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Why Normalise by $h$ in the Partition Function for Classical Harmonic Oscillator?

The $N$ is not really the Planck's constant $h$. It is denoted as such because that was the convention. This has to do with the history of the subject. Statistical mechanics, in its classical form was ...
Soham Mitra's user avatar
2 votes

Power-series expansion in coupling/Planck constant

Feynman diagram techniques are inherently perturbative. Diagramatically, in a theory like scalar $\phi^3$ or QED with a photon and an electron where you have one vertex, an expansion in the coupling ...
Andrew's user avatar
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Why does $\hbar$ appear twice in the axioms of QM?

The Schrodinger equation is not at all a postulate of quantum mechanics, it is derived directly by the other axioms, so it inherits the constant from the Commutation relations. See Sakurai for a full ...
LolloBoldo's user avatar
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-1 votes

Why does $\hbar$ appear twice in the axioms of QM?

$\hbar$ (or $h$) is scattered in many places over the whole QM subject, including well-known Planck law : $$\large {\displaystyle B_{\nu }(\nu ,T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }...
Agnius Vasiliauskas's user avatar
7 votes

How does the Planck constant enter into the uncertainty principle?

Notice that the kernel of the transform from position to momentum representations is $$ e^{- i x p / \hbar}. $$ Thus, comparing with your definition of Fourier transform, you need to have $$ \xi = ...
Aschkal's user avatar
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