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

+ 2 like - 0 dislike
363 views

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 $Z$ 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 $\phi^r(x)$ ?

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 $\phi^r$.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.