0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $n\ne 2$.)

$\endgroup$
  • $\begingroup$ What's "energy dimension -1"? Do you mean $(energy \, dimension)^{-1}$? $\endgroup$ – badjohn Nov 9 '18 at 16:03
  • $\begingroup$ @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 '18 at 16:33
0
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

$E=h\nu$, 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.

$\endgroup$
  • 2
    $\begingroup$ This answer could use a lot more elaboration. $\endgroup$ – user191954 Nov 9 '18 at 12:25
  • $\begingroup$ @Chair short and correct is my pref $\endgroup$ – my2cts Nov 9 '18 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.