Understanding the Math behind meson decay computations

[JD_PM]

Homework Statement:: I am studying how to compute the meson-decay amplitude worked out in Tong's notes (pages 55 and 56; I attached the PDF).
Relevant Equations:: $$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \phi (x) |i>$$

The initial and final states are given by

$$|i > = \sqrt{2 E_{\vec p}} a_{\vec p}^{\dagger} |0>$$

$$|f > = \sqrt{4 E_{\vec q_1}E_{\vec q_2}} b_{\vec q_1}^{\dagger}c_{\vec q_2}^{\dagger} |0>$$

Using Dyson formula's expansion we get that

$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \phi (x) |i>$$

We first expand out $\phi \sim a + a^{\dagger}$ using the following formula

Doing so we get

$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} a_{\vec k} a_{\vec p}^{\dagger} e^{-ikx} |0> = -ig<f|\int d^4 x \psi^{\dagger} (x) \psi (x) e^{-ipx} |0>$$

I do not understand why the $a_{\vec k}e^{ipx}$ term vanishes.

Tong justifies it as follows: when $a^{\dagger}$ hits |i> we get a two meson state, which has 'zero overlap' with $<f|$ and 'and there’s nothing in the $\psi$ and $\psi^{\dagger}$ operators that lie between them to change this fact'.

What does he mean with 'zero overlap' with $<f|$?

He uses the same argument to justify the final solution:

With this thread I am aimed at understanding and getting the provided solution.

Thank you.

PS: The amplitude can be computed by formula (3.26), The unitary matrix is given by formula (3.23), where the $H_{int}$ is given by (3.25)

 

[nrqed]

Homework Statement:: I am studying how to compute the meson-decay amplitude worked out in Tong's notes (pages 55 and 56; I attached the PDF).
Relevant Equations:: $$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \phi (x) |i>$$

The initial and final states are given by

$$|i > = \sqrt{2 E_{\vec p}} a_{\vec p}^{\dagger} |0>$$

$$|f > = \sqrt{4 E_{\vec q_1}E_{\vec q_2}} b_{\vec q_1}^{\dagger}c_{\vec q_2}^{\dagger} |0>$$

Using Dyson formula's expansion we get that

$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \phi (x) |i>$$

We first expand out $\phi \sim a + a^{\dagger}$ using the following formula

View attachment 260586
Doing so we get

$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} a_{\vec k} a_{\vec p}^{\dagger} e^{-ikx} |0> = -ig<f|\int d^4 x \psi^{\dagger} (x) \psi (x) e^{-ipx} |0>$$

I do not understand why the $a_{\vec k}e^{ipx}$ term vanishes.

Tong justifies it as follows: when $a^{\dagger}$ hits |i> we get a two meson state, which has 'zero overlap' with $<f|$ and 'and there’s nothing in the $\psi$ and $\psi^{\dagger}$ operators that lie between them to change this fact'.

What does he mean with 'zero overlap' with $<f|$?

He uses the same argument to justify the final solution:View attachment 260583

With this thread I am aimed at understanding and getting the provided solution.

Thank you.

PS: The amplitude can be computed by formula (3.26), The unitary matrix is given by formula (3.23), where the $H_{int}$ is given by (3.25)

When people say there is "no overlap" between two states, they mean that $\langle a | b \rangle = 0$. So here he is saying that when the state will be applied to $\langle f|$, the result will be zero.

 

[nrqed]

Homework Statement:: I am studying how to compute the meson-decay amplitude worked out in Tong's notes (pages 55 and 56; I attached the PDF).
Relevant Equations:: $$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \phi (x) |i>$$

The initial and final states are given by

$$|i > = \sqrt{2 E_{\vec p}} a_{\vec p}^{\dagger} |0>$$

$$|f > = \sqrt{4 E_{\vec q_1}E_{\vec q_2}} b_{\vec q_1}^{\dagger}c_{\vec q_2}^{\dagger} |0>$$

Using Dyson formula's expansion we get that

$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \phi (x) |i>$$

We first expand out $\phi \sim a + a^{\dagger}$ using the following formula

View attachment 260586
Doing so we get

$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} a_{\vec k} a_{\vec p}^{\dagger} e^{-ikx} |0> = -ig<f|\int d^4 x \psi^{\dagger} (x) \psi (x) e^{-ipx} |0>$$

I do not understand why the $a_{\vec k}e^{ipx}$ term vanishes.

To help a bit more, first let me add this. First, note that you meant that $a_{\vec k}^{\dagger} \, e^{ipx}$ vanishes. Write this term and look at all the annihilation and creation operators sandwiched between the vacuum states. What do you observe?

 

[JD_PM]

To help a bit more, first let me add this. First, note that you meant that $a_{\vec k}^{\dagger} \, e^{ipx}$ vanishes. Write this term and look at all the annihilation and creation operators sandwiched between the vacuum states. What do you observe?

Yes you are right, I meant to ask 'why does $a_{\vec k}^{\dagger} \, e^{ipx}$ vanish?'

Let's write it explicitly and see why it does. We have:

$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} \Big(a_{\vec k}e^{-ikx}+a_{\vec k}^{\dagger}e^{ikx} \Big) a_{\vec p}^{\dagger}|0>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} \Big(a_{\vec k}a_{\vec p}^{\dagger}e^{-ikx}|0>+a_{\vec k}^{\dagger}a_{\vec p}^{\dagger}e^{ikx}|0> \Big)$$

Mmm but I still do not see why the following term is zero

$$ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} a_{\vec k}^{\dagger}a_{\vec p}^{\dagger}e^{ikx}|0>=0$$

 

[nrqed]

Yes you are right, I meant to ask 'why does $a_{\vec k}^{\dagger} \, e^{ipx}$ vanish?'

Let's write it explicitly and see why it does. We have:

$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} \Big(a_{\vec k}e^{-ikx}+a_{\vec k}^{\dagger}e^{ikx} \Big) a_{\vec p}^{\dagger}|0>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} \Big(a_{\vec k}a_{\vec p}^{\dagger}e^{-ikx}|0>+a_{\vec k}^{\dagger}a_{\vec p}^{\dagger}e^{ikx}|0> \Big)$$

Mmm but I still do not see why the following term is zero

$$ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} a_{\vec k}^{\dagger}a_{\vec p}^{\dagger}e^{ikx}|0>=0$$

Two $\phi$ particles are created out of the vacuum. Do theyget annihilated before you apply the vacuum bra?

 

[JD_PM]

Two $\phi$ particles are created out of the vacuum. Do they get annihilated before you apply the vacuum bra?

I'd say yes but I do not know why... could you please give me a hint? :)

 

[nrqed]

I'd say yes but I do not know why... could you please give me a hint? :)

Sure. If you have two creation operators acting on the vacuum, you need two annihilation operators for the same particle type in order to have a non zero result. So you need two operators $a_p, a_p'$ somewhere in your expression to compensate for the two creation operators. Do you have these?

 

[JD_PM]

Sure. If you have two creation operators acting on the vacuum, you need two annihilation operators for the same particle type in order to have a non zero result. So you need two operators $a_p, a_p'$ somewhere in your expression to compensate for the two creation operators. Do you have these?

Oh so the reason is that, as there are no $a_p, a_p'$ compensating $a_p^{\dagger}, a_p'^{\dagger}$, then that term has to be zero. Alright thanks!

How could we prove that statement though? Through commutation relations?

I am going to try again and see if I get the final answer (it will be tricky though )

 

[nrqed]

Oh so the reason is that, as there are no $a_p, a_p'$ compensating $a_p^{\dagger}, a_p'^{\dagger}$, then that term has to be zero. Alright thanks!

How could we prove that statement though? Through commutation relations?

I am going to try again and see if I get the final answer (it will be tricky though )

It follows from the fact that states with different number of particles (of each type) are orthogonal. For example, if $| \alpha \rangle$ is a state containing two particles of type $\phi$, in other words if it is proportional to

$$ |\alpha \rangle \propto a_k^\dagger a^\dagger_{k'} |0 \rangle $$

then we automatically have

$$ \langle 0 | \alpha \rangle = 0. $$

Good luck with the rest of your calculation. If you get stuck on something, don't hesitate to ask!

 

[JD_PM]

Please let me show all my work and then add some comments.

So we were dealing with the leading order in g

$$\langle f|S|i\rangle=-ig \langle f| \int d^4 x \psi^{\dagger} (x) \psi (x) \phi (x) |i \rangle \ \ \ \ (1)$$

I'll go step by step

1) We plug the mode expansion for $\phi$ and the given definition for $|i\rangle$ into (1) to get

$\langle f|S|i \rangle =-ig \langle f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} \Big(a_{\vec k}a_{\vec p}^{\dagger}e^{-ikx}|0 \rangle +a_{\vec k}^{\dagger}a_{\vec p}^{\dagger}e^{ikx}|0 \rangle \Big)$

Thanks to your explanation at #9 I see that we have $\langle 0 |a_{\vec k}^{\dagger}a_{\vec p}^{\dagger}| 0 \rangle=0$. So we end up with

$$\langle f|S|i\rangle=-ig \langle f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} a_{\vec k}a_{\vec p}^{\dagger}e^{-ikx}|0\rangle \ \ \ \ (2)$$

2) Use the commutation relation $[a_{\vec k}, a_{\vec p}^{\dagger}] = \delta^3 (\vec p - \vec k)$

Based on the commutation relation we know that $a_{\vec k}a_{\vec p}^{\dagger}=\delta^3 (\vec p - \vec k)+a_{\vec p}^{\dagger}a_{\vec k}$. Plugging it into (2) we get

$$\langle f|S|i\rangle=-ig \langle f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} \Big( \delta^3 (\vec p - \vec k)+a_{\vec p}^{\dagger}a_{\vec k} \Big)e^{-ikx}|0\rangle $$

We know that, by definition, we have $a_{\vec k}|0\rangle = 0$. Thus (2) of course becomes

$$\langle f|S|i\rangle=-ig \langle f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} \delta^3 (\vec p - \vec k) e^{-ikx}|0\rangle$$

Based on the famous sifting property of the Dirac delta function (i.e. $\int f(t) \delta (t-T) dt = f(T)$), we compute the following integral

$$\int \frac{d^3 k}{\sqrt{2E_{\vec k}}} \delta^3 (\vec p - \vec k) e^{-ikx} = \frac{(2\pi)^3}{\sqrt{2E_{\vec p}}} e^{-ipx}$$

Thus we end up with

$$\langle f|S|i\rangle=-ig \langle f| \int d^4 x \psi^{\dagger} (x) \psi (x) e^{-ipx}|0\rangle \ \ \ \ (3)$$

3) Recall the mode expansion of the complex field

$$\psi = \int \frac{d^3 k_1}{(2 \pi)^3} \frac{1}{\sqrt{2E_{\vec k_1}}}\Big( b_{\vec k_1}e^{i k_1 x}+c_{\vec k_1}^{\dagger}e^{-i k_1 x}\Big)$$

$$\psi^{\dagger} = \int \frac{d^3 k_2}{(2 \pi)^3} \frac{1}{\sqrt{2E_{\vec k_2}}}\Big( b_{\vec k_2}^{\dagger}e^{-i k_2 x}+c_{\vec k_2}e^{i k_2 x}\Big)$$

4) Plug the mode expansion for $\psi$, $\psi^{\dagger}$ and $\langle f|= \langle 0| \sqrt{4 E_{\vec q_1} E_{\vec q_2}}c_{\vec q_2} b_{\vec q_1}$ into (3)

$\langle f|S|i\rangle=-ig \langle 0|\iiint d^4 x \frac{d^3 k_2 d^3 k_1}{(2 \pi)^6} \frac{\sqrt{4 E_{\vec q_1} E_{\vec q_2}}}{\sqrt{4 E_{\vec k_2} E_{\vec k_1}}}c_{\vec q_2} b_{\vec q_1}\Big( b_{\vec k_2}^{\dagger}e^{-i k_2 x}+c_{\vec k_2}e^{i k_2 x}\Big) \Big( b_{\vec k_1}e^{i k_1 x}+c_{\vec k_1}^{\dagger}e^{-i k_1 x}\Big) e^{-ipx}|0\rangle \ \ \ \ (4)$

The only non-zero term is $\langle 0| c_{\vec q_2} b_{\vec q_1} b_{\vec k_2}^{\dagger} c_{\vec k_1}^{\dagger} | 0 \rangle$. Thus (4) becomes

$$\langle f|S|i\rangle=-ig \langle 0|\iiint d^4 x \frac{d^3 k_2 d^3 k_1}{(2 \pi)^6} \frac{\sqrt{4 E_{\vec q_1} E_{\vec q_2}}}{\sqrt{4 E_{\vec k_2} E_{\vec k_1}}}c_{\vec q_2} b_{\vec q_1} b_{\vec k_2}^{\dagger}c_{\vec k_1}^{\dagger}|0\rangle e^{-i(k_2+k_1+p)x} \ \ \ \ (5)$$

5) Use the commutation relation $[b_{\vec q_1}, b_{\vec k_2}^{\dagger}] = \delta^3 (\vec k_2 - \vec q_1)$

Based on the commutation relation we know that $b_{\vec q_1} b_{\vec k_2}^{\dagger}=\delta^3 (\vec k_2 - \vec q_1)+b_{\vec k_2}^{\dagger}b_{\vec q_1}$. Plugging it into (5) we get

$$\langle f|S|i\rangle=-ig \langle 0|\iiint d^4 x \frac{d^3 k_2 d^3 k_1}{(2 \pi)^6} \frac{\sqrt{4 E_{\vec q_1} E_{\vec q_2}}}{\sqrt{4 E_{\vec k_2} E_{\vec k_1}}} \delta^3 (\vec k_2 - \vec q_1) c_{\vec q_2} c_{\vec k_1}^{\dagger}|0\rangle e^{-i(k_2+k_1+p)x}$$

Where I've used the fact that $b_{\vec q_1} |0\rangle =0$

Based on the famous sifting property of the Dirac delta function (i.e. $\int f(t) \delta (t-T) dt = f(T)$), we compute the following integral

$$\int \frac{d^3 k_2}{\sqrt{2E_{\vec k_2}}} \delta^3 (\vec k_2 - \vec q_1) e^{-i k_2 x} = \frac{(2\pi)^3}{\sqrt{2E_{\vec q_1}}} e^{i q_1 x}$$

6) Use the commutation relation $[c_{\vec q_2}, c_{\vec k_1}^{\dagger}] = \delta^3 (\vec k_1 - \vec q_2)$

Based on the commutation relation we know that $c_{\vec q_2} c_{\vec k_1}^{\dagger}=\delta^3 (\vec k_1 - \vec q_2)+c_{\vec k_1}^{\dagger}c_{\vec q_2}$.

At this point we proceed the same way as 5): we use the fact that $c_{\vec q_2}|0\rangle=0$ and we solve the integral

$$\int \frac{d^3 k_1}{\sqrt{2E_{\vec k_1}}} \delta^3 (\vec k_1 - \vec q_2) e^{-i k_1 x} = \frac{(2\pi)^3}{\sqrt{2E_{\vec q_2}}} e^{i q_2 x}$$

7) Use 5) and 6) to solve equation (5)

$$\langle f|S|i\rangle=-ig \langle 0|\int d^4 x e^{i(q_1+q_2-p)x}|0\rangle = -ig (2\pi)^4 \delta^4 (q_1+q_2-p)$$

 

[JD_PM]

I think I got it right but, to be honest, it feels like I cheated; I already knew the answer and based on it I got the the following integral:

$$\int \frac{d^3 k_2}{\sqrt{2E_{\vec k_2}}} \delta^3 (\vec k_2 - \vec q_1) e^{-i k_2 x} = \frac{(2\pi)^3}{\sqrt{2E_{\vec q_1}}} e^{i q_1 x}$$

Particularly, I do not see why the sign of the exponent changes after solving the integral.