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.


2 Answers 2


(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.

  • 2
    $\begingroup$ Thanks! Your answer makes me realize that I have been confusing time translation with time evolution. Time translation means merely adjusting the observer's clock in the "passive" viewpoint, or rearranging our experiments so that everything happens in a later time in the "active" viewpoint, where as time evolution is linked to the dynamics. Since time translation is nothing different from rotation or spatial translation, it is allowed to modify Heisenberg-picture-states as well as operators. $\endgroup$
    – Kaius
    Commented Feb 2, 2016 at 11:44
  • $\begingroup$ Seems it's on the right track, but won't the generator of time translation be the free particle Hamiltonian, that is, $H_0 = (p^0)_1 + (p^0)_2 + \cdots$ where $(p^0)_i$ is the time component of the $i$-th particle's momentum operator, rather than $H$, the "full" Hamiltonian with interaction included? $\endgroup$
    – Kaius
    Commented Feb 2, 2016 at 11:54
  • 1
    $\begingroup$ The generator of time translation is the full Hamiltonian $H$. Even though the interaction is included here, $E = p_{1}^{0} + p_{2}^{0} + \cdots$ because we consider in and out states, which are noninteracting in the limit $t \to \mp \infty$. More precisely, a wave packet constructed from in(out) states will consist of particles that are infinitely far away from one another at $t \to -\infty$ ($+\infty$). $\endgroup$
    – higgsss
    Commented Feb 2, 2016 at 15:04

Just one comment to the higgsss answer.

Formally from the Wigner theorem we have that if there exist time shift symmetry, for which the scalar product of quantum mechanical rays is conserved, $$ \tag 1 |\langle \psi (t)|\kappa (t)\rangle| = |\langle \psi{'}(t+\tau)| \kappa{'}(t+\tau) \rangle|, $$ then the symmetry transformation acts on $|\psi\rangle$ as the linear unitary operator (or antilinear antiunitary, which isn't realized for time shift symmetry), $$ U = e^{i\hat{H}t} $$ $\hat{H}$ has the formal sense of time translation generator, and the physical sense of the energy of system. The statement of the theorem doesn't depend on details of system (i.e., it may be the set of free non-interacting one-particle states, or such interacting system), except the property $(1)$.

The interpretation of time translation may be active or passive. From passive point of view, there are many observers, which are related to each other by the symmetry transformation, and describe the same system. Here we see nothing but the symmetry transformation, not the evolution.

From an active point of view, however, there is single observer, but the system itself undergoes the time symmetry transformation (i.e., is changed in time). By the other words, due to active point of view, transformed by the symmetry transformation state $$ |\psi{'}\rangle \equiv |\psi(t)\rangle $$ is the state which appeared from the state $$ |\psi\rangle = |\psi (0)\rangle $$ By the Wigner theorem and active point of view we have that $$ |\psi (t)\rangle = e^{-iHt}|\psi (0)\rangle , $$ i.e., the Schrodinger equation.


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