- #1
hideelo
- 91
- 15
I am trying to follow modern QFT by Tom Banks and I am having an issue with literally the first equation.
He claims that beginning from ## |p_1 , p_2, ... , p_k> \: = \: a^\dagger (p_1) a^\dagger (p_2) \cdots a^\dagger (p_k)|0> ## with the commutation relation ##[a (p),a^\dagger (q)]_\pm \: = \: \delta^3(p-q)##, (+for fermions, - for bosons) I can reproduce the following equation
$$<p_1 , p_2, ... , p_k|q_1 , q_2, ... , q_l> \: = \: \frac{\delta_{kl}}{k!}\sum_\sigma (-1)^{S\sigma}\delta^3(p_1-q_{\sigma(1)}) \cdots \delta^3(p_k-q_{\sigma(k)})$$Where S = 1 for fermions and 0 for bosons. I can show that for the 1 particle state I get ##<p|q> \: = \: \delta^3(p-q)##, but as soon as I get to the two particle state my result differs from his.His formula tells me that for the two particle state I should get:
$$<sr|pq> \: = \: \frac{1}{2} \left( \delta^3(p-s)\delta^3(r-q) \mp \delta^3(r-p) \delta^3(q-s) \right)$$
But I am getting
$$<sr|pq> \: = \: \delta^3(p-s)\delta^3(r-q) \mp \delta^3(r-p) \delta^3(q-s) $$
missing the factor of ##\frac{1}{2}##
$$<sr|pq> \: = \:<r|a(s)a^\dagger(p)|q>\: = \:<r|[a(s),a^\dagger(p)]_\pm |q> \mp <r|a^\dagger(p)a(s)|q> \:= $$
$$ \delta^3(p-s)<r|q> \mp <r|a^\dagger(p)a(s)|q> \: = \: \delta^3(p-s)\delta^3(r-q) \mp <r|a^\dagger(p)a(s)a^\dagger(q)|0> \: = $$
$$ \delta^3(p-s)\delta^3(r-q) \mp <r|a^\dagger(p)[a(s),a^\dagger(q)]_\pm |0> \pm <r|a^\dagger(p)a^\dagger(q)a(s) |0> \: = $$
$$ \delta^3(p-s)\delta^3(r-q) \mp <r|a^\dagger(p)|0> \delta^3(q-s) \: = \: \delta^3(p-s)\delta^3(r-q) \mp <r|p> \delta^3(q-s) \: = $$
$$ \delta^3(p-s)\delta^3(r-q) \mp \delta^3(r-p) \delta^3(q-s)$$
Whats going wrong?
He claims that beginning from ## |p_1 , p_2, ... , p_k> \: = \: a^\dagger (p_1) a^\dagger (p_2) \cdots a^\dagger (p_k)|0> ## with the commutation relation ##[a (p),a^\dagger (q)]_\pm \: = \: \delta^3(p-q)##, (+for fermions, - for bosons) I can reproduce the following equation
$$<p_1 , p_2, ... , p_k|q_1 , q_2, ... , q_l> \: = \: \frac{\delta_{kl}}{k!}\sum_\sigma (-1)^{S\sigma}\delta^3(p_1-q_{\sigma(1)}) \cdots \delta^3(p_k-q_{\sigma(k)})$$Where S = 1 for fermions and 0 for bosons. I can show that for the 1 particle state I get ##<p|q> \: = \: \delta^3(p-q)##, but as soon as I get to the two particle state my result differs from his.His formula tells me that for the two particle state I should get:
$$<sr|pq> \: = \: \frac{1}{2} \left( \delta^3(p-s)\delta^3(r-q) \mp \delta^3(r-p) \delta^3(q-s) \right)$$
But I am getting
$$<sr|pq> \: = \: \delta^3(p-s)\delta^3(r-q) \mp \delta^3(r-p) \delta^3(q-s) $$
missing the factor of ##\frac{1}{2}##
$$<sr|pq> \: = \:<r|a(s)a^\dagger(p)|q>\: = \:<r|[a(s),a^\dagger(p)]_\pm |q> \mp <r|a^\dagger(p)a(s)|q> \:= $$
$$ \delta^3(p-s)<r|q> \mp <r|a^\dagger(p)a(s)|q> \: = \: \delta^3(p-s)\delta^3(r-q) \mp <r|a^\dagger(p)a(s)a^\dagger(q)|0> \: = $$
$$ \delta^3(p-s)\delta^3(r-q) \mp <r|a^\dagger(p)[a(s),a^\dagger(q)]_\pm |0> \pm <r|a^\dagger(p)a^\dagger(q)a(s) |0> \: = $$
$$ \delta^3(p-s)\delta^3(r-q) \mp <r|a^\dagger(p)|0> \delta^3(q-s) \: = \: \delta^3(p-s)\delta^3(r-q) \mp <r|p> \delta^3(q-s) \: = $$
$$ \delta^3(p-s)\delta^3(r-q) \mp \delta^3(r-p) \delta^3(q-s)$$
Whats going wrong?