It's probably more complicated to first try to find the velocity of particle A in B's frame and then to use that velocity to calculate the energy. There's a more direct way that makes the problem a bit simpler.
Recall, E = M c^2 \gamma
So, to measure the energy of A in B's frame, we only need...
Well, you have the wavefunction for the ground state from the first part, correct? The following is a very important idea in QM:
The allowed wavefunctions are by definition eigenfunctions of the Hamiltonian operator H. There is a theorem that says that eigenfunctions of a hermitian operator...
I think it you have the rules for phi^4 theory, the rules for phi^3 are nearly identical, and the only difference is the type of diagrams that you are allowed to draw. The initial and final state wave functions will be exactly the same and the propagators will be exactly the same. If you're...
The above poster went into some of the math.
In modern atomic theory, the electron isn't in anyone place at a time. Rather, we can't ever say for sure where the electron actually IS. The best we can do is determine where it is very likely to be, or where it spends most of its time. Thus...
Yeah, he covers it in chapters 9,10 and lists Feynman rules. If you're just starting out with QFT, I think Srednicki may be a bit awkward in his notation since he includes a lot of factors early on that anticipate renormalization (there are a lot of Z's floating around that may confuse you at...
I'm not an expert, but since no one else is answering, I'll throw in my two cents:
For this theory, it's clear what Feynman diagrams will look like. You can draw lines and you can draw three lines meeting at a point. So, we can pretty easily see what diagrams contribute at a particular level...
You are correct, V = 0 in the region that we're interested in. The kinetic energy operator is given by
T = - \frac{ \hbar^2}{2m}\frac{d^2}{dx^2}
So, if we know what the total energy operator is (it's just the kinetic energy operator since there is no potential) we can find it's expectation...
To be specific, here's what I had in mind when thinking about the Lorentz transformation:
Consider frame S' which is attached to the back of the train. Frame S is attached to the ground, and frame S' moves with velocity v1 with respect to frame S. The two origins for these frames coincide at...
iii: In QM, energy is described by an energy operator known as the Hamiltonian. I would start by writing out the Hamiltonian operator. The next step is to recall the formula for the expectation value of an operator in terms of a given wavefunction.
ii: A wavefunction is normalized when $\int \psi (x) ^* \psi (x) dx $ = 1, so to calculate N, you need to integrate the given wavefunction from negative infinity to infinity (which is effectively the integral from -a to a), equate that to 1, and solve for N.
Really, I think the best thing to do is to use the full Lorentz Transformation. The second part of the problem is tricky because the relative positions of the frames change as a function of time. Here's a hint:
Assume that all frames start at the same position at time t = 0 when the runner...
In the second part, be sure that the time dilation takes place between the frame of the runner and the frame of a person standing on the ground. The relative velocity between those two frames isn't v1 or v2, so you'll have to use velocity addition.
The overall point is that the fact that space is curved means that it is impossible and meaningless to compare velocities of things that aren't adjacent (or close enough to adjacent such that the space locally is to a good approximation Minkowski).