- #1
physichu
- 30
- 1
What is the value of ##\left\langle {{\bf{p}}|{\bf{p}}} \right\rangle ## when ##a_{\bf{k}}^ + k\left| 0 \right\rangle = \sqrt {2{E_{\bf{p}}}} \left| 0 \right\rangle ##? (like in Peskin)
I suppose that ##\left\langle {\bf{k}} \right|{a_{\bf{k}}}{\bf{P}}a_{\bf{k}}^ + \left| {\bf{k}} \right\rangle = {\bf{k}}##.
So
##\eqalign{
& \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + a_{\bf{p}}^{}a_{\bf{k}}^ + \left| 0 \right\rangle } = \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + \left\{ {\left[ {a_{\bf{p}}^{},a_{\bf{k}}^ + } \right] + \overbrace {a_{\bf{k}}^ + a_{\bf{p}}^{}}^{ = 0}} \right\}\left| 0 \right\rangle } = \cr
& = \int {{d^3}p{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + \left| 0 \right\rangle \delta \left( {{\bf{k}} - {\bf{p}}} \right)} = {\bf{k}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{k}}^ + \left| 0 \right\rangle = {\bf{k}} \cr} ##.
This implies ##\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{k}}^ + \left| 0 \right\rangle = 1##
and
##\left\langle {{\bf{p}}|{\bf{p}}} \right\rangle = {1 \over {2{E_{\bf{p}}}}}##
am I right?
I suppose that ##\left\langle {\bf{k}} \right|{a_{\bf{k}}}{\bf{P}}a_{\bf{k}}^ + \left| {\bf{k}} \right\rangle = {\bf{k}}##.
So
##\eqalign{
& \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + a_{\bf{p}}^{}a_{\bf{k}}^ + \left| 0 \right\rangle } = \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + \left\{ {\left[ {a_{\bf{p}}^{},a_{\bf{k}}^ + } \right] + \overbrace {a_{\bf{k}}^ + a_{\bf{p}}^{}}^{ = 0}} \right\}\left| 0 \right\rangle } = \cr
& = \int {{d^3}p{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + \left| 0 \right\rangle \delta \left( {{\bf{k}} - {\bf{p}}} \right)} = {\bf{k}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{k}}^ + \left| 0 \right\rangle = {\bf{k}} \cr} ##.
This implies ##\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{k}}^ + \left| 0 \right\rangle = 1##
and
##\left\langle {{\bf{p}}|{\bf{p}}} \right\rangle = {1 \over {2{E_{\bf{p}}}}}##
am I right?