Checking that a given potential has a ground state

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SUMMARY

The discussion centers on verifying that the potential \( V(x) = -\frac{h^2 a^2}{m}\text{sech}^2(ax) \) has a ground state represented by the wavefunction \( \psi_0(x) = A\text{sech}(ax) \). Participants emphasize the necessity of substituting \( \psi_0 \) into the time-independent Schrödinger equation to confirm it satisfies the equation, which indicates that it is indeed a valid ground state solution. Additionally, the relationship between the number of nodes in the wavefunction and energy levels is highlighted, with the consensus that the ground state in one dimension has no nodes.

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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 Schrödinger equation and show that it is a solution, but surely that just shows that it satisfies the Schrödinger 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 Schrödinger 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

$$-\frac{h^2}{2m} \frac{d^2\psi_0}{dx^2} + V\psi_0 = E \psi_0$$
 
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hmparticle9 said:
I spent all day on this problem when I was right in the beginning: just jam it into the Schrödinger 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.
 
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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 Schrödinger equation and show that it is a solution, but surely that just shows that it satisfies the Schrödinger 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. :)
 

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