Recent content by dark_matter_is_neat

I'm trying to calculate the scattering cross section between a charged particle and a particle with a magnetic moment. I believe the interaction term in this case would be ##\mu \cdot (\vec{v} \times \frac{e\vec{r}}{4 \pi r^{3}})## although I'm a bit confused how to properly treat it in this...

Yes I want the interaction term associated with the magnetic moment of the neutral particle interacting with the magnetic field produced by the moving charged particle. The neutral particle mass may be much smaller, much larger, or comparable to the moving charge mass, so I'm interested in...

So could I just treat the velocity as not an operator, if I want to calculate this without QFT (Just keep treating ##\vec{B} = \frac{\vec{v} \times \vec{r}}{4*\pi*r^{3}}## as having ##\vec{v}## as a parameter and not an operator)?

The wavefunctions refer to the two-particle systems (for the most simple case we can treat the two particles as starting and ending as free particles, so the initial and final wavefunctions are just products of plane-wave states). Neither of the particles are fixed. The neutral particle of...

I am interested in calculating the scattering cross section for the 2-to-2 scattering of a moving charge (Such as proton) with a 0-charge particle with a magnetic moment, So I want the interaction term associated with the moving proton's magnetic field treated using quantum mechanically, so I...

What I'm interested in is how to treat this magnetic field when expressed in term of operators. Classically ##\vec{B} = \frac{q\vec{v} \times \vec{r}}{4*\pi*r^3}## but in quantum mechanics ##\vec{v} = \vec{p}/m = -i*\vec{ \nabla }/m##, so what I'm wondering is the magnetic field treated as...

The interacting particle is a spin 1/2 fermion with an intrinsic magnetic moment but no charge. Its mostly just an extraneous detail for my question, since this is just going to lead to an interaction term of form ##-\vec{\mu} \cdot \vec{B}## what I'm mainly concerned with is the form of ##\vec{B}##

I'm wanting to calculate the interaction term of a magnetic moment with the magnetic field of a moving charged particle but I've confused myself about how to treat the magnetic field in quantum mechanics. In classical mechanics the magnetic field is just ##\frac{q*\vec{v} \times...

Since the problem is really just asking about the temperature dependence of the heat capacity I don't really need to consider the pre-factors, So rewriting ##k_{x}## and ##k_{y}## in terms of k as if in polar coordinates, ##E = k\sqrt{a^{2}cos^{2}(\phi) + b^{2}sin^{2}(\phi)}##. So E is...

For me the part of the problem that is giving me issues is obtaining the density of states, since typically how you would calculate D(E) as D(E) = ##\frac{A}{2 \pi} *k*\frac{dk}{dE}## but this shouldn't work since this assumes angular symmetry in k space which this dispersion relationship...

Well since the light source is laser, there is a continuous stream of photons. Since the laser has a power P and the energy of a photon is ##\frac{hc}{\lambda}##, the number of incident photons per second is ##\frac{P \lambda}{hc}##. The momentum per photon is ##\frac{h}{\lambda}##, so...

So, when the light is being reflected by the cube's interior the force is always just ##\frac{2Pcos(45)}{c}## Since ##\frac{2P}{c}## is the rate of photon momentum being deposited, but it is deposited at a angle of 45 degrees from the horizontal so the horizontal force applied is ##\frac{2P...

I'm not really sure what to do, clearly I need to get an equation of motion for the box and then solve for the final momentum. I can solve for the rate at which photons hit the box by dividing the power, P, by the energy of photon which would be ##\frac{P \lambda}{hc}## so multiplying this by...

So since it looks like I made a formatting error and for some reason I can't edit my post anymore, here is the section with messed up formatting fixed: ##\gamma = \frac{1}{\sqrt{1 - v_{CM}^{2}}}## and ##\alpha = v_{CM} \gamma## when using natural units. So So rearranging the expression for E'...

So from particle 1 scattering in all directions with equal probability in the CM frame, I believe that that means probability of finding particle 1 in an angular range ##d\theta## is just ##\frac{d \theta}{2 \pi}##. Let P(E') be the probability density of E', so from the probability of finding...