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In quantum mechanics, one requires that $$\frac{d}{dt}\int_{-\infty}^\infty\left|\psi(x,t)\right|^2dx=0$$ in order for normalization to be independent of time.

In general, is it true that for any function $f\in L^2(\mathbb R^2)$, $$\frac{d}{dy}\int_{-\infty}^\infty \left|f(x,y)\right|^2dx=0?$$

2 Answers 2

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No, it's not! Consider for example

$$f(x,y) := g(y) \cdot \left( \frac{1}{\sqrt{2\pi}} \cdot \exp \left(- \frac{x^2}{2} \right) \right)^{\frac{1}{2}}$$

where $g \in L^1$ is an arbritary function. Then $$\int_{-\infty}^\infty |f(x,y)|^2 \, dx = |g(y)|^2 \cdot \int_{\mathbb{R}}\frac{1}{\sqrt{2\pi}} \cdot \exp \left(- \frac{x^2}{2} \right) \, dx = |g(y)|^2$$ since the (remaining) integrand is the density of the normal distribution $N(0,1)$.

You could choose for example $g(y) := e^{-{y^2}}$, then

$$\frac{d}{dy} \int_{-\infty}^\infty |f(x,y)|^2 \, dx = \frac{d}{dy} e^{-2y^2} = -4y \cdot e^{-2y^2} \not= 0$$

The equality $\frac{d}{dy} \int_{-\infty}^\infty |f(x,y)|^2 \, dx=0$ holds iff $\int_{-\infty}^\infty |f(x,y)|^2 \, dx =$ constant.

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No of course, unless $$\int_{-\infty}^\infty \left\lvert f(x, y)\right\rvert^2\, dx=\text{constant.}$$