# Why only the first derivative of the wave function must be continuous?

Let's say we have a piece-wise differentiable periodic potential. For example, like a zigzag. Let's assume that we know eigenvalues and eigenfunctions on each interval, $$\Psi_{1, 2}$$ on the left and $$\Psi_{3, 4}$$ on the right. How do we find the general solution?

For a periodic array of finite potential wells I would say that the wave function and its derivative must be continuous at the boundaries of each pit, but the second derivative can have discontinuity because the potential has a discontinuity. However, it is not the case for a continuous potential. Do I also require that the second derivatives are equal? But that implies more equations than I have eigenfunctions for a given energy.

In case of a potential like in picture, the functions on the left are

$$\Psi_{left}(x) = C_{1} \Psi_{1}(x) + C_{2} \Psi_{2}(x),\\ \Psi_{right}(x) = C_{3} \Psi_{3}(x) + C_{4} \Psi_{4}(x).$$

What I should do? Obliviously, the functions are continuous, so

$$C_{1} \Psi_{1}(0) + C_{2} \Psi_{2}(0) = C_{3} \Psi_{3}(0) + C_{4} \Psi_{4}(0)$$

Their derivatives also must be continuous:

$$C_{1} \Psi_{1}^{\prime}(0) + C_{2} \Psi_{2}^{\prime}(0) = C_{3} \Psi_{3}^{\prime}(0) + C_{4} \Psi_{4}^{\prime}(0)$$

But what is my next step? Do I also require second derivatives to be continuous,

$$C_{1} \Psi_{1}^{\prime\prime}(0) + C_{2} \Psi_{2}^{\prime\prime}(0) = C_{3} \Psi_{3}^{\prime\prime}(0) + C_{4} \Psi_{4}^{\prime\prime}(0) ?$$

But in this case, when I apply Bloch theorem, do I also apply constraints on the function derivatives?

$$\left[C_{1} \Psi_{1}(-L/2) + C_{2} \Psi_{2}(-L/2)\right] e^{ikL} = C_{3} \Psi_{3}(L/2) + C_{4} \Psi_{4}(L/2)$$ ...and the same for the first and the second derivative? I would end up with 6 equations on 4 coefficients in this case. However, I cannot allow second derivative to have discontinuity anywhere because the potential doesn't have discontinuities. What am I doing wrong?

• Jun 18 at 22:42

You don't have to enforce the continuity of $$\psi''$$. The time-independent Schrödinger equation is

$$\psi''(x) = - \frac{2m\left[E-V(x)\right]}{\hbar^2}\psi(x).$$

Since the potential $$V$$ is continuous, if $$\psi$$ is continuous, $$\psi''$$ will automatically be continuous. In other words, you could impose the continuity of $$\psi''$$, but this boundary condition won't give you an independent equation: it will be the same equation as the one you get by imposing the continuity of $$\psi$$.

• Thank you, that's it! As long as the functions satisfy the Schrödinger equation in left and right intervals, the condition on continuity of the second derivatives should be just the condition on the function itself multiplied by a constant. Jun 19 at 3:22

Ignoring Bloch's theorem for the moment, my starting point would be to take the time independent Schrödinger equation, $$-{\hbar^2\over2m}{\mathrm d^2\psi\over d x^2} + U(x)\psi(x)= E\psi(x)$$, then I would observe that $$U''(x)$$ is some sort of Dirac comb. So this means that the second derivative of the Schrödinger equation will have naked $$\delta(x-x_0)$$ terms in it.

When we have a differential equation that has naked $$\delta$$-terms in it, we usually soak those up entirely with the highest derivative we are taking, which in this case would be $$\psi^{(4)}(x).$$ This means that $$\psi'''(x)$$ will be discontinuous at the edges by a determined amount, but $$\psi'',\psi',$$ and $$\psi$$ will all be continuous.

Then we get to the difficult part, which is trying to figure out how Bloch's theorem fits in, and I don't think it alters this basic story, just changes the mass term or something and thereby changes the exact size of the discontinuities, probably.