Confusion Regarding Relativity in QFT

Question

I've been looking further into QFT after my QM and GR courses this past semester so I can better understand Carroll's section from Geometry and Spacetime, "QFT in Curved Spacetime". I'm fundamentally struggling to reconcile with how/why the Schrödinger equation is still valid in its formalism. The Schrodinger equation, the way I see it cannot be lorentz-invariant because it contains two space derivatives but just one time derivative:$$i\hbar \frac{d}{dt}|\Psi(t)\rangle = \hat{H}|\Psi(t)\rangle$$ $$H=\int d^3x \text{ }\frac{1}{2}\pi^2+\frac{1}{2}(\nabla\phi)^2+V(\phi) = \int d^3x\text{ }T^{00}$$

Where $T^{00}$ is the 00 component of the stress energy tensor constructed from the lagrangian $\mathcal{L}=\frac{1}{2}\dot{\phi}^2-\frac{1}{2}(\nabla\phi)^2-V(\phi)$ and $\pi=\frac{\partial\mathcal L}{\partial \dot\phi}$

After integrating the $\nabla \phi$ term by parts we get the familiar notion of a hamiltonian with two spatial derivatives as in non-relativistic QM (neglecting the boundary term),$$H=\int d^3x \text{ }\frac{1}{2}\pi^2-\frac{1}{2}\phi(\nabla^2\phi)+V(\phi)$$

I really don't understand how to reconcile this. I've heard explanations that the Schrödinger equation describes evolution in the hilbert space not "real" space - but still does this not assume some sort of global ticking clock on the hilbert space? Another explanation I've heard is that the field operators are lorentz-invariant since they obey klein-gordon/dirac equations, but again why is this enough? Perhaps this is a matter of just misunderstanding the formalism but it seems strange that the equation supposedly modeling the "time evolution" of a quantum state in QFT isn't even in terms of the state/particle's proper time.

Any thoughts or clarifications are much appreciated.

P.S. if you have any QFT resource recommendations for an ambitious undergrad I would also be very appreciative.

Answer 1

The answer is that by using Hamiltonian formalism, you have explicitly pick out time, and you are not treating space and time on equal footing. And that's why by simply looking at Hamiltonian it is not straightforward to tell whether this system has Lorentz symmetry, since you have explicitly treat time specially in your mathematical formalism (notice: nothing change in physics, it is just a change of mathematical tool).

To see why Hamiltonian treat time specially, remember that we need canonical momenta of a degree of freedom to construct hamiltonian:

$$H=\pi^i \dot{\phi}^i-L$$

where $\phi^i$ is the degree of the freedom and the $\pi^i$ is the corresponding canonoical momenta, given by $$\pi^i=\frac{\partial L}{\partial \dot{\phi}^i}$$

You can see that both $\phi$ and $\pi$ are dependent on time, meaning that you have explicitly pick out time in Hamiltonian formalism, therefore, the space part of the Hamiltonian and the time part of the Hamiltonian will look differently. As a result ,taking the derivative of time on LHS does not violate Lorentz symmetry, since Hamiltonian formalism intrinsically treat time differently.

On the contrary, Lagrangian formalism treat space and time on equal footing, and that's why in modern science, we usually prefer to use Lagrangian formalism rather than Hamiltonian.

Regarding recommendation, see this post. Personally I think David Tong's lecture note is the best introductory lecture note you can find on QFT, which can be accessed here.

Answer 2

Oh, you are falling prey to yet another one of those ambiguous terms in physics.

Some horrible people decided to call $$\tag1\mathrm i\hslash\frac{\partial\ }{\partial t}\psi=\hat H\psi$$ as the Schrödinger equation. This is horrible because the Dirac equation would then also be a Schrödinger equation because it looks like this too, just that its $\psi$ is a 4-component Dirac-spinor and its Hamiltonian operator $\hat H$ is a complicated thing with the Dirac alpha and beta matrices, or equivalently, Dirac gamma matrices.

In fact, by introducing an auxiliary variable, you could cast the Klein-Gordon equation to look like this too. It is usually called the Dirac form of the Klein-Gordon equations.

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There is nobody claiming that the Schrödinger equation that looks like $$\tag2\mathrm i\hslash\frac{\partial\ }{\partial t}\psi=-\frac{\hslash^2}{2m}\vec\nabla\cdot\vec\nabla\psi+\hat V\psi$$ can ever be compatible with the Special Theory of Relativity. It is incompatible, for the reason that you correctly pointed out, the single time derivative v.s. double space derivatives.

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It is very good that you have learnt QM and GR. However, QFT is such a tremendous beast that you should totally give up any notion of curved spacetimes and instead just try to learn it in its simplest form. Indeed, do not resist gauge-fixing onto Coulomb gauge, because while gauge-invariance and so on are important to QFT, the topic is so difficult, that one really wants to have the simplicity of only having transverse waves. Merely counting degrees of freedom is already a challenging task in QFT.

Answer 3

The Schrodinger equation, the way I see it cannot be lorentz-invariant because it contains two space derivatives but just one time derivative:$$i\hbar \frac{d}{dt}|\Psi(t)\rangle = \hat{H}|\Psi(t)\rangle$$

The equation you have written above contains one time derivative and zero space derivatives.

If you choose to consider a non-relativistic Hamiltonian $$ \hat H = \frac{1}{2m}\hat{p}^2 + V(\hat x)\;,\tag{1} $$ then the theory is not Lorentz invariant... but that was your choice.

If you write the Schrödinger equation in the position basis, where $\hat p \to -i\hbar\frac{d}{dx}$, and you choose to use Eq. (1) as your Hamiltonian, then the Schrödinger equation looks like: $$ i\hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x,t)}{\partial x^2}+V(x)\Psi(x,t)\;. \tag{2} $$ Eq. (2) above does contain two spatial derivatives and one time derivative. But, again, this was seemingly your choice to work with a non-relativistic Hamiltonian.