Understanding the Math behind meson decay computations

In summary: Ah, I see where my mistake was. I was considering the wrong term, I was looking at the term with the creation operator in front of the vacuum state instead of the annihilation operator.So, let's try again. We have:$$<f|S|i>=-ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} \Big(a_{\vec k}^{\dagger}e^{ikx}+a_{\vec k}e^{-ikx} \
  • #1
JD_PM
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Homework Statement:: I am studying how to compute the meson-decay amplitude worked out in Tong's notes (pages 55 and 56; I attached the PDF).
Relevant Equations:: $$<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

Captura de pantalla (1021).png

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:
Captura de pantalla (1020).png


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)
 

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  • #2
JD_PM said:
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.
 
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  • #3
JD_PM said:
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?
 
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  • #4
nrqed said:
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

$$ig<f| \int d^4 x \psi^{\dagger} (x) \psi (x) \int \frac{d^3 k \sqrt{2E_{\vec p}}}{(2 \pi)^3 \sqrt{2E_{\vec k}}} a_{\vec k}^{\dagger}a_{\vec p}^{\dagger}e^{ikx}|0>=0$$
 
  • #5
JD_PM said:
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?
 
  • #6
nrqed said:
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? :)
 
  • #7
JD_PM said:
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?
 
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  • #8
nrqed said:
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 o_O)

Captura de pantalla (1020).png
 
  • #9
JD_PM said:
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 o_O)

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!
 
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  • #10
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)

$$\langle f|S|i\rangle=-ig \langle 0|\int d^4 x e^{i(q_1+q_2-p)x}|0\rangle = -ig (2\pi)^4 \delta^4 (q_1+q_2-p)$$
 
  • #11
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.
 

Related to Understanding the Math behind meson decay computations

1. What is the purpose of understanding the math behind meson decay computations?

The purpose of understanding the math behind meson decay computations is to gain a deeper understanding of the fundamental principles of particle physics and the behavior of subatomic particles. It also allows scientists to make more accurate predictions and interpretations of experimental data.

2. What are the key equations and concepts involved in meson decay computations?

The key equations and concepts involved in meson decay computations include the Standard Model of particle physics, which describes the fundamental particles and their interactions, as well as the principles of quantum mechanics and special relativity. Specific equations used in meson decay computations include the Feynman diagrams and the equations for calculating decay rates and branching fractions.

3. How do scientists use meson decay computations in their research?

Scientists use meson decay computations in their research to study the properties and behavior of mesons, which are unstable particles that decay into other particles. By understanding the math behind these computations, scientists can make predictions about the decay patterns and lifetimes of different mesons, which can then be compared to experimental data to test the validity of the theories.

4. What are some challenges in understanding the math behind meson decay computations?

One of the main challenges in understanding the math behind meson decay computations is the complexity of the equations and concepts involved. These computations require a strong background in mathematics and physics, and can be difficult to grasp without proper training and education. Additionally, the unpredictable nature of subatomic particles and their interactions can make it challenging to accurately predict decay patterns and lifetimes.

5. How does understanding the math behind meson decay computations contribute to advancements in science and technology?

Understanding the math behind meson decay computations contributes to advancements in science and technology by providing a deeper understanding of the fundamental principles of particle physics. This knowledge can then be applied to other areas of research, such as nuclear energy, medical imaging, and particle accelerators. Additionally, the development of new mathematical and computational techniques for meson decay computations can have broader applications in other fields of science and technology.

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