How does one compute the Fourier-Transform of the Dirac-Hamiltonian?

[PhysicsRock]

Greentings,
I've dealt with Quantum Theory a lot, but there's one thing I don't quite understand.

When deriving the Fermion-Propagator, say $S_F$, all the authors I've read from made use of the Fourier-Transform. Basically, it always goes like
$$
\begin{align}
H_D S_F(x-y) &= (i \hbar \gamma^\mu \partial_\mu - mc) \int \tilde{S}_F(p) \cdot e^{i p (x - y) / \hbar} \, \frac{d^4p}{(2 \pi)^4} \notag \\
&= \int (\gamma^\mu p_\mu - mc) \tilde{S}_F(p) e^{i p (x - y) / \hbar} \, \frac{d^4p}{(2 \pi)^4} \notag
\end{align}
$$

After that, the Dirac-Delta is used in Integral-representation and by comparing coefficiants, a momentum-space Propagator can be written as $\tilde{S}_F(p) = \frac{\gamma^\mu p_\mu + mc}{p^2 - m^2 c^2}$. What I don't get is how in line two of the above equation the Hamiltonian can be dragged into the integral, especially in it's momentum-space form, which I also can't find a derivation for. I have attempted it in the following way

$$
\begin{align}
\tilde{H}_D &= \int (i \hbar \gamma^\mu \partial_\mu - mc) e^{-i p x / \hbar} \, d^4x \notag \\
&= \int (i \hbar \gamma^\mu (-i \frac{p_\mu}{\hbar}) - mc) e^{-i p x / \hbar} \, d^4x \notag \\
&= \int (\gamma^\mu p_\mu - mc) e^{i p x / \hbar} \, d^4x \notag
\end{align}
$$

Obviously, this already displays what has been used above, however, within an integral, so I cannot just assume $\tilde{H}_D = (\gamma^\mu p_\mu - mc)$.

So, can someone please explain to me how this is valid? I can't seem to understand it myself and I cannot find anything on the Internet either that would make sense out of this.

Thank you in advance and have a great day everyone :)

 

[Demystifier]

The operator you are studying is not called Hamiltonian. It's called Dirac operator. Leaving that aside, here is a hint. Don't start from the propagator, start from the Dirac field $\psi(x)$. Write down the Dirac equation for it and then perform a Fourier transform.

 

[PhysicsRock]

The operator you are studying is not called Hamiltonian. It's called Dirac operator. Leaving that aside, here is a hint. Don't start from the propagator, start from the Dirac field $\psi(x)$. Write down the Dirac equation for it and then perform a Fourier transform.

Wouldn't that lead to the convolution theorem?

 

[Demystifier]

Wouldn't that lead to the convolution theorem?

Convolution involves a product of two functions. The Dirac equation (unlike the propagator) does not contain such a product, so no, you will not get the convolution theorem if you know what you are doing.

 

[PhysicsRock]

Convolution involves a product of two functions. The Dirac equation (unlike the propagator) does not contain such a product, so no, you will not get the convolution theorem if you know what you are doing.

That is quite literally the problem, I have no idea what I'm doing or rather, how I'm supposed to do it. I would appreciate it a lot if authors would provide more detailed derivations and more steps leading to the final result.

However, coming back to what you said earlier, namely "Write down the Dirac equation for it and then perform a Fourier transform". Do you mean to imply that I can take the Fourier-Transform of the individual components ($H_D$ and $\Psi(x)$) individually?

Thanks for your replies by the way!

 

[samalkhaiat]

Greentings,
I've dealt with Quantum Theory a lot, but there's one thing I don't quite understand.

Do you know how to solve differential equations by Green’s function methods?

1) K-G propagator: Recall that the FREE KG field satisfies \left( \partial^{2} + m^{2}\right) \phi (x) = 0 , \ \ \ \mbox{where} \ \partial^{2} = \partial_{\mu}\partial^{\mu}. \ \ (1) Now let us modify the above FREE KG equation to include a source J(x), i.e., we write the following equation for the interacting KG field \varphi (x): \left( \partial^{2} + m^{2}\right) \varphi (x) = J(x) \equiv \int d^{4}y \ \delta^{4}(x - y) J(y) . The most general (formal) solution of the interacting KG field is obtained by applying the inverse operator (\partial^{2} + m^{2})^{-1} to the above equation and taking into account the FREE equation (1): \varphi (x) = \phi (x) + \int d^{4}y \big[ (\partial^{2} + m^{2})^{-1} \delta^{4}(x - y) \big] \ J(y) . This means that the bracketed object under the integral is the KG propagator, let us call it - \Delta (x - y). Then, for our formal solution, we have \varphi (x) = \phi (x) - \int d^{4}y \ \Delta (x - y) \ J(y) , with (the KG Green function) \Delta (x - y) satisfying \left( \partial^{2} + m^{2}\right) \Delta (x - y) = - \delta^{4}(x - y) . \ \ \ \ \ (2) So, our task is reduced to solving (2). This is done by Fourier transforming the propagator \Delta (x - y) = \frac{1}{(2 \pi)^{4}} \int d^{4}k \ e^{- i k(x - y)} \ \Delta_{F}(k) . \ \ \ (3) Now, substitute (3) in (2); use the identity (\partial^{2} + m^{2}) e^{- i k(x - y)} = (- k^{2} + m^{2}) e^{- i k(x - y)}, and the integral representation of the delta function \delta^{4}(x - y) = \frac{1}{(2 \pi)^{4}} \int d^{4}k \ e^{- i k(x - y)} . Doing that gives you \int d^{4}k \ e^{- i k(x - y)} = \int d^{4}k \ e^{- i k(x - y)} \ (k^{2} - m^{2}) \Delta_{F}(k) . This in turn gives you \Delta_{F}(k) = \frac{1}{k^{2} - m^{2}} . Substituting this in (3), you get the space-time KG propagator \Delta (x) = \frac{1}{(2 \pi)^{4}} \int d^{4}k \frac{e^{- ikx}}{k^{2} - m^{2}} \equiv - (\partial^{2} + m^{2})^{-1} \delta^{4}(x) .

2) The Dirac propagator: Repeat the above!

3) Given \mathcal{L} = - \frac{1}{4} \int d^{4}x \ F_{\mu\nu}F^{\mu\nu}, with F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}. Show that the field A_{\mu} has NO propagator.

 

[PhysicsRock]

Do you know how to solve differential equations by Green’s function methods?

Hello,
yes, I am familiar with Green's functions.

Thank you so much for the help. I think, I got it now. I will write down my approach for the Fermion-Propagator below. If you want to, feel free to check it and correct me, if I'm wrong anywhere. If you don't wish to check it, that's totally fine as well.

Before showing my solution, I want to ask a quick question in advance. When you proposed the problem of proving there is no propagator for the what I assume to be the Photon-Field, you gave the Lagrangian as, quote, "$\mathcal{L} = - \frac{1}{4} \int d^4x \, F_{\mu\nu} F^{\mu\nu}$". What confuses me, is the integral sign. To me, the Maxwell-Lagrangian should be this expression, simply without the integral. What you provided here is the Action of the System I would assume, or am I wrong here?

Alright, now to my solution. I start by writing down the equation defining the Propagator in coordinate space as
$$
\begin{align}
(i \hbar \gamma^\mu \partial_\mu - mc) S_F(x - y) = \delta^4(x - y)
\end{align}
$$

I then write down the Propagator as a Fourier-Transform like

$$
\begin{align}
S_F(x-y) = \frac{1}{(2 \pi)^4} \int d^4p \, \tilde{S}_F(p) \, e^{-ip(x - y) / \hbar}
\end{align}
$$

Following that, I note that

$$
\begin{align}
(i \hbar \gamma^\mu \partial_\mu - mc) \, e^{-ip(x-y)/\hbar} = (\gamma^\mu p_\mu - mc) \, e^{-ip(x-y)/\hbar}
\end{align}
$$

Next, I substitute the Fourier-Transform into the defining equation and by dragging the Dirac-Operator inside the integral, giving me

$$
\begin{align}
(i \hbar \gamma^\mu \partial_\mu - mc) \, \int \frac{d^4p}{(2 \pi)^4} \, \tilde{S}_F(p) \, e^{-ip(x-y)/\hbar}
= \int \frac{d^4p}{(2 \pi)^4} \, (\gamma^\mu p_\mu - mc) \tilde{S}_F(p) e^{-ip(x-y)/\hbar}
\end{align}
$$

I then make use of the Integral Representation of the Dirac-Delta, which leads to

$$
\begin{align}
(\gamma^\mu p_\mu - mc) \tilde{S}_F(p) = \mathbb{I}
\end{align}
$$

Where $\mathbb{I}$ denotes the identity matrix in $\mathbb{R}^4$. Expanding on both sides with $(\gamma^\mu p_\mu + mc)$ yields

$$
\begin{align}
(p^2 - m^2c^2)\tilde{S}_F(p) = \gamma^\mu p_\mu + mc
\end{align}
$$

Where $(\gamma^\mu p_\mu)(\gamma^\mu p_\mu) = p^2$ follows from the properties of the Gamma-matrices. The final result is then obtained by dividing both sides with $(p^2 - m^2 c^2)$,
so I get

$$
\begin{align}
\tilde{S}_F(p) = \frac{\gamma^\mu p_\mu + mc}{p^2 - m^2 c^2}
\end{align}
$$

Are there any mistakes in my derivation? If so, please let me know. Thanks again for your help, have a good day.