When Griffiths derives the time-independent Schrodinger equation he divides both sides of the Schrodinger equation by $\psi$. I take this as a tacit assumption that $\psi\neq0$ when we intend to solve the time-independent Schrodinger equation for $\psi$.

Later, when solving the infinite square well, one of the boundary conditions assumes $\psi(0)=\psi(a)=0$.

I would argue that if we assume $\psi=0$ anywhere in the domain in which we intend to solve for psi, then we cannot use a time-independent Schrodinger equation that has been derived by assuming $\psi\neq0$.

And in the case of the the infinite square well, the solutions clearly have zero values in the domain $0\leq x\leq a$: $$\psi_n(x)= \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi}{a} x\right) \;.$$

I'm wondering if the derivation of the time-independent Schrodinger equation does not actually depend on dividing by $\psi$, but that Griffiths used a lazy trick to let the reader more easily see, read, or follow the derivation to the final equation.

So, why are we allowed to divide by $\psi$ when $\psi=0$?

  • 5
    $\begingroup$ If you are ever in doubt about the solution to a differential equation, you can always just plug it back in and check that it works by hand. Don't right this fact off as trivial. it means that often a very good guess (and there are many, not wholly rigorous, methods to come up with a very good guess) can be turned into rigorous solution pretty easily. $\endgroup$ Jun 14, 2017 at 14:39
  • 3
    $\begingroup$ What @BySymmetry says is spot on. When trying to solve ODE explicitly, we mathematicians are happy to use whatever method, as long as the results we plug in at the end really works. There are rigorous results that say that if a solution works, and you have a certain number of parameters, then that's the general solution. Can't argue with that. $\endgroup$
    – Ryan Unger
    Jun 14, 2017 at 15:09

2 Answers 2


This is a standard "issue" when one solves differential equation with a separation of variable. To be rigorous, the case $\psi=0$ should be handled separately. However, you can check that $\psi''=0$ when $\psi=0$, and that $\psi''/\psi$ is well behaved there, so you are in fact allowed to divide by $\psi$ on the whole interval.

But don't say that to a mathematician.

EDIT : Since this answer has some success, I would like to improve it by showing how one can avoid the division by $\psi$. For this, let's assume that after separation of variables (i.e. after plugging the Ansatz $\psi(x,t)=T(t) X(x))$, we obtain the following PDE $$ a T'(t)X(x)=b\, T(t) X''(x),\quad\quad (1) $$ where $a$ and $b$ are some non-zero constants. Now assume that there exist $x_0$ and $t_0$ such that $a T'(t_0)X(x_0)=b T(t_0) X''(x_0)\neq0$. Then there is a constant $\lambda\neq 0$ such that $$ X''(x)=\lambda X(x),\\ T'(t)=\frac ba\lambda \, T(t). $$ (The case $\lambda=0$ corresponds to the case where there is no such point $(x_0,t_0)$, which is equivalent to say that $X''(x)=0$ and $T'(t)=0$.)

Proof: Plugging $t=t_0$ in (1) gives $$ X''(x)=\frac{a T'(t_0)}{b\, T(t_0)}X(x), $$ since $b T(t_0)$ does not vanish from our hypothesis. Calling $\lambda=\frac{a T'(t_0)}{b T(t_0)}$ gives the ODE for $X(x)$. Note that we also have that $\lambda=X''(x_0)/X(x_0)$ which is also a well-defined ratio. Next, plugging $x=x_0$ in (1) gives directly the ODE for T(t).

This method is trivially generalized to other PDE that can be solved by separation of variables.


Remainder to make the argument easier to understand for those who don't have Griffith at hand. He separates $t$ and $x$ (page 20)

$$\Psi(x,t)=\psi(x)f(t) \tag{2.1}$$

then Schrödinger equation becomes

$$i\hbar\, \psi\frac{df}{dt} = -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}\,f+V\psi\, f$$

and then he divides both side by $\psi\, f$.

The mathematically rigorous way to do this sort of things is to postulate that $\psi(x)$ is non-zero for a least one $x$, then since $\psi$ is assumed to be continuous, it is non-zero on some open interval $]x_1, x_2[$. On that interval, we can then divide by $\psi(x)$ and solve the differential equation. Then we can make the interval as big as possible, and it may end up that $\lim_{x->x_2^-}\psi(x)=0$ then. The reasoning is still entirely rigorous in $]x_1, x_2[$. But then usually what will happen is twofold

  • to the right of $x_2$, we have another interval $]x_2, x_3[$ on which $\psi(x)$ is non-zero. So now we can put the two pieces together: $\psi$ is defined everywhere but at $x_2$; and
  • $\lim_{x->x_2^+} \psi(x)=0=\lim_{x->x_2^-}\psi(x)$, so we can define $\psi(x_2)$ to be 0 by continuity.

Troubles appear when the last condition on limit from the left and from the right does not hold.


This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .