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amjad-sh
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If we set ##\hbar=1##, do we need in return to change the value of the wave vector K to be equal to the momentum p, as ##\hbar## K=p? or make p equal to K ?
Thanks in advance
Thanks in advance
Last edited:
amjad-sh said:If we set ##\hbar=1##, do we need in return to change the value of the wave vector K to be equal to the momentum p, as ##\hbar K=p##? or make p equal to K ?
I am asking this because in bulk metal,for example, the fermi wave vector has a quantity near ##10^9 m^{-1}## while the momentum of an electron should have a much less quantity, since it is multiplied by the Plancks constant.PeterDonis said:No, you don't need to "change" the wave vector. If you use units where ##\hbar = 1##, then the wave vector already is the same as the momentum, without having to change anything.
amjad-sh said:the momentum of an electron should have a much less quantity, since it is multiplied by the Plancks constant.
amjad-sh said:when we use ##\hbar## in natural units, what value should we insert for K?
then the answer is ##p=10^9##?PeterDonis said:This should be obvious from the equation p=ℏKp=ℏKp = \hbar K. That relation is true regardless of what units you choose for ℏℏ\hbar.
Yes.amjad-sh said:then the answer is ##p=10^9##?
Well, to be sure:amjad-sh said:then the answer is ##p=10^9##?
or p and k should have a different value from that we used in dimensional analysis?
amjad-sh said:or p and k should have a different value from that we used in dimensional analysis?
can we say for example that K=[ value of p in SI units]##kgms^{-1}##, like inverse of what @Gaussian97 said? and if no why not?PeterDonis said:If you are using units where ##\hbar = 1##, dimensional analysis says that ##p## and ##k## have the same units.
amjad-sh said:can we say for example that K=[ value of p in SI units]kgms−1kgms^{-1}, like inverse of what @Gaussian97 said? and if no why not?
I'm talking about using ##\hbar=1##.PeterDonis said:If you're using SI units, you're not using units in which ##\hbar = 1##. Which system of units do you want to talk about? You seem to be mixing them up.
By deciding to use ##\hbar=1## one does not decide that the magnitude of ##k## is the same as in SI units. To determine this magnitude one must make an additional choice. This additional choice may be taken in many different ways; for instance so that ##k## has the same magnitude as in SI, or alternatively so that ##p## has the same magnitude as in SI. This freedom of choice comes from the fact that the unit of the Planck constant isamjad-sh said:when we changed to ##\hbar=1##,why we decided for the magnitude and unit of k to be the same in SI units? why we didn't play the game reversely by letting the magnitude and unit of p in ##\hbar=1## unit be the same in the SI unit, and so the unit and magnitude of k will change this time to be equal to p?
amjad-sh said:when we changed to ##\hbar=1##,why we decided for the magnitude and unit of k to be the same in SI units?
Actually, as I explained in #14, by choosing ##\hbar = 1## one chooses an infinite class of systems of units different from SI.PeterDonis said:By choosing ##\hbar = 1## you are choosing a different system of units from SI units.
hbar, also written as ℏ, is the reduced Planck constant, which is equal to the Planck constant (h) divided by 2π. It is a fundamental constant in quantum mechanics and is used to relate the momentum and wave vector of a particle.
In quantum mechanics, the momentum (p) of a particle is related to its wave vector (k) by the equation p = ℏk. This means that the momentum of a particle is directly proportional to its wave vector, with hbar acting as a conversion factor.
hbar is important because it allows us to connect the classical concept of momentum with the wave-like behavior of particles in quantum mechanics. It also allows us to make predictions about the behavior of particles based on their wave properties.
To calculate the momentum of a particle using hbar, we use the equation p = ℏk, where p is the momentum and k is the wave vector. We can also rearrange this equation to solve for the wave vector, k = p/ℏ, or for the Planck constant, h = p/k.
Yes, hbar can be used to solve for the wave vector of a particle. As mentioned in the previous answer, we can rearrange the equation p = ℏk to solve for k, which represents the wave vector. This allows us to calculate the wave properties of a particle based on its momentum.