General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Question
Chapter 28, Problem 46E
(a)
To determine
The Schrӧdinger equation for the particle.
(b)
To determine
The boundary conditions for the wave function.
(c)
To determine
Whether the given wave function is the satisfactory solution for the Schrӧdinger equation.
(d)
To determine
The energy corresponding to the give wave function.
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A function of the form e^−gx2 is a solution of the Schrodinger equation for the harmonic oscillator, provided that g is chosen correctly. In this problem you will find the correct form of g.
(a) Start by substituting Ψ = e^−gx2 into the left-hand side of the Schrodinger equation for the harmonic oscillator and evaluating the second derivative.
(b) You will find that in general the resulting expression is not of the form constant × Ψ, implying that Ψ is not a solution to the equation. However, by choosing the value of g such that the terms in x^2 cancel one another, a solution is obtained. Find the required form of g and hence the corresponding energy.
(c) Confirm that the function so obtained is indeed the ground state of the harmonic oscillator and has the correct energy.
Consider the Schrodinger equation for a one-dimensional linear harmonic oscillator:
-(hbar2/2m) * d2ψ/dx2 + (kx2/2)*ψ(x) = Eψ(x)
Substitute the wavefunction ψ(x) = e-(x^2)/(ξ^2) and find ξ and E required to satisfy the Schrodinger equation. [Hint: First calculate the second derivative of ψ(x), then substitute ψ(x) and ψ′′(x). After this substitution, there will be an overall factor of e-(x^2)/(ξ^2) on both sides of the equation which canbe an canceled out. Then, gather all terms which depend on x into one coefficient multiplying x2. This coefficient must be zero because the equation must be satisfied for any x, and equating it with zero yields the expression for ξ. Finally, the remaining x-independent part of the equation determines the eigenvalue for energy E associated with this solution.]
Consider the 1D time-independent Schrodinger equation
ħ² ď²
2m dr²
with the potential
where to is a parameter.
(a) Show that
V(x) =
+V(x)] = Ev
v
is a solution of the Schrodinger equation.
ħ²
mx²
sech²
1 = A sech
x
xo
(₁)
Chapter 28 Solutions
General Physics, 2nd Edition
Ch. 28 - Prob. 1RQCh. 28 - Prob. 2RQCh. 28 - Prob. 3RQCh. 28 - Prob. 4RQCh. 28 - Prob. 5RQCh. 28 - Prob. 6RQCh. 28 - Prob. 7RQCh. 28 - Prob. 8RQCh. 28 - Prob. 9RQCh. 28 - Prob. 10RQ
Ch. 28 - Prob. 1ECh. 28 - Prob. 2ECh. 28 - Prob. 3ECh. 28 - Prob. 4ECh. 28 - Prob. 5ECh. 28 - Prob. 6ECh. 28 - Prob. 7ECh. 28 - Prob. 8ECh. 28 - Prob. 9ECh. 28 - Prob. 10ECh. 28 - Prob. 11ECh. 28 - Prob. 12ECh. 28 - Prob. 13ECh. 28 - Prob. 14ECh. 28 - Prob. 15ECh. 28 - Prob. 16ECh. 28 - Prob. 17ECh. 28 - Prob. 18ECh. 28 - Prob. 19ECh. 28 - Prob. 20ECh. 28 - Prob. 21ECh. 28 - Prob. 22ECh. 28 - Prob. 23ECh. 28 - Prob. 24ECh. 28 - Prob. 25ECh. 28 - Prob. 26ECh. 28 - Prob. 27ECh. 28 - Prob. 28ECh. 28 - Prob. 29ECh. 28 - Prob. 30ECh. 28 - Prob. 31ECh. 28 - Prob. 32ECh. 28 - Prob. 33ECh. 28 - Prob. 34ECh. 28 - Prob. 35ECh. 28 - Prob. 36ECh. 28 - Prob. 37ECh. 28 - Prob. 38ECh. 28 - Prob. 39ECh. 28 - Prob. 40ECh. 28 - Prob. 41ECh. 28 - Prob. 42ECh. 28 - Prob. 43ECh. 28 - Prob. 44ECh. 28 - Prob. 45ECh. 28 - Prob. 46ECh. 28 - Prob. 47ECh. 28 - Prob. 48ECh. 28 - Prob. 49ECh. 28 - Prob. 50ECh. 28 - Prob. 51ECh. 28 - Prob. 52ECh. 28 - Prob. 53ECh. 28 - Prob. 54E
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