In the Schrödinger wave equation, where does the $8\pi^2 m/h$ come from?

Where $m$ is the mass and $h$ is Planck's constant

I understand the variables... but I'm unsure of the application of $8\pi^2 m/h$ as it relates to mass. Apologies if this is a Modern Physics question, as I have not taken it yet. I usually have seen explanations for the rest of the function, but this part is usually considered as self-evident. I figure it's from a more elementary Physics course, but I've yet to find it's origin.

Here is the source:

$$\frac{d^2\psi}{dx^2} - \frac{8\pi^2m}{h}(E-V)\psi = 0$$ where $m$ is the mass of the particle and $V$ the expression for the potential energy. This is a one-dimensional equation, independent of time. To solve this equation one has to, first of all, define the appropriate expression for the potential energy $V$, which will depend on the problem studied. When this expression is inserted into the Schrodinger wave equation, the differential equation so obtained can be solved to find $\Psi$ and $E$. In three dimensions the Schrodinger equation becomes [...]

(From Biophysics, by V. Pattabui and N. Gautham (Kluwer, 2002).)


1 Answer 1


This is a typo in the book, together with an extremely-far-from-the-usual presentation.

Serious QM texts will generally present the Schrödinger equation in the form $$ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} +(V-E)\psi = 0, $$ where the factor of $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi$ acting on the wavefunction comes from the kinetic-energy term $\frac{\hat{p}^2}{2m}\psi$ and the identification $\hat p = -i\hbar \frac{d}{dx}$. (For more information on those, see any serious QM textbook.)

To get from there to the form in your textbook, you need to

  • expand the reduced Planck's constant as $\hbar = h/2\pi$ explicitly (despite the fact that no serious physics treatment this side of the 1950s will do that),
  • push all the constants, including the sign, from the derivative and onto the potential term (for no clear reason and no clear gain),
  • get the sign of the potential-energy term wrong, and
  • get the denominator wrong by skipping the missing square in front of the $h$.

If you do want to write the Schrödinger equation in that form, then the correct way to write it is $$ \frac{d^2\psi}{dx^2} + \frac{8\pi^2m}{h^2}(E-V)\psi = 0. $$ Though if you really want to learn QM, I would recommend looking for an alternative resource - none of the above inspires me with any confidence in that text.


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