Physics Problems & Solutions: Normalization

Showing posts with label Normalization. Show all posts

Quantum Mechanics - Infinite Potential Well

_{See Problem 51

The eigenfunctions satisfy the condition

_₀∫^{^a}ψ_{_n}*(x) ψ_{_l}(x) dx = δ_nl

δ_nl = 1 if n = l, otherwise δ_nl = 0. This is a statement that the eigenfunctions are

A. Solutions to the Schrodinger equation

B. Orthonormal
C. Bounded

D. Linearly dependent
E. Symmetric

(GR9277 #52)

Solution:

Orthonormality = orthogonal and normal

_₀∫^{^a}ψ_{_n}*(x) ψ_{_l}(x) dx = 〈ψ_{_n}|ψ_{_l}〉= δ_nl

Orthogonal, δ_nl = 0

Normal, δ_nl = 1 if n = l

Answer: B}

Quantum Mechanics - Schrodinger Equation

The wave function ψ(x) = A exp (−b²x²/2), where A and b are real constants, is a normalized eigenfunction of the Schrodinger equation for a particle of mass M and energy E in a one dimensional potential V(x) such that V(x) = 0 at x = 0. Which of the following is correct?

A. V = ħ²b⁴/2M
B. V = ħ²b⁴x²/2M
C. V = ħ²b⁶x⁴/2M
D. E = ħ²b²(1 − b²x²)
E. E = ħ²b⁴/2M

(GR8677 #18)

Solution:

Schrodinger Equation: $-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)\psi(x)=E\psi(x)$

$\begin{aligned} \frac{\partial\psi}{\partial x}&=\frac{\partial}{\partial x}e^{-b^2x^2/2}=-b^2x\psi \\\frac{\partial^{2}\psi}{\partial x^{2}}&=-b^2\frac{\partial}{\partial x}x\psi=-b^2\left( x\frac{\partial\psi}{\partial x}+\psi\frac{\partial x}{\partial x}\right ) \\\frac{\partial^{2}\psi}{\partial x^{2}}&=-b^2(-b^2x^2\psi+\psi)=(b^4x^2-b^2)\psi\end{aligned}$

Schrodinger Equation: $-\frac{\hbar^{2}}{2m}(b^4x^2-b^2)\psi(x)+V(x)\psi(x)=E\psi(x)$

To find E → V(x = 0) = 0:

$\\-\frac{\hbar^{2}}{2m}(0-b^2)+0=E \\\rightarrow E=\frac{\hbar^{2}b^2}{2m}$

To find V(x):
$\\-\frac{\hbar^{2}}{2m}(b^4x^2-b^2)+V(x)=E \\\\-\frac{\hbar^{2}b^4x^2}{2m}+\frac{\hbar^{2}b^2}{2m}+V(x)=E \\\\-\frac{\hbar^{2}b^4x^2}{2m}+E+V(x)=E \\\\\rightarrow V(x)=\frac{\hbar^{2}b^4x^2}{2m}$

Answer: B

Quantum Mechanics - Normalization of Wavefunction

Eigenfunction for a rigid dumbbell rotating about its center have a φ dependence of the form ψ(φ) = Ae^imφ, where m is a quantum number and A is a constant. Which of the following values of A will properly normalize the eigenfunction?

A. √(2π)
B. 2π
C. (2π)²
D. 1/√(2π)
E. 1/(2π)

(GR8677 #28)

Solution:
Normalized Wavefunction: $\int |\psi(x)|^2dx=\int \psi(x)\psi^*(x)dx=1$
$\\\psi(\phi)=Ae^{im\phi} \\\psi^*(\phi)=Ae^{-im\phi}$

$\\\int_{0}^{2\pi} |\psi(\phi)|^2d\phi=\int_{0}^{2\pi} |A|^2d\phi=1 \\|A|^2\int_{0}^{2\pi}d\phi=|A|^2 2\pi=1\rightarrow A = \frac{1}{\sqrt{2\pi}}$

Answer: D