Checking that a given potential has a ground state

  • Thread starter Thread starter hmparticle9
  • Start date Start date
  • Tags Tags
    Quantum mechanics
hmparticle9
Messages
151
Reaction score
26
Homework Statement
Checking that a given potential has a ground state
Relevant Equations
$$V(x) = -\frac{h^2 a^2}{m}\text{sech}^2(ax)$$
In my book the following potential is given:

$$V(x) = -\frac{h^2 a^2}{m}\text{sech}^2(ax)$$

then the task is: Check that this potential has the ground state ##\psi_0(x) = A\text{sech}(ax)##.

I first thought to put it in the time-independent Schrodinger equation and show that it is a solution, but surely that just shows that it satisfies the Schrodinger equation. The question is to show that the potential has a ground state equal to ##\psi_0##. I spent all day on this problem when I was right in the beginning: just jam it into the Schrodinger equation and out comes its ground state energy times itself. But by doing that are we actually showing that the GROUND STATE is given by ##\psi_0##? That is, are the solutions to this problem incorrect? Is there something missing?

Just to clarify what the solutions are saying: they show that ##\psi_0## satisfies

$$-\frac{h^2}{2m} \frac{d^2\psi_0}{dx^2} + V\psi_0 = E \psi_0$$
 
Last edited:
Physics news on Phys.org
hmparticle9 said:
I spent all day on this problem when I was right in the beginning: just jam it into the Schrodinger equation and out comes its ground state energy times itself. But by doing that are we actually showing that the GROUND STATE is given by ##\psi_0##? That is, are the solutions to this problem incorrect? Is there something missing?
The number of nodes of the wavefunction is related to the energy level.
 
  • Love
Likes hmparticle9
https://en.wikipedia.org/wiki/Ground_state

Says here that in one-dimension it can be proven that the ground state has no nodes. Thanks for putting me on the right track :)
 
hmparticle9 said:
https://en.wikipedia.org/wiki/Ground_state

Says here that in one-dimension it can be proven that the ground state has no nodes. Thanks for putting me on the right track :)
That's what it says, but do you have an intuitive idea why that might be the case? What is the connection between energy and number of nodes?
hmparticle9 said:
I first thought to put it in the time-independent Schrodinger equation and show that it is a solution, but surely that just shows that it satisfies the Schrodinger equation.
Surely it does. Did you do it?
 
The connection between energy and number of nodes is that the energy increases as the number of nodes increases?

Yes I did it. :)
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top