Feynman Rules for Crossing Lines in Green's Function Diagrams?

[latentcorpse]

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

Consider the Feynman rules for Green's Functions given at the top of p79 in these notes.

Now let us consider the diagram given in the example on p78.
Take for example the 2nd diagram in the sum i.e. the cross one where x1 is joined to x4 and x2 is joined to x3 and these two lines cross over each other.

Suppose I wanted to apply the Feynman rules to this diagram:

I am not sure if the point where they cross is a vertex or not? I'm going to assume that it must be otherwise that diagram would be the same as the 1st one in the sum (but with vertices relabelled). So let us label the vertex with the spacetime position y.

The Feynman rules then tell us that this Feynman diagram contributes

$-i \lambda \int d^4y \int \frac{d^4k}{(2 \pi)^4} \frac{i e^{-ik \cdot (x_1-x_4)}}{k^2-m^2+i \epsilon} \int \frac{d^4p}{(2 \pi)^4} \frac{ie^{-ip \cdot (x_2 - x_3)}}{p^2-m^2+i \epsilon}$

Is this correct? Can it be simplified? It looks pretty messy?

Presumably when we added the other contributions from all the other diagrams, we'd end up with a nice final expression for $G^{(4)}(x_1 , \dots , x_4)$?

Thanks.

 

[fzero]

If the point y is a vertex, then we need to write propagators from each point $$x_i$$ to $$y$$. Defining the momenta to point inwards, we could write

$$-i \lambda \int d^4y \prod_i \Delta_F(y,x_i).$$

Note that the role of the integration over y is to enforce momentum conservation through the integral

$$\int d^4y e^{-iy\sum_i k_i} \sim \delta^4\left(\sum_i k_i\right).$$

Amplitudes look a bit simpler in momentum space, but as things go, this one is pretty tame.

 

[latentcorpse]

If the point y is a vertex, then we need to write propagators from each point $$x_i$$ to $$y$$. Defining the momenta to point inwards, we could write

$$-i \lambda \int d^4y \prod_i \Delta_F(y,x_i).$$

Note that the role of the integration over y is to enforce momentum conservation through the integral

$$\int d^4y e^{-iy\sum_i k_i} \sim \delta^4\left(\sum_i k_i\right).$$

Amplitudes look a bit simpler in momentum space, but as things go, this one is pretty tame.

I see. Thanks. Can you just confirm that for that diagram we were talking about, there is ACTUALLY a vertex at y, yes?

 

[fzero]

I see. Thanks. Can you just confirm that for that diagram we were talking about, there is ACTUALLY a vertex at y, yes?

Yes, there's a vertex in that diagram. The diagram without a vertex that connects $$x_1\leftrightarrow x_4$$ and $$x_2\leftrightarrow x_3$$ is part of the "2 Similar" diagrams in the preceeding term.

 

[latentcorpse]

Yes, there's a vertex in that diagram. The diagram without a vertex that connects $$x_1\leftrightarrow x_4$$ and $$x_2\leftrightarrow x_3$$ is part of the "2 Similar" diagrams in the preceeding term.

Thanks. I do have one or two other questions about later in these notes though.

(i) On p82, he calculates in (4.9) the matrix $M^{12}$ using the formula (4.8).

Now I tried to replicate this basic calculation and got the elements the wrong way round. I have traced this back to me taking the $\mu$ index to label the rows and the $\nu$ index to label the columns. However, to get the right answer, it seems it must be the other way around. I don't know how we were expected to know this though as he doesn't explain what the indices label and I thought that by convention the first index was for rows and the second for columns, no?

Have I made a mistake here or something?This is then used in (4.29) to get the matrix exponential he has written down. However, he hasn't put in the 1/2 I don't think so I think (4.29) should be
$\text{exp } \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{\phi^3}{2} & 0 \\ 0 & - \frac{\phi^3}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
If we then substitute $\phi^3=2 \pi$
we get
$\text{exp } \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & \pi & 0 \\ 0 & - \pi & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
Why is that equal to 1?

(ii) Under (4.53) on p90 he says that $\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu = \frac{1}{2} \{ \gamma^\mu , \gamma^\nu \} \partial_\mu \partial_\nu$
I cannot see why this is true! If I expand out that commutator I get $\frac{1}{2} ( \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu )$ but the $\gamma$ matrices don't commute so surely that identity is wrong?

Thanks a lot.

 

[fzero]

It looks like he's using the opposite convention for the matrix form or just made typos that propagated through the notes.

For the 2nd question, the 1/2 is accounted for by the sum over $$\rho,\sigma$$. You get $$\phi^3$$, not $$\phi^3/2$$.

 

[latentcorpse]

It looks like he's using the opposite convention for the matrix form or just made typos that propagated through the notes.

For the 2nd question, the 1/2 is accounted for by the sum over $$\rho,\sigma$$. You get $$\phi^3$$, not $$\phi^3/2$$.

Hmmm. Sorry I don't see it. How does that sum give us a factor of 2 to cancel the 1/2?

 

[fzero]

Hmmm. Sorry I don't see it. How does that sum give us a factor of 2 to cancel the 1/2?

$$\Omega_{\rho\sigma} \mathcal{M}^{\rho\sigma} = \Omega_{12} \mathcal{M}^{12} +\Omega_{21} \mathcal{M}^{21}$$

 

[latentcorpse]

$$\Omega_{\rho\sigma} \mathcal{M}^{\rho\sigma} = \Omega_{12} \mathcal{M}^{12} +\Omega_{21} \mathcal{M}^{21}$$

Ok. So then we get to the matrix

$\text{exp } \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 2 \pi & 0 \\ 0 & - 2 \pi & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
How can we show that this matrix exponential is 1?

Also, under (4.53) on p90 he says that $\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu = \frac{1}{2} \{ \gamma^\mu , \gamma^\nu \} \partial_\mu \partial_\nu$
I cannot see why this is true! If I expand out that commutator I get $\frac{1}{2} ( \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu )$ but the $\gamma$ matrices don't commute so surely that identity is wrong?

And then finally in (4.62) on p91, he calculates the lagrangian. However, I find that
$L=\psi^\dagger \gamma^0 ( i \gamma^\mu \partial_\mu - m) \psi = i \begin{pmatrix} u_+^\dagger \\ u_-^\dagger \end{pmatrix} \gamma^0 \gamma^\mu \partial_\mu \begin{pmatrix} u_+ & u_- \end{pmatrix} - m \begin{pmatrix} u_+^\dagger \\ u_-^\dagger \end{pmatrix} \gamma^0 \begin{pmatrix} u_+ & u_- \end{pmatrix}$
which doesn't look like it's going to give the right answer. In particular there are factors of $\gamma^0$ floating about in awkward places...

Thanks.

 

[fzero]

Ok. So then we get to the matrix

$\text{exp } \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 2 \pi & 0 \\ 0 & - 2 \pi & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
How can we show that this matrix exponential is 1?

$$\exp \begin{pmatrix} 0 & \phi \\ -\phi & 0 \end{pmatrix} = \begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{pmatrix}$$

Also, under (4.53) on p90 he says that $\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu = \frac{1}{2} \{ \gamma^\mu , \gamma^\nu \} \partial_\mu \partial_\nu$
I cannot see why this is true! If I expand out that commutator I get $\frac{1}{2} ( \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu )$ but the $\gamma$ matrices don't commute so surely that identity is wrong?

$$\partial_\mu \partial_\nu$$ is symmetric.

And then finally in (4.62) on p91, he calculates the lagrangian. However, I find that
$L=\psi^\dagger \gamma^0 ( i \gamma^\mu \partial_\mu - m) \psi = i \begin{pmatrix} u_+^\dagger \\ u_-^\dagger \end{pmatrix} \gamma^0 \gamma^\mu \partial_\mu \begin{pmatrix} u_+ & u_- \end{pmatrix} - m \begin{pmatrix} u_+^\dagger \\ u_-^\dagger \end{pmatrix} \gamma^0 \begin{pmatrix} u_+ & u_- \end{pmatrix}$
which doesn't look like it's going to give the right answer. In particular there are factors of $\gamma^0$ floating about in awkward places...

Thanks.

$$\bar{\psi}$$ has a factor of $$\gamma^0$$, see (4.40).

 

[latentcorpse]

$$\exp \begin{pmatrix} 0 & \phi \\ -\phi & 0 \end{pmatrix} = \begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{pmatrix}$$



$$\partial_\mu \partial_\nu$$ is symmetric.



$$\bar{\psi}$$ has a factor of $$\gamma^0$$, see (4.40).

Haven't I put the $\gamma^0$ in though in my previous post?

 

[latentcorpse]

Actually I have now figured out how that works.

What about in (4.117), why does $\sqrt{ p \cdot \sigma} = \sqrt{ E + p^3}$ here?
Whereas in (4.115) it was equal to $\sqrt{E - p^3}$

And in (4.118), why does this operator give us 1/2 when it acts on (4.116)? Thanks.

 

[fzero]

Actually I have now figured out how that works.

What about in (4.117), why does $\sqrt{ p \cdot \sigma} = \sqrt{ E + p^3}$ here?
Whereas in (4.115) it was equal to $\sqrt{E - p^3}$

(4.115) has the spin up state, while (4.117) is spin down. Alternatively, you can just write out the matrices and compute.

And in (4.118), why does this operator give us 1/2 when it acts on (4.116)? Thanks.

I can't find a definition for $$\hat{p}_i$$, but I assume it's just the unit vector, so $$\hat{p}_i = (0,0,1)$$. Then $$\sigma^3$$ appears in the operator and $$h=1/2$$ follows by direct computation.

 

[latentcorpse]

(4.115) has the spin up state, while (4.117) is spin down. Alternatively, you can just write out the matrices and compute.

I don't follow this still.

in (4.115) we have $\sqrt{p \cdot \sigma} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \sqrt{ E-p^3} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ using the definition of 4 vector products in Minkowski space $p \cdot \sigma =p^0 \sigma_0 - p^i \sigma_i$

I don't see how this calculation would be any different for (4.117)?

 

[fzero]

Because

$\sigma^3 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \neq \sigma^3 \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$

 

[latentcorpse]

Because

$\sigma^3 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \neq \sigma^3 \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$

Ok. That makes sense.

What about the bit inbetween eqns (4.126) and (4.127). He claims that $(p \cdot \bar{\sigma})(p' \cdot \sigma) = (p_0 + p_i \sigma^i)(p_0-p_i \sigma_i)$

I disagree. I find:

$p \cdot \bar{\sigma} = p_\mu \bar{\sigma}^\mu = p_0 \bar{\sigma}^0 + p_i \bar{\sigma}^i = p_0 - p_i \sigma^i$

since$\bar{\sigma}^i=-\sigma^i$

and then

$p' \cdot {\sigma} = p'_\mu \sigma^\mu = p'_0 {\sigma}^0 + p'_i \sigma}^i = p_0 - p_i \sigma^i$

since $p'^i=-p^i$

Is that right?

 

[fzero]

Your calculation is correct. He's made a mistake in carrying out the multiplication in the 2nd line of (4.126). He should have $$\sqrt{(p\cdot\sigma)(p'\cdot\sigma)}$$ in the first term and $$\sqrt{(p\cdot\bar{\sigma})(p'\cdot\bar{\sigma})}$$ in the second. Both of those expressions give $$p_0^2 - \vec{p}^2$$ under the square root.

 

[latentcorpse]

Your calculation is correct. He's made a mistake in carrying out the multiplication in the 2nd line of (4.126). He should have $$\sqrt{(p\cdot\sigma)(p'\cdot\sigma)}$$ in the first term and $$\sqrt{(p\cdot\bar{\sigma})(p'\cdot\bar{\sigma})}$$ in the second. Both of those expressions give $$p_0^2 - \vec{p}^2$$ under the square root.

Yes. I was doing it in my head. I guess if I'd written it out, I would have seen that he hadn't carried the factors through properly from the matrix multiplication.

And could you have a look at (5.22)? Here I find that $\vec{\alpha}=\gamma^0 \vec{\gamma}$
i.e. there should be no minus sign.

Since the Dirac equation gives

$( i \gamma^\mu \partial_\mu - m ) \psi =0$
$(i \gamma^0 \partial_t + i \gamma^i \partial_i - m) \psi=0$
$i \frac{\partial \psi}{\partial t} = - i \gamma^0 \gamma^i \partial_i \psi+ m \gamma^0 \psi$
$i \frac{\partial \psi}{\partial t} = - \vec{\alpha} \cdot \vec{\nabla} \psi + m \beta \psi$

So we would find that $\vec{\alpha}= \gamma^0 \vec{\gamma}$, no?

Also, under (5.34), I don't understand the reasoning for that minus sign being there?

And in the paragraph where he's trying to explain why the minus sign is there he claims that $\{ \psi(x) , \bar{\psi}(y) \} = 0$. But we showed in (5.14) that $\{ \psi(x) , \psi^\dagger(y) \} = \delta_{\alpha \beta} \delta^{(3)}(x-y)$ as a result of fermionic quantisation. Wouldn't this imply there is a non trivial commutation relation for $\{ \psi(x) , \bar{\psi}(y) \}$ since $\bar{\psi} = \psi^\dagger \gamma^0$ after all?

 

[fzero]

Yes. I was doing it in my head. I guess if I'd written it out, I would have seen that he hadn't carried the factors through properly from the matrix multiplication.

And could you have a look at (5.22)? Here I find that $\vec{\alpha}=\gamma^0 \vec{\gamma}$
i.e. there should be no minus sign.

Since the Dirac equation gives

$( i \gamma^\mu \partial_\mu - m ) \psi =0$
$(i \gamma^0 \partial_t + i \gamma^i \partial_i - m) \psi=0$
$i \frac{\partial \psi}{\partial t} = - i \gamma^0 \gamma^i \partial_i \psi+ m \gamma^0 \psi$
$i \frac{\partial \psi}{\partial t} = - \vec{\alpha} \cdot \vec{\nabla} \psi + m \beta \psi$

So we would find that $\vec{\alpha}= \gamma^0 \vec{\gamma}$, no?

$$i \gamma^\mu \partial_\mu - m = i \gamma^0 \partial_t - i \gamma^i \partial_i -m$$

Also, under (5.34), I don't understand the reasoning for that minus sign being there?

And in the paragraph where he's trying to explain why the minus sign is there he claims that $\{ \psi(x) , \bar{\psi}(y) \} = 0$. But we showed in (5.14) that $\{ \psi(x) , \psi^\dagger(y) \} = \delta_{\alpha \beta} \delta^{(3)}(x-y)$ as a result of fermionic quantisation. Wouldn't this imply there is a non trivial commutation relation for $\{ \psi(x) , \bar{\psi}(y) \}$ since $\bar{\psi} = \psi^\dagger \gamma^0$ after all?

Those should be equal time commutation relations, note that 5.14 depends on the spatial variables only. Besides, time and normal ordering are definitions of the product that subtract off anything you'd have gotten from commutation relations. Just the overall sign must be kept track of for fermions.

 

[latentcorpse]

$$i \gamma^\mu \partial_\mu - m = i \gamma^0 \partial_t - i \gamma^i \partial_i -m$$

Surely not. $a^\mu b_\mu = a^0 b_0 + a^i b_i$
We only get a minus sign when we include the metric as follows:
$a^\mu b_\mu = \eta_{\mu \nu} a^\mu b^\nu = a^0 b^0 - a^i b^i$
No?

Those should be equal time commutation relations, note that 5.14 depends on the spatial variables only. Besides, time and normal ordering are definitions of the product that subtract off anything you'd have gotten from commutation relations. Just the overall sign must be kept track of for fermions.

So what's the actual reason for the minus sign then?

 

[fzero]

Surely not. $a^\mu b_\mu = a^0 b_0 + a^i b_i$
We only get a minus sign when we include the metric as follows:
$a^\mu b_\mu = \eta_{\mu \nu} a^\mu b^\nu = a^0 b^0 - a^i b^i$
No?

You're right, I was a bit too fast. $$p_\mu = (p_0, - \vec{p})$$, but $$\partial_\mu = (\partial_t,\partial_i)$$. I don't believe that the minus sign affects anything else said in that section though.

So what's the actual reason for the minus sign then?

The time-ordered product still has to respect the signs of the anticommutation relation.

 

[latentcorpse]

You're right, I was a bit too fast. $$p_\mu = (p_0, - \vec{p})$$, but $$\partial_\mu = (\partial_t,\partial_i)$$. I don't believe that the minus sign affects anything else said in that section though.

This won't change what I did though since $p_\mu$ doesn't appear in this calculation does it?

The time-ordered product still has to respect the signs of the anticommutation relation.

Why does $\{ \psi(x) , \bar{\psi}(y) \} = 0$ though? Can't we pull out a factor of $\gamma^0$ and then have $\gamma^0 \{ \psi(x) , \bar{\psi}(y) \}$ which would certainly bee non-zero?

Also, I noticed something that's bugging me. Take for example, (5.40), he says that he is expanding to second order and so he gets a factor of $\frac{(-i \lambda)^2}{2}$ as you would expect from the exponential in Dyson's formula. But didn't we show earlier in (3.23) that the factor of $\frac{1}{2}$ isn't necessary?

And lastly, on p120, I'm getting very confused trying to source the origin of these minus signs. Take for example the equation beneath (5.48), why does

$\langle 0 | c^{r'}_{\vec{q'}} b^{s'}_{\vec{p'}} {c^m_{\vec{l_1}}}^\dagger {b^n_{\vec{l_2}}}^\dagger$
rearrange to give us the minus sign - I have been playing about with this using the anticommutation relations and cannot for the life of me get it to work out right!

Thanks a lot.

 

[latentcorpse]

may have sorted it all except the cbcb bit at the end there...