hmparticle9
- 151
- 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
$$-\frac{h^2}{2m} \frac{d^2\psi_0}{dx^2} + V\psi_0 = E \psi_0$$
$$V(x) = -\frac{h^2 a^2}{m}\text{sech}^2(ax)$$
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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