- #1

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Is it true to say that the integral over all volume of ∇ψ where ψ is a scalar function of position and time is just ψ ?

Thanks

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- #1

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Is it true to say that the integral over all volume of ∇ψ where ψ is a scalar function of position and time is just ψ ?

Thanks

- #2

BvU

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No

- #3

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When I integrate by parts I need to know the integral of ∇ψ. I assumed it was ψ . Can you tell me what it is ?

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- #4

BvU

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Do you realize that, if ##\ \psi(\vec r, t)\ ## is a scalar, then ##\ \nabla \psi(\vec r, t)\ ## is a vector ? Meaning that ##\ \displaystyle \int \psi^* \,\nabla \psi(\vec r, t)\ dV ## is also a vector and can therefore never be equal to the scalar ##\psi## ?

- #5

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[QUOTE="

Do you realize that, if ##\ \psi(\vec r, t)\ ## is a scalar, then ##\ \nabla \psi(\vec r, t)\ ## is a vector ? Meaning that ##\ \displaystyle \int \psi^* \,\nabla \psi(\vec r, t)\ dV ## is also a vector and can therefore never be equal to the scalar ##\psi## ?[/QUOTE]

Yes I understand that ψ is a scalar and ∇ψ is a vector but to integrate by parts I need to know the integral of ∇ψ and assuming it is just ψ does end up giving me the correct answer in the end. How do I integrate ∇ψ over all volume ?

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- #6

Chestermiller

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What is ##\psi^*## supposed to be? The complex conjugate of ##\psi##?

- #7

BvU

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Yes

- #8

BvU

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Well, if it has three components, maybe you might want to do it one component at the time ... ?showing that the expectation value of momentum is real

Another thing that comes to mind is Gauss' theorem -- probably a much better path to explore in the case you have already seen things like probability density ( ##\rho = \psi^* \psi## ) -- which is clearly real, and its conservation requirement ## \left ( \ \displaystyle {\partial \rho \over \partial t } + \vec \nabla\cdot\vec j = 0 \right ) ## .

Is this homework ? What does your textbook say ?

Doesn't sound good at all to me, but I can be mistaken. Could you elaborate ? (Note that you don't integrate ##\nabla\psi## but ##\psi^*\nabla\psi##...)to integrate by parts I need to know the integral of ∇ψ and assuming it is just ψ does end up giving me the correct answer

Normally integration by parts just lets you write (very useful) things like ##\displaystyle {\int \bigl [ \psi^* \nabla \psi - (\nabla \psi^*)\psi\bigr ] \ dV = 2 \int \psi^* \nabla \psi \; dV}\ \ ## -- where you use that ##\psi^*\psi=0## at infinity.

- #9

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- #10

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- #11

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Obviously in 1-d its the integral of dψ/dx which is just ψ. Does the 3-d version not have a similar form ?

- #12

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$$\nabla \psi = (\partial\psi/\partial x,\partial\psi/\partial y, \partial\psi/\partial z)=\frac{\partial\psi}{\partial x} e_x +\frac{\partial\psi}{\partial y} e_y+\frac{\partial\psi}{\partial z} e_z$$, where ##e_x##, ##e_y## and ##e_z## are the basis vectors. So, you can do the integration term by term.

- #13

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Thanks. It just all looks so messy. I was hoping that a simple elegant way existed to perform the integration by parts as it would in 1-d

$$\nabla \psi = (\partial\psi/\partial x,\partial\psi/\partial y, \partial\psi/\partial z)=\frac{\partial\psi}{\partial x} e_x +\frac{\partial\psi}{\partial y} e_y+\frac{\partial\psi}{\partial z} e_z$$, where ##e_x##, ##e_y## and ##e_z## are the basis vectors. So, you can do the integration term by term.

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