To me it looks like all he did was to sum up/down electrons for a spin-free system.
Looks valid if you are interested in "pure" charge density and want to forget about spin.
Nothing wrong with it.
This is NOT homework.
I am trying to solve a generalized circuit ( each node is a 4-component generalization) and matrices are 4x4 tensors.
And I am trying to write them down -- making common nodes "zero" current, ending up with a matrix as in what's shown in the link...
Yes - I can do the 2x2 proof I guess.
Because any 2x2 Hermitian matrix can be written as:
H=\begin{bmatrix}
a & c -i \ d \\
c + i \ d & b
\end{bmatrix}
where a,b,c,d are all real numbers.
Then H can be uniquely defined in terms of Pauli matrices:
\frac{1}{2}\left[ (a+b) \ I_{2\times...
No - B is the best answer because
Force is defined as the net rate of change of momentum.
\sum_i \vec{F_i}=0 = \frac{d \vec{\ p}}{dt}
However, the momentum being constant does not mean that the speed has to be the same, nor the Energy to be the same.
Mathematically you can see it like...
I posted this question over at the QM page,
https://www.physicsforums.com/showthread.php?t=714076
but I realized I am really looking for a
hard Mathematical proof ...
A description of a numerical way of proving this would also be very helpful for me.
or a reference covering the...
Hi , Thank you for the responses ... However, I still don't understand it from a matrix point of view.
Let's take N = 2 , and have a 4x4 H matrix ... can one prove that my representation will always cover the full space ?
I didn't follow it from the Dirac notation,
Many thanks for responses.
Hi,
Wasn't sure if I should post this to Linear Algebra or here.
My question is really simple:
Can a 2N by 2N random, and Hermitian Matrix ( Hamiltonian ) be always written as:
H = A \otimes I_{2\times 2} + B \otimes \sigma_x + C \otimes \sigma_y + D \otimes \sigma_z
where A,B,C,D are all...
In 2D- as you correctly pointed out the error is proportional to kf^2 ...
But N is the electron number as you point out.
in 2D at low temperatures
kf = sqrt ( 2 pi N )
that is
kf ^ 2 is proportional to Electron Density ...
Then the error indeed goes as 1/N.
Pick a practical PDE, like one of those already suggested ( say, Poisson Equation in 2D)
and try to solve it numerically.
Yes, numerical attempt will be like designing an experiment and you will learn a lot along the way.
Once you solve it numerically, in order to make sure you did...
Seebeck coefficient is a function of energy; just like conductivity. Therefore when you want the total Seebeck coefficient, it has to be weighted by their energy dependent conductivities.
Fredrik, I agree with you %99 percent. No question. Whether or not we can measure S^2 , we can discuss separately. I can think of, admitedly hypothetical, spin-functional devices to do the deed but that's another point.
I think I have to agree that I was too dismissive, theories certainly...
Yeah I think I have made my point. I do research professionally, so I don't worry about accuracy when I post on PF in my free time.
To tell beginners that to understand spin, they need to dig their heads in a physics book is far from inaccurate, it's very misleading.
The term "spin" can be...