# Beginner question on constraints of a wave function in quantum mechanics

I am working through Griffiths’ Introduction to Quantum Mechanics. In chapter 1, he attempts to impose a condition such that $$\frac{d}{dt}\int_{-\infty}^\infty\left|\psi(x,t)\right|^2dx=0$$ so that the normalization of a solution to the Schrödinger equation is independent of time. He derives that $$\frac{d}{dt}\int_{-\infty}^\infty\left|\psi(x,t)\right|^2dx=\frac{i\hbar}{2m}\left.\left(\psi^\star\frac{\partial\psi}{\partial x}-\frac{\partial\psi^\star}{\partial x}\psi\right)\right|_{-\infty}^\infty.$$ Griffiths then concludes that a wave function must satisfy $$\psi\rightarrow 0\qquad\textrm{as}\qquad x\rightarrow\pm\infty.$$

1. Is this condition really enough?
2. For example, are there square-integrable functions $\psi$ such that $$\psi\rightarrow 0,\frac{\partial\psi^\star}{\partial x}\rightarrow\infty\qquad\textrm{as}\qquad x\rightarrow\pm\infty?$$
3. What would be a necessary condition to impose on $\psi$ to ensure that the above quantity is zero?
4. The question Must the derivative of the wave function at infinity be zero? suggests that having a function with compact support is sufficient. Is this necessary?
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I don't think you are describing Griffith's argument correctly. He does not claim that $\psi \rightarrow 0$ as $x \rightarrow \infty$ as a consequence of the time independence of the normalization of $\psi$. He claims $\psi \rightarrow 0$ as a consequence of $\psi$ being normalizable. However he then has a horrible footnote warning about good mathematicians with pathological counterexamples and saying that in physics the wave function always goes to zero at infinity. Except of course he later uses a formalism for scattering problems where $\psi(x,t)$ behaves as $e^{ikx}$ at infinity. ;) –  Jeff Harvey Dec 24 '12 at 18:10

1. No this is not enough. You were given counterexamples on the web site you mention in 4.

2. Yes, there are such functions, for example $(\sin x^3)/x$

3. It is hard to tell what is necessary, besides the trivial condition $\psi\psi'\to 0$.

4. Compact support is sufficient but not necessary.

5. You wrote the formula incorrectly: your RHS is $0$.

6. Physicists frequently do not state their conditions precisely, you should accept this when you read physics literature.

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@5. I corrected the typo. –  james Dec 24 '12 at 17:31