Solving for hbar: Momentum and Wave Vector Relationship

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In summary, when using units where ##\hbar=1##, the wave vector K and the momentum p have the same units and magnitude. However, the choice of which units to use for the Planck constant allows for different magnitudes for K and p, and this choice does not affect the symbolic expressions used in theoretical physics.
  • #1
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
 
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  • #2
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 ?

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.
 
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  • #3
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.
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.
But when we use ##\hbar## in natural units, what value should we insert for K?
 
  • #4
amjad-sh said:
the momentum of an electron should have a much less quantity, since it is multiplied by the Plancks constant.

Not if you are using units where ##\hbar = 1##; "much less" implicitly assumes that Planck's constant has a very small value, but ##1## is not a very small value.

amjad-sh said:
when we use ##\hbar## in natural units, what value should we insert for K?

This should be obvious from the equation ##p = \hbar K##. That relation is true regardless of what units you choose for ##\hbar##.
 
  • #5
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.
then the answer is ##p=10^9##?
or p and k should have a different value from that we used in dimensional analysis?
 
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  • #6
amjad-sh said:
then the answer is ##p=10^9##?
Yes.
 
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  • #7
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?
Well, to be sure:
If you are asking if ##p=10^9\text{kgms}^{-1}## then the answer is absolutely no.
If you are asking if ##p=10^9\text{m}^{-1}## then the answer is yes. But it's rare to use the unit of ##m^{-1}## when you set ##\hbar=1##
 
  • #8
If you want to go back from natural units, where ##\hbar=c=k_{\text{B}}=1## to the usually unhandy SI units in the realm of microphysics you only need ##\hbar c=0.197 \text{GeV} \, \text{fm}##. This is usually handy in HEP physics. In other fields like atomic physics other units may be more convenient. I've seen textbooks, where they used units with ##c=\alpha_{\text{em}}## :-)).
 
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  • #9
amjad-sh said:
or p and k should have a different value from that we used in dimensional analysis?

If you are using units where ##\hbar = 1##, dimensional analysis says that ##p## and ##k## have the same units.
 
  • #10
PeterDonis said:
If you are using units where ##\hbar = 1##, dimensional analysis says that ##p## and ##k## have the same units.
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?
 
  • #11
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?

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.
 
  • #12
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.
I'm talking about using ##\hbar=1##.
If we use ##\hbar =1##, p and k have the same unit and same magnitude.
p and k in SI units have different magnitudes and units.
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?
I don't know if you got my point.
 
  • #13
Well the new SI in some sense does right this. It defines values of the fundamental constants, including Planck's constant. It's only not choosing the value ##\hbar=1## which indeed is pretty convenient in (theoretical) microscopic physics but pretty unhandy in everyday life, where we deal with macroscopic objects.
 
  • #14
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? 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?
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 is
$$[\hbar]=kg\,m^2\,s^{-1}$$
so when you set ##\hbar=1## you can choose to change the units of mass ##kg##, or you can choose to change the units of length ##m##, or you can choose to change the units of time ##s##, or you can choose to make a combination of such choices. There is an infinite number of combinations. But in the actual literature you will never see what the actual choice is. That's because, in practice, the units ##\hbar=1## are used only for dealing with general symbolic formulas, and never for writing down explicit numbers (*), so in practice no choice is made at all. When physicists want to calculate the actual numbers (magnitudes, as you call them), they first return their final formula to some of the standard units such as SI and then express the numbers in those standard units.

(*) An exception is when physicists use units ##\hbar=c=G_N=k_B=1## (standing for the Planck constant, speed of light, Newton gravitational constant and Boltzmann constant, respectively), in which case all physical quantities can be considered as "dimensionless", so one can write down explicit numbers without carrying about units. For an amusing paper see also https://pdfs.semanticscholar.org/a8...33.1234800079.1563437390-736102029.1555316195
 
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  • #15
It's usually a convention in specific physics communities. In my own, relativistic heavy-ion/high-energy particle physics, we usually use natural units with ##\hbar=c=k_{\text{B}}=1##, i.e., we set the modified Planck action, the speed of light, and Boltzmann's constant to 1. Then we usually give masses, energies, momenta, temperatures, chemical potentials in GeV and lengths and time in ##\text{fm}=10^{-15} \;\text{m}##. If you have things like energy density, it's given in GeV/fm^3. Of course, you can easily convert it to GeV^4 using ##\hbar c=0.197 \; \text{GeV} \, \text{fm}##.
 
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  • #16
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?

You didn't. By choosing ##\hbar = 1## you are choosing a different system of units from SI units. You do not appear to grasp this simple fact.
 
  • #17
PeterDonis said:
By choosing ##\hbar = 1## you are choosing a different system of units from SI units.
Actually, as I explained in #14, by choosing ##\hbar = 1## one chooses an infinite class of systems of units different from SI.
 
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Related to Solving for hbar: Momentum and Wave Vector Relationship

1. What is hbar in the context of momentum and wave vector?

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.

2. How is momentum related to wave vector in the context of hbar?

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.

3. Why is hbar important in solving for momentum and wave vector?

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.

4. How is hbar used to calculate the momentum of a particle?

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.

5. Can hbar be used to solve for the wave vector of a particle?

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.

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