Consider the quantized real scalar field acting on the vacuum state $\vert 0 \rangle $. We can interpret the state $\phi(\textbf{x})\vert 0 \rangle $ (defined in the Schrodinger picture at $t=0$) as a particle created at $(t=0,\textbf{x})$.

Peskin and Schroeder say that the state $\phi(x)\vert 0 \rangle $ is particle prepared at spacetime point $x$. I see how this works in the Heisenberg picture. But I want to work in the Schrodinger picture to convince myself of whats going on. So I time evolve the state $\phi(\textbf{x})\vert 0 \rangle $ by acting with the time evolution operator $U=e^{-iHt}$. However this time evolution gives an incorrect expression, in particular instead of having $e^{iEt}$ which is what I need, I have a $e^{-iEt}$ term which doesn't coincide with the Heisenberg picture result (or the result you get from just solving the Klein Gordon equation).

What's the problem?

Edit: A more in depth explanation;

We have $\phi(\textbf{x})\vert 0 \rangle = \displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\textbf{p}}}e^{-i\textbf{p}.\textbf{x}}a^\dagger_\textbf{p} \vert 0 \rangle$. In the Heisenberg picture we can show that at time $t$ we have $\phi(x)\vert 0 \rangle = e^{iHt}\phi(\textbf{x})e^{-iHt}\vert 0 \rangle$.

Expanding $e^{-iHt}$ as a Taylor series and since the state $\vert 0 \rangle$ is such that $H\vert 0 \rangle=0$ we have $e^{-iHt}\vert 0 \rangle=\vert 0 \rangle$. This gives us $\phi(x)\vert 0 \rangle = e^{iHt}\phi(\textbf{x})\vert 0 \rangle$. Using the commutation relations between the ladder operators and $H$ we obtain the correct result.

The problem is that if I take the state $\phi(\textbf{x})\vert 0 \rangle$ and time evolve with the time evolution operator $U=e^{-iHt}$ we obtain $\phi(x)\vert 0 \rangle = e^{-iHt}\phi(\textbf{x})\vert 0 \rangle$. Comparing our expressions we have $e^{-iHt}=e^{iHt}$ implying $t=0$ or $H=0$, the former coinciding with the fact that the pictures agree at $t=0$.

Why am I not obtaining the same result in the Schrodinger picture?

  • $\begingroup$ Why don't the results coincide? It seems like you've done something wrong. The equivalency of Heisenberg and Schrodinger picture is proved in Quantum Mechanics. $\endgroup$ – Prof. Legolasov Jul 18 '15 at 20:55
  • $\begingroup$ "What's the problem?" I also don't see that. Explain more precise where your results differ. $\endgroup$ – Nontriviality Jul 18 '15 at 22:28
  • $\begingroup$ @Nontriviality I added more explanation $\endgroup$ – Okazaki Jul 19 '15 at 9:52

This problem arises because you can not compare states in the Heisenberg picture and states in the Schrödinger picture (except for $t=0$) because they are physically different objects.

The only thing that coincides and can be compare are matrix elements or expectation values. If you have a look at expectation values, you will see (quite trivially) that there is no difference.

  • $\begingroup$ Ah so things corellation functions would be the same? I could compute $\langle 0 \vert \phi(y)\phi(x)\vert 0 \rangle$ with $y^0=t_2$, $x^0=t_1$ and it would be equal to $\langle 0 \vert \phi(\textbf{y})U^{\dagger}(t_2)U(t_1)\phi(\textbf{x})\vert 0 \rangle$ ? $\endgroup$ – Okazaki Jul 19 '15 at 12:29
  • $\begingroup$ It should work. Compare the time dependent field expansion and how the time evolution operators act (use commut relations like $[H,a]$). $\endgroup$ – Nontriviality Jul 21 '15 at 8:42

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