# The total derivative of a wavefunction

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## Main Question or Discussion Point

I have read that the integral of d3x ∇(ψ*ψ) is zero because the total derivative vanishes if ψ is normalizable.
Does this mean that the integral of d3x ∇(ψ*ψ) is ψ*ψ evaluated at the limits where ψ is zero ?
Thanks

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Perhaps Gauss' theorem for gradients is useful here: https://math.stackexchange.com/questions/1958252/gauss-divergence-theorem-for-volume-integral-of-a-gradient-field If $\psi^{\dagger} \psi$ is localized in extent, this says the result will be zero. $\\$ Additional item: The most use I have previously had for this gradient form of Gauss' law is to prove Archimededes' theorem, but every so often, it comes in handy. (Note: Force per unit volume in a fluid is $f_v=-\nabla p$. In addition, gravitational force per unit volume $f_g=-\delta g \hat{z}$, where $\delta$ is the density, in mass per unit volume. At equilibrium, the sum of the forces is zero. Archimedes' principle follows with a volume integral of this. Buoyant force=$-\int p \, \hat{n} dA=\int \delta g \, d^3 x$ ).

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I have read that the integral of d3x ∇(ψ*ψ) is zero because the total derivative vanishes if ψ is normalizable.
Does this mean that the integral of d3x ∇(ψ*ψ) is ψ*ψ evaluated at the limits where ψ is zero ?

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Perhaps. If $\int \psi^{\dagger} \psi \, d^3 x=1$ over a finite volume, than yes, that explanation would work. Otherwise, it is really necessary to go the complete vector calculus route. The complete vector calculus route is really the preferred one. The gradient is similar to, but, (in my opinion), is not exactly a differential. $\\$ In one dimension, with a differential you get $\int\limits_{-\infty}^{+\infty} f'(x) \, dx= \int\limits_{-\infty}^{+\infty} d f(x)= f(+\infty)-f(-\infty)$. With vector calculus and a volume integral of a gradient, you get a surface integral.

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The integral over all space of ψ*ψ does equal 1 as the wavefunction is normalisable but is it correct to say that the integral of
d3x ∇(ψ*ψ) is ψ*ψ evaluated at the limits of space ?

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The integral over all space of ψ*ψ does equal 1 as the wavefunction is normalisable but is it correct to say that the integral of
d3x ∇(ψ*ψ) is ψ*ψ evaluated at the limits of space ?
No. It becomes a surface integral of $\int \psi^{\dagger} \psi \, \hat{n} dA$ at $r=+\infty$.

dyn
No. It becomes a surface integral of $\int \psi^{\dagger} \psi \, \hat{n} dA$ at $r=+\infty$.
And that is because of Gauss' theorem for gradients and the fact that the wavefunction tends to zero as x,y.z all tend to ± infinity ?
If using spherical coordinates then ψ→0 as r→ infinity but what about the other limit r=0 ?

Yes, to your first question. And $r=0$ doesn't matter. Only the surface integral at $r=+\infty$.