Sorry if this is a naive question, but I am new to QFT. In the treatment of scattering in section 3.1 of The quantum theory of fields, vol.1, Weinberg first presented the general transformation rule for several non-interacting particles under the action of a Poincare group element $(\Lambda, a)$ (p.108, eq.(3.1.1)):
$$
U(\Lambda, a)\Psi_{p_1,~\sigma_1,~n_1;~p_2,~\sigma_2,~n_2;~\cdots}
=\exp\left( -ia_\mu((\Lambda p_1)^\mu + (\Lambda p_2)^\mu+\cdots) \right)
\times\sqrt{\displaystyle{\frac{(\Lambda p_1)^0(\Lambda p_2)^0\cdots}{p_1^0p_2^0\cdots}}}\sum_{\sigma'_1~\sigma'_2~\cdots}
D_{\sigma'_1~\sigma_1}^{(j_1)}\left( W(\Lambda,p_1) \right)
D_{\sigma'_2~\sigma_2}^{(j_2)}\left( W(\Lambda,p_2) \right)
\cdots \times \Psi_{\Lambda p_1,~\sigma'_1,~n_1;~\Lambda p_2,~\sigma'_2,~n_2;~\cdots}.
$$
Here $\Lambda$ is an arbitrary homogeneous Lorentz transformation and
$a$ is a space-time translation applied after $\Lambda$. The labels
$p_1,~\sigma_1,~n_1;~p_2,~\sigma_2,~n_2;~\cdots$ represent the states
of particles, with the first particle having momentum $p_1$, spin $\sigma_1$, charge $n_1$ and so on. the $D$'s are Wigner rotation matrices not directly relevant to the present question.
I am pretty confident in understanding this equation since it follows directly from the previous chapter. However, I get confused when Weinberg
says that $U(\Lambda, a) = \exp(iH\tau)$ if we set
$\Lambda^{\mu}_{~~\nu} = \delta^{~\mu}_{~~~\nu}$ and $a^{\mu}\sim(0,0,0,\tau)$ (the fourth component being time).
As I understand it, $H$ in this chapter no longer denotes the free-particle
Hamiltonian as in Chapter 2, but the 'total' Hamiltonian with interaction included. This is most evidently seen from his equation (3.1.8). Therefore,
the claim $U(\Lambda, a) = \exp(iH\tau)$ is simply a statement that Hamiltonian generates time evolution, which follows from, well, the Schroedinger equation. (The lack of minus sign in the exponential being
due to the 'passive' view we are taking.) But I really doubt whether this is the correct understanding since Weinberg
have made no explicit mention of the Schroedinger equation, or time evolution of any kind up to this point in the book.
What made me even more confused is his statement in the middle paragraph on page 109:
To maintain manifest Lorentz invariance, in the formalism we are using here, state-vectors do not change with time — a state vector $\Psi$ describes the whole spacetime history of a system of particles. (This is known as the Heisenberg picture …)
Now in the Heisenberg picture, time evolution is carried out by operators rather than the state vector. Then how come that time evolution by $\tau$
will result in $\exp(iH\tau)$ being acted on the state vector? Also I
cannot get the point of using Heisenberg picture in maintaining manifest Lorentz invariance.
To sum up, here are my main questions:
(1) Does the statement 'if we set
$\Lambda^{\mu}_{~~\nu} = \delta^{~\mu}_{~~~\nu}$ and $a^{\mu}\sim(0,0,0,\tau)$, then $U(\Lambda, a) = \exp(iH\tau)$ ' involve an implicit application of the Schroedinger equation, or some time evolution equation of a similar kind?
(2) How does the fact that we are using Heisenberg picture comply with
the change of state vector under this special choice of $U(\Lambda, a)$ ?
(3) Why does the application of Heisenberg picture allow us to see manifest Lorentz invariance? How can I see it?
I would really appreciate it if anyone could possibly offer me some hints or insight, or simply point out where I have gone wrong.
Best Answer
(1) The operator $U(\Lambda,a)$ is a unitary "rotation" in the Hilbert space corresponding to an inhomogeneous Lorentz transformation of the spacetime coordinates. When $U(\Lambda,a)=\exp(iH\tau)$, it is an operator that adjusts the clock forward by $\tau$. Conceptually this is not a physical time evolution of the system.
(2) A unitary rotation $U$ in the Hilbert space transforms operators as well as states, for the same reason as a rotation in $\mathbb{R}^{3}$ transforms vectors as well as matrices acting on vectors. That is, we have $\hat{O}\Psi$ $\rightarrow$ $\hat{O}^{\prime}\Psi^{\prime}$, where $\hat{O}^{\prime} = U\hat{O}U^{-1}$ and $\Psi^{\prime} = U\Psi$. This is true in both the Heisenberg and Schrödinger pictures.
(3) Someone else may have a better answer, but as far as the scattering theory itself is concerned, I don't see an advantage of the Heisenberg picture over the Schödinger picture. However, once we know that we are working with a quantum field theory, it is more natural to use the Heisenberg picture because it treats space and time on an equal footing. That is, in this picture, quantum field operators are functions of spacetime while states have no spacetime dependence at all. On the other hand, in the Schödinger picture, quantum fields depend only on spatial coordinates while states depend only on time.