Quantum Engi Q and A_Part_10

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

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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. ...
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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.

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