Yukawa Potential Homework: Sketch, Schrodinger Eq & Perturbation Theory

You should get:E_0^{(1)}=-\frac{\lambda}{2}E_0Yes, the potential should look like that.By the way, you are very close with your Schrodinger equation, but it's just a couple of typos (I think you just forgot the squared part in the second term). It should be:E\psi (r,\theta,\phi)=\left[-\frac{\hbar^{2}}{2m_{e}r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial}{\partial r}\right)-\frac{\hbar^{2}}{2m_{e}r^{2}}\
  • #36
Hart said:
Got a bit confused now, this is what I have noted down in addition to the previous post (I think this is more correct):

[tex]\left<\psi V_{\gamma}(r)\psi\right> = \int^{\infty}_{0}\left[-\left(\frac{1}{\pi r a^{3}}\right)e^{-r\left(\frac{2}{a}-\gamma\right)}\right]dr[/tex]

so then get this:

[tex]=\left[\frac{1}{\pi a^{3}}\left(\frac{2}{a}-\gamma\right)e^{-r\left(\frac{2}{a}-\gamma\right)}\right]^{\infty}_{0}[/tex]

and after input limits:

[tex]=\frac{1}{\pi a^{3}}\left(\frac{2}{a}-\gamma\right)[/tex]

.. any good?

You are still leaving out constants, like charge. Also, you completely changed the solution. Not sure what happened but now things are worse.

Go back and carefully step through everything on paper that way you can keep track of everything.
 
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  • #37
These are my exact full calculations:

[tex]V_{\gamma}(r) = -\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}[/tex]

[tex]\left<\psi(r)|V_{\gamma}(r)|\psi(r)\right> = \left<\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right> = 4\pi \int^{\infty}_{0}\psi(r)V_{\gamma}(r)\psi(r)r^{2}dr[/tex]

[tex]=\left(-\frac{4\pi q^{2}}{4\pi \epsilon_{0}}\right)\int^{\infty}_{0}r e^{-\gamma r}\psi(r)\psi(r) dr[/tex]

[tex]=\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\int^{\infty}_{0}r e^{-\left(\frac{2}{a}+\gamma\right)r} dr[/tex]

[tex]=\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(-\frac{4 \pi}{\frac{2}{a}+\gamma}\right)[/tex]

[tex]=\frac{4 q^{2}}{\epsilon_{0}a^{3}\left(\frac{2}{a}+\gamma\right)}[/tex]

.. I'm hoping this is along the right lines, though I can't see that I've missed anything else now, but I'm unsure what I need to do further now.
 
Last edited:
  • #38
The solution to the integral in the 2nd to last line is incorrect. Scroll back a page and see how we solved it.
 
  • #39
Hart said:
so..

[tex]\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left(\left(-\frac{r}{\alpha}-\frac{1}{\alpha^{2}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right)\right|^{\infty}_{0}[/tex]

??

And then inputting the limits:

[tex]\implies = \left(0-\frac{1}{\alpha^{2}}\right)[/tex]

??

Therefore:

[tex]\implies = \left(-\frac{4 \pi }{\alpha^{2}}\right)[/tex]

.. better? :wink:

nickjer said:
Yes, now put back in all the constants. And it should look very similar to the ground state energy of a hydrogen atom using the coulomb potential.

.. that's what I've used, obviously with the value of [itex]\alpha[/itex] substituted.

.. OH wait, it's [itex]\alpha^{2}[/itex] not [itex]\alpha[/itex], so then the result should be:

[tex]=\frac{4 q^{2}}{\epsilon_{0}a^{3}\left(\frac{2}{a}+\gamma\right)^{2}}[/tex]

correct now?
 
  • #40
Then where did the 4*pi come from since you used it in line 2 already. And why isn't the denominator squared?
 
  • #41
Also, once you finished this, you will need to do the same for the coulomb potential. Since you need to do

[tex]<\psi_0|V_\lambda-V|\psi_0>=<\psi_0|V_\lambda|\psi_0>-<\psi_0|V|\psi_0>[/tex]

But that will be much easier to solve, since we solved it basically already. You would replace [tex]\alpha[/tex] from your previous integral with just [tex]2/a_0[/tex]
 
  • #42
so it should be:

[tex]\left<\psi(r)|V_{\gamma}(r)|\psi(r)\right> = \left<\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right> =\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(-\frac{1}{\left(\frac{2}{a}+\gamma}\right)^{2}\right)[/tex]

better?
 
  • #43
Hart said:
.. of course! what a silly mistake there! :grumpy:

so..

[tex]\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left(\left(-\frac{r}{\alpha}-\frac{1}{\alpha^{2}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right)\right|^{\infty}_{0}[/tex]

??

And then inputting the limits:

[tex]\implies = \left(0-\frac{1}{\alpha^{2}}\right)[/tex]

??

Therefore:

[tex]\implies = \left(-\frac{4 \pi }{\alpha^{2}}\right)[/tex]

.. better? :wink:

Forgot to mention, the solution is positive. You subtract the last limit of r=0, making it positive.
 
  • #44
erm.. yep. so should be:

[tex]\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left(\left(\frac{r}{\left(\frac{2}{a}+\gamma\right)}-\frac{1}{\left(\frac{2}{a}+\gamma\right)^{2}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right)\right|^{\infty}_{0} = \frac{4\pi}{\left(\frac{2}{a}+\gamma\right)^{2}}[/tex]

right?

but that doesn't help simplify this:

[tex]\left<\psi(r)|V_{\gamma}(r)|\psi(r)\right> = \left<\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right> =\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(\frac{1}{\left(\frac{2}{a}+\gamma}\right)^{2}\right)[/tex]

does it?
 
  • #45
There is no 4*Pi in the first line, since you already pulled out the angular integrals. The last line is now correct. To simplify it, I would recommend pulling out the 2/a from the denominator so it more resembles the ground state of the coulomb potential.

Now go back and solve this again for just the coulomb potential. You will notice a similarity.
 
  • #46
um, don't get how I can rearrange that how you said :confused:

.. also, so to get the final answer I need also caclulate this:

[tex]\int_{0}^{\infty} r e^{-\left(\frac{2}{a_{0}}\right)r}dr = \left(\left(-\frac{a_{0}r}{2}-\frac{a_{0}^{2}}{4}\right)e^{-\left(\frac{a_{0}}{2}\right)r}\right)\right|^{\infty}_{0} = \left(\frac{a_{0}r}{2}+\frac{a_{0}^{2}}{4}\right)[/tex]

Then take this away from what we're currently working out, so that the final answer is:

[tex]<\psi_0|V_\lambda-V|\psi_0>=<\psi_0|V_\lambda|\psi_0>-<\psi_0|V|\psi_0>[/tex]

correct??
 
  • #47
That is correct (in theory). But you solved the integral wrong. It is the exact same integral as before but without the gamma.
 
  • #48
..huh?
 
  • #49
nickjer said:
You have:

[tex]\int r e^{-\alpha r}dr[/tex]

Use,
[tex]r = u[/tex]
[tex]e^{-\alpha r} dr = dv[/tex]

That gives,
[tex]dr = du[/tex]
[tex]\frac{-1}{\alpha} e^{-\alpha r} = v[/tex]

Use,
[tex]\int u dv = uv - \int v du[/tex]
[tex]\int r e^{-\alpha r}dr = \frac{-r}{\alpha} e^{-\alpha r}+\int \frac{1}{\alpha} e^{-\alpha r}dr
= \frac{-r}{\alpha} e^{-\alpha r}-\frac{1}{\alpha^2}e^{-\alpha r}[/tex]

I left out the bounds for simplicity, but you should include them. I also used alpha to make my work easier.

The same as this but now alpha = 2/a and not 2/a+gamma.
 
  • #50
.. so you mean for the second part it will be this:

[tex]\int_{0}^{\infty} r e^{-\left(\frac{2}{a}\right)r}dr = \left(\frac{r}{\left(\frac{2}{a}\right)}-\frac{1}{\left(\frac{2}{a})^{2}}\right)\right|^{\infty}_{0} = \frac{4\pi}{\left(\frac{2}{a}\right)^{2}} = \pi a^{2}[/tex]

??

Going back to the first bit, how do I rearrange that result better then? I didn't get what you said about pulling the [itex]\frac{2}{a}[/itex] term out?!
 
<h2>1. What is the Yukawa Potential?</h2><p>The Yukawa Potential is a mathematical function used to describe the interaction between two particles, such as an electron and a proton, in quantum mechanics. It is named after the Japanese physicist Hideki Yukawa, who first proposed its use in 1935.</p><h2>2. How is the Yukawa Potential represented graphically?</h2><p>The Yukawa Potential is typically represented as a potential energy curve, with the distance between the two particles on the x-axis and the potential energy on the y-axis. The curve starts at a high value when the particles are close together, decreases rapidly, and eventually approaches zero as the particles move further apart.</p><h2>3. What is the Schrodinger Equation and how is it related to the Yukawa Potential?</h2><p>The Schrodinger Equation is a fundamental equation in quantum mechanics that describes how the wave function of a quantum system changes over time. The Yukawa Potential is one of the terms in the Schrodinger Equation, representing the potential energy of the system.</p><h2>4. How is Perturbation Theory used in solving problems involving the Yukawa Potential?</h2><p>Perturbation Theory is a mathematical technique used to approximate solutions to complex problems by breaking them down into simpler, solvable parts. In the case of the Yukawa Potential, Perturbation Theory is used to find an approximate solution to the Schrodinger Equation by treating the potential energy term as a small perturbation to the overall system.</p><h2>5. What are some real-world applications of the Yukawa Potential?</h2><p>The Yukawa Potential has been used in various fields of physics, including nuclear physics, particle physics, and condensed matter physics. It has also been applied to study the behavior of atoms and molecules, as well as the interactions between subatomic particles in particle accelerators. Additionally, the Yukawa Potential has been used in astrophysics to model the gravitational interaction between galaxies and dark matter.</p>

1. What is the Yukawa Potential?

The Yukawa Potential is a mathematical function used to describe the interaction between two particles, such as an electron and a proton, in quantum mechanics. It is named after the Japanese physicist Hideki Yukawa, who first proposed its use in 1935.

2. How is the Yukawa Potential represented graphically?

The Yukawa Potential is typically represented as a potential energy curve, with the distance between the two particles on the x-axis and the potential energy on the y-axis. The curve starts at a high value when the particles are close together, decreases rapidly, and eventually approaches zero as the particles move further apart.

3. What is the Schrodinger Equation and how is it related to the Yukawa Potential?

The Schrodinger Equation is a fundamental equation in quantum mechanics that describes how the wave function of a quantum system changes over time. The Yukawa Potential is one of the terms in the Schrodinger Equation, representing the potential energy of the system.

4. How is Perturbation Theory used in solving problems involving the Yukawa Potential?

Perturbation Theory is a mathematical technique used to approximate solutions to complex problems by breaking them down into simpler, solvable parts. In the case of the Yukawa Potential, Perturbation Theory is used to find an approximate solution to the Schrodinger Equation by treating the potential energy term as a small perturbation to the overall system.

5. What are some real-world applications of the Yukawa Potential?

The Yukawa Potential has been used in various fields of physics, including nuclear physics, particle physics, and condensed matter physics. It has also been applied to study the behavior of atoms and molecules, as well as the interactions between subatomic particles in particle accelerators. Additionally, the Yukawa Potential has been used in astrophysics to model the gravitational interaction between galaxies and dark matter.

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