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In most cases, people discuss the effective action or the effective potential in the convention $\hbar=1$. Occasionally, we see the expression at the 1-loop order as $$\Gamma[\phi]=S[\phi]+\frac{i\hbar}{2}\text{Tr}\log[\partial^2+V''[\phi]]+{\cal O}(\hbar^2)\;,\tag{12}$$ see, e.g., eq. (12) in A. Zee's QFT textbook on pp. 239.

I wonder, does the $\hbar$ really play a role in the effective potential? In this case, it seems that any quantum corrections to the classical potential is very small, suppressed by $\hbar$. But in many discussions, the effective potential can differ the classical potential dramatically.

I currently have a discrepancy with my colleague about the physical role of the $\hbar$ appearing in the above formula. My colleague thinks that, the $\hbar$ appearing in the second term is only for a bookkeeping purpose and is not physical and argues that it can be cancelled somehow while I disagree. To me, perhaps the intuitive way to understand my colleague's opinion is that the $\hbar$ appears like a total factor in front of the action and any rescaling the units could change the magnitude of the $\hbar$ but the physics shall not be affected. Any opinions?

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  • $\begingroup$ @Qmechanic Thank you very much for the link. Actually, I am very familiar with the derivation of the effective action with or without the $\hbar$. But I currently have a discrepancy with my colleague about the physical role of the $\hbar$ appearing in the above formula. My colleague thinks that, the $\hbar$ appearing in the second term is only for a bookkeeping purpose and is not physical and argues that it can be cancelled somehow while I disagree. Could you tell me your opinion? $\endgroup$ – Wein Eld Jun 30 '18 at 11:12
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    $\begingroup$ Related physics.stackexchange.com/q/300643 ? $\endgroup$ – SRS Jun 30 '18 at 12:57
  • $\begingroup$ Are you and your colleague unequivocally unconflicted about the classical limit and the semiclassical expansion in plain, dull, QM? If not, multiplexing the problem by throwing in infinite degrees of freedom might not get you on the same page. $\endgroup$ – Cosmas Zachos Jun 30 '18 at 15:22
  • $\begingroup$ @CosmasZachos The problem is that, I cannot understand (and cannot agree) the opinion on "ℏ is unphysical in the above formula". Could you please simply give me a positive or negative answer to that such that I can make sure whether I really miss something very fundamental? $\endgroup$ – Wein Eld Jun 30 '18 at 16:10
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After reading the comments, I don't think that $\hbar$ is unphysical here in general. In particular, the quantum effective action should also encompass some of the classical non-linearities of the theory. This should stem from the fact that in the presence of massives particles, the Klein-Gordon equation and hence the Feynman propagator has also hidden inside a factor of $\hbar$ in it see e.g. wikipedia. So the $\hbar$ should come up as a compensator to extract some classical mass-dependent non linearities of the theory.

I say 'should' everywhere because I'm not sure about the quantum effective action really. What I'm sure about is that loop computations in, say, QED (or gravity), do encompass classical pieces and mix different orders in $\hbar$ (when massive particles are being exchanged at least). I see no reason why this would not directly translate to the vertices you get in your quantum effective action.

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