Suppose one is to find the stationary states of a particle in an infinite cubic well. Inside the box the time independent SE is:
- \frac{\hbar}{2m} \big( \frac{\partial ^2 \psi}{\partial x ^2 } + \frac{\partial ^2 \psi}{\partial z ^2 } + \frac{\partial ^2 \psi}{\partial z ^2 } \big)= E\psi...
I can't get the following limit to work:
lim(X->0) 1/(ix)*(exp(imx) - 1 ) = m
I'm sorry for the poor notation. I tried expanding the exponential, and L'hopitals rule and combinations of these approaches, but i can't get it to work out. Any help is much appreciated!
Consider a coaxial cable with inner radius a and outer radius b. The potential difference between the inner and outer conducter is V. How can it be shown that the electric field at a distance r from the inner conducter is given by:
E = \frac{V}{r ln\frac{b}{a}}
Any help would be much...
Homework Statement
Im am considering a covariant differential:
D_\mu H = ( partial_\mu + \frac{1}{2} i g \tau_j W_{j\mu} + ig B_\mu ) H
H is an isospiner, \tau_j are the pauli spin matrices, \partial_\mu is the four-gradient \frac{\partial}{\partial x^\mu} and W_{j \mu} and B_\mu are...
Let me just post the solution, u are very close anyway:
Let E be an eigenvector of H with eigenvalue e:
H E = e E
Then we have:
H^4 E = e H^3 E = e^2 H^2 E = e^3 H E = e^4 E.
You simply let one H act on E in each step.
Since you have H^4 = 1 we have:
H^4 E = E = e^4 E.
So all you need to...
I think you can do it simple like this:
Let E be an eigenvector of H with eigenvalue e:
H E = e E
Then H^4 E = e H^3 E = e^2 H^2 E = ...
And use H^4 E = E since H^4 = 1.
This gives you an general equation for the eigenvalues. Now when H is hermitian the eigenvalues must be real...
If the box extend all through the well, the problem is just the same as for V=0. If u just have a bumb in the middle of the well, you can use perturbation theory.
Oh i think i found my error. What Ben wrote must be transformation between covariants and contravariants? i thought it was Lorentz transformations. Well, if I'm right here is what i get. In my notation: x^{\mu} = contravariant x_{\mu} = covariant g_{\mu \nu} = metric.
Connection between...
"Also, the question only asks whether taking the contravariant gradient of a scalar function results in a contravariant 4-vector. It doesn't ask you anything about transforming from one inertia system to another."
Hmm.. I think I'm getting confused. The contravariant gradient is:
\frac{\partial...
Thanks for your replies. Now, here is what I've tried:
I consider the Lorentz transformation:
\bar{x}^{\mu} = \Lambda ^{\mu} _{\nu} x^{\nu}.
Differentiating i obtain:
\Lambda ^{\mu} _{\nu} = \frac{\partial \bar{x} ^{\mu}}{\partial x^{\nu}}.
Acting upon the transformation with the...