The total derivative of a wavefunction

[dyn]

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

 

[Charles Link]

Perhaps Gauss' theorem for gradients is useful here: https://math.stackexchange.com/ques...eorem-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=$-\int p \, \hat{n} dA=\int \delta g \, d^3 x$ ).

 

[dyn]

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 for your reply. I read the link but is it possible to explain this example using the total derivative explanation ?

 

[Charles Link]

Thanks for your reply. I read the link but is it possible to explain this example using the total derivative explanation ?

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.

 

[dyn]

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 ?

 

[Charles Link]

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 ?

 

[Charles Link]

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 $r=+\infty$.