Checking that a given potential has a ground state

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The discussion revolves around verifying whether the potential V(x) = -h²a²/m sech²(ax) has a ground state represented by the wavefunction ψ₀(x) = A sech(ax). Participants debate the validity of simply inserting ψ₀ into the time-independent Schrödinger equation to confirm it as a ground state, questioning if this method adequately demonstrates that ψ₀ is indeed the ground state. They note that in one-dimensional systems, the ground state typically has no nodes, which is linked to its energy level. The conversation highlights the relationship between the number of nodes in a wavefunction and energy levels, emphasizing that energy increases with more nodes. Ultimately, the participants seek a deeper understanding of the implications of their findings regarding the ground state.
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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 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

$$-\frac{h^2}{2m} \frac{d^2\psi_0}{dx^2} + V\psi_0 = E \psi_0$$
 
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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 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.
 
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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. :)
 
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