Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
3rd Edition
ISBN: 9781107189638
Author: Griffiths, David J., Schroeter, Darrell F.
Publisher: Cambridge University Press
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Chapter 8, Problem 8.16P

(a)

To determine

An upped bound on the ground state of the infinite square well using the function ψ(x)=Ax(ax) for 0<x<a, otherwise 0.

(a)

Expert Solution
Check Mark

Answer to Problem 8.16P

An upped bound on the ground state of the infinite square well using the function ψ(x)=Ax(ax) for 0<x<a, otherwise 0 is Egs10π2π222ma2.

Explanation of Solution

Given,

ψ(x)=Ax(ax)

Normalize the above wave function,

1=0a|ψ(x)|2dx=0aA2x2(ax)2dx=A2a501u2(1u)2du=A2a530

Solving further for A2

A2=30a5        (I)

The expectation value of the Hamiltonian is

H=0aψ*(x)[22md2ψdx2]dx=0aψ*(x)[22m(2A)]dx=2A2m0ax[ax]dx=202ma5a301u[1u]du

Solving further,

H=52ma2

Therefore,

Egs10π2π222ma2

Conclusion:

An upped bound on the ground state of the infinite square well using the function ψ(x)=Ax(ax) for 0<x<a, otherwise 0 is Egs10π2π222ma2.

(b)

To determine

The optimal value of p and the best bound on the ground state energy.

(b)

Expert Solution
Check Mark

Answer to Problem 8.16P

The optimal value of p is p=2+64 and the best bound on the ground state energy is Egs5+26π2π222ma2.

Explanation of Solution

Given,

ψ(x)=A[x(ax)]p

Where, p is a real number.

Normalize the above wave function,

1=0a|ψ(x)|2dx=0aA2x2p(ax)2pdx=A2a4p+101|u(1u)|2pdu        (II)

The expectation value of the Hamiltonian is

H=12m0aψ*[22md2ψdx2]dx=22m0a(dψdx)*dψdxdx=2A22m01p2(a2x)2[x(ax)]2p2du=p222ma201(12u)2[u(1u)]2p2du01|u(1u)|2pdu        (III)

Solving the integral in the numerator separately for simplicity,

01(12u)2[u(1u)]2p2du=12p101(12u)ddu[u(1u)]2p1du=12p1[0+201[u(1u)]2p1du]

Substitute the above equation in equation (III),

H=p222ma212p1[201[u(1u)]2p1du]01|u(1u)|2pdu=2p22p122ma201[u(1u)]2p1du01|u(1u)|2pdu=2p(4p+1)2p122ma2

(Using example Schaum’s 18.24)

Differentiate the above equation with respect to p

dHdp=022ma216p216p2(2p1)2=0

Solving the above quadratic equation and the positive root is the solution that diverges at 0 to a.

p=2+64

Substituting the above relation in Equation 8.1

Egs5+26π2π222ma2

Conclusion:

Thus, the optimal value of p is p=2+64 and the best bound on the ground state energy is Egs5+26π2π222ma2.

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