Schwartz's Quantum field theory (12.9)

[Plantation]

Homework Statement

$$[a_{\vec{p_1} s_1 n}, a^{\dagger}_{\vec{p_2}s_2 n}]= (2 \pi)^3 \delta^{3}(\vec{p_1}-\vec{p_2}) \delta_{s_1, s_2} . $$

Relevant Equations

$$ \langle \vec{p_1} | \vec{p_2} \rangle = 2 \omega_1 ( 2\pi)^3 \delta^{3}(\vec{p_1}-\vec{p_2}) $$

I am reading the Schwartz's quantum field theory, p.207 and stuck at some calculation.

In the page, he states that for identical particles,

$$ | \cdots s_1 \vec{p_1}n \cdots s_2 \vec{p_2} n \rangle = \alpha | \cdots s_2 \vec{p_2}n \cdots s_1 \vec{p_1}n \cdots \rangle, \tag{12.5}$$

where $\alpha = e^{i\phi}$ for some real $\phi$.

From this, he argues that for boson case we obtain

$$a^{\dagger}_{\vec{p_1} s_1 n} a^{\dagger}_{\vec{p_2}s_2 n} | \psi \rangle = a^{\dagger}_{\vec{p_2}s_2 n} a^{\dagger}_{\vec{p_1} s_1 n} | \psi \rangle \tag{12.7}$$ for all $|\psi\rangle$ (I don't know why this is true from (12.5)) so that

$$ [a^{\dagger}_{\vec{p_1} s_1 n}, a^{\dagger}_{\vec{p_2}s_2 n}] = [a_{\vec{p_1} s_1 n}, a_{\vec{p_2}s_2 n}] =0 \tag{12.8}$$

And he saids that since $\langle \vec{p_1} | \vec{p_2} \rangle = 2 \omega_1 ( 2\pi)^3 \delta^{3}(\vec{p_1}-\vec{p_2})$, we can use same argument to show that

$$[a_{\vec{p_1} s_1 n}, a^{\dagger}_{\vec{p_2}s_2 n}]= (2 \pi)^3 \delta^{3}(\vec{p_1}-\vec{p_2}) \delta_{s_1, s_2} . \tag{12.9} $$

Q. And why this is true? How can we use the formula for $\langle \vec{p_1} | \vec{p_2} \rangle$ ? What should I catch?

 

[vanhees71]

He has a somewhat confusing normalization convention (in fact almost any author uses another more or less confusing convention; at the end it doesn't matter). He defines the momentum-spin eigenstates as normalized as
$$\langle \vec{p},s |\vec{p}',s' \rangle=(2 \pi)^3 \delta^{(3)} 2 \omega_{\vec{p}} (\vec{p}-\vec{p}') \delta_{s s'}$$
and the corresponding creation operator by
$$\hat{a}^{\dagger}(\vec{p},s) |\Omega \rangle=\frac{1}{\sqrt{2 \omega_{\vec{p}}}} |\vec{p},\vec{s} \rangle.$$
Adjoining this gives
$$\langle \Omega|\hat{a}(\vec{p},s) = \frac{1}{\sqrt{2 \omega_{\vec{p}}}} \langle \vec{p},s|.$$
From this you get for bosons
$$\langle \Omega |[\hat{a}(\vec{p},s), \hat{a}^{\dagger}(\vec{p}',s')]|\Omega = \langle \Omega |\hat{a}(\vec{p},s) \hat{a}^{\dagger}(\vec{p}',s') \Omega= \frac{1}{\sqrt{2 \omega_{\vec{p}}}} \frac{1}{\sqrt{2 \omega_{\vec{p}'}}} 2 \omega_{\vec{p}} (2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}') \delta_{ss'}=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}') \delta_{ss'}.$$

 

[Plantation]

He has a somewhat confusing normalization convention (in fact almost any author uses another more or less confusing convention; at the end it doesn't matter). He defines the momentum-spin eigenstates as normalized as
$$\langle \vec{p},s |\vec{p}',s' \rangle=(2 \pi)^3 2 \omega_{\vec{p}} \delta^{(3)} (\vec{p}-\vec{p}') \delta_{s s'}$$
and the corresponding creation operator by
$$\hat{a}^{\dagger}(\vec{p},s) |\Omega \rangle=\frac{1}{\sqrt{2 \omega_{\vec{p}}}} |\vec{p},\vec{s} \rangle.$$
Adjoining this gives
$$\langle \Omega|\hat{a}(\vec{p},s) = \frac{1}{\sqrt{2 \omega_{\vec{p}}}} \langle \vec{p},s|.$$
From this you get for bosons
$$\langle \Omega |[\hat{a}(\vec{p},s), \hat{a}^{\dagger}(\vec{p}',s')]|\Omega \rangle = \langle \Omega |\hat{a}(\vec{p},s) \hat{a}^{\dagger}(\vec{p}',s') | \Omega \rangle= \frac{1}{\sqrt{2 \omega_{\vec{p}}}} \frac{1}{\sqrt{2 \omega_{\vec{p}'}}} 2 \omega_{\vec{p}} (2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}') \delta_{ss'}=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}') \delta_{ss'}.$$

Thanks.

Question 1 : Why the final equality is true? ; i.e., why $\frac{1}{\sqrt{2 \omega_{\vec{p}}}} \frac{1}{\sqrt{2 \omega_{\vec{p}'}}} 2 \omega_{\vec{p}} =1$ ? Perhaps,

$$\langle \vec{p},s |\vec{p}',s' \rangle=(2 \pi)^3 2 \sqrt{\omega_{\vec{p}}} \sqrt{\omega_{\vec{p'}}} \delta^{(3)} (\vec{p}-\vec{p}') \delta_{s s'}$$

is more correct normalization condition ? ; i.e., the author (Schwartz) made mistake ?

Question 2 : Is it really enough to show only $$\langle \Omega |[\hat{a}(\vec{p},s), \hat{a}^{\dagger}(\vec{p}',s')]|\Omega \rangle = (2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}') \delta_{ss'} $$ to show the (12.9) : $$[a_{\vec{p_1} s_1 n}, a^{\dagger}_{\vec{p_2}s_2 n}]= (2 \pi)^3 \delta^{3}(\vec{p_1}-\vec{p_2}) \delta_{s_1, s_2} . $$ in my question, without having to show the statement for all $|\psi\rangle$ instead of $\Omega$ (ground state?) ?

 

[vanhees71]

Ad 1: It's together with the $\delta$ distribution, because $\delta^{(3)}(\vec{p}-\vec{p}')=0$ for $\vec{p} \neq \vec{p}'$. The final result indeed is
$$\langle \vec{p},s|\vec{p}',s' \rangle=(2 \pi)^3 2 \omega_{\vec{p}} \delta^{(3)}(\vec{p}-\vec{p}') \delta_{ss'}.$$
The advantage of this normalization convention is that this scalar product between momentum-spin eigenstates gives a Lorentz scalar.

Ad 2: It's enough, because from the canonical equal-time commutation relations, which you postulate, e.g., in the canonical-quantization argument of field quantization, it's clear that $[\hat{a}_{\vec{p}_1 s_2 n},\hat{a}_{\vec{p}_2 s_2 n}] \propto \hat{1}$.

 

[haushofer]

Ah, the confusion one gets when implicitly implying Dieac delta's and then forgetting about them :P