Is Griffiths Quantum 2.51 the Ground State Solution?

[NeoDevin]

[SOLVED] Griffiths Quantum 2.51

This is problem 2.51 from Griffiths Introduction to Quantum Mechanics, 2nd ed. p89.

Homework Statement
Consider the potential

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

where $a$ is a positive constant, and "sech" stands for the hyperbolic secant.

a) Graph this potential.

b) Check that this potential has the ground state

$$\psi_0(x) = A sech(a x)$$

and find its energy. Normalize $\psi_0$, and sketch its graph.

c) Show that the function

$$\psi_k(x) = A\left(\frac{i k - a tanh(a x)}{i k + a}\right)e^{i k x}$$

(where $k = \sqrt{2 m E}/\hbar$ as usual) solves the Schr\ddot{o}dinger equation for any (positive) energy $E$. Since $tanh z \rightarrow -1$ as $z \rightarrow -\infty$,

$$\psi_k(x) \approx A e^{i k x}$$, for large negative x.

This represents, then, a wave coming in from the left with no accompanying reflected wave (i.e., no term exp(-ikx)). What is the asymptotic form of $\psi_k(x)$ at large positive $x$? What are R and T, for this potential? Comment: This is a famous example of a reflectionless potential - every incident particle, regardless of its energy, passes right through.

The attempt at a solution

Part a is easy, just draw the graph, intersecting the y-axis at -\frac{\hbar^2 a^2}{m}.

Part b is a little more difficult. I can show that it is a solution, but I'm not sure how to guarantee that it's the ground state. With that potential, I get that

$$\hat H\psi_0 = -\frac{\hbar^2 a^2}{2 m} \psi_0$$

If this result was the same as the minimum potential, I could say for certain that it's the ground state.

And for the normalization constant I get

$$A = \sqrt{\frac{a}{2}}$$

Part c I don't have much clue for. I tried just putting the hamiltonian into mathematica with that wave function, to see what it gives me, but it didn't give me anything that I can see looks like the RHS of the equation, any suggestions on how to go about part c would be appreciated.

 

[Avodyne]

Part b: do you recall any paricular property that the ground state wave function always has for any potential (in one dimension), and that no excited-state wave function has?

Part c: plug in and grind should work to show that it is a solution.

 

[NeoDevin]

Part b: do you recall any paricular property that the ground state wave function always has for any potential (in one dimension), and that no excited-state wave function has?

The lowest energy?

Part c: plug in and grind should work to show that it is a solution.

Ok, I'll try it again and write back here later/tomorrow.

 

[NeoDevin]

The fact that it has no nodes guarantees that it's the ground state?

 

[Avodyne]

The fact that it has no nodes guarantees that it's the ground state?

Bingo!