Schrodinger Equation in Three Dimensions Flashcards
Schrodinger Equation in 3D
Time Dependent
iℏ ∂ψ(|r,t)/∂t= -ℏ²/2m ∇²ψ(|r,t) + V(|r,t)ψ(|r,t)
Schrodinger Equation in 3D
Probability Density
ψ(|r,t) |²
Schrodinger Equation in 3D
Normalisation
∫ | ψ(|r,t) |² dV = 1
Schrodinger Equation in 3D
Time Independent
- ℏ²/2m ∇²ψ(|r) + V(|r)ψ(|r) = E ψ(|r)
- i.e. V=V(|r) => ψ(|r,t) = ψ(|r) * e^(-itE/ℏ)
3D Infinite Potential Well
Description
- consider a box with no potential inside (V=0) and infinite potential outside (V=∞)
- solve the Schrodinger equation inside the box and then impose the condition that solutions must be zero at the walls of the box, i.e standing waves in the box
3D Infinite Potential Well
Proof
-the potential is time independent so the Schrodinger equation takes the form:
-ℏ²/2m ∂²ψ(|r)/∂x² - ℏ²/2m ∂²ψ(|r)/∂y² - ℏ²/2m ∂²ψ(|r)/∂z²
= E ψ(|r)
-different coordinates appear independently so seek solution of the form:
ψ(|r) = F(x)G(y)H(z)
-sub into the Schrodinger equation
-this gives rise to 3 equations:
d²F/dx² * 1/F + FCx = 0
=> Enx = nx²ℏ²π²/2mLx² , nx = 1,2,3,…
Fnx(x) = [2/Lx]sin(nxπx/Lx)
-and similarly for y and z
-total energy:
E = Enx + Eny + Enz
= ℏ²π²/2m * [nx²/Lx² + ny²/Ly² + nz²/Lz²]
-total wave function:
ψ(|r) = F(x)G(y)H(z)
= √[8/LxLyLz]sin(nxπx/Lx)sin(nyπy/Ly)sin(nzπ*z/Lz)
Degeneracy of the Energy Spectrum
1D
-in the 1D case:
En = ℏ²π²n²/2mL²
-for each En, there exists a unique wave function:
φ(x) = √[2/L] * sin(πnx/L)
-one wave function per energy level, no degeneracy
Degeneracy of the Energy Spectrum
3D
- different combinations of nx, ny and nz can have the same energy but different wave functions
- in the 3D case, SOME energy states are degenerate, this is due to symmetry of the well
- when the symmetry breaks, degeneracy is removed
Schrodinger Equation for a Central Potential
[^K+^U] ψ(|r) = Eψ(|r)
Hydrogen Atom
^K
^K = -ℏ²/2m [∂²/∂x² + ∂²/∂y² + ∂²/∂z²]
Hydrogen Atom
|^L²
|^L² = -ℏ²[1/sinθ ∂/∂θ(sinθ∂/∂θ) + 1/sin²θ ∂²/∂φ²]
Hydrogen Atom
Schrodinger Equation
- ℏ²/2m [1/r ∂/∂r (r² ∂/∂r) - 1/ℏ²r² |^L]*ψ(|r) + U(r)ψ(|r) = Eψ(|r)
- > where U(r) = -e²/(4πϵo*r)
Angular Momentum
Classical vs Quantum
-in classical physics: |L = |r x |p -in quantum mechanics: |^L = |^r x |^p ^Lx = ^y^pz - ^z^py ^Ly = ^z^px - ^x^pz ^Lz = ^x^py - ^y^px
Can the three quantum angular momentum operators ^Lx, ^Ly, ^Lz be measured simultaneously?
-operators can be measure simultaneously IF [^A,^B]=0
-but:
[^Lx , ^Ly] = -iℏ^Lz
[^Ly , ^Lz] = -iℏ^Lx
[^Lz , ^Lx] = -iℏ^Ly
-therefore we cannot measure all three components of angular momentum at the same time
Eigenvalue Problem of ^|L
-define ^|L² = Lx² + Ly² + Lz²
=>
[^Lz , ^|L²] = 0
[^Ly , ^|L²] = 0
[^Lx , ^|L²] = 0
-this means we can pick one component e.g. ^Lz and diagonalise it together with ^|L²
-the common eigenvectors of these two operators are:
{ |l*ml⟩ }
-corresponding eigenvalues:
^|L² |l*ml⟩ = ℏ²l(l+1) |l*ml⟩
^Lz |l*ml⟩ = ℏm |l*ml⟩
-therefore we need two numbers to label the eigenvalues and eigenvectors of these operators, l and ml
