# QFT Dyson series: why are we solving the Schrodinger equation?

In quantum field theory, the solution of the time evolution operator of the Schrodinger equation (in the interaction picture) is given by Dyson's series, which is used to calculate the S-matrix. Why we are still working with the non-relativistic Schrodinger equation (instead of the Klein-Gordon equation)?

Source: David Tong's notes; page 50. (http://www.damtp.cam.ac.uk/user/dt281/qft.html)

• Do the notes say $\hat H$ or $-\frac{\hbar}{2m}\nabla^2+V$? Dec 10, 2013 at 16:06
• @NickKidman They say $\hat{H}$. I've edited to original post with the link. Dec 10, 2013 at 16:07

"Schroedinger equation" unfortunately is a bit ambiguous word. It could refer to $$i\hbar\frac{d \psi(t)}{dt} = H_t\psi(t) \tag{1}$$ but also to a more precise form like this: $$i\hbar\frac{d \psi(t)}{dt} = \left(-\frac{1}{2m}{\bf P}^2 + V_t\right)\psi(t) \:.$$ The former version does not depend on the quantum physical system you are dealing with. So, in particular, it holds for relativistic quantum systems, provided you insert the appropriate Hamiltonian operator therein. If you are studying the quantized Klein Gordon field, there is a corresponding Hilbert space and a corresponding Hamiltonian operator for the quantum states of that system.

In a more abstract version (1) can be used to define the evolution operator of the theory. Assuming $$\psi_t = U(t,t_0) \psi_{t_0} \tag{2}$$ where now there are no restrictions on the used vectors, since the domain of $U(t.t_0)$ is the whole Hilbert space, differently form the domain of $H_t$ which is only a dense subspace in general. Indeed, inserting (2) in (1) and exploiting the fact that $\psi_{t_0}$ is arbitrary, we obtain the equation for $U(t,t_0)$: $$-i\hbar\frac{d}{dt}U(t,t_0) = H_t U(t,t_0) \tag{3}$$ that at least formally, gives rise to the Heisenberg picture version of Dayson's series as the solution: $$U(t,t_0) = T\left[ e^{-\frac{i}{\hbar} \int_{t_0}^t H_\tau d\tau}\right]\:,$$ (in QFT is more appropriate to deal with the so called interaction picture so that the in the analogue of (3) uses the only interaction part of the Hamiltonian operators appears in place of $H_t$).

In QFT (at least referring to the non-interacting case) you have to make a distinction between two levels: one-particle situation and arbitrarily many particles case associated with the quantum field.

Let us focus on the former case. There is the Hilbert space ${\cal H}$ of a single particle, for instance that of a scalar boson field $\phi$ verifying Klein-Gordon equation. The Hilbert space is again $L^2(R^3, d^3x)$ position representation or, equivalently, $L^2(R^3, d^3p)$ momentum representation, and there are some other natural representations, for instance the momentum representation but using a covariant measure $dp^3/p_0$. The relation between momentum and position representation is more complicated (one should use the so called Newton-Wigner representation of position observable) and usually is not discussed in textbooks.

Dealing with Schroedinger picture in a fixed inertial reference frame, time evolution for a single particle is always described by the abstract Schroedinger equation (1), where now (I henceforth assume $c= \hbar=1$ for the sake of simplicity): $$K := P_0 =\sqrt{{\bf P}^2 + m^2I}\:,$$ where $m$ is the mass of the particle and ${\bf P}$ the 3-momentum operator.

A quantized quantum field, however, involves states including an arbitrary number of particles. So one has to pass from the one-particle Hilbert space to the whole Hilbert space of the (free) theory which is called (symmetrized in the boson case) Fock space:

$${\cal F}_+({\cal H}) = C \oplus{\cal H} \oplus ({\cal H} \otimes {\cal H})_s \oplus ({\cal H} \otimes {\cal H} \otimes {\cal H})_s \oplus ({\cal H} \otimes {\cal H}\otimes{\cal H} \otimes {\cal H})_s \oplus \cdots$$

Above $_s$ indicates that one has only to consider the completely symmetric subspace of ${\cal H}\otimes \cdots \otimes {\cal H}$, and $\oplus$ is the ortogonal direct sum of Hilbert spaces. For instance, the vectors in $({\cal H} \otimes {\cal H})_s$ are those describing a system of two identical particles associated with the quantum field. The Hamiltonian operator of the complete system of particles is therefore given by the sum of the Hamiltonian operators of each subsystem:

$$H = 0 \oplus K \oplus (K\otimes I + I \otimes K) \oplus \cdots$$

In this juncture, the quantum field $\hat{\phi}(t,x)$ is pictured in terms of certain operators working between (arbitrary) couples of subspaces ${\cal H}\otimes \cdots \otimes {\cal H}$, adding and removing particles. I will not enter into details because I should introduce too many notions. I just say that the (Hamiltonian) evolution equation of the quantum field operator constructed this way is, again, given by the Hamiltonian $H$ introduced above:

$$\frac{\partial}{\partial t} \hat{\phi}(t,x) = i[ H, \hat{\phi}(t,x)]\:.$$

That is nothing but the standard evolution equation in Heisenberg picture for any quantum observable. Just a final comment is worth. Writing that equation in terms of $\hat{\phi}$ and its conjugated momentum $\hat{\pi}$, and adding the analogous equation of motion for the latter, one obtains a system of equations that is completely equivalent to the Lagrangian equations of motion for the field, I mean Klein-Gordon equation.

• Wow nice answer; thank you. Could you tell me what book/texts you have used to get this information? Dec 11, 2013 at 12:53
• Well, I think every rigorous QFT book would be OK. Streater-Wightman's textbook for instance ("PCT, Spin and Statistics, and All That") Dec 11, 2013 at 13:40

The Schrödinger equation you use in non-relativistic quantum mechanics describes the evolution of the wave-function for a single particle, or at least, a fixed number. So you can think of it as being the "wave equation" for a one particle wave-function. Nice, neat interpretation. Also incomplete.

But the Schrödinger can also be viewed as the defining equation for time-evolution in Quantum mechanics, being something that you can derive from a path integral and a Lagrangian.

So in relativistic QFT (or any QFT, for that matter), our "canonical variables" are no longer the position of a particle, but the value of a field $\phi$, and these variables have an extra dangling label "$x^\mu$. So in principle you could define a wave-function (actually, a wave-functional) $\Psi[\phi]$ and a Hamiltonian operator that acts on this type of Hilbert space, and Schrödinger equation that gives it's time evolution.

It's a very curious conspiracy that the classical equations of motion for the Dirac field give us a fairly consistent "single particle quantum mechanics". It somehow inherits it's quantumness from QFT, while being fully "classical". I think it has to do with the presence of $\hbar$ in the Lagrangian, but I don't recall the details.

• More precisely, in quantum mechanics, we have a "wavefunction" $\Psi(x;t)$, while, in QFT, we have a "wavefunction" $\Psi(\phi; x,t)$. The semi-colon notation is here to insist on the separation of fields ($X$ in Quantum mechanics and $\phi$ in QFT), and parameters ($t$ in Quantum mechanics and $x,t$ in QFT) Dec 10, 2013 at 18:02
• +1, If you decide you want to look at amplitudes which eat up phase factors, then you're in quantum territory and the group theory admits spins. I think you mystify the Dirac equation a little - I couldn't decide beforehand if I'd expect a fairly consistent "single particle quantum mechanics" or not. But before one can settle it, it would be good if you consider it to be fully "classical" or if it has quantumness. Dec 10, 2013 at 18:10
• It is not necessary to use wavefunction (representation of state vector) here. It is the time-evolution of state in Schrodinger picture. Like Heisenberg picture used in Eq. (2.77) in Tong's note. Dec 10, 2013 at 19:14
• @Hunter : "But what exactly does the "wave function" for QFT represent?" The same thing that it represents in QM. In QM, you cannot speak about a precise trajectory of the particle $x(t)$, you can only speak of a probability amplitude $\Psi(x,t)$. In QFT, you cannot speak about a precise field $\phi(x,t)$, you can only speak about a probability amplitude $\Psi(\phi;x,t)$, that you may write also, in some sense, $\Psi(\phi(x);t)$ Dec 10, 2013 at 19:24
• Perhaps it would be helpful for the OP to emphasize in the response the fact that as with any quantum system, the state of the system in quantum field theory is still described by a state in Hilbert space (not of wave functionals) and that the Shrodinger equation is the equation of time evolution for the Hilbert state of the system. I personally feel that talk of wave functionals and path integral quantization can be confusing when discussing the Shrodinger equation since it's most directly thought of as an equation governing time evolution in canonical quantization. Dec 11, 2013 at 1:57