I recently read a bit about the Schrodinger picture in QFT and wavefunctionals, see e.g. Polchinski's String Theory lectures, and I wanted to ask if the intuitive understanding of QFT I got is "right". I understood that a state in quantum field theory is very similar to the state in usual QM: There, a particle may be at any position in the configuration space, and the wavefunction give's a probability amplitude to every such configuration. In QFT, a field can be in any classical state $\phi(x)$, and these are weighted with the amplitude of the wave-functional $\Psi[\phi(x)]$. I also read that a QFT state may be written as $\int D\phi \Psi[\phi(x)]|\phi(x)\rangle$, where the $|\phi(x)\rangle$ are similar to position eigenstates in QM states which correspond to classical field configurations. But how do I construct these eigenstates $|\phi(x)\rangle$? Does something like $$|\phi(\vec x,t)\rangle=\int d^3x \phi(\vec x,t) \Phi(\vec x) |0\rangle$$ work, where $\Phi(x)$ shall be the Schrodinger field operator which creates a particle localized at position $\vec x$?
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1The short answer is yes. However, the integral you wrote is more like a symbol, as it makes no mathematical sense. The details can be found in Streater Wightman. – Phoenix87 Sep 30 '17 at 14:27
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1Yep! The eigenstates are just basis states, however. Much like $\left| x \right>$ for different values of $x$ in QM. There isn't a formula for them. You could formally expand $\left|0\right>$ as a wavefunctional though. – Prof. Legolasov Sep 30 '17 at 16:09
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You might, or might not, be interested in this, or this one. – Cosmas Zachos Oct 28 '17 at 23:28