# Need help with units conversion in time dependent Schrodinger equations

#### Doniyor

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 $$\displaystyle a_0$$?

#### studiot

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.

#### Doniyor

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:

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