Recent content by student111

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...

doh! I differentiated the whole part instead of nominator and denominator separately.. Thx a lot

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'hospital's 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...

Great. Thx a lot.

Can a quadratic form always be diagonalised by a rotation?? Thx in advance

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...

exactly, i mistyped. Nonhermitian: 1, -1, i, -i. And only 1,-1 if it is hermitian.

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.

Ok. Thanks a lot for your help and inspiration.

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...