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I'm reading the chapter Uses of Instantons from Sidney Coleman's Aspects of Symmetry. In the introduction to the chapter, page 265, he considered the traditional $\phi^4$theory described by the Lagrangian $$\mathscr{L}=\frac{1}{2}(\partial_\mu\phi)^2-m^2\phi^2-g^2\phi^4.\tag{1}$$ Next, he says that

For classical physics, $g$ is an irrelevant parameter.

To substantiate this claim, he made a change of variables $\phi^\prime=g\phi$ so that the Lagrangian $\mathscr{L}$ in (1) becomes $$\mathscr{L}=\frac{1}{g^2}\Big[\frac{1}{2}(\partial_\mu\phi^\prime)^2-\frac{1}{2}m^2\phi^{\prime 2}-\phi^{\prime 4}.\Big]\tag{2}$$

All I can see from (2) is that the Euler-Lagrange equation of motion for $\phi^\prime$ do not depend upon $g$. Since $g$ is absent from the field equation, if one could solve for $\phi^\prime$, one can also solve for $\phi$ irrespective of the value of $g$. As I understand, this is what he meant by $g$ being an irrelevant parameter.

Coleman argues that although $g$ is irrelevant in classical field theory, it's not in the corresponding quantum theory. He argues that since there is a new constant $\hbar$, and the correct object to consider $$\frac{\mathscr{L}}{\hbar}=\frac{1}{g^2\hbar}\Big[\frac{1}{2}(\partial_\mu\phi^\prime)^2-\frac{1}{2}m^2\phi^{\prime 2}-\phi^{\prime 4}\Big]\tag{3}$$ which appears in the path-integral $Z=\int D\phi e^{iS[\phi]/\hbar}$. This is used to claim that $g^2\hbar$ is relevant. I don't understand why.

Can we not absorb $g^2\hbar$ into $\phi^\prime$? I don't understand how does the appearance of $\hbar$ makes $g$ relevant. Can someone help?

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    $\begingroup$ By "can we have an exact solution" do you mean "does a solution exist, regardless of our ability to write it in closed form"? Or do you mean "does a solution exist that we can write down in a non-self-referential way"? $\endgroup$ – probably_someone Aug 4 '17 at 8:31
  • $\begingroup$ @probably_someone I don't really understand how to interpret Coleman's point. I'm not quite sure whether this is the right question to ask to understand Coleman's point. $\endgroup$ – SRS Aug 4 '17 at 8:38
  • $\begingroup$ Though I'm missing some context here, your interpretation seems reasonable. My question to you, to try to understand the source of the confusion, is: what do you mean by an "exact solution"? $\endgroup$ – probably_someone Aug 4 '17 at 8:42
  • $\begingroup$ @probably_someone Can we always work out the solution the EOM resulting from (2) given the necessary boundary conditions? At least in principle? But again, I don't know whether I'm asking the right question at all. $\endgroup$ – SRS Aug 4 '17 at 8:47
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    $\begingroup$ @AccidentalFourierTransform I have changed the question because I think this one captures my confusion better. $\endgroup$ – SRS Aug 4 '17 at 9:37
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The basic point is that in the path integral $Z=\int \text{D}\phi e^{\text{i}S[\phi]/\hbar}$, the value of the action is relevant, whereas in the classical theory the action is just there to be extremised, i.e. to derive the equtions of motion. Hence, in the quantum theory, a rescaling of the action as in $S\to g^{-2} S$ changes the theory.

Note that the rescaled action does not simply imply a rescaling of the path integral by a constant factor, but rather a change in the "interference of different paths".

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Consider a much simpler case showing the same confusion – a free Newtonian particle with Lagrangian $$\mathcal{L} = \frac{m}{2} \dot{x}^2. $$

Classically, $m$ is completely irrelevant. Indeed, the equation of motion is $\ddot{x} = 0$.

Quantum-mechanically, $m$ is relevant. To understand that, just consider the commutation relation

$$ [x, \dot{x}] = i \hbar m^{-1}. $$

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    $\begingroup$ This is not very convincing; after all, I could argue that $[x,p]$ is mass independent. $\endgroup$ – Javier Aug 4 '17 at 9:53
  • $\begingroup$ Coleman has some path-integral argument in mind and $g$ is like the quartic coupling of a quartic oscillator (not any Lagrangian). What is wrong with the commutation relation? It's the usual one $[x,p]=i\hbar$. Do I miss something? @Solenodon Paradoxus $\endgroup$ – SRS Aug 4 '17 at 9:54
  • $\begingroup$ @SRS my point was that $m$ shows up in the commutator. The path integral argument that Coleman is making is trivial: in path integral formalism the classical action appears in the exponential with a $i\hbar^{-1}$ coefficient. Adding $g^{-2}$ is equivalent to tweaking the $\hbar$. $\endgroup$ – Prof. Legolasov Aug 4 '17 at 9:58

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