I'll start by claiming that the action of the Schrödinger equation is the following:

$$\int \frac {h^2} {2m}\psi^{*'} \psi^{'} + \psi^* (V-E) \psi dx$$

To verify this, let's take the variation of it with respect to a small variation in $\psi^*$ and equate that to zero (note that $\psi^*$ and $\psi$ are considered independent):

$$\delta \int \frac {h^2} {2m} \psi^{*'} \psi^{'} + \psi^* (V-E) \psi dx = 0$$

$$\int \frac {h^2} {2m}\delta \psi^{*'} \psi^{'} + \delta \psi^* (V-E) \psi dx = 0$$

Integrating by parts:

$$\frac {h^2} {2m} \psi^{'} \delta \psi^{*}\Big{|}^b_a - \frac {h^2} {2m} \int \psi^{''} \delta \psi^* dx + \int \delta \psi^* (V-E) \psi dx = 0$$

Since the variation $\delta \psi^*$ is an arbitrarily small function that goes to zero at the end points of $[a,b]$, let's make the added assumption that it's first and second derivatives also go to zero at the end points:

$$\int \Big[- \frac {h^2} {2m} \psi^{''} + (V-E) \psi \Big] \delta \psi^* dx = 0$$

Again, since the variation $\delta \psi^*$ is arbitrary (but small), and zero at the end points, the fundamental theorem of variational calculus tells us that:

$$- \frac {h^2} {2m} \psi^{''} + (V-E) \psi = 0$$

So far so good. Now if in the step where we did integration by parts, we changed the order, we would have got the following:

$$\frac {h^2} {2m} \delta \psi^{*'} \psi\Big{|}^b_a - \frac {h^2} {2m} \int \delta \psi^{*''} \psi dx + \int \delta \psi^* (V-E) \psi dx = 0$$

Our assumption about the derivatives of the variation in $\psi^*$ means that:

$$\int \Big[ -\frac {h^2} {2m} \delta \psi^{*''} + \delta \psi^* (V-E) \Big] \psi dx = 0$$

Now what we have in the square brackets is a function which is zero at the endpoints of $[a,b]$, and arbitrary in the middle (though small):

So it seems that the fundamental theorem of variations calculus implies:

$$\psi = 0$$

This obviously is untrue, as $\psi$ wasn't any particular wave function, and could be non-zero. So where am I going wrong.

We know that the action is correct; something we proved. It's just that changing the order of functions while integrating by parts gives us a dubious result. It might also be that since we limited the derivatives of $\delta \psi^*$ to zero at the end points, the fundamental theorem used might not be valid, but I am not sure.

  • $\begingroup$ Hints to the question (v1): Write out all manipulations explicitly. Identify precisely how and where you intend to use the fundamental theorem of variations. Scrutinize in particular the validity of your above sentence: the square bracket is arbitrary in the middle. It is not. $\endgroup$ – Qmechanic Aug 2 '16 at 20:09
  • $\begingroup$ @Qmechanic: If you are trying to say that the integral is simply $\int \delta S^* \psi dx$, where $S^*$ is just the complex conjugate of Schrödinger equation, and equal to zero, then I claim (and I might be wrong) that $\delta S^* \neq 0$. I say so because the variation in $\psi^*$ is just a small function we add to $\psi^*$, and it doesn't have to satisfy the Schrödinger equation, hence it can be non-zero. $\endgroup$ – Sidd Aug 2 '16 at 20:27

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