I apologize if this seems like a simple question, but I have been agonizing over it recently. In nonrelativistic quantum mechanics, a plane wave of the form $e^{\pm i\vec p\cdot \vec x}$ is called outgoing (incoming for the minus sign). However, in QFT (particularly in the context of the mode expansion) we have terms of the form $$\hat a(p)e^{-ip\cdot x}+\hat a^\dagger(p)e^{ip\cdot x} $$ where $p$ and $x$ are now the usual momentum and position 4-vectors. Books such as Lancaster and Blundell's Quantum Field Theory for the Gifted Amateur say that the plane wave $e^{-ip\cdot x}$ is an incoming wave, and $e^{ip\cdot x}$ is an outgoing wave. However, using just a simple definition of the 4-vector inner product, the former becomes $e^{i(\vec p\cdot \vec x -Et)}$, which would seem to be outgoing, not incoming! Doing the same thing for the wave $e^{ip\cdot x}$ I conclude that it must be incoming, not outgoing. So my question is, in quantum field theory how do we define incoming and outgoing waves? Am I wrong about the nonrelativistic definition?

Edit: I used the mostly minus metric signature in this question.

  • $\begingroup$ Your understanding of what is called incoming/outgoing in basic QM is incorrect - see, e.g., their definition in this answer $\endgroup$
    – Roger V.
    Jul 6, 2022 at 12:40
  • $\begingroup$ @RogerVadim In 3d scattering is it not true that a wave of the form $e^{ik\cdot r}$ ($e^{-ik\cdot r}$) is outgoing (incoming)? $\endgroup$ Jul 6, 2022 at 14:43
  • $\begingroup$ You probably mean spherically symmetric rather than any 3d case? However in the OP you are referring to a plane wave $\endgroup$
    – Roger V.
    Jul 6, 2022 at 15:41
  • $\begingroup$ I think you are probably right. In the general case how do we define incoming and outgoing if not by looking at the sign of the momenta? $\endgroup$ Jul 6, 2022 at 16:25
  • $\begingroup$ Actually, we are looking at the flux - whether it flows towards or away from the scattering target. And we mean here the flux of the free space solutions, which can be plane wave, spherical waves or something else, and often a mixture. $\endgroup$
    – Roger V.
    Jul 6, 2022 at 16:47

1 Answer 1


The notion of position space wavefunctions in Quantum Field Theory appears and reveals its usefulness in the context of the LSZ reduction formula. First I wish to emphasize that it has a different nature than in non-relativistic Quantum Mechanics. The reason is that there is no useful generalization of the position operators $X^i$ of non-relativistic QM to QFT. In the absence of those, we don't really have the position basis $|x\rangle$ and without it there is no sense in which we can construct an object $\psi(x)=\langle x|\psi\rangle$ which represents "the probability amplitude for a particle to be found at $x$".

Instead, what happens is that we have one inner product on the space of solutions to the linearized field equations which allows us to extract from in/out fields the creation and annihilation operators for one-particle states. Let me be precise with a scalar field. In/out fields are defined by $$\phi_{\rm in/out}(x)=\int \dfrac{d^3 p}{(2\pi)^3 2p^0}\left(a_{\rm in/out}(p)e^{ipx}+a_{\rm in/out}^\dagger(p)e^{-ipx}\right)\tag{1}$$ where I use $(-,+,+,+)$ signature. The in/out fields are related to the interacting fields $\phi(x)$ in the asymptotic regions of spacetime as $$\phi(x)\to \sqrt{Z}\phi_{\rm in/out}(x),\quad \text{as $t\to \pm \infty$}\tag{2}$$ where $Z$ is a quantity called wavefunction renormalization. Now a scattering amplitude can be evaluated as $$\langle \text{out}|\text{in}\rangle=\langle 0|a_{\rm out}(p_1')\cdots a_{\rm out}(p_n')a^\dagger_{\rm in}(p_1)\cdots a_{\rm in}^\dagger(p_m)|0\rangle \tag{3}.$$

The idea of LSZ reduction is to invert (1) to write $a_{\rm out}(p_i')$ and $a_{\rm in}^\dagger(p_i)$ in terms of $\phi_{\rm in/out}(x)$ and then use (2) to relate to the interacting field. One effectively relates an ${\cal S}$-matrix element to correlation functions in a field theory. To see this discussion in more detail (and in particular understand why that $\sqrt{Z}$ appears in (2)) I suggest the book by Itzykson and Zuber.

Now the relevant point to your question is: how do we invert (1)? The answer is that there is an inner product on the space of solutions to the KG equation, known as KG inner product, defined by $$(\phi,\psi)_{KG}=-i\int_{\Sigma}d\Sigma n^\mu \left(\phi^\ast \partial_\mu \psi-\psi\partial_\mu \phi^\ast\right)\tag{4}$$

whre $\Sigma$ is a so-called Cauchy slice (a generalization of the surfaces of constant Minkowski time). Using (4) one may show that

\begin{eqnarray} (e^{ipx},e^{iqx})_{KG} &=&(2\pi)^3(2\omega_p)\delta^{(3)}(\vec{p}-\vec{q}),\tag{5.1}\\ (e^{-ipx},e^{-iqx})_{KG} &=&-(2\pi)^3(2\omega_p)\delta^{(3)}(\vec{p}-\vec{q}),\tag{5.2}\\ (e^{ipx},e^{-iqx})_{KG} &=&0\tag{5.3}, \end{eqnarray}

where we assume that $p^0>0$ and $q^0>0$. These equations are orthogonality relations which allow you to extract the creation and annihilation operators from the in/out fields. In particular observe that:

\begin{eqnarray} (e^{ipx},\phi_{\rm in/out})_{KG} &=& a_{\rm in/out}(p)\tag{6.1},\\ (e^{-ipx},\phi_{\rm in/out})_{KG} &=& -a_{\rm in/out}^\dagger(p)\tag{6.2}. \end{eqnarray}

Therefore to evaluate (3) we need $a_{\rm in}^\dagger(p)=-(e^{-ipx},\phi_{\rm in})_{KG}$ and $a_{\rm out}(p)=(e^{ipx},\phi_{\rm out})_{KG}$. One then eventually chooses the Cauchy slice $\Sigma$ at $t\to \pm \infty$ to use (2) to write these as integrals involving the interacting field $\phi(x)$ and converts these into integrals over the whole spacetime. In the end one finds the LSZ formula which is what everyone uses to evaluate scattering amplitudes in practice $$\langle \text{out}|\text{in}\rangle = \prod_{i=1}^{m}\prod_{j=1}^n\int d^4x_{i}d^4x'_j e^{-ip_i\cdot x_i}e^{ip_j'\cdot x_j}\langle 0|T\{\phi(x_1)\cdots \phi(x_m)\phi(x_1')\cdots \phi(x_n')\}|0\rangle_{A,C}\tag{7}$$

where the $A,C$ subscript means that we have the connected component of the correlator with external legs removed.

With this we can finally answer your question: in QFT position space wavefunctions appear as the solutions to the linearized field equations upon which we project the in/out quantum fields to extract creation and annihilation operators of asymptotic particle states. When we use these to evaluate a scattering amplitude, to get incoming particles we must project on $e^{-ipx}$ and to get outgoing particles we must project on $e^{ipx}$. The final result is the LSZ formula (7) in which the amplitude is obtained by integrating the connected amputated time-ordered correlator against the external wavefunctions. In particular the difference in sign of the exponentials dictates which are the incoming ones and which are the outgoing ones for the reasons that have been explained.

  • $\begingroup$ Thanks for the answer. So (in the mostly minus signature) in order to get incoming particles we project onto $e^{ip\cdot x}$ and to get outgoing particles we project onto $e^{-ip\cdot x}$? And we get such conventions for incoming/outgoing waves because of the way we do the mode expansion and from how we derive the LSZ reduction formula? $\endgroup$ Jul 6, 2022 at 19:13
  • 1
    $\begingroup$ Yes, that's right. I shall update the answer later today to the mostly minus signature. But the whole point is that to get creation or annihilation operators you project on plane waves with different signs, and on the scattering amplitude outgoing particles come from annihilation operators and incoming one from creation operators. This in turn associates outgoing and incoming particles with plane waves with different signs. $\endgroup$
    – Gold
    Jul 6, 2022 at 19:40
  • $\begingroup$ Great! Thanks for helping me to understand $\endgroup$ Jul 6, 2022 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.