# Where is the Feynman Green's function in quantum mechanics?

In quantum field theory, the Feynman/time ordered Green's function takes the form $$D_F(p) \sim \frac{1}{p^2 - m^2 + i \epsilon}$$ and the $i \epsilon$ reflects the fact that the Green's function is not unique, but must be fixed by boundary conditions. Different choices of boundary conditions yield different placements of the $i \epsilon$, such as $$\frac{1}{(p_0-i\epsilon + \mathbf{p})^2 - m^2}, \quad \frac{1}{(p_0+i\epsilon + \mathbf{p})^2 - m^2}, \quad \frac{1}{p^2 - m^2 - i \epsilon}$$ which correspond to retarded, advanced, and "anti-time ordered". The placement of the $i \epsilon$ determines things like the allowed direction of Wick rotation.

Quantum mechanics is simply a quantum field theory in zero spatial dimensions, so everything here should also apply to quantum mechanics. But the correspondence is a bit obscure. At the undergraduate level, Green's functions just don't come up in quantum mechanics, because we usually talk about the propagator $$K(x_f, t_f, x_i, t_i) = \langle x_f | U(t_f, t_i) | x_i \rangle$$ instead (see here, here for propagators vs. Green's functions, which are often confused). At this level it's not as useful to talk about a Green's function, because we don't add source terms to the Schrodinger equation like we would for fields.

In a second course on quantum mechanics, we learn about scattering theory, where Green's functions are useful. But some of them are missing. The four choices in quantum field theory come from the fact that there are both positive and negative frequency solutions to relativistic wave equations like the Klein-Gordan equation, but in quantum mechanics this doesn't hold: for a positive Hamiltonian the Schrodinger equation has only positive frequency/energy solutions. So it looks like there are only two Green's functions, the retarded and advanced one, which are indeed the two commonly encountered (see here). These correspond to adding an infinitesimal damping forward and backward in time.

Are there four independent Green's functions in quantum mechanics or just two? What happened to the Feynman Green's function? Can one define it, and if so what is it good for?

Edit: see here for a very closely related question.

• The part about Green's functions vs. propagators is confusing - you seem to think that there is a difference between the two in quantum mechanics but that we only talk about the propagator, when the answer to the very question you link as reference says that Green's function and propagator are the same in quantum mechanics. Could you be more explicit about your notion of "Green's function" here? Feb 4, 2018 at 11:15
• @ACuriousMind I should've been more clear since there are some differing conventions. By propagator, I mean the solution to the homogeneous equation. By Green's function, I mean the inhomogeneous response to a delta function driving. Feb 4, 2018 at 11:29
• These are not the same. For instance, the propagator is always nonzero for positive and negative time differences. The Green's function, on the other hand, is not even defined without boundary conditions; for instance it can be either zero for negative time differences (retarded) or zero for positive time differences (advanced) or neither. Duhamel's principle says that the propagator times $\Theta(t)$ is the retarded Green's function, but the propagator by itself is not equal to any Green's funtion. Feb 4, 2018 at 11:30
• You're missing a $t_i$ on the right hand side of the third expression. Feb 4, 2018 at 13:49
• I disagree with your last comment. In this case, even Wikipedia got it sort of right.
– user178876
Feb 4, 2018 at 16:03

Where is the pole prescription in quantum mechanics?

The pole prescription comes about when you solve time-dependent wave equations via Fourier transforms. In quantum mechanics that would take the time-dependent Schrödinger equation to the time-independent Schrödinger equation. However the corresponding time-dependent Green's functions do not converge when you Fourier transform them, which is how the $i\epsilon$ prescription comes about. For more detail see my answer to this question, which was also linked by the OP as related.

Are there four independent Green's functions in quantum mechanics or just two?

In quantum mechanics, a time-dependent Green function is a solution of

$$(i \frac{\partial}{\partial t} - H ) G(t) = \mathbb{I}\delta(t) .$$

The solution is not unique, therefore there are as many Green's functions as boundary conditions (BCs) you can think up. A particularly useful one is the BC

$$G^+(t) = 0 \quad \textrm{for } t<0,$$

since it gives a solution to the initial value problem via

$$\psi(t) = iG^+(t-t_0)\psi(t_0) \quad \textrm{for } t>t_0.$$

Note that this is exactly the same in relativistic quantum mechanics/quantum field theory, only that the corresponding wave equation differs and you have to Fourier transform a few more dimensions. E.g. for the Klein-Gordon equation, which was used as an example in the question, we have

$$(\partial^2 + m^2) \psi(t) = 0,$$

such that

$$(\partial^2 + m^2) G(t) = \mathbb{I}\delta(t).$$

Imposing the same boundary condition as above and manipulating a little bit gives the Green function in Fourier space as $$G(p) \propto \frac{1}{p^2 - m^2 + i\epsilon}.$$

Note that this is unique now (details), the sign of $i\epsilon$ has implicitly been chosen by the boundary condition of the time-dependent Green function. I can never remember if that is called the advanced/retarded/Feynman Green's function and I think the terms also differ in the literature (e.g. in scattering theory you would call this the retarded Green function, but in QFT it seems to be the Feynman Green function). Either way it is the one that is useful for solving initial value problems.

Is the quantum mechanical path integral secretly related to the Feynman Green's function, or is it indifferent?

• Hi, I enjoyed this answer but when I revisited this question I realized I was very confused and mixed up two separate things! (i.e. the purpose of the $i \epsilon$, which you explain here, and the specific use of it in QM to get the Feynman Green's function) I edited my question to be more correct, but in the process it made your answer less relevant. But still, +1 for the explanation. Apr 15, 2018 at 17:44
I'm not sure if this was your question, but there s an easy interpretation of the Feynman propagator and the $$i \epsilon$$ prescription in quantum mechanics. It just comes from time ordering the two point function. However here, the two "points" are just points in time, not space time, because we are doing QM and not QFT.
Consider the 1D quantum harmonic oscillator. The ground state is $$| 0 \rangle$$. The time ordered two-point function is just $$G(t, t') = \langle 0 | \mathbb{T} \{ \hat q(t) \hat q(t') \} | 0 \rangle.$$ There's a lot of ways we could evaluate $$G(t, 0)$$. We could, for example, unleash the full wrath of the path integral here. We would then introduce an $$i \epsilon$$ and evaluate the path integral from $$T = -\infty$$ to $$T = + \infty$$ in order to get a vacuum to vacuum correlation function. However, that's completely overkill here. We can just use the algebra of creation and annihilation operators. The only wrinkles are that 1. we have to time evolve by $$|t - t'|$$ because we want a time ordered product and 2. we will implicitly subtract off the ground state energy $$\hbar \omega / 2$$.
\begin{align*} G(t, t') &= \langle{0}| \hat q e^{-i \hat H |t - t'| / \hbar}\hat q|{0}\rangle \\ &= \frac{\hbar}{2 m \omega }\langle{0}| (\hat a + \hat a^\dagger) e^{-i \hat H |t - t'| / \hbar} (\hat a + \hat a^\dagger) |0 \rangle \\ &= \frac{\hbar}{2 m \omega }\langle{0}| \hat a e^{-i \hat H |t - t'| / \hbar} \hat a^\dagger |{0}\rangle \\ &= \frac{\hbar}{2 m \omega }\langle{1}| e^{-i \hat H |t - t'| / \hbar} |{1}\rangle \\ &= \frac{\hbar}{2 m \omega} e^{-i \omega |t - t'|} \end{align*} This is the Feynman green function for the quantum harmonic oscillator. Note that at no step did we fool around with an $$i \epsilon$$. However, its hidden in the absolute-value bars. We could also write it using the Heavyside step function, which some people like to do.