Integration of the gradient of a vector

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dyn
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Hi.
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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  • #2
BvU
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No
 
  • #3
dyn
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Thanks. I'm trying to integrate over all volume ψ*∇ψ
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 ?
Thanks
 
  • #4
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Not so well-versed in telepathy, I'm afraid. So a bit of context might be helpful. My initial two-letter answer was more a feeble attempt at irony. But if I have to guess: you are in introductory quantum mechanics and looking at probability density and probability current density ?

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
dyn
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Yes your guess is correct. It is QM and showing that the expectation value of momentum is real.
[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 ?
Thanks
 
  • #6
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What is ##\psi^*## supposed to be? The complex conjugate of ##\psi##?
 
  • #7
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Yes
 
  • #8
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showing that the expectation value of momentum is real
Well, if it has three components, maybe you might want to do it one component at the time ... :rolleyes: ?


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 ?

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
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##...)

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
dyn
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I am just trying to follow some lecture notes I found online and to be honest , I don't like them. As part of the integration by parts I need to know what the integral of ∇ψ is but that doesn't seem to be very straightforward
 
  • #10
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Presumably, it would help to post the link to the lecture notes. If you really need to integrate ##\nabla\Psi##, my suggestion would be to do it component-wise.
 
  • #11
dyn
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To integrate ψ*(∇ψ)over all volume by parts I need to know what the integral of ∇ψ is. Does the integral of ∇ψ not have a simple form ?
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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Yes, in cartesian coordinates you have
$$\nabla \psi = (\partial\psi/\partial x,\partial\psi/\partial y, \partial\psi/\partial z)=\frac{\partial\psi}{\partial x} e_x +\frac{\partial\psi}{\partial y} e_y+\frac{\partial\psi}{\partial z} e_z$$, where ##e_x##, ##e_y## and ##e_z## are the basis vectors. So, you can do the integration term by term.
 
  • #13
dyn
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Yes, in cartesian coordinates you have
$$\nabla \psi = (\partial\psi/\partial x,\partial\psi/\partial y, \partial\psi/\partial z)=\frac{\partial\psi}{\partial x} e_x +\frac{\partial\psi}{\partial y} e_y+\frac{\partial\psi}{\partial z} e_z$$, where ##e_x##, ##e_y## and ##e_z## are the basis vectors. So, you can do the integration term by term.
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
 
  • #14
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some lecture notes I found online
o_O Have a link for us ?
 

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