Quantum Question: Why k^2<0 in QFT?

[latentcorpse]

In the following online notes,

http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf

at the very top of page 60, I don't follow the two line argument as to why $k^2<0$ - can anybody explain it to me?

Thanks.

 

[betel]

He is using the metric with signature +---. So as $$k= (0,k_1,k_2,k_3)$$ obviously $$k^2=k_\mu k^\mu=-\sum_i k_i^2 <0$$.

 

[latentcorpse]

He is using the metric with signature +---. So as $$k= (0,k_1,k_2,k_3)$$ obviously $$k^2=k_\mu k^\mu=-\sum_i k_i^2 <0$$.

ok. but why is $k_i=p_i=p'_i$ in the first place?

 

[betel]

This follows from the delta functions in 3.51 which ensure 4-momentum conservation. From one delta function you get for the zero component of k
$$k_0=\sqrt{m^2+\vec p_1^2}-\sqrt{m^2+\vec p'_1^2}$$. From the other delta function you have $$k_0=\sqrt{m^2+\vec p_2'^2}-\sqrt{m^2+\vec p_2^2}$$.
Now in the center of mass frame $$\vec p_1 = \vec p_2$$ and the same for the primed quantities.
From the fact that you need both delta functions to be satiesfied for the product to be nonzero you get.
$$k_0=\sqrt{m^2+\vec p_1^2}-\sqrt{m^2+\vec p'_1^2}=k_0=\sqrt{m^2+\vec p'_2^2}-\sqrt{m^2+\vec p_2^2}$$
Inserting the center of mass relation this gives
$$k_0=\sqrt{m^2+\vec p_1^2}-\sqrt{m^2+\vec p'_1^2}=k_0=\sqrt{m^2+\vec p'_1^2}-\sqrt{m^2+\vec p_1^2}$$
And thus
$$\sqrt{m^2+\vec p_1^2}=\sqrt{m^2+\vec p'_1^2}$$
Therefore you know the absolute value of the momentum has to be conserved, so the meson only carries spatial momentum, i.e. its zero component is zero.

 

[latentcorpse]

This follows from the delta functions in 3.51 which ensure 4-momentum conservation. From one delta function you get for the zero component of k
$$k_0=\sqrt{m^2+\vec p_1^2}-\sqrt{m^2+\vec p'_1^2}$$. From the other delta function you have $$k_0=\sqrt{m^2+\vec p_2'^2}-\sqrt{m^2+\vec p_2^2}$$.
Now in the center of mass frame $$\vec p_1 = \vec p_2$$ and the same for the primed quantities.
From the fact that you need both delta functions to be satiesfied for the product to be nonzero you get.
$$k_0=\sqrt{m^2+\vec p_1^2}-\sqrt{m^2+\vec p'_1^2}=k_0=\sqrt{m^2+\vec p'_2^2}-\sqrt{m^2+\vec p_2^2}$$
Inserting the center of mass relation this gives
$$k_0=\sqrt{m^2+\vec p_1^2}-\sqrt{m^2+\vec p'_1^2}=k_0=\sqrt{m^2+\vec p'_1^2}-\sqrt{m^2+\vec p_1^2}$$
And thus
$$\sqrt{m^2+\vec p_1^2}=\sqrt{m^2+\vec p'_1^2}$$
Therefore you know the absolute value of the momentum has to be conserved, so the meson only carries spatial momentum, i.e. its zero component is zero.

hi there.

clearly I am either being an idiot or missing something here. How do the delta functions imply
$$k_0=\sqrt{m^2+\vec p_1^2}-\sqrt{m^2+\vec p'_1^2}$$ and $$k_0=\sqrt{m^2+\vec p_2'^2}-\sqrt{m^2+\vec p_2^2}$$?

 

[betel]

You are not an idiot, just not used to four momenta.
The four momentum is given by $$p^\mu=(E,\vec p) = (\sqrt{m^2+\vec p^2},\vec p)$$.
The delta functions imply $$k^\mu-p'_1^\mu+p_1^\mu=0, k^\mu-p_2^\mu-p'_2^\mu=0$$ if you take the first components of this you get the equations for the energy.

 

[betel]

Oh, and I have set c=1, if that is confusing you.

 

[latentcorpse]

Oh, and I have set c=1, if that is confusing you.

ok. I'll work through that later on - I think I get what you're saying though.

What about in equation 4.76,

why does the parity transformation take psi to $\gamma^0 \psi ( - \vec{x} , t )$.

In particular, where did that $\gamma^0$ come from?

 

[betel]

If you take a look at the definition of $$\psi$$ and write both $$\psi$$ and $$\gamma^0$$ as matrices you will see why it has to appear if you want to get the transformation behaviour of parity.

 

[latentcorpse]

If you take a look at the definition of $$\psi$$ and write both $$\psi$$ and $$\gamma^0$$ as matrices you will see why it has to appear if you want to get the transformation behaviour of parity.

Yeah thanks! That way we can switch the $\psi_+$ and $\psi_-$

Can you follow the calculation on p108? There's a lot of questionable minus signs in my opinion!

 

[betel]

starting where?

 

[latentcorpse]

starting where?

well, looking at eqn 5.8, i find something wrong here

$\pi \dot{\psi} - L = i \psi^\dagger \dot{\psi} - \bar{\psi} ( i \gamma^\mu \partial_\mu - m ) \psi = i \bar{\psi} \gamma^0 \partial_0 \psi - i \bar{\psi} \gamma^\mu \partial_\mu \psi + m \bar{\psi} \psi = i \bar{\psi} \gamma^0 \partial_0 \psi - i \bar{\psi} \gamma^0 \partial_0 \psi + i \bar{\psi} \gamma^i \partial_i \psi + m \bar{\psi} \psi = \bar{\psi} ( i \gamma^i \partial_i \psi + m \bar{\psi} \psi$

but he has a minus sign inside those brackets?

 

[betel]

You splitted the d slash wrong.
$$\gamma^\mu\partial_\mu=\gamma^0\partial_0+\gamma^i\partial_i$$
The relative minus sign will only appear if you explicitly write the matrices. SO his result is correct.

 

[latentcorpse]

You splitted the d slash wrong.
$$\gamma^\mu\partial_\mu=\gamma^0\partial_0+\gamma^i\partial_i$$
The relative minus sign will only appear if you explicitly write the matrices. SO his result is correct.

sorry. could you elaborate a bit more please. I'm still a little confused.

is the scalar product of two four vectors not defined to have a minus sign?

 

[betel]

The notation with the mus just means summing, nothing else. The minus only appears if you write everythings in terms of only lower indices components or only upper.
$$p^\mu p_\mu = p^0 p_0 + p^ip_i = p_0 p_0 - p_i p_i = p^0 p^0 - p^i p^i$$

 

[latentcorpse]

The notation with the mus just means summing, nothing else. The minus only appears if you write everythings in terms of only lower indices components or only upper.
$$p^\mu p_\mu = p^0 p_0 + p^ip_i = p_0 p_0 - p_i p_i = p^0 p^0 - p^i p^i$$

ok. on the next page he talks about the "small subtlety with the minus signs in deriving this equation"

is this the same idea again?

 

[betel]

Yes.

 

[latentcorpse]

Yes.

so when i write

$\vec{a} \cdot \vec{b}$ this is not equal to $a_i b^i$?

It is in fact equal to $\Sigma_i a^i b^i = \Sigma_i a_i b_i = - a_i b^i = - a^i b_i$ since we have to raise/lower accordingly with a metric that has Lorentzian signature?

If that's correct, my next concern is in the paragraph at the top of page 109 - he says the $-c^\dagger c$ term is disasterous since our Hamiltonian is not bounded delow.
This doesn't make any sense to me or the explanation about producing $c^dagger$ particles! We also have a $b^\dagger b$ term - this has the annihilation operator on the right aswell. What's going on here? Is it the minus sign that's causing the problems? And if so, why?

Thanks

 

[betel]

The scalar product is correct

The minus sign is important. This means, that each of this particles contributes with a negative energy. So every time you create a c particle you get a certain amount of energy. For the b particles you have to pay energy to get your particle.

So with both c and b particles you can produce as many particles as you want, i.e. you vacuum will rapidly decay to a firestorm of particles - you vacuum is unstable.

 

[latentcorpse]

The scalar product is correct

The minus sign is important. This means, that each of this particles contributes with a negative energy. So every time you create a c particle you get a certain amount of energy. For the b particles you have to pay energy to get your particle.

So with both c and b particles you can produce as many particles as you want, i.e. you vacuum will rapidly decay to a firestorm of particles - you vacuum is unstable.

I don't really follow sorry.

So if I make a c particle, I produce a negative amount of energy (i.e. energy is given off) and if I produce a b particle, I take in a certain amount of energy (i.e. i produce a positive amount of energy). So we are saying that this is all fine and well until we consider what happens when we take our system and produce loads and loads of c particles - this produces an infinitely large amount of negative energy, correct?

What's the problem with that? Is it unphysical? If so, why?

 

[latentcorpse]

The scalar product is correct

The minus sign is important. This means, that each of this particles contributes with a negative energy. So every time you create a c particle you get a certain amount of energy. For the b particles you have to pay energy to get your particle.

So with both c and b particles you can produce as many particles as you want, i.e. you vacuum will rapidly decay to a firestorm of particles - you vacuum is unstable.

Hi. Sorry to load you with questions but you're helping me loads here lol!

So I am ok for the next wee while up until eqn 5.34. I don't understand where this minus sign comes from in the case of $y^0 >x^0$?

And then the Wick's Theorem this leads to in 5.36 at teh bottom of the page is exactly the same as the one earlier for bosons in eqn 3.43 even though we now have a difference in that "bosonic operators commute inside T, fermionic operators anti-commute". Surely this statement describes a fundamental difference so why doesn't Wick's Theorem change?

 

[betel]

The total energy is conserved. So usually it costs energy to produce a particle, this is why there cannot suddenly appear particles in vacuum and stay there. This means if you have a vacuum without particles, it is stable forever, there vacuum will always be the same.

Now the c particles are different. Producing a c particle does not cost energy but gives energy. So if you produce a c particle you get enough energy to produce e.g. a b particle.

Now your empty vacuum will spontaneously produce pairs of b and c particles and rapidly decay, thus it is unstable.

 

[latentcorpse]

The total energy is conserved. So usually it costs energy to produce a particle, this is why there cannot suddenly appear particles in vacuum and stay there. This means if you have a vacuum without particles, it is stable forever, there vacuum will always be the same.

Now the c particles are different. Producing a c particle does not cost energy but gives energy. So if you produce a c particle you get enough energy to produce e.g. a b particle.

Now your empty vacuum will spontaneously produce pairs of b and c particles and rapidly decay, thus it is unstable.

Thanks for that!

Do you have any ideas about this one:


So I am ok for the next wee while up until eqn 5.34. I don't understand where this minus sign comes from in the case of $y^0 >x^0$?

And then the Wick's Theorem this leads to in 5.36 at teh bottom of the page is exactly the same as the one earlier for bosons in eqn 3.43 even though we now have a difference in that "bosonic operators commute inside T, fermionic operators anti-commute". Surely this statement describes a fundamental difference so why doesn't Wick's Theorem change?

 

[betel]

Well, the minus sign here appears because you have to commute two fermionic operators and those have different commutation rules.

And as the same behaviour holds for T and normal ordering is seems sensible that Wicks theorem goes through unchanged, although you should check explicitly.

 

[latentcorpse]

Well, the minus sign here appears because you have to commute two fermionic operators and those have different commutation rules.

And as the same behaviour holds for T and normal ordering is seems sensible that Wicks theorem goes through unchanged, although you should check explicitly.

to check expilicitly do you mean do this:

$C \bar{\psi}(y) \psi(x) = T ( \bar{\psi}(y) \psi(x)) - : \bar{\psi}(y) \psi(x) : = - T ( \psi(x) \bar{\psi}(y)) + : \psi(x) \bar{\psi}(y) : = - \left( T ( \psi(x) \bar{\psi}(y)) - : \psi(x) \bar{\psi}(y) : \right) = - C \psi(x) \bar{\psi}(y)$

where I've used C to denote contraction since I didn't know how to do those bracket-y things above the psi terms!



And then on page 121, the circular loop diagram. It says we pick up an overall minus and justifies it with all that contraction stuff at the bottom. I don't get how that shows anything - let alone where the minus comes from?

Cheers again!

 

[betel]

Well it's more like that you can write Time ordering as Normal ordering plus all possible contractions. And for this the proof is exactly the same as in the bosonic case.

In the example you move the $$\psi(y)$$ from the right to the left, on the sway swapping it three times with other fermionic operators. Each time you pick up a minus sign. You have to do this, because only if all contracted operators are next to each other you can replace them by propagators.

 

[latentcorpse]

Well it's more like that you can write Time ordering as Normal ordering plus all possible contractions. And for this the proof is exactly the same as in the bosonic case.

In the example you move the $$\psi(y)$$ from the right to the left, on the sway swapping it three times with other fermionic operators. Each time you pick up a minus sign. You have to do this, because only if all contracted operators are next to each other you can replace them by propagators.

Ok. I get that now. But in the example on p121, what do these terms have to do with that circular loop? I don't follow the connection?

Thanks.

 

[betel]

The fermionic part of that diagram, i.e. the loop is given by to the 4 factors of psi. At each point x and y you absorb and create one particle. Combined this gives the propagators as shown.

 

[latentcorpse]

The fermionic part of that diagram, i.e. the loop is given by to the 4 factors of psi. At each point x and y you absorb and create one particle. Combined this gives the propagators as shown.

so why do we contract them in that particular way? why do we not contract the x's together and the y's together

 

[betel]

You are right, this contraction has to be included. But this will gove a so called vacuum, bubble or reducible diagram. iF you consider it in this specific diagram you will have the bosons going straight through plus two separate circles which correponds to fermions created and annihilated again. But first of all, this diagram will give an infinite contribution and second it would also happen in vacuum and would give an infinite contribution to the vacuum. So the practical view is to simply ignore such diagrams. It can also be shown using the partition function that this treatment is correct. The bubble diagram give no physical measurable effect.

 

[latentcorpse]

You are right, this contraction has to be included. But this will gove a so called vacuum, bubble or reducible diagram. iF you consider it in this specific diagram you will have the bosons going straight through plus two separate circles which correponds to fermions created and annihilated again. But first of all, this diagram will give an infinite contribution and second it would also happen in vacuum and would give an infinite contribution to the vacuum. So the practical view is to simply ignore such diagrams. It can also be shown using the partition function that this treatment is correct. The bubble diagram give no physical measurable effect.

Hi, I have one or two other small things in these notes that I was wondering if you could help me with.

In the 1st line of eqn 5.6, this only works if we assume that $u$ and $u^dagger$ commute, but it doesn't say anywhere in the notes that this is true!

For example, if we expand the commutator, we will get something like $b b^\dagger u u^\dagger - b^\dagger b u^\dagger u$ and in order for this to equal $[b, b^\dagger] u u^\dagger$ then the $u$ and $u^\dagger$ must commute. Why is this?

And secondly, in 4.128, why doesn it not make sense to consider something like $\Sigma_{r,s} u^r(\vec{p}) \bar{u}^s(\vec{p})$? What are these indices r and s and what do they mean? Are these the spinor indices?

Thanks again!

 

[betel]

The us are no operators but spinors. nevertheless there is a difference which one comes first, similar to vectors and their transpose. In this case this is taken care of by the indices which pick out exactly one component and sums over them. Similiar to writing maxtrix vector product using the indices, then you can commute them.

Yes, the r,s are spinor indices. And you only consider the one sum, because this one appears in the diagrams.

 

[latentcorpse]

The us are no operators but spinors. nevertheless there is a difference which one comes first, similar to vectors and their transpose. In this case this is taken care of by the indices which pick out exactly one component and sums over them. Similiar to writing maxtrix vector product using the indices, then you can commute them.

Yes, the r,s are spinor indices. And you only consider the one sum, because this one appears in the diagrams.

Hmmmm... I don't see how the indices work to tell us which way round the spinors should come?

Could you write it out or elaborate perhaps?

Thanks!

 

[betel]

He was a bit sloppy in the notation. He takes the commutator of the field spinor with its hermitian conjugate. You should notice that this is not well defined, as the first is a 4x4 matrix, whereas the second one is a scalar. What he really wanted to compute was the commutator of one component of each, as in eq. 5.3. The spinors will then commute because you have only simple numbers as explained in more detail below. So to be correct you should put an index alpha or beta on the u or v.
The rest should go through unchanged (I haven't checked) and you would have to take the alpha-beta component of all the $$(\not p-m)\gamm^0$$ combinations at the end. If I should explain in more detail pls let me know.

In standard linear algebra you can write e.g. the scalar product using the vectors.
$$(\vec v,\vec u)=\vec v^\dagger \vec u\neq \vec u\vec v^\dagger$$
Here the order is important.
Using indices the order does not matter
$$(\vec v,\vec u)=\sum_i v_i^*u_i = \sum_i u_i v_i^*$$
In exactly the same way it works here.