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EE201/MSE207 Lecture 9 Quantum mechanics in three dimensions - - PowerPoint PPT Presentation
EE201/MSE207 Lecture 9 Quantum mechanics in three dimensions - - PowerPoint PPT Presentation
EE201/MSE207 Lecture 9 Quantum mechanics in three dimensions (still only one particle) Natural generalization (, , , ) Schrdinger = equation: usually no t -dependence 2 + 2 + 2 2
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Separation of variables in Cartesian coordinates
Simplification if π
π = π
1 π¦ + π 2 π§ + π 3 π¨ ;
then 3D TISE can be replaced with three 1D equations
β β2 2π π2π ππ¦2 + π2π ππ§2 + π2π ππ¨2 + π π π = πΉ π
TISE Look for (assume)
π π = π1 π¦ π2 π§ π3 π¨
Divide TISE by π, then
β β2 2π π2π1 π¦ ππ¦2 π1 π¦ + π2π2 π§ ππ§2 π2 π§ + π2π3 π¨ ππ¨2 π3 π¨ + π
1 π¦ + π 2 π§ + π 3(π¨) = πΉ
Then three equations, with πΉ = πΉ1 + πΉ2 + πΉ3
β β2 2π π2π1 π¦ ππ¦2 π1 π¦ + π
1 π¦ = πΉ1
and two similar equations for π§ and π¨ (not in the textbook)
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Simplification if π π = π
1 π¦ + π 2 π§ + π 3 π¨
(cont.)
Rewrite as usual
β β2 2π π2π1 π¦ ππ¦2 + π
1 π¦ π1 π¦ = πΉ1 π1 π¦
β β2 2π π2π2 π§ ππ§2 + π
2 π§ π2 π§ = πΉ2 π2 π§
β β2 2π π2π3 π¨ ππ¨2 + π
3 π¨ π3 π¨ = πΉ3 π3 π¨
πΉ = πΉ1 + πΉ2 + πΉ3 π π = π1 π¦ π2 π§ π3(π¨)
Each equation has many solutions
ππ,π,π π = π1,π π¦ π2,π π§ π3,π(π¨)
Energy
πΉ = πΉπ¦,π + πΉπ§,π + πΉπ¨,π
(replaced 1,2,3 with π¦, π§, π¨) General solution
Ξ¨ π , π’ =
π, π, π
ππ,π,π π1,π π¦ π2,π π§ π3,π(π¨) exp βπ πΉπ¦,π + πΉπ§,π + πΉπ¨,π β π’
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Examples
Unfortunately, not many examples when this trick is useful Semiconductor quantum well, quantum wire, quantum dot
(terminology for semiconductor structures is slightly different than in QM)
quantum well (QW), 2D electron gas (2DEG) electrons do not move in z-direction, free motion in x and y z x quantum wire (QWi), 1D electrons electrons move only in x-direction, restricted along y and z quantum dot (QD), 0D electrons motion is restricted in all direction (x, y, and z) Only the first case (QW) can be truly represented as π
1 π¦ + π 2 π§ + π 3 π¨ ;
however, other cases can also be treated in this way approximately
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Semiconductor Quantum Well
z π π = π π¨ = 0 + 0 + π
3 π¨
z
(finite depth QW along z)
Wavefunctions
π(π¦, π§, π¨) = ππ π¨ ππππ¦π¦ππππ§π§ 1
2π
- r 1
2πβ
πΉ = πΉπ + β2ππ¦
2
2π + β2ππ§
2
2π
If infinite depth, a π π = 0, 0 β€ π¨ β€ π β,
- therwise
then
π(π¦, π§, π¨) =
2 π sin ππ π π¨
ππππ¦π¦ππππ§π§ 1
2π
- r 1
2πβ
πΉ = π2π2β2 2ππ2 + β2ππ¦
2
2π + β2ππ§
2
2π
π
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Rectangular Quantum Wire
x If finite depth in y and z directions, then we cannot use this trick. However, it works for infinite depth. Assume π π¦, π§, π¨ = 0, if 0 β€ π¨ β€ π 0 β€ π§ β€ π β,
- therwise
π π¦, π§, π¨ = 2 π 2 π sin ππ¨π π π¨ sin ππ§π π π§ ππππ¦π¦ 1 2π
- r
1 2πβ
If not rectangular and/or finite depth, then still 2+1 dimensions
πΉ = ππ¨
2 π2β2
2ππ2 + ππ§
2 π2β2
2ππ2 + β2ππ¦
2
2π
π π
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Rectangular (cuboid) Quantum Dot
Again need to assume infinite depth π π¦, π§, π¨ = 0, if 0 β€ π¨ β€ π 0 β€ π§ β€ π 0 β€ π¦ β€ π β,
- therwise
π π¦, π§, π¨ = 2 π 2 π 2 π sin ππ¨π π π¨ sin ππ§π π π§ sin ππ¦π π π¦ πΉ = ππ¨
2
π2 + ππ§
2
π2 + ππ¦
2
π2 π2β2 2π
Degeneracy if π, π, and π are equal or commensurate. In semiconductors π is effective mass. π π π
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Another example: 3D oscillator (e.g., atom in a lattice)
π π = 1 2 πππ¦
2π¦2 + 1
2 πππ§
2π§2 + 1
2 πππ¨
2π¨2
πΉππ¦,ππ§,ππ¨ = ππ¦ +
1 2 βππ¦ + ππ§ + 1 2 βππ§+ ππ¨ + 1
2 βππ¨
Again, degeneracy if ππ¦, ππ§, or ππ¨ are equal or commensurate.
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Spherically symmetric potential (similar trick)
π π = π( π )
Most important for atoms Then it is natural to look for
π π , π, π = π π π(π, π)
where π , π, π are spherical coordinates TISE β β2 2π πΌ2π + π π π = πΉ π Rewriting Laplacian in spherical coordinates β β2 2π 1 π 2 π ππ π 2 ππ ππ + 1 π 2 sin π π ππ sin π ππ ππ + 1 π 2 sin2 π π2π ππ2 + π π π = πΉ π Divide by π = ππ and multiply by β2ππ 2/β2 1 π π ππ π 2 ππ ππ β 2ππ 2 β2 π π β πΉ + 1 π 1 sin π π ππ sin π ππ ππ + 1 sin2 π π2π ππ2 = 0
π (π + 1) βπ (π + 1)
(so far just a notation) const const combine
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Assume π π, π = Ξ π Ξ¦(π), again separation of variables 1 Ξ(π) 1 sin π π ππ sin π πΞ ππ + π π + 1 sin2 π + 1 Ξ¦(π) π2Ξ¦ ππ2 = 0
βπ2 π2
const const
Ξ¦ π = ππππ, π = 0, Β±1, Β±2, β¦
(since should be periodic with 2π) This is why π is integer.
Ξ π = π΅ π
π π(cos π)
Associated Legendre function This is why π is integer. π = 0, 1, 2, β¦ (integer) π = βπ, βπ + 1, β¦ 0, β¦ π β 1, π
π: anguar momentum quantum number
(azimuthal q. n. , orbital q. n. )
π: magnetic quantum number π
π π π, π = Ξπ π π Ξ¦π(π)
are called spherical harmonics
These function are the same for any spherically symmetric potential π(π ).
Spherical harmonics π
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Radial function π
π = π π π(π, π)
Let us introduce π£ π = π π π , for this function the equation is
β β2 2π π2π£ ππ 2 + π π + β2 2π π π + 1 π 2 π£ = πΉ π£
centrifugal term So, the equation for π£(π ) is similar to 1D TISE, but with the centrifugal term. It has some solutions, depending on π (orbital q.n.) and π (solution index). Corresponding energy: πΉπ,π . Overall, 3 quantum numbers: π, π, π. However, energy depends only on π and π.
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Hydrogen atom
π π = β π2 4ππ0 1 π
Consider only bound states (since atom) ο πΉ < 0 Effective potential
β π2 4ππ0 1 π + β2 2π π π + 1 π 2
Accidentally, for this potential πΉπ,π is highly degenerate
πΉπ = β π 2β2 π2 4ππ0
2
1 π2 = πΉ1 π2 π = 1, 2, 3, β¦ πΉ1 = β13.6 eV (ground state)
π = 1, 2, 3, β¦ π = 0, 1, β¦ π β 1 π = 0, Β±1, Β±2, β¦ Β± π
Total degeneracy: π=0
πβ1(2π + 1) = π2
Ground state: π100 π , π, π =
1 ππ3 πβπ /π π = 4ππ0β2 ππ2 = 0.53 β«
Bohr radius
(almost same theory for dopant levels and excitons) principal azimuthal (ang.mom.) magnetic
π(π )
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