Using the normalisation condition in 3D

In summary: This is because the total Hamiltonian for an atom is ##\hat{H} = \frac{-\hbar^2 \nabla_R^2}{2M} - \frac{\hbar^2 \nabla^2}{2\mu} - Ae^{\frac{-r}{a}}## and since ##\vec{R}## is the position vector of the center of mass, it follows that ##\vec{R} = (m_p+m_n)\vec{r}## where ##m_p+m_n## is the total mass of the atom. Therefore, the wavefunction of the center of mass is just a plane wave that travels in the
  • #1
Kara386
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Homework Statement


The Hamiltonian for an atom of deuteron is
##\hat{H} = \frac{-\hbar^2 \nabla_R^2}{2M} - \frac{\hbar^2 \nabla^2}{2\mu} - Ae^{\frac{-r}{a}}##
Where ##\nabla_R## is the differential operator for the centre of mass co-ordinates ##R = \frac{m_p\vec{r_p} + m_n\vec{r_n}}{M}## and ##\nabla## is the differential operator for the difference co-ordinates ##r_p## and ##r_n##. ##M## is the total mass ##m_p+m_n## and ##\mu## is the reduced mass ##\frac{m_pm_n}{m_p+m_n}##.

Approximate the ground state wavefunction with:
##\phi(\vec{r}) = ce^{\frac{-\alpha r}{2a}}## where c and ##\alpha## are real and positive constants.
1. Use the normalisation condition to find c in terms of a.

2. Find an expression for the expectation value ##<\phi|\hat H|\phi>##.
If there's a better way of doing Braket notation in latex, please let me know! :)

Homework Equations

The Attempt at a Solution


My first question is about 1. So the question has ##\phi## as a function of the vector ##\vec{r}##, but the r in the exponential is not a vector. So in what way is ##\phi## actually a function of ##\vec{r}##? I'm unclear on how that works, in terms of when I'm calculating the expectation value, am I integrating with respect to ##d^3\vec{r}##? Or just ##dr##, since there is no ##\vec{r}## dependence. If I do need to integrate w.r.t. ##d^3\vec{r}##, how do I do that? I have a feeling it involves spherical co-ordinates. Probably a ##4\pi r^2dr##. But I'd appreciate a good explanation as to why; I've had a look online and can't find one!

For the second question, the most horrible integral I've ever encountered appeared. The same question applies here, really: do I integrate ##d^3\vec{r}##? Or just ##dr##? Either way, I have no idea how to actually calculate ##\nabla_R^2## when both ##r## and ##\vec{r}## depend on ##R##. I really have no idea what ##\nabla_R^2 ## applied to ##\phi## actually means, in terms of how to calculate those derivatives!

That's a long question, most of it stating things I don't know, so thank you for taking the time to read it! And thanks for any help. :)
 
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  • #2
Kara386 said:
If there's a better way of doing Braket notation in latex, please let me know! :)
Use \rangle, ##\rangle## or \langle, ##\langle##.
Kara386 said:
My first question is about 1. So the question has ϕϕ\phi as a function of the vector ⃗rr→\vec{r}, but the r in the exponential is not a vector. So in what way is ϕϕ\phi actually a function of ⃗rr→\vec{r}?
A notation of the form ##\phi(\vec{r})## can also be written as ##\phi(r, \theta,\varphi)##, therefore if only ##r## is present in the argument, that means ##\phi## is independent of ##\theta## and ##\varphi##.
Kara386 said:
I'm unclear on how that works, in terms of when I'm calculating the expectation value, am I integrating with respect to d3⃗rd3r→d^3\vec{r}? Or just drdrdr, since there is no ⃗rr→\vec{r} dependence. If I do need to integrate w.r.t. d3⃗rd3r→d^3\vec{r}, how do I do that?
You are integrating with respect to ##d\vec{r}^3##. It's just the notation for the volume element in spherical coordinate. So, find out the expression for the volume element in this coordinate system.
Kara386 said:
But I'd appreciate a good explanation as to why; I've had a look online and can't find one!
You can as well integrate in Cartesian coordinate, but since the given form of the wavefunction is easier to be integrated in the spherical coordinate, you better do it this way.
Kara386 said:
Either way, I have no idea how to actually calculate ∇2R∇R2\nabla_R^2 when both rrr and ⃗rr→\vec{r} depend on RRR. I really have no idea what ∇2R∇R2\nabla_R^2 applied to ϕϕ\phi actually means, in terms of how to calculate those derivatives!
No, ##\vec{r}## and ##\vec{R}## are independent variables. The question is rather strange actually because it asks you to calculate the expectation value of the total energy but it gives you only the wavefunction of the relative position. In this case, I would suggest that you define your own wavefunction for the center of mass. Note that the total wavefunction should look like ##\psi(\vec{R},\vec{r}) = \gamma(\vec{R})\phi(\vec{r})## where ##\gamma(\vec{R})## is the wavefunction of the center of mass. To find this wavefunction, note that the total Hamiltonian suggest that it should be that of a plane wave.
 
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Related to Using the normalisation condition in 3D

1. What is the normalisation condition in 3D?

The normalisation condition in 3D refers to the mathematical concept of normalisation, which is used to ensure that a vector or function has a magnitude of 1. In 3-dimensional space, this means that the square of the vector's length (or the magnitude of the function) plus the square of its width plus the square of its height should equal 1.

2. Why is the normalisation condition important in 3D?

The normalisation condition is important in 3D because it ensures that vectors and functions are properly scaled and do not become infinitely large or small. This is essential in many scientific and mathematical applications, as it allows for more accurate and meaningful calculations and interpretations.

3. How is the normalisation condition used in 3D?

The normalisation condition is used in 3D by dividing the vector or function by its length, width, and height, or by dividing each component of the vector by the square root of the sum of its squares. This ensures that the resulting vector or function has a magnitude of 1 and satisfies the normalisation condition.

4. What are the benefits of using the normalisation condition in 3D?

Using the normalisation condition in 3D has several benefits, including simplifying calculations, ensuring accuracy, and allowing for easier interpretation of results. It also helps to avoid errors and inconsistencies that may arise from vectors or functions with varying magnitudes.

5. Are there any limitations to using the normalisation condition in 3D?

While the normalisation condition is a useful mathematical concept, it may not always be applicable or necessary in certain situations. For example, in some cases, the normalisation condition may result in loss of information or oversimplification of a problem. It is important for scientists to carefully consider the context and purpose of their calculations before applying the normalisation condition in 3D.

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