Recent content by tommy01

Hi everybody, I encountered this question as an example of what a google applicant is asked: You are shrunk to the height of a 2p coin and thrown into a blender. Your mass is reduced so that your density is the same as usual. The blades start moving in 60 seconds. What do you do? It is...

Thanks a lot ...

Hi togehter. I encountered the following problem: The timeordering for fermionic fields (here Dirac field) is defined to be (Peskin; Maggiore, ...): T \Psi(x)\bar{\Psi}(y)= \Psi(x)\bar{\Psi}(y) \ldots x^0>y^0 = -\bar{\Psi}(y)\Psi(x) \ldots y^0>x^0 where \Psi(x) is a Dirac...

Thanks again. But that's what i meant. It is unlikely but not impossible. And a very poor lower bound doesn't justify to call \Delta p the typical momentum. It seems as if one could conclude the mean value (which i assume is the typical value) from the standard deviation. But there's no...

Thanks, but i think it's meaningless to give a lower bound and speak of typical momenta. The uncertainty can be a very small fraction of the actual momentum or vice versa, so i can't believe they meant it this way.

They say: "The uncertainty principle gives an estimate for the typical electron momenta when they are confined to such a linear domain." Then they give the above formula. But all they get from the uncertainty principle is the uncertainty \Delta p which is not the typical momentum. In my...

Sorry for bumping, but i can't believe that nobody who did some lectures on particle physics ever encountered this (or a similar misuse of the uncertainty principle) problem. I would be very glad for an answer. Greetings.

They conclude that the momentum of the electron is \approx \hbar/\Delta x , so it surely isn't zero. They want to calculate the typical momentum of an electron in an atom, it would not be very helpful to chooce such a reference frame, because the answer would be that the typical momentum is...

Hi together ... In many textbooks on particle physics i encounter - at least in my mind - a misuse of the Heisenberg uncertainty principle. For completeness we talk about \Delta p \Delta x \geq \hbar/2 For example they state that the size of an atom is of the order of a few Angstroms...

Hi. The whole thing is rather sloppy and mathematically not very well justified by Dirac. If you have an operator A in the Hilbert space the eigenvalues and eigenkets of the operator satisfy A |a\rangle = a|a\rangle where a is an eigenevalue if you have a continuum of eigenvalues in an (real)...

O.k. so everything in the exponent multiplied by t except the -i is the frequency \omega. As E=\hbar \omega the negative energy comes from the negative frequency. Mhh i think i probably got it although there are many cases i remember where the time dependence was \exp(i \omega t) and no one...

my point is: you say that there are negative energy solutions because, as you showed, the general solution bears a plus AND a minus sign in the exponent. but it seems to me it would be the same to say the energy can be imaginary because there is an i in the exponent. i simply don't get the...

thank you. i also heard this statement, but it's the same story i think. only this time the sign is carried over to the paramter t instead of the energy. i'll try to make my point clear: although the solutions are \Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x})) you don't...

yes, but as i mentioned earlier, why does this sign carry over to the energy? why don't say the energy is positive but the functional dependence, to get linearly independent solutions, is exp(- ...) respectively exp(+ ...)? thanks for your help and please excuse my problems understanding that...

Thanks. Thats a linear superposition of the fundamental solutions i cited above, but as you wrote \omega and thefore the energy is positive and there is no need for negative energy states called antiparticles. Or did i misunderstand something? edit: Back when they dealt with the Klein-Gordon...