Need help with units conversion in time dependent Schrodinger equations

Feb 2020
3
0
Uzbekistan
I have a time-dependent Schrodinger equation for an electron in 1D:
\(\displaystyle i\hbar\frac{\partial \psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m_e}\frac{\partial^2\psi(x,t)}{\partial x^2}+V(x)\psi(x,t).\)

I don't want to "carry" constant prefactors and want to deal with
\(\displaystyle i\frac{\partial \psi(x,t)}{\partial t}=-\frac{1}{2}\frac{\partial^2\psi(x,t)}{\partial x^2}+V(x)\psi(x,t),\)
by setting up my units as \(\displaystyle \hbar=m_e=1.\)

Then, after finding the solution (for example, numerically) I want to return back to physical units.

My question is: "How to convert length and time back to physical (for example, SI) units? I know, that I should scale by some factor, e.g. Bohr radius for the length. But, how to prove that?"

In other words, how much meters in SI units correspond to \(\displaystyle x=1\) in my units? How to derive this scaling factor?

You may say, multiply it by some number \(\displaystyle a_0\), but I need the answer, why exactly this number \(\displaystyle a_0\)?

Thank you in advance for your answer!
 
Apr 2015
1,227
356
Somerset, England
You only need units for mass (M) , length (L) and time (T) for this.

Well the mass of the electron has only one dimensional unit so to start with that


\(\displaystyle {m_e} = 9.1*{10^{ - 31}}kg\)

So


\(\displaystyle 1kg = \frac{{{m_e}}}{{9.1}}*{10^{31}}\quad {m_e}units\)

hbar has dimensions \(\displaystyle M{L^2}{T^{ - 2}}\)

and a value in ISO of

\(\displaystyle \hbar = 1*{10^{ - 34}}kgmetre{s^2}/\sec ond\)

Replacing kg by \(\displaystyle {m_e}units\) we have


\(\displaystyle \hbar = 1*{10^{ - 34}}*\frac{{{m_e}}}{{9.1}}*{10^{31}}{m_e}metre{s^2}/\sec ond\)

since length and time are independent there are many ways you can adjust hbar to unity so having pointed the way, I will leave that to you.
 
Feb 2020
3
0
Uzbekistan
You only need units for mass (M) , length (L) and time (T) for this.

Well the mass of the electron has only one dimensional unit so to start with that


\(\displaystyle {m_e} = 9.1*{10^{ - 31}}kg\)

So


\(\displaystyle 1kg = \frac{{{m_e}}}{{9.1}}*{10^{31}}\quad {m_e}units\)

hbar has dimensions \(\displaystyle M{L^2}{T^{ - 2}}\)

and a value in ISO of

\(\displaystyle \hbar = 1*{10^{ - 34}}kgmetre{s^2}/\sec ond\)

Replacing kg by \(\displaystyle {m_e}units\) we have


\(\displaystyle \hbar = 1*{10^{ - 34}}*\frac{{{m_e}}}{{9.1}}*{10^{31}}{m_e}metre{s^2}/\sec ond\)

since length and time are independent there are many ways you can adjust hbar to unity so having pointed the way, I will leave that to you.
Thanks for the answer. Now I understand how to derive that.

Continuing the discussion, I'd like to measure the length in Bohr radius \(\displaystyle a_0=0.5*10^{-10} m\). So, 1 meter is then \(\displaystyle 1m=a_0*2*10^10\)

We can substitute it into your \(\displaystyle \hbar\) formula to get

\(\displaystyle \hbar = 1*{10^{ - 34}}*\frac{{{m_e}}}{{9.1}}*{10^{31}}{m_e}a_0^2*4*10^{20}/second\).

From the formula above we can find what is 1s in our new units:

\(\displaystyle 1 s=\frac{1*10^{-34}}{9.1}*10^{31}*4*10^{20}*\frac{a_0^2 m_e}{\hbar}=0.46*10^{17} *\frac{a_0^2 m_e}{\hbar}\).