Green's functions in QFT for the gifted amateur

[Amentia]

Hello,

I am reading the book QFT for the gifted amateur and I have a question concerning how to go from the wave function picture to the Green's function as defined by equations (16.13) and (16.18) at page 147.

$\phi(x,t_{x}) = \int dy G^{+}(x,t_{x},y,t_{y})\phi(y,t_{y})$

$G^{+}(x,t_{x},y,t_{y}) = \theta(t_{x}-t_{y}) \langle x(t_{x})|y(t_{y})\rangle$

I tried to write $\phi(x,t_{x})$ as a function of $\phi(y,t_{y})$ by introducing an integral form for the identity operator:

$\phi(x,t_{x}) = \langle x|\hat{U}(t_{x})|\phi)\rangle$
$\phi(x,t_{x}) = \int dy \langle x|\hat{U}(t_{x})|y\rangle\langle y|\phi\rangle$
$\phi(x,t_{x}) = \int dy \langle x|\hat{U}(t_{x})\hat{U}(-t_{y})\hat{U}(t_{y})|y\rangle\langle y|\hat{U}(-t_{y})\hat{U}(t_{y})|\phi\rangle$
$\phi(x,t_{x}) = \int dy \langle x|\hat{U}(t_{x}-t_{y})\left[\hat{U}(t_{y})|y\rangle\langle y|\hat{U}^{\dagger}(t_{y})\right]\hat{U}(t_{y})|\phi\rangle$

Now I assume the jacobian of the unitary matrix U (time evolution operator) is 1 so that I rewrite the integral:
$\phi(x,t_{x}) = \int dy \langle x|\hat{U}(t_{x}-t_{y})|y\rangle\langle y|\hat{U}(t_{y})|\phi\rangle$
$\phi(x,t_{x}) = \int dy \langle x|\hat{U}(t_{x}-t_{y})|y\rangle\phi(y,t_{y})$
$\phi(x,t_{x}) = \int dy \langle x(t_{x})|y(t_{y})\rangle\phi(y,t_{y})$

So by identification, I get:

$G^{+}(x,t_{x},y,t_{y}) = \langle x(t_{x})|y(t_{y})\rangle$

and my question is: where does the Heaviside function $\theta(t_{x}-t_{y})$ come from in the definition? I do not obtain it from this derivation, unless it was necessary at some point to introduce it in the integral and I failed to see it.

 

[George Jones]

I am reading the book QFT for the gifted amateur and I have a question concerning how to go from the wave function picture to the Green's function as defined by equations (16.13) and (16.18) at page 147.

For $G^+$, what does the $+$ mean? Look at the last full paragraph on page 146.

 

[Amentia]

Thank you for your answer, I understood that it is introduced so that particles do not travel backward in time. But in the wavefunction representation, there is no Heaviside function, so does that mean that in quantum mechanics the particle can actually go backward in time? The books seems to say that the two equations are equivalent but they also say at some point that the propagator is more general than the wavefunctions because it cares about the starting point of the particle in spacetime. So this is what confuses me, I guess.

 

[Amentia]

$\phi(x,t_{x}) = \int dy G^{+}(x,t_{x},y,t_{y})\phi(y,t_{y})$

$G^{+}(x,t_{x},y,t_{y}) = \theta(t_{x}-t_{y}) \langle x(t_{x})|y(t_{y})\rangle$

So by identification, I get:
$G^{+}(x,t_{x},y,t_{y}) = \langle x(t_{x})|y(t_{y})\rangle$

Here I could also say:
$G^{+}(x,t_{x},y,t_{y}) = \langle x(t_{x}-t_{y})|y(0)\rangle$

So a particle started at position y at time 0 and ends at position x at time $t_{x}-t_{y}$. But this time can then be negative in the wavefunction representation or do we need the Heaviside function somewhere when using the time-evolution operator?

 

[vanhees71]

The Heaviside function is there to define the retarded Green's function. It's also important to say what you want to calculate. For the time evolution of states usually you don't need the retarded but the time-oredered Green's function. So I'm not sure that your question can be answered seriously without more context about the problem you like to solve. I don't have the textbook you quote at hand. So I can't check.

 

[Demystifier]

But in the wavefunction representation, there is no Heaviside function, so does that mean that in quantum mechanics the particle can actually go backward in time?

In a sense, yes. As long as quantum mechanics is viewed as a microscopic theory, it is invariant under the time inversion. However, when the effect of measurement is taken into account, then some macroscopic effects are taken into account too. Macroscopic physics in general, and measurement process in particular, are not invariant under the time inversion. This fact can be taken into account by replacing the Green function $G=G^+ +G^-$ with $G^+$. It is really a trick, a shortcut that allows you to describe a macroscopic effect by dealing only with microscopic degrees, without messing with macroscopic degrees explicitly.

 

[vanhees71]

The problem is that we have to guess what is treated in the quoted passage of the textbook. For non-relativistic QT the propagator for the evolution of the wave function is identical with the retarded one. In 2nd quantization that's because in non-relativistic QM the field operator mode decomposition is a pure annihilation field, and thus the time-ordered is the same as the retarded propagator. This is no longer true in relativistic local QFTs, where all field operators' mode decompositions are the sum of both creation and annhilation operators, and thus the time ordered Green's functions are not the same as the retarded ones.

In all QTs, no matter whether relativistic or non-relativistic the retarded propgator occurs in linear-response theory and thus, e.g., the Kubo relation to calculate transport coefficients is always to be used with the retarded (and only with the retarded!) propagator.

We have to wait, which problem is discussed in the OP, before we can definitely answer the question.

 

[Amentia]

Hello, thank you for your answers.

The Heaviside function is there to define the retarded Green's function. It's also important to say what you want to calculate. For the time evolution of states usually you don't need the retarded but the time-oredered Green's function. So I'm not sure that your question can be answered seriously without more context about the problem you like to solve. I don't have the textbook you quote at hand. So I can't check.

What I want to calculate is very general. It is simply about going from the wave function to the Green's function picture, so just solving the equation in the first message. In the book, this is almost the first definition of the Green's function, just after $\hat{L} G(t,u) = \delta (t-u)$, used to make a link between Green's function and quantum mechanics by defining a new quantity called a propagator. So supposedly it does not involve any advanced quantum field theory calculations (no quantum fields, no second quantization yet, I think it starts at the next chapter). How would you introduce this time-ordered Green's function in my calculation?

In a sense, yes. As long as quantum mechanics is viewed as a microscopic theory, it is invariant under the time inversion. However, when the effect of measurement is taken into account, then some macroscopic effects are taken into account too. Macroscopic physics in general, and measurement process in particular, are not invariant under the time inversion. This fact can be taken into account by replacing the Green function $G=G^+ +G^-$ with $G^+$. It is really a trick, a shortcut that allows you to describe a macroscopic effect by dealing only with microscopic degrees, without messing with macroscopic degrees explicitly.

Maybe it is the answer to my question. If introducing the retarded Green's function is a trick and does not come naturally from the wave function picture, then it means my calculation to go from one representation to the other is correct and that the integral is somehow cut to positive times to make the theory consistent later on...

----------------------------

If you really need the full section of the book to answer, I could copy it there later if this is allowed.

 

[vanhees71]

There's no "Green's function picture" in QM. Since you talk about wave functions, you must talk about non-relativistic quantum theory. Then the time-ordered propagator is the same as the retarded one.

It cannot be so difficult to briefly state the problem. I don't think we need the complete section of the book.

 

[Demystifier]

In 2nd quantization that's because in non-relativistic QM the field operator mode decomposition is a pure annihilation field

Well, if the operator $\psi$ is a pure annihilation field, then $\psi^{\dagger}$ is a pure creation field. And nobody prevents you to define a new non-relativistic field $\phi=\psi+\psi^{\dagger}$.

 

[Amentia]

So in the book, they say: the amplitude $\langle x(t_{x})|y(t_{y}\rangle$ is known as a propagator. (...) Propagators for single particles have a neat mathematical property: they are the Green's function of the equation of motion of the particle. Then they define the general equation for Green's function with the delta function and give a few examples. After this they recall the Schrodinger equation in 1 dimension and say: " Why might the Green's function of the Schrodinger equation be useful, and what interpretation might such a function have? The real beauty of a Green's function is the property" from the previous example, namely that:

$\phi(x,t_{x}) = \int dy G^{+}(x,t_{x},y,t_{y})\phi(y,t_{y})$

Then they say that the Green's function propagates the particle and explain that the retarded Green's function is there to prevent particles going back in time. Finally they add that we can interpret $\phi(y,t_{y})$ as the amplitude to find a particle at $(y,t_y)$ and $\phi(x,t_{x})$ as the amplitude to find a particle at $(x,t_x)$. It follows that the propagator is the probability amplitude that a particle in state y at time $t_y$ ends up in a state x at time $t_x$. So the Green's function may be written:

$G^{+}(x,t_{x},y,t_{y}) = \theta(t_{x}-t_{y}) \langle x(t_{x})|y(t_{y})\rangle$There is nothing more to my problem. I simply want to start from $\phi(x,t_{x}) = \langle x|\hat{U}(t_{x})|\phi)\rangle$ and derive the expression for the propagator that they give in the book. And my method is to introduce a closure relation to make an integral appear so that it looks like the first equation. And finally I hope to write it in a way that makes possible the identification of the expression of the Green's function in the integral. But as you have seen, the Heaviside function did not appear in my calculation and that was my question. Did I need to introduce it somewhere or is my calculation correct but the Heaviside function is introduced manually because it is needed to make quantum field theory consistent? Which would mean the two representations are not really equivalent. I know they also say in the book that the propagator actually contains more information than the wave function but this is not explicitly stated in going from one equation to another.

 

[vanhees71]

Well, if the operator $\psi$ is a pure annihilation field, then $\psi^{\dagger}$ is a pure creation field. And nobody prevents you to define a new non-relativistic field $\phi=\psi+\psi^{\dagger}$.

I talk about standard QFT where the field operators obey standard canonical commutation relations and transform locally under Galilei or Poincare transformations. In non-relativistic QT this standard field operator is a pure annihilation operator, and that's why
$$\mathrm{i} G(t,x)=\langle \Omega|T \psi(t,x) \psi^{\dagger}(0,0)|\Omega \rangle$$
is in fact the retarded propgator in non-relaivistic QFT.

 

[Demystifier]

But as you have seen, the Heaviside function did not appear in my calculation and that was my question. Did I need to introduce it somewhere or is my calculation correct but the Heaviside function is introduced manually because it is needed to make quantum field theory consistent?

Mathematically, it is not necessary to introduce the Heaviside function. It is there for physical reasons. Some equations in physics have more than one solution, but some of those solutions do not describe the world that we see.