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  What's the difference between quantum mechanical operator $\Phi^r(x)$ and its corresponding quantum fields $\phi^r(x)$ ?

+ 2 like - 0 dislike

Consider a quantum field theory with action $I[\phi]$, and suppose we 'turn on' a set of classical currents $J_r(x)$ coupled to the fields $\phi^r(x)$ of the theory. The complete vacuum vacuum amplitude in the presence of these currents is then

\(Z[J]\equiv\left< \text{VAC, out}|\text{VAC, in}\right>_J\\ =\int \Bigg[\prod_{s,y}d\phi^s(y)\Bigg]\text{exp}\bigg(iI[\phi]+i\int d^4x\phi^r(x)J_r(x)+\epsilon \text{ terms}\Bigg).\)

The Feynman rules for calculating $Z[J]$ are just the same as for calculating the vacuum-vacuum amplitude $Z[0]$ in the absence of the external current, except that the Feynman diagrams now contain vertices of a new kind, to which a $single$ $\phi^r$-line is attached. Such a vertex labelled with a coordinate x contributes a 'coupling' factor $iJ_r(x)$ to the integrand of the position-space Feynman amplitude. Equivalently, we could say that in the expansion of $Z[J]$ in powers of $J$, the coefficient of the term proportional to $iJ_r(x)iJ_s(y)\cdots$ is just the sum of diagrams with external lines (including propagators) corresponding to the fields $\phi^r(x), \phi^s(y,)$ etc. In particular, the first derivative gives the vacuum matrix element of the quantum mechanical operator $\Phi^r(x)$ corresponding to $\phi^r(x)$:

\(\Bigg[\frac{\delta}{\delta J_r(y)}Z[J]\Bigg]_{J=0}=\int \Bigg[\prod_{r,x}d\phi^r(x)\Bigg]\phi^r(y)\text{exp}\{iI[\phi]+\epsilon \text{ terms}\}\\ =i\left< \text{VAC, out}|\Phi^r(x)|\text{VAC, in}\right>_{J=0}.\)

What's the difference between quantum mechanical operator $\Phi^r(x)$ and its corresponding quantum fields $\phi^r(x)$ ?

asked Jun 1, 2014 in Theoretical Physics by coolcty (125 points) [ revision history ]
edited Jun 1, 2014 by coolcty

1 Answer

+ 2 like - 0 dislike

What should be called a quantum field is $\Phi^r$ : it is field with values in the operators of the theory, i.e. for every $x$, $\Phi^r(x)$ is an operator acting on the Hilbert space of the theory. The field $\phi^r$ is a classical field : for every $x$, $\phi^r(x)$ is a number (if $\phi^r$ is a scalar field, or more generally an element of some representation of the Lorentz group). As clear from the formulas in the question, the path integral is an integral over the space of classical fields $\phi^r$.

answered Jun 1, 2014 by 40227 (5,140 points) [ no revision ]

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