- #1

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^{3}x ∇(ψ

^{*}ψ) is zero because the total derivative vanishes if ψ is normalizable.

Does this mean that the integral of d

^{3}x ∇(ψ

^{*}ψ) is ψ

^{*}ψ evaluated at the limits where ψ is zero ?

Thanks

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- Thread starter dyn
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- #1

- 764

- 57

Does this mean that the integral of d

Thanks

- #2

- 5,453

- 2,765

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 ## ).

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- #3

- 764

- 57

Thanks for your reply. I read the link but is it possible to explain this example using the total derivative explanation ?I have read that the integral of d^{3}x ∇(ψ^{*}ψ) is zero because the total derivative vanishes if ψ is normalizable.

Does this mean that the integral of d^{3}x ∇(ψ^{*}ψ) is ψ^{*}ψ evaluated at the limits where ψ is zero ?

- #4

- 5,453

- 2,765

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

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- #5

- 764

- 57

d

- #6

- 5,453

- 2,765

No. It becomes a surface integral of ## \int \psi^{\dagger} \psi \, \hat{n} dA ## at ## r=+\infty ##.^{*}ψ does equal 1 as the wavefunction is normalisable but is it correct to say that the integral of

d^{3}x ∇(ψ^{*}ψ) is ψ^{*}ψ evaluated at the limits of space ?

- #7

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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 ?No. It becomes a surface integral of ## \int \psi^{\dagger} \psi \, \hat{n} dA ## at ## r=+\infty ##.

If using spherical coordinates then ψ→0 as r→ infinity but what about the other limit r=0 ?

- #8

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- 2,765

Yes, to your first question. And ## r=0 ## doesn't matter. Only the surface integral 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 ?

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