A I'm getting the wrong inner product of Fock space

[hideelo]

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?

 

[Jilang]

Are you happy where the k! comes from? ( As that seems to be the difference)

 

[hideelo]

Are you happy where the k! comes from? ( As that seems to be the difference)

No, it's the k! that's bothering me

 

[Jilang]

Isnt it there to avoid double counting?

 

[hideelo]

Isnt it there to avoid double counting?

That's what I thought, but when I worked it out I found no double counting

 

[Jilang]

Maybe I’m not understanding your notation, but which operator in your first line annihilates the q state?

 

[hideelo]

The convention I am using (and I think its standard) is that $a^\dagger(q)$ creates a state with momentum q and annihilates a costate with momentum q i.e.

$$a^\dagger(q)|0> = |q>$$
and
$$<q|a^\dagger(q) = <0|$$

Similarly $a(q)$ annihilates a state with momentum q and creates a costate with momentum q i.e.

$$a(q)|q> = |0>$$
and
$$<0|a(q) = <q|$$

 

[Jilang]

Great thanks for confirming that. So wouldn’t you need four operators for 2 particles? Two to annihilate the two original states and two to create the new ones?

 

[hideelo]

Great thanks for confirming that. So wouldn’t you need four operators for 2 particles? Two to annihilate the two original states and two to create the new ones?

In general yes, but what I'm doing is in only "pulling out" one creation operator. So while it's true that $|pq> =a^\dagger(p) a^\dagger(q)|0>$ I can then let the $a^\dagger(q)|0> =|q>$ which gives me what I have up there $|pq> a^\dagger(p) |q>$

I think that was OK to do...

 

[Jilang]

Quantum Field Theory for the Gifted Amateur by Lancaster and Blundell doesn’t have the 1/2, which is interesting! In that text the factors only appear when calculating the wave-function. So something is wrong with one of them...

 

[Demystifier]

Whats going wrong?

Banks is probably wrong.

First, because S. Weinberg, The Quanum Theory of Fields I, Eq. (4.1.6) confirms your result and contradicts the Banks's result.

Second, because the Banks's book is generally bad: https://www.physicsforums.com/threads/bad-books.693040/

Besides, I have shown that one very influential paper by Banks et al was deeply wrong: https://arxiv.org/abs/1502.04324

 

[vanhees71]

For fermions there shouldn't be any factorials since each single-particle state can only be occupied with at most one particle. For bosons you have factorials for normalizing the states which occur only when a single-particle state is occupied by more than one particle. In the infinite-volume limit this leads to trouble. So you should keep all the momenta (labeled with $p$ and $q$ in your formula) general. The proper expression comes out right due to the $\delta$ distributions. So in the standard meaning of the creation and annihilation operators there should be indeed no $k!$ in the formula in the OP. So you are right with your calculation.

For a very nice and careful treatment of the operator formalism, see

Fetter, Walecka, Quantum theory of many-particle systems

For this calculational technique it doesn't play a role whether you consider relativistic or non-relativistic QFT (Fetter&Walecka is non-relativistic).