Quantum Engi Q and A_Part_10

# Quantum Engi Q and A_Part_10 - 28 3.5.9 3.6 CHAPTER 3....

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 28 3.5.9 3.6 CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICS Quantum conﬁnement The Harmonic Oscillator 3.6.1 The Hamiltonian 3.6.2 Solution using separation of variables h0 (x) = h1 (x) = h2 (x) = h3 (x) = h4 (x) = 1 (πℓ2 )1/4 e−ξ 2ξ 2 /2 e−ξ 2ξ 2 − 1 e−ξ e−ξ 2 /2 (4πℓ2 )1/4 (9πℓ2 )1/4 4ξ 4 − 12ξ 2 + 3 (576πℓ2 )1/4 c m ℓ= 2 /2 2ξ 3 − 3ξ ω= 2 /2 (4πℓ2 )1/4 h ¯ mω ξ= e−ξ x ℓ 2 /2 Table 3.1: First few one-dimensional eigenfunctions of the harmonic oscillator. 3.6. THE HARMONIC OSCILLATOR 3.6.2.1 29 Solution harmb-a Question: Write out the ground state energy. Answer: Taking the generic expression 2nx + 2ny + 2nz + 3 hω ¯ 2 and substituting the lowest possible value, 0, for each of nx , ny , and nz , you get the ground state energy 3 ¯ E000 = hω 2 Enx ny nz = 3.6.2.2 Solution harmb-b Question: Write out the ground state wave function fully. Answer: Taking the generic expression ψnx ny nz = hnx (x)hny (y )hnz (z ) and substituting nx = ny = nz = 0, you get the ground state eigenfunction ψ000 = h0 (x)h0 (y )h0 (z ). Now substitute for h0 from table 3.1: ψ000 = 1 (πℓ2 )3/4 e−x 2 /2ℓ2 e−y 2 /2ℓ2 e−z 2 /2ℓ2 where the constant ℓ is as given in table 3.1. You can multiply out the exponentials: ψ000 = 3.6.2.3 1 (πℓ2 )3/4 e−(x 2 +y 2 +z 2 )/2ℓ2 . Solution harmb-c Question: Write out the energy E100 . Answer: Taking the generic expression 2nx + 2ny + 2nz + 3 hω ¯ 2 and substituting nx = 1, ny = nz = 0, you get Enx ny nz = E100 = 5 hω ¯ 2 30 3.6.2.4 CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICS Solution harmb-d Question: Write out the eigenstate ψ100 fully. Answer: Taking the generic expression ψnx ny nz = hnx (x)hny (y )hnz (z ) and substituting nx = 1, ny = nz = 0, you get ψ100 = h1 (x)h0 (y )h0 (z ). Now substitute for h0 and h1 from table 3.1: √ 2x/ℓ −x2 /2ℓ2 −y2 /2ℓ2 −z2 /2ℓ2 e e e ψ100 = (πℓ2 )3/4 where the constant ℓ is as given in table 3.1. You can multiply out the exponentials: ψ100 = 3.6.3 √ 2x/ℓ (πℓ2 )3/4 e−(x 2 +y 2 +z 2 )/2ℓ2 . Discussion of the eigenvalues 9 hω ¯ 2 7 hω ¯ 2 5 hω ¯ 2 3 hω ¯ 2 nx = ny = nz = nx = ny = nz = nx = ny = nz = 3 0 0 2 0 0 1 0 0 nx = ny = nz = 0 3 0 0 2 0 0 1 0 0 0 3 0 0 2 0 0 1 2 1 0 1 1 0 0 2 1 1 0 1 10211 01021 22101 0 1 1 0 0 0 0 Figure 3.3: The energy spectrum of the harmonic oscillator. ...
View Full Document

## This note was uploaded on 11/13/2011 for the course PHY 4458 taught by Professor Garvin during the Fall '11 term at University of Florida.

Ask a homework question - tutors are online