I thought there was a separate wave equation for the E and B fields: the wave equations for them are decoupled. How is it possible to discern causality from decoupled equations?
I think that's correct. Energy is conserved. Momentum changes direction, both in the direction of the slit and perpendicular to the slit, but the magnitude of the momentum is the same. I'm not completely sure though.
Can renormalization of QED really be interpreted as a dielectric shielding of the vacuum by electron/positron pairs that appear and disappear out of the vacuum?
I understand that's what the Feynman diagram for the QED vertex suggests, since it's the internal fermion lines that interact with a...
Should one view correlation functions as:
(1) \; \langle T{\phi(x)\phi(y)}\rangle
or
(2) \; \langle T{\phi(x)\phi(y)}\rangle - \langle \phi(x)\rangle \langle\phi(y)}\rangle
with the second term being zero?
(2) makes more sense as it really measures whether the two fields are...
The electron self-energy occurs in classical electrodynamics where if you calculate the potential of an electron at the point of the electron, you get infinity.
In classical electrodynamics this is ignored, but can be calculated in QED.
In QED the Green's function G(x,y) which represents...
Re: definition of "indistinct" in pauli exclusion principle
The u and d quarks can be transformed into each other under flavor isospin SU(2).
Are there separate creation operators for u and d quarks? Or is there just one creation operator, with isospin a label just like spin or momentum...
O I see what you're saying. So for the charged KG equation:
\Box \phi + m^2 \phi=0
and it's conjugate equation, you can add a term to get a new equation:
\Box \phi + m^2 \phi+\partial^\mu(\phi^*\partial_\mu\phi-\phi \partial_\mu \phi^*)=0
whose solution is the KG-field, since the term in...
oh, I think you and I made a slight error. It's not:
P(A|B) = P(A)P(B)
but
P(A \cap B)=P(A)P(B)
If you wanted to use conditional probability, then the equation would be:
P(A|B) = P(A)
Anyways, I don't think this changes the concept, just the notation:
P( (x=a) \cap...
oh yes, that is almost by definition. If you have P(x,y), and P(x|y=a)=P(x,a)=f(x)g(a), then it must be true that P(x,y)=f(x)g(y), or that the state can be written:
|\psi>=|\sqrt{f}>\otimes|\sqrt{g}>
I was under the impression that discussion was over the free, non-charged KG equation.
The free, charged KG equation has a U(1) symmetry but not a gauge symmetry.
The solution to the EOMs of free KG fields are unique, so I don't believe there is any freedom to impose gauge conditions.
Doesn't it depend on which gauge you choose? What if I chose the axial gauge: A3=0 ?
Would:
\Box A^\mu-\partial^\mu \partial_\nu A^\nu=0
reduce to the Klein-Gordan equation?
Re: definition of "indistinct" in pauli exclusion principle
What makes bound electrons special?
Say I know that a spin up electron is at x=2, and a spin down electron is at x=7. Then would I have to antisymmetrize the wavefunction to be:
|2,\uparrow \rangle \otimes |7,\downarrow \rangle -...
If that were true, then any function in 2 variables can be decomposed as a product of two functions in one variable.
But such a function would have the property that it has no zeroes, or infinitely many zeroes, since if h(x0,y0)= f(x0)g(y0)=0, then either f(x0)=0 or g(y0)=0, which implies...
I'm not that familiar with the spin 3/2 equation.
But does the electromagnetic potential obey the Klein-Gordan equation? I don't think it does unless you choose the Lorenz gauge or the Coloumb gauge.
addendum:
it goes something like this:
\Box A^\mu-\partial^\mu \partial_\nu A^\nu=0
The...
definition of "indistinct" in pauli exclusion principle
I'm a little confused about what constitutes a distinct particle.
For example, a muon is not an electron as they've got different masses. So the wavefunction for the electron/muon system does not have to be antisymmetric (although it can...
Spin 3/2 obeys the Rarita-Schwinger equation. The spin 3/2 field has a mixture of vector and spinor indices.
The spin 2 field should obey the Einstein equation.
Yeah so this seems right. Take one particle in 2-dimensions, and say you want the two dimensions to be uncorrelated, say:
(|x_1>+\pi|x_2>)\otimes (|y_1>+\lambda|y_2>)
The amplitude to be in x2 is pi times more likely than to be in x1, and y2 lambda times more likely than to be in y1...
I've never read that particular book, but I'm a big fan on his lectures in physics, volume I.
I vaguely recall Peskin and Schroeder saying that the propagator is the amplitude also, although I've never read their book.
It kind of makes sense on a conversational level. Propagators are joined at...
Yes, I got confused because you used the notation delta E, so I thought it meant change in energy or difference in energy levels. It could be the case that your reference for the energy is 0, but for half integral spins that reference is not achievable.
Anyways, to get the neutron magnetic...
Not every state can be represented by a direct product. Do states that can be written as a direct product have anything special about them?
It seems that states that can be written as a direct product lose correlation between between each individual state. More specifically, stronger than...
I'm not sure what you're asking.
Shouldn't
\Delta E = -\mu \cdot B
be
\Delta E = 2 \mu \cdot B
?
The g-factor is related to energy in that the energy is proportional to the g-factor. If the g-factor is zero, and assuming there is no orbital angular momentum...
Since there are computers that can do simple Fourier transforms, I thought I'd experiment with some simple ones ( http://www.wolframalpha.com/input/?i=fourierx[%281%29%2F%28x^2%2B1%29] and http://www.wolframalpha.com/input/?i=fourier[%28-x^2%29%2F%28x^2%2B1%29] ):
\frac{1}{\sqrt{2\pi}} \int...
Right, I messed up on that one. But what if I had this one instead:
g(k)=.5(1+\frac{k_{0}^2}{\vec{k}^2})
This does not introduce a pole in the complex k_{0}-plane.
The idea is that two transform functions, one with g(k)=1 and one with
g(k)=.5(1+\frac{k_{0}^2}{\vec{k}^2})
, give...
If I choose g(k)=1, they are the same. But what if I choose g(k)=1+(k_{0}^2-\vec{k}^2) . Will the two different g(k)'s give the same result, since eventually (k_{0}^2=\vec{k}^2) when integrating over ko?
I chose g(k)=.5(1+\frac{\vec{k}^2}{k_{0}^2}) , which when you set (k_{0}^2=\vec{k}^2) ...
Well, say g(k)=\frac{k_{0}^2}{\vec{k}^2} , or g(k)=1+(k_{0}^2-{\vec{k}^2) . These shouldn't affect the pole when integrating just over k_0. So it seems that you should be able to set:
g(k)=\frac{k_{0}^2}{\vec{k}^2}=1
and
g(k)=1+(k_{0}^2-{\vec{k}^2)=1
from the very beginning.
But if this...
The basic idea I'm trying to get at is this:
D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}g(k)}{k^2-m^2+i\epsilon }
The integral over k0 is done by taking the residues at k_0=\pm \sqrt{\vec{k}^2+m^2} to get:
D(x,y)=-i\theta(x_o-y_o)\int \frac{d^3k}{2...
Because of the k02 term in the denominator, that is a pole of order 2 when integrating over k0.
So what you do is multiply by the integrand by k02, and take one derivative, and set k0=0 to get your residue. In this case this will result in zero.
Another way to see it is to note that the...
thanks. you used -+++, but the gist of what you did is correct. changing to +---:
0.5 + 0.5 \frac{\vec{k}^{2}}{k^{2}_{0}} = 0.5 \left( 1 + \frac{-k^{2} + k^{2}_{0}}{k^{2}_{0}}\right) = 1-0.5 \frac{k^{2}}{k^{2}_{0}}
D(x, y) =\int{\frac{d^{4}k}{(2 \pi)^{4}} \, \frac{e^{i k (x -...