If I express a Hamiltonian $H$ in units of Hz by dividing the energy terms in the Hamiltonian by hbar $\tilde{H}=\dfrac{H}{\hbar}$ which means you set $\hbar =1$. Then what are the units of time? Also let's say you wanted to calculate the evolution of a system with $U=e^{i/\hbar H t}$ between $t = 1$ to $t=10$ s. What numbers do you have to use for $t$ when $\tilde{U}=e^{i \tilde{H} t}$. Do you have to multiply $t$ with $\hbar$ to obtain the same results?
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$\begingroup$ this wiki article has tables and formulae en.wikipedia.org/wiki/Natural_units $\endgroup$– anna vNov 9, 2018 at 11:59
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3 Answers
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It depends what else, if anything, you set to $1$. If nothing, time has energy dimension $-1$. If $c=1$, we can also say time has mass dimension $-1$. (You could even say the momentum dimension is $-1$, but no-one does.) In the convention $c=\hbar=1$, known as natural units, mass/energy dimensions are often just called dimensions, so time has dimension $-1$. If we also take $G=1$ as in Planck units, time can be nondimensionalised, with the Planck time given by $\sqrt{G\hbar/c^5}$. (Well, in our $4$-dimensional spacetime, anyway; the analogous result in $n$-dimensional spacetime, with one time dimension, is $(G\hbar/c^{n+1})^{1/(n-2)}$ provided $n\ne 2$.)
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$\begingroup$ What's "energy dimension -1"? Do you mean $(energy \, dimension)^{-1}$? $\endgroup$– badjohnNov 9, 2018 at 16:03
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$\begingroup$ @badjohn The dimension $E^{-1}$ can just be called dimension $-1$ if everything has to be of the form $E^p$, but the prefix energy clarifies that it's powers of energy we're using. $\endgroup$– J.G.Nov 9, 2018 at 16:33
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If frequency is in Hz then assuming you are using a coherent set of units time must be in s. Setting Planck's constant to 1 will fix either your unit of mass or your unit of length, but you still have complete freedom to choose the other one.
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$E=h\nu$, so the units of frequency and energy become identical if you set h=1. The answer is therefore the inverse of whatever energy unit you have. You still have a lot of freedom to fix this.
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2$\begingroup$ This answer could use a lot more elaboration. $\endgroup$– user191954Nov 9, 2018 at 12:25
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