Integration of the gradient of a vector

In summary: I don't have a link. But I found this video on YouTube that might help:In summary, this conversation is about integrating over all volume of ∇ψ where ψ is a scalar function of position and time. The integral of ∇ψ is ψ*∇ψ, but to integrate by parts you need to know the integral of ∇ψ. The expectation value of momentum is real and can be found using Gauss' theorem.
  • #1
dyn
773
61
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
No
 
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Likes Chestermiller
  • #3
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
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
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
What is ##\psi^*## supposed to be? The complex conjugate of ##\psi##?
 
  • #7
Yes
 
  • #8
dyn said:
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 ?

dyn said:
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 let's 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
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
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
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
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
eys_physics said:
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
dyn said:
some lecture notes I found online
o_O Have a link for us ?
 

1. What is the gradient of a vector?

The gradient of a vector is a mathematical operation that represents the rate of change of a scalar field in a particular direction. It is a vector that points in the direction of the steepest increase of the scalar field and its magnitude represents the rate of change.

2. How is the gradient of a vector calculated?

The gradient of a vector is calculated by taking the partial derivatives of the components of the vector with respect to the variables in the scalar field. These partial derivatives are then combined to form a vector, which is the gradient of the original vector.

3. What is the significance of the gradient of a vector?

The gradient of a vector is important in many fields of science and engineering, as it helps us understand the behavior of scalar fields and their rates of change. It is used in various applications such as physics, meteorology, and computer graphics.

4. What is the relationship between the gradient of a vector and its direction?

The gradient of a vector always points in the direction of the steepest increase of the scalar field. This means that the direction of the gradient vector is perpendicular to the level curves of the scalar field at that point.

5. Can the gradient of a vector have a negative magnitude?

Yes, the magnitude of the gradient of a vector can be negative if the scalar field is decreasing in the direction of the gradient. This indicates a decrease in the rate of change of the scalar field in that direction.

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