# Normalization Conditions of Wave functions

1. Jan 8, 2012

### fscman

I am currently reading through Griffiths Quantum Mechanics textbook, and on page 14, Griffiths proves that
$\frac{d}{dt}\int_{-\infty}^{\infty} |\Psi(x,t)|^2 \, dx = \left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}$
and he claims that the right hand side evaluates to zero since $\Psi$ must be normalizable and hence $\lim_{x \to \infty} \Psi(x,t) = 0$.
My question is whether $\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}$ must also equal zero. I ask this because:

Suppose we take the function $\Psi (x,t)=\frac{1}{x} \sin(x^9)$. Then, we can calculate $\int_{-\infty}^{\infty} \Psi^2 \, dx = 1.14...$ so $\Psi$ IS normalizable. We can find that $\frac{\partial \Psi}{\partial x}=\frac{1}{x^2}\left(9x^9 \cos(x^9)-\sin(x^9)\right)=9x^7 \cos(x^9)-\frac{\sin(x^9)}{x^2}$,
so if we take $\Psi^* \frac{\partial \Psi}{\partial x}=9 x^6 \sin(x^9) \cos (x^9) - \frac{\sin^2(x^9)}{x^3}$, we find that this expression does not approach 0 as x approaches infinity (which renders Griffiths' argument using the fact $\lim_{x \to \infty} \Psi(x,t) = 0$ to explain why the original integral false; there are functions that tend to zero as x approaches infinity but the derivative can grow arbitrarily large). Of course, in this case, $\left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}$ happens to vanish since $\Psi^*=\Psi$, but if we choose some general complex function, then I would guess that $\left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}$ might not vanish if we do not set $\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}$ equal to zero. However, I have never seen this condition in quantum mechanics textbooks. Is there something wrong with my analysis, or is $\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}=0$ an implicit assumption physicists make? Thanks.

Last edited: Jan 8, 2012
2. Jan 9, 2012

### Fredrik

Staff Emeritus
I don't have a complete answer, just a few observations.
• Physicists are really sloppy with these things, so they'll say that $\psi$ must go to 0, even though it may not be true.
• As you have noted, it isn't true. To prove that $\psi\to 0$, you will need other assumptions than than square-integrability.
• We are also assuming that $\psi$ is a solution of the Schrödinger equation. So $\psi$ is differentiable, and therefore continuous.
• The right-hand side you're working with is 0 if $\psi\to 0$ and $\partial\psi/\partial x$ is bounded.
I suspect that it's possible to prove that if $\psi$ is square-integrable, continuous and has bounded first derivatives, then $\psi\to 0$. I also suspect that there's something very "unphysical" about wavefunctions with unbounded partial derivatives. I expect them to have infinite energy or something like that.

3. Jan 9, 2012

### tom.stoer

You are right!

Physicists sometimes believe that normalization of a wave function requires that it approaches zero at infinity, i.e. lim x→±∞ ψ(x) = 0. This assumption is wrong - there are well-known counter examples. e.g.:

ψ(x) = 1 for x in [n,n+1/n2] for all integers n
ψ(x) = 0 elsewhere

lim x→±∞ ψ(x) does not exist, but of course the infinite sum for the normalization converges which means that ψ(x) is normalizable. You can even construct wave functions with increasing height of the peaks for x→±∞

You may exclude those wave functions based on "physical reasoning" but in pure maths i.e. for the L2 Hilbert space they are perfectly valid.

4. Jan 9, 2012

### Fredrik

Staff Emeritus
Here's a nice example of a continuous square-integrable function that doesn't go to zero as the variable goes to infinity, from this article.

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5. Jan 9, 2012

### fscman

Thanks for your replies! I will look more into the mathematical side of L2 spaces and read the article. I believe the "physical reality" argument for why ψ approaches 0 as x approaches infinity, and I just realized that from the time independent Schrodinger equation, if we require the potential energy to be bounded at infinity, then $\frac{\partial^2 \Phi}{\partial x^2}$ must approach 0 as x approaches infinity if the original function ψ behaves that way. Then, by mathematical arguments involving the mean value theorem, I believe that implies $\lim_{x \to \infty} \frac{\partial \Phi}{\partial x}=0$ also.