De-dimensionalizing/rescaling units

[MaestroBach]

TL;DR Summary

I need to set certain values to 1, then rescale the rest of my units around that, but am unsure of how to do this in general.

Sorry if this is the wrong place to put this- I had no idea what category my question should fall under.

Does anyone possibly have good resources or good explanations on how to re-scale units? Something with concrete examples instead of just abstract explanations? For example, something like if I set hbar = 1, how would I manipulate my other units such as energy/what units would they end up as? I don't know if de-dimensionalizing is the correct name for what I want to do- my professor calls it rescaling units.

Google wasn't very helpful, and I got confused by wikipedia's page on nondimensionalization.

I would appreciate any help!

 

[jedishrfu]

It would help a lot if you provided some context here.

okay so you’re wanting to set certain physical constants to 1 or 0 in order to simplify an equation.

I think this is called natural units:

https://en.wikipedia.org/wiki/Natural_units

and this discussion on Physics Stack Exchange

https://physics.stackexchange.com/questions/168983/setting-constants-equal-to-1-conditions

 

[MaestroBach]

It would help a lot if you provided some context here.

okay so you’re wanting to set certain physical constants to 1 or 0 in order to simplify an equation.

I think this is called natural units:

https://en.wikipedia.org/wiki/Natural_units

and this discussion on Physics Stack Exchange

https://physics.stackexchange.com/questions/168983/setting-constants-equal-to-1-conditions

For context, something I'm doing is just solving schrodinger's equations for particles in boxes, but it's suggested that I set hbar = 1, in order to make numbers easier to handle/look at when plotting. However, I just don't get how setting hbar = 1 affects my other units/variables- I don't understand how the units of energy or displacement are affected by setting hbar = 1, if that makes sense.

 

[Gaussian97]

Well, it's easy, start with the Schrödinger equation written like
$$i\hbar \frac{\partial}{\partial t} \Psi = \left[-\frac{\hbar^2}{2m}\nabla^2+V\right]\Psi$$
What you can do first, to simplify this equation is to define new variables, call them $\hat{\vec{x}}=\frac{\vec{x}}{\hbar}$ and $\hat{t}=\frac{t}{\hbar}$, notice that trivially, because $\hbar$ is constant
$$\hbar\frac{\partial}{\partial t}=\hbar \frac{\partial}{\partial(\hbar\hat{t})}=\frac{\partial}{\partial \hat{t}}, \qquad \hbar^2\nabla_{x}^2= \nabla_{\hat{x}}^2$$
Where obviously $\nabla_{x}^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ and $\nabla_{\hat{x}}^2=\frac{\partial^2}{\partial \hat{x}^2}+\frac{\partial^2}{\partial \hat{y}^2}+\frac{\partial^2}{\partial \hat{z}^2}$
Then Schrödinger equation reads
$$i \frac{\partial}{\partial \hat{t}} \Psi = \left[-\frac{1}{2m}\nabla_{\hat{x}}^2+V\right]\Psi$$
Also, you define the wave functions with a scalar product given by
$$\int_{-\infty}^{\infty} \Psi^*\Phi d^3x$$
We can also redefine the wave functions with $\hat{\Psi}=\frac{\Psi}{\hbar^{3/2}}$, then the scalar product becomes
$$\int_{-\infty}^{\infty} \hat{\Psi}^*\hat{\Phi} d^3\hat{x}$$

Note that the Schrödinger equation also has the same form
$$i \frac{\partial}{\partial \hat{t}} \hat{\Psi} = \left[-\frac{1}{2m}\nabla_{\hat{x}}^2+V\right]\hat{\Psi}$$
What physicist usually mean with 'take $\hbar=1$' is to do this substitutions and then we simply drop all the $\hat{ }$ from the equations.

By the way, note that these substitutions are not unique, I've started with $\hat{x}=\frac{x}{\hbar}$, but anything fulfilling $\hbar\frac{\hat{x}^2}{\hat{t}}=\frac{x^2}{t}$ will do the job as well. The motivation doing in my way (aside that may seem more natural) is that $\hat{x}$ and $\hat{t}$ have units like $\text{kg}^\alpha (\text{ms}^{-1})^\beta$.

 

[MaestroBach]

Well, it's easy, start with the Schrödinger equation written like
$$i\hbar \frac{\partial}{\partial t} \Psi = \left[-\frac{\hbar^2}{2m}\nabla^2+V\right]\Psi$$
What you can do first, to simplify this equation is to define new variables, call them $\hat{\vec{x}}=\frac{\vec{x}}{\hbar}$ and $\hat{t}=\frac{t}{\hbar}$, notice that trivially, because $\hbar$ is constant
$$\hbar\frac{\partial}{\partial t}=\hbar \frac{\partial}{\partial(\hbar\hat{t})}=\frac{\partial}{\partial \hat{t}}, \qquad \hbar^2\nabla_{x}^2= \nabla_{\hat{x}}^2$$
Where obviously $\nabla_{x}^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ and $\nabla_{\hat{x}}^2=\frac{\partial^2}{\partial \hat{x}^2}+\frac{\partial^2}{\partial \hat{y}^2}+\frac{\partial^2}{\partial \hat{z}^2}$
Then Schrödinger equation reads
$$i \frac{\partial}{\partial \hat{t}} \Psi = \left[-\frac{1}{2m}\nabla_{\hat{x}}^2+V\right]\Psi$$
Also, you define the wave functions with a scalar product given by
$$\int_{-\infty}^{\infty} \Psi^*\Phi d^3x$$
We can also redefine the wave functions with $\hat{\Psi}=\frac{\Psi}{\hbar^{3/2}}$, then the scalar product becomes
$$\int_{-\infty}^{\infty} \hat{\Psi}^*\hat{\Phi} d^3\hat{x}$$

Note that the Schrödinger equation also has the same form
$$i \frac{\partial}{\partial \hat{t}} \hat{\Psi} = \left[-\frac{1}{2m}\nabla_{\hat{x}}^2+V\right]\hat{\Psi}$$
What physicist usually mean with 'take $\hbar=1$' is to do this substitutions and then we simply drop all the $\hat{ }$ from the equations.

By the way, note that these substitutions are not unique, I've started with $\hat{x}=\frac{x}{\hbar}$, but anything fulfilling $\hbar\frac{\hat{x}^2}{\hat{t}}=\frac{x^2}{t}$ will do the job as well. The motivation doing in my way (aside that may seem more natural) is that $\hat{x}$ and $\hat{t}$ have units like $\text{kg}^\alpha (\text{ms}^{-1})^\beta$.

This made a ton of sense! Thank you so much!

Couple questions:

When you were redefining $\hat{\Psi}$, where did the $\hbar^{3/2}$ come from?

Also, given that I'm now using $\hat{\vec{x}}$, does that mean that the units of my displacement are whatever my displacement units originally were (such as meters) divided by $\hbar$? I guess I'm confused as to how you get the $\text{kg}^\alpha (\text{ms}^{-1})^\beta$ units

I've seen people take $c = 1$ in tandem with $\hbar = 1$, but does that mess anything up, given that setting $\hbar = 1$ specified the units of displacement and time, but $c = 1$ would set up a different set of units?

 

[Gaussian97]

When you were redefining $\hat{\Psi}$, where did the $\hbar^{3/2}$ come from?

Well, simply because you want your new equation to look exactly the same as the old ones (because in this way you can throw away the $\hat{}$ without any change), since you have originally
$$\int_{-\infty}^\infty \Psi^*\Phi d^3x$$
With the new variable $x=\hbar\hat{x}$ you have
$$\int_{-\infty}^\infty \hbar^3\Psi^*\Phi d^3\hat{x} = \int_{-\infty}^\infty \hat{\Psi}^*\hat{\Phi} d^3\hat{x}\Longrightarrow \hat{\Psi}^*\hat{\Phi}=\hbar^3 \Psi^*\Phi $$
Because you want this to be true for all the functions $\Psi$ and $\Phi$ you must choose the change
$$\hat{\Psi}=\hbar^{3/2}\Psi$$.

Also, given that I'm now using $\hat{\vec{x}}$, does that mean that the units of my displacement are whatever my displacement units originally were (such as meters) divided by $\hbar$? I guess I'm confused as to how you get the $\text{kg}^\alpha (\text{ms}^{-1})^\beta$ units

Yes, the units of $x$ are $\text{m}$ and the units of $\hbar$ are $\text{kgm}^2\text{s}^{-1}$ in SI, so therefore the units of $\hat{x}$ are
$$[\hat{x}]=\frac{[x]}{[\hbar]}=\frac{\text{m}}{\text{kgm}^2\text{s}^{-1}}=\text{kg}^{-1}(\text{ms}^{-1})^{-1}$$
And a similar thing for $\hat{t}$.

I've seen people take $c = 1$ in tandem with $\hbar = 1$, but does that mess anything up, given that setting $\hbar = 1$ specified the units of displacement and time, but $c = 1$ would set up a different set of units?

As I said the change is not unique, because $\hbar$ and $c$ are linearly independent magnitudes you can always choose such a transformation so that all the $\hbar$ and $c$ of your equations disappear (as we physicists say, we can put $\hbar=c=1$). Actually that's the reason why I did the change I did, because now adding the condition $c=1$ is trivial (note that all the $\hat{ }$ variables are proportional to units of velocity).

 

[MaestroBach]

Well, simply because you want your new equation to look exactly the same as the old ones (because in this way you can throw away the $\hat{}$ without any change), since you have originally
$$\int_{-\infty}^\infty \Psi^*\Phi d^3x$$
With the new variable $x=\hbar\hat{x}$ you have
$$\int_{-\infty}^\infty \hbar^3\Psi^*\Phi d^3\hat{x} = \int_{-\infty}^\infty \hat{\Psi}^*\hat{\Phi} d^3\hat{x}\Longrightarrow \hat{\Psi}^*\hat{\Phi}=\hbar^3 \Psi^*\Phi $$
Because you want this to be true for all the functions $\Psi$ and $\Phi$ you must choose the change
$$\hat{\Psi}=\hbar^{3/2}\Psi$$.
Yes, the units of $x$ are $\text{m}$ and the units of $\hbar$ are $\text{kgm}^2\text{s}^{-1}$ in SI, so therefore the units of $\hat{x}$ are
$$[\hat{x}]=\frac{[x]}{[\hbar]}=\frac{\text{m}}{\text{kgm}^2\text{s}^{-1}}=\text{kg}^{-1}(\text{ms}^{-1})^{-1}$$
And a similar thing for $\hat{t}$.As I said the change is not unique, because $\hbar$ and $c$ are linearly independent magnitudes you can always choose such a transformation so that all the $\hbar$ and $c$ of your equations disappear (as we physicists say, we can put $\hbar=c=1$). Actually that's the reason why I did the change I did, because now adding the condition $c=1$ is trivial (note that all the $\hat{ }$ variables are proportional to units of velocity).

Really, really appreciate your help, thank you!