Is Griffiths Quantum 2.51 the Ground State Solution?

  • Thread starter Thread starter NeoDevin
  • Start date Start date
  • Tags Tags
    Griffiths Quantum
AI Thread Summary
The discussion centers on problem 2.51 from Griffiths' "Introduction to Quantum Mechanics," focusing on a specific potential and its ground state solution. The potential V(x) is defined as V(x) = -(\hbar^2 a^2/m)sech^2(ax), and the ground state wave function is proposed as ψ_0(x) = A sech(ax). Participants confirm that the absence of nodes in the wave function indicates it is indeed the ground state, and the energy associated with this state is calculated as -(\hbar^2 a^2/2m). Additionally, the discussion touches on the reflectionless nature of the potential, emphasizing that all incident particles pass through without reflection.
NeoDevin
Messages
334
Reaction score
2
[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.
 
Last edited:
Physics news on Phys.org
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.
 
Avodyne said:
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?

Avodyne said:
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.
 
The fact that it has no nodes guarantees that it's the ground state?
 
NeoDevin said:
The fact that it has no nodes guarantees that it's the ground state?

Bingo!
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top