6
$\begingroup$

In Weinberg's QFT Page 109, he defines the "in" and "out" states as

the 'in' and 'out' states* $\Psi_{\alpha}^{+}$ and $\Psi_{x}^{-}$ will be found to contain the particles described by the label $\alpha$ if observations are made at $t \rightarrow-\infty$ or $t \rightarrow+\infty$, respectively.

And then he claims that

Note how this definition is framed. To maintain manitest 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, in distinction with the Schrödinger picture, where the operators are constant and the states change with time.) Thus we do not say that $\Psi_{\alpha} \pm$ are the limits at $t \rightarrow \mp \infty$ of a time-dependent state-vector $\Psi(t)$

However, implicit in the definition of the states is a choice of the inertial frame from which the observer views the system; different observers see equivalent state-vectors, but not the same state-vector. In particular, suppose that a standard observer $\mathcal{O}$ sets his or her clock so that $t=0$ is at some time during the collision process, while some other observer $\mathcal{O}^{\prime}$ at rest with respect to the first uses a clock set so that $t^{\prime}=0$ is at a time $t=\tau ;$ that is, the two observers' time coordinates are related by $t^{\prime}=t-\tau .$ Then if $\mathcal{O}$ sees the system to be in a state $\Psi, \mathcal{O}^{\prime}$ will see the system in a state $U(1,-\tau) \Psi=\exp (-i H \tau) \Psi .$ Thus the appearance

Now my question is: as we are talking about a state-vector $\Psi$ in the Heisenberg picture, which do not evolve over time. Why does the state vector change under the change of observers with different setting of time.

$\endgroup$
5
  • 1
    $\begingroup$ The operators which correspond to the measurement of a particular quantity change with time. For instance, the operator that measures 4-momentum at time $t$ is $P_\mu(t) = e^{iHt}P_\mu e^{-iHt}$ where $P_\mu$ is the operator that measures 4-momentum at time $t=0$. Effectively, this is the same thing as saying that an observer at time $t$ will measure the system in the state $e^{-iHt} \Psi$. $\endgroup$
    – Prahar
    Nov 4 '20 at 12:05
  • $\begingroup$ @Prahar So why doesn't he just use the Schodinger picture to say that the state-vector $\Psi$ is dependent on time, rather than using Heisenberg picture and arguing about the change of inertial frame(which is what confuses me). $\endgroup$
    – Jiahao Fan
    Nov 4 '20 at 12:14
  • $\begingroup$ In this part of the discussion, it's true that the Schrodinger picture is more useful. However, the Heisenberg picture is more convenient for most of the discussions later and my guess would be that Weinberg did not want to keep moving between pictures. He wanted to have the entire discussion in the Heinsenberg picture throughout the book because that is better to do in the long run (as far as his book is concerned). $\endgroup$
    – Prahar
    Nov 4 '20 at 12:16
  • $\begingroup$ @Prahar From my understanding, your argument "this is the same thing as saying that an observer at time $t$ will measure the system in the state $e^{-iHt} \Psi$ " is actually changing from Heisenberg picture to Schodinger picture since you are talking about the change of state vectors. $\endgroup$
    – Jiahao Fan
    Nov 4 '20 at 12:22
  • $\begingroup$ Yes. That's a good way to put it. $\endgroup$
    – Prahar
    Nov 4 '20 at 12:25
3
$\begingroup$

I think I got the point.

In Heisenberg's picture, the state-vectors do not change according to the Schodinger's equation governing the time evolution of the state. Since different pictures are defined in how operators and state vectors are changing with $\textbf{time evolution equation}$.

But the state vectors $\textbf{do}$ change under symmetry transformations such as Lorentz transformation. And one Lorentz transformation is "time translation", which coincides with time evolution operator of the schdinger equation but the physical meaning is different.

Now let's do a "change of inertial frame of observing the system", what we are doing here is doing a "Lorentz transformation" rather than doing a "time evolution", so the state vectors do change and it changes with the same way of time evolution incidentally.

$\endgroup$
1
+50
$\begingroup$

a state-vector $\Psi$ describes the whole spacetime history of a system of particles.

Think of $\Psi$ heuristically as corresponding to the worldlines of a system of particles. (Obviously there are no actual worldlines, because we're doing quantum mechanics.) The worldlines describe the entire spacetime history of the particles, but the worldlines are not Lorentz invariant. Different observers in different inertial frames will see different worldlines. Of course this is just the classical picture, but the basic idea that the state of the system is not Lorentz invariant carries over to the quantum case.

$\endgroup$
4
  • $\begingroup$ Thanks for your answer! But why does he use Heisenberg picture in the first place? Why not just use Schodinger picture, which explicitly tells the time evolution? $\endgroup$
    – Jiahao Fan
    Oct 14 '20 at 5:14
  • $\begingroup$ Because in relativistic field theory it is more convenient to make the operators $\psi(x,t)$ time dependent, so that $t$ is on the same footing as $x$.. $\endgroup$
    – mike stone
    Nov 4 '20 at 12:02
  • $\begingroup$ @mikestone Well $\Psi(x)$ is the state vector here so that it does not change with time... $\endgroup$
    – Jiahao Fan
    Nov 4 '20 at 12:09
  • $\begingroup$ The state vector $|\Psi\rangle$ is usually not a function of position $x$. The "in" state vector is a function $|{\bf k}_1,{\bf k}_2,\ldots, {\bf k}_N\rangle$ of the the momenta ${\bf k}_i$ of the asymptotically far-separated incoming particles. More precisely one makes wavepackets out of the plane-wave $|{\bf k}_1,{\bf k}_2,\ldots, {\bf k}_N\rangle$ so the that the particles are well separated in the distant past. Those packets can depend on where the particles were long ago but the plane wave states are the ones that used for Lorantz invariance arguments. $\endgroup$
    – mike stone
    Nov 4 '20 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.