Recent content by physichu

Puting a minus in front of the momentum in the field expansion gives ##\phi \left( {\bf{x}} \right) = \int {{d^3}\tilde p} \left( {{a_{\bf{p}}}{e^{i{\bf{p}} \cdot {\bf{x}}}} + a_{\bf{p}}^ + {e^{ - i{\bf{p}} \cdot {\bf{x}}}}} \right){\rm{ }}\phi \left( {\bf{x}} \right) = \int {{d^3}\tilde...

My Hero :) Just so I understand, I was supposed to get to that by myself?

Whether the tension in a pendulum is less than mg at some point in its swing.

I think you miss an ##i## in the ##{p_0}## term.

Not necessarily, it's depands on ##{v_o}##.

I think you get an extra i from the ##{p^0}## integral ##\eqalign{ & {p^0} = ip_E^0 \cr & \downarrow \cr & dp_{}^0 = idp_E^0 \cr} ##.

After that corection I will say that it's depand on ##{v_0}##.

OOPSS. The first equation should be ##{1 \over 2}mv_0^2 = {1 \over 2}m{v^2} + mgr\left( {1 - \cos \theta } \right)##

look at the scetch Let's say that the length of the swinging part of the rope is ##r## and we give the right mass a ##{v_0}## speed. Then when it's in the ##\theta## angle position. it will have a speed ##v## such that ##mgr\left( {1 - \cos \theta } \right) + {1 \over 2}mv_0^2 = {1 \over...

I have a problem with that equation. I understand (dont know if I'm right) that ##p = - M##. But than, isn't ##g\left( { - {{{p^2}} \over {{M^2}}}} \right)## just equal ##g\left( { - 1} \right)##? And my bigest problem: in 12.66 ##\left[ {p{\partial \over {\partial p}} - \beta \left( \lambda...

where do the other exponential and the ##ipr## denominator come from?

Hello :) I don't get this integral (Peskin & Schroeder P. 27 ) ##\int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{1 \over {{E_{\bf{p}}}}}{e^{i{\bf{p}} \cdot {\bf{r}}}}} = {{2\pi } \over {{{\left( {2\pi } \right)}^3}}}\int\limits_0^\infty {dp{{{p^2}} \over {2{E_{\bf{p}}}}}{{{e^{ipr}} -...

But... Isn't ##\mathop {\lim }\limits_{a \to \infty } \exp \{ - {{\left| {{{\bf{k}}^2} - {{\bf{p}}^2}} \right|} \over a}\} /b = \delta \left( {\bf{p}} \right)## so ##\eqalign{ & \mathop {\lim }\limits_{a \to \infty } \left| \phi \right\rangle = {1 \over {\sqrt {{{\left( {2\pi }...

O. K. I THINK I GOT IT! Following Peskin equation (4.65) ##\left| \phi \right\rangle = \int {{{{d^3}k} \over {{{\left( {2\pi } \right)}^3}}}{1 \over {\sqrt {2{E_{\bf{k}}}} }}\phi \left( {\bf{k}} \right)\left| {\bf{k}} \right\rangle } = \int {{{{d^3}k} \over {{{\left( {2\pi }...

Let me see if i got it right: from answer to my question i understand that ##a_{\bf{k}}^ + \left| 0 \right\rangle ## is not a physical state, so it is O.K. that ##\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{k}}^+ \left| 0 \right\rangle ## have no meaning. Is it right to say that...