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I’m studying QFT and I learnt that we didn’t change the Schrödinger equation into relativistic one but Klein-Gordon and Dirac equation are now the equation of motion of our fields operator. But our states (or wave function) are evolving with Schrödinger equation yet, I cannot understand how can we deal with this. For instance, in the $S$-matrix formalism our asymptotic states evolve with Schrödinger equation but we want to describe a relativistic process. So my doubt is in which way do we have to interpret the Schrödinger equation when we’re doing QFT.

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  • $\begingroup$ I am quite certain that you have misunderstood the terminology, and it is not that you are using Schrödinger's equation in QFT. After all, if you say that you are using Klein-Gordon and Dirac equations of motion for your field operators, that means you are not using Schrödinger's equation, even though you may hear it being called so. $\endgroup$ Commented May 30, 2023 at 9:43
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    $\begingroup$ If you really want a Schrödinger equation, you can always write one for the fields. It's a bit cumbersome since it is now a "wavefunctional" (takes a field and outputs a complex number), but this is doable for fields with quadratic Lagrangians (non interacting). $\endgroup$
    – LPZ
    Commented May 30, 2023 at 10:01
  • $\begingroup$ No no my question is what’s the meaning of $S = T \exp( -i \int_\inf^\inf dt H)$, is it relativistic or not in QFT. Because I thought it was just the formal solution of Shröedinger equation and hence non-relativistic. $\endgroup$ Commented May 30, 2023 at 10:12
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2 Answers 2

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The Schrödinger equation is just $$ H\psi = \mathrm{i}\partial_t \psi$$ and nothing about this is "non-relativistic" or "relativistic" - it's just that relativistically, the $t$ there is a coordinate time and hence once you use a Hamiltonian you are not writing stuff down in a manifestly Lorentz covariant way, but you've instead chosen and fixed some particular coordinate system. This is somewhat "ugly" from an abstract viewpoint where we'd like to keep everything Lorentz covariant, but it's not somehow wrong or forbidden.

The Schrödinger equation is just the statement that the Hamiltonian is the generator of time translation in the same sense that momentum is the generator of spatial translations, see also this answer of mine for more on just viewing the Schrödinger equation as a special case of the relation between transformations and generators formulated by Stone's theorem.

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  • $\begingroup$ Ok thank you, you centered the point. Now I have another question, when I’m doing QFT and I have my field with their eq. of motion, where do we have the “relativity”? Is it right to think, classically, a field (e.g. a KG field) as a wave with constant velocity equals to c, as in Maxwell theory.? $\endgroup$ Commented May 30, 2023 at 12:51
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    $\begingroup$ @michaelpasqui What do you mean "where"? We're doing relativity by saying that the Lorentz transformations are our symmetries, and the Klein-Gordon equation is a relativistic equation of motion. That we might temporarily break manifest Lorentz covariance during a derivation does not mean we have given up the idea of Lorentz covariance entirely - the end results will not contain some specific choice for time and Hamiltonian! $\endgroup$
    – ACuriousMind
    Commented May 30, 2023 at 12:54
  • $\begingroup$ I mean that at the end of the story this theory has to describe a relativistic particles in “motion” and that motion has to be relativistic. Is the passage from saying that the dynamics of the field is relativistic to saying that the motion of the particles is it that confuse me. But maybe my problem is in the definition of particle in quantum field theory. $\endgroup$ Commented May 30, 2023 at 13:02
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    $\begingroup$ @michaelpasqui This is a quantum theory - there is no "motion" in the classical sense because there are no definite positions (even less than there are in ordinary non-relativistic QM). If you are not satisfied with "relativistic QFT is relativistic by having Lorentz symmetry and using the Minkowksi metric", you need to ask a more specific question where you detail what you expect a "relativistic theory" to be and why. $\endgroup$
    – ACuriousMind
    Commented May 30, 2023 at 13:25
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  • The Schrödinger equation

The Schrödinger equation is the evolution equation for states in a quantum theory : $$i\hbar \frac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle$$ Whether this describes a relativistic system or not has nothing to do with the Schrödinger equation, and everything to do with the Hamiltonian. More specifically, whether there is a faithful unitary representation of the Poincaré group on states whose generators commute with the Hamiltonian.

  • A non relativistic quantum system

The Hamiltonian for a free non-relativistic particle is : $$H = \frac{\hat p^2}{2m}$$ It acts on the Hilbert space $\mathcal H = L^2(\mathbb R)$, which contains wave functions $\psi$, with $\hat p = i\hbar \frac{\partial}{\partial x}$. This is indeed a non-relativistic system.

  • A relativistic quantum system

For a massive relativistic particle, there is a field representation, which is the easiest to work with in the Lagrangian formalism (with lagrangian density $\mathcal L = \frac12 \partial_\mu \varphi\partial^\mu\varphi + \frac 12 m^2 \varphi^2$). There is also a Fock space representation, which I use below.

In this case, the Hilbert space in the (bosonic) Fock space : $$\mathcal F = \bigoplus_{n= 0}^{+\infty} S^n\mathfrak h$$ on the $1$-particle Hilbert space $\mathfrak h = L^2(\mathbb R^d)$. The Hamiltonian is : $$H = \int \text d^d \vec p\sqrt{\vec p^2 + m^2}{a_\vec p}^\dagger{a_\vec p}$$ where the creation/annihilation operators are normalized such that $[a_\vec p, {a_\vec q}^\dagger] = \delta^{(d)}(\vec p-\vec q)$.

States evolve according to the Schrödinger equation ! For example, the $1$-particle state $|\psi(t=0)\rangle = |\vec p\rangle = {a_\vec p}^\dagger|0\rangle$ evolves according to : $$|\psi(t=0)\rangle = e^{-iHt} |\psi(0)\rangle = e^{-i\sqrt{p^2 +m^2}}|\psi(0)\rangle$$

To check that this is indeed a relativistic system, we need to find hermitian generators for the Poincaré group $P^\mu, M^{\mu\nu}$ (with $P^0 = H$) and make sure that they satisfy the Poincaré algebra.

To do this, for notational convenience, write $p^0 = \sqrt{\vec p^2 +m^2}$ (making each $d$-vector $\vec p$ into a $d+1$ vector $p$) and $\frac{\partial}{\partial p_0} = 0$. Then set : \begin{align} P_\mu &= \int \text{d}^d \vec{p} {a_\vec p}^\dagger p^\mu a_{\vec{p}} \\ M_{\mu\nu} &= i\int\text d^d \vec p {a_\vec p}^\dagger \mathcal M_{\mu\nu}(p) a_\vec p \end{align}

where $\mathcal M_{\mu\nu}(p) = i\left(p_\mu \frac{\partial}{\partial p^\nu} - p_\nu \frac{\partial}{\partial p^\mu}\right)$. These are indeed Hermitian operators and we can compute their commutators. We can check that : \begin{align} [\mathcal M_{\mu\nu}(p),p_\sigma] &= i(p_{\mu}\eta_{\mu\sigma}- p_\nu \eta_{\mu\sigma}) \\ [\mathcal M_{\mu\nu}(p),\mathcal M_{\alpha\beta}(p)] &= i\left(\eta_{\nu\alpha}\mathcal M_{\mu\beta}(p) + \eta_{\mu\beta}\mathcal M_{\nu\alpha}(p) - \eta_{\mu\alpha}\mathcal M_{\nu\beta}(p) - \eta_{\mu\beta}\mathcal M_{\mu\alpha}(p)\right) \end{align}

With this in hand, we compute the algebra : \begin{align} [M_{\mu\nu},P_\sigma] &=\int\text d^d \vec p\int\text d^d \vec q \left[ {a_\vec p}^\dagger \mathcal M_{\mu\nu}(p) a_\vec p, {a_\vec q}^\dagger q_\sigma a_\vec q\right] \\ &= \int\text d^d \vec p\int\text d^d \vec q\left({a_\vec p}^\dagger \mathcal M_{\mu\nu}(p) \delta^{(d)}(\vec p - \vec q)q_\sigma a_\vec q - {a_\vec q}^\dagger q_\sigma \delta^{(d)}(\vec p-\vec q) \mathcal M_{\mu\nu}(p)a_\vec p \right)\\ &= \int\text d^d p {a_\vec p}^\dagger [\mathcal M_{\mu\nu}(p),p_\sigma]a_p\\ &= \int\text d^d p {a_\vec p}^\dagger i(p_\mu\eta_{\nu\sigma} -p_\nu\eta_{\mu\sigma})a_p \\ &= i(P_\mu\eta_{\nu\sigma} -P_\nu\eta_{\mu\sigma}) \end{align} and likewise we would find : $$[M_{\mu\nu}, M_{\alpha\beta}] = i\left(\eta_{\nu \alpha} M_{\mu\beta} + \eta_{\mu\beta} M_{\nu \alpha} - \eta_{\nu \beta} M_{\mu\alpha} - \eta_{\mu\alpha} M_{\nu\beta}\right)$$

To recap, we have a relativistic quantum system whose states obey evolve in time according to the Schrödinger equation. If you were to define field (distributional) operators and switch to the Heisenberg picture, you would find that those quantum field obey the Klein-Gordon equation.

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