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In QFT, the Klein-Gordon equation is solved with the field operator $\hat \psi(x)$/$\hat \psi^\dagger(x)$ in the Heisenberg picture, and (as I understand it) gives the evolution of a single on-mass-shell particle/antiparticle.

My question is, could we solve it in the Schrodinger picture of the field configurations instead? What would this look like? Would it be unhelpful? How would it deal with positive frequency solutions?

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In this case, the state changes with time as

$$\Psi\left[\phi_{2}, t_{2}\right]=\int \mathcal{D} \phi_{1} \mathcal{S}\left[\phi_{2}, t_{2} ; \phi_{1}, t_{1}\right] \Psi\left[\phi_{1}, t_{1}\right]$$ here $$\mathcal{S}\left[\phi_{2}, t_{2} ; \phi_{1}, t_{1}\right]=\left\langle\phi_{2}\left|e^{-i H\left(t_{2}-t_{1}\right) / \hbar}\right| \phi_{1}\right\rangle$$

where $H$ is constant. Since $H$ is constant in this picture we don't need to think about the time ordering in the unitary operator.

And the field operator is constant in time defined by (Note instead of 4 momentum and 4 position vector we have 3d vectors)

$$\psi (x)=\int \left.{\frac {\mathrm {d} ^{3}p}{(2\pi )^{3}}}{\frac {1}{2E(\mathbf {p} )}}\left(A(\mathbf {p} )e^{i\vec{p}\cdot \vec{x}}+B(\mathbf {p} )e^{-i\vec{p}\cdot \vec{x}}\right)\right|_{p^{0}=+E(\mathbf {p} )}$$

The S matrix of both pictures is equal.

$$\langle f|S| i\rangle_{\text {H}}=\langle f ; \infty \mid i ;-\infty\rangle_{\text {S}}$$

Read about Schrödinger functional for more information.

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  • $\begingroup$ How does this deal with the positive frequency solutions that the Klein Gordon equation will permit? $\endgroup$
    – Alex Gower
    Mar 25 at 18:25
  • $\begingroup$ @AlexGower I think for the negative energy solution $\mathcal{S}\left[\phi_{2}, t_{2} ; \phi_{1}, t_{1}\right]=\left\langle\phi_{2}\left|e^{i H\left(t_{2}-t_{1}\right) / \hbar}\right| \phi_{1}\right\rangle$ since the filed is time independent. $\endgroup$ Mar 25 at 18:30
  • $\begingroup$ What I mean is. QFT textbooks often introduce the KGE then say the object which solves it cannot be a wavefunction since the KGE permits positive and negative energy solutions. For the heisenberg field operator this is fine since the negative energy solutions accompany annihilation operators. How is it fine here? $\endgroup$
    – Alex Gower
    Mar 25 at 18:33
  • $\begingroup$ @AlexGower negative energy solutions will come when you do relativistic quantum mechanics (which is a description of reality less fundamental than QFT but more fundamental than QM). In QFT we never get negative energy solutions. The solutions with negative energy in RQM become positive energy solutions that move backward in time in QFT. $\endgroup$ Mar 25 at 18:38
  • $\begingroup$ @AlexGower but the definitions of different pictures are the same in both RQM and QFT. $\endgroup$ Mar 25 at 18:39

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