Why is the Hamiltonian in QFT the generator of time evolution?

Question

In non-relativistic Quantum Mechanics one can derive that the time translation operator that acts on quantum states is given (in natural units) by \begin{equation} e^{-iHt}, \end{equation} where $H$ is the Hamiltonian operator. This shows that the Hamiltonian is indeed the generator of time translations.

In Quantum Field Theory (QFT), the Hamiltonian also seems to be the generator of time translations. (I had a lecture about it this week.) Time evolution in the Schrödinger picture is now given by \begin{equation} \psi(\vec{x},0)|0\rangle\to e^{-iHt}\psi(\vec{x},0)|0\rangle \tag{1} \end{equation} for a free scalar field $\psi$, say. Or, in the Heisenberg picture, we can display the time evolution by \begin{equation} \psi(\vec{x},t) = e^{-iHt}\psi(\vec{x},0)e^{iHt} \tag{2} \end{equation} which is actually more general then (1). My question is as follows.

How does one derive that time evolution in QFT is given by (1) or (2)? I know that $H$ is the conserved charge corresponding to time translation, so an answer might begin from this fact. But if the answer states that the conserved charge of a symmetry is always the generator of the symmetry, I would appreciate a proof/derivation of that.

Answer 1

Your picture isn't quite right. In QFT what was the wave function gets promoted to observable operator, and $\mathbf{x}$ gets demoted to parameter on the same level as $t$. The time evolution of an operator is not given by $\operatorname{e}^{-iHt}$, that's the evolution of a state vector. Operators evolve, in the Heisenberg picture, according to: $$\psi(t,\mathbf{x}) = \operatorname{e}^{-iHt} \psi(0,\mathbf{x}) \operatorname{e}^{iHt}.$$ You can go further than that, though, and add in the space translation generators to get: $$\psi(t,\mathbf{x}) = \operatorname{e}^{-i P_\mu x^\mu} \psi(0) \operatorname{e}^{i P_\mu x^\mu},$$ in the $(+,-,-,-)$ signature metric.

What's going on here is that most treatments of QFT elid over the state vector necessary for a Schrodinger treatment. That state vector still obeys a Schrodinger type equation, it just has to be cast in terms of functional analysis instead of ordinary calculus.

As an example, the free real scalar field has Lagrangian density: $$ \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{m^2}{2} \phi^2.$$ The momentum canonically conjugate to $\phi$ is $\pi \equiv \frac{\partial\mathcal{L}}{\partial \dot{\phi}} = \dot{\phi}$. This produces a Hamiltonian in the usual way: $$H = \int \operatorname{d}^3 x \left[\frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{m^2}{2} \phi^2\right].$$ The fields are now promoted to operators that obey the equal time canonical commutation relations, $[\phi(\mathbf{x}), \pi(\mathbf{y})] = i \delta(\mathbf{x}-\mathbf{y})$. The state vector now has to assign a probability density, per unit function space volume ($[\mathcal{D} \phi]$), to every distinct configuration the field can take at any given time. This is known as the wave functional, denoted $\Psi[\phi]$. The canonical commutation relations imply that $\Psi$ obeys a Schrodinger type equation: $$\int \operatorname{d}^3 x \left[-\frac{1}{2} \frac{\delta^2 \Psi}{\delta \phi(\mathbf{x})^2} + \frac{1}{2}(\nabla\phi)^2 \Psi + \frac{m^2}{2} \phi^2 \Psi \right] = i \frac{\partial \Psi}{\partial t}.$$ This equation is, of course, just the simple harmonic oscillator for which we can construct raising and lower operators in the usual way (after changing to Fourier space). The ground state is given by the Gaussian functional: $$\Psi_0[\phi] \propto \exp\left(-\frac{1}{2}\int \operatorname{d}^3k \left[[\phi(k)]^2 \sqrt{\mathbf{k}^2 + m^2}\right]\right), $$ with excited states built using raising operators, $a^\dagger(\mathbf{k}) = \sqrt[4]{\frac{k^2 + m^2}{4}}\left[\phi(\mathbf{k}) - \frac{i}{\sqrt{k^2 + m^2}} \pi(\mathbf{k})\right]$, in the usual way.

I can only speculate that QFT isn't taught this way in most textbooks for two reasons. First, QFT is primarily used for calculating scattering amplitudes, and other formalisms are easier to get results from. Second, the infinities that plague QFT, requiring renormalization, could be even more difficult to manage in this formalism. This 1996 paper by Long and Shore is one example of professionals using this formalism.

Answer 2

The easy reason is because that's how it works for waves. The basis of quantum mechanics is to use the same framework used for waves for matter as well. Planck showed that light quanta have energies given by h$\nu$, so the evolution of waves is given by $e^{-\it iE/\hbar\space t}$. We bring that to the matter particles and voila.

Answer 3

Consider a system in state $|\psi\rangle$. Alice uses a set of basis states $\langle a|$. The state of the system has components $\langle a|\psi\rangle$ in Alice's frame of reference. Bob uses basis states $\langle a'|$ which are related to Alice's basis by a unitary coordinate transformation $\langle a'|=\langle a|\hat{U}$. In Bob's reference frame, the system $|\psi\rangle$ has components $\langle a'|\psi\rangle=\langle a|\hat{U}|\psi\rangle$. We can think of the system as being in the fixed state $|\psi\rangle$ and Alice uses "axes" $\langle a|$ and Bob uses "rotated axes" $\langle a'|=\langle a|\hat{U}$. This is the passive viewpoint. Alternatively, Bob can think of his components $\langle a'|\psi\rangle=\langle a|\hat{U}|\psi\rangle=\langle a|(\hat{U}|\psi\rangle)$ as being the result of the state changing from $|\psi\rangle$ to $|\psi'\rangle=\hat{U}|\psi\rangle$ with respect to the fixed basis $\langle a|$. This is the active viewpoint. Both points of view are equivalent.

Let's use the active viewpoint to see how an operator $\hat{O}$ transforms. Alice prepares a system in a state $|\psi\rangle$. Bob sees this system in state $|\psi'\rangle =\hat{U}|\psi\rangle$. Alice acts on the state with an operator $\hat{O}$ to produce $\hat{O}|\psi\rangle$. Bob sees the new state as $\hat{U}(\hat{O}|\psi\rangle)=\hat{U}\hat{O}\hat{U}^{-1}\hat{U}|\psi\rangle=(\hat{U}\hat{O}\hat{U}^{-1})\hat{U}|\psi\rangle=(\hat{U}\hat{O}\hat{U}^{-1})|\psi'\rangle$. In other words, Alice's operator $\hat{O}$ appears to Bob as the operator $\hat{O}'=\hat{U}\hat{O}\hat{U}^{-1}$.

A unitary coordinate transformation obeys $\hat{U}^{\dagger}=\hat{U}^{-1}$. This implies that an infinitesimal unitary coordinate transformation can be written $\hat{U}=1-i\epsilon \hat{G}$ where $\epsilon$ is an infinitesimal number and $\hat{G}$ is Hermitian (proof: $\hat{U}^{\dagger}=(1-i\epsilon\hat{G})^{\dagger}=1+i\epsilon\hat{G}=\hat(U)^{-1}$). The sign of $\epsilon$ is one's own convention. A finite unitary transformation is made by stacking up $N$ small transformations. $\hat{U}=(1-i\epsilon\hat{G})^{N}=e^{-iN\epsilon\hat{G}}=e^{-i\tau\hat{G}}$ where the finite parameter is $\tau=N\epsilon$. A state now transforms (actively) as $|\psi'\rangle=e^{-i\tau\hat{G}}|\psi\rangle$ and an operator transforms (actively) as $\hat{O}'=e^{-i\tau\hat{G}}\hat{O}e^{i\tau\hat{G}}$. In quantum mechanics, the unitary operators correspond to canonical transformations in classical mechanics. The Hermitian operators $\hat{G}$ correspond to the generators of canonical transformations in classical mechanics. In classical mechanics, the Hamiltonian function $H$ is the generator of time translations, so the unitary coordinate transformation corresponding to time translation is $\hat{U}=e^{-it\hat{H}}$.