a quantum leap in math by michael smith. introduction to quantum mechanics niels bohr erwin...
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A QUANTUM LEAP IN A QUANTUM LEAP IN MATHMATH
By Michael SmithBy Michael Smith
Introduction to Quantum MechanicsIntroduction to Quantum Mechanics
Niels BohrNiels Bohr
Erwin SchrErwin Schröödingerdinger
Werner HeisenbergWerner Heisenberg
Bohr’s Model of AtomBohr’s Model of Atom
Max Planck and quanta of energyMax Planck and quanta of energy E = hυE = hυ
Bohr wanted to explain how electrons orbit Bohr wanted to explain how electrons orbit nucleusnucleus TheorizedTheorized orbits of electrons quantizedorbits of electrons quantized
Bohr’s Model of AtomBohr’s Model of Atom
Force associated with the charged particle = to Force associated with the charged particle = to centripetal force of a rotating electroncentripetal force of a rotating electron
Angular momentum constant, quantized, and Angular momentum constant, quantized, and related to Planck’s constantrelated to Planck’s constant
2 2
20
Ze mv =
4πε r r
nhv =
2mrπ
Bohr’s Model of AtomBohr’s Model of Atom
Using some algebra Bohr found radiusUsing some algebra Bohr found radius
Plugging in appropriate values first 3 orbits of Plugging in appropriate values first 3 orbits of Hydrogen atom are 0.529 Å, 2.116 Å, and Hydrogen atom are 0.529 Å, 2.116 Å, and 4.761Å which matched experimental data4.761Å which matched experimental data
2 20
2 2
n h 4πεr =
4π mZe
Bohr’s Model of AtomBohr’s Model of Atom
Total Energy of SystemTotal Energy of System
Using a little algebra and formula for radius Bohr foundUsing a little algebra and formula for radius Bohr found
First 3 energy levels for hydrogen calculated to be -13.6 eV, -First 3 energy levels for hydrogen calculated to be -13.6 eV, -3.4 eV, and -1.5 eV agreed with experimental data3.4 eV, and -1.5 eV agreed with experimental data
22
0
1 ZeE = mv -
2 4πε r
2 4
2 2 20
1 4π meE = -
2 h (4πε ) n
SchrSchröödinger’s Equationdinger’s Equation
de Broglie proposed that matter, like light, could de Broglie proposed that matter, like light, could possess properties of both particles and wavespossess properties of both particles and wavesAlso said energy equations developed by Einstein and Also said energy equations developed by Einstein and Planck were equal Planck were equal
From this stated that wavelength could be determined From this stated that wavelength could be determined by knowing a particle’s mass and velocityby knowing a particle’s mass and velocity
2E = hυ = mc
hλ =
mv
SchrSchröödinger’s Equationdinger’s Equation
SchrSchröödinger convinced by 2 colleagues (Henri and dinger convinced by 2 colleagues (Henri and DeBye) to come up with wave equation to explain de DeBye) to come up with wave equation to explain de Broglie’s conceptBroglie’s concept
To verify applied it to hydrogen atom putting it in To verify applied it to hydrogen atom putting it in spherical coordinatesspherical coordinates
2 2 2 2
2 2 2 2
-hEψ = + + + V ψ
8π m x y z
22
(r)2 2 2
1 1 2msinθ (r ) + (sinθ ) + ψ + (E - V )ψ = 0
r sinθ r r θ θ sinθ φ
SchrSchröödinger’s Equationdinger’s Equation
SchrSchröödinger equation made up of angular dinger equation made up of angular component (component (θθ, , φφ) and radial component (r)) and radial component (r)
Solving for each component, the wave Solving for each component, the wave function for hydrogen at ground statefunction for hydrogen at ground state
32
-ra
1000
1 1Ψ = e
aπ
SchrSchröödinger’s Equationdinger’s Equation
Only radial component at ground state, SchrOnly radial component at ground state, Schröödinger dinger equation for hydrogenequation for hydrogen
Solving givesSolving gives
or -13.6 eV which agrees with Bohr’s model and or -13.6 eV which agrees with Bohr’s model and experimental data experimental data
3 32 2
-r -r2 a a(r)2 2
0 0
1 d d 1 1 2m 1 1(r e ) + (E - V ) e = 0
r dr dr a aπ π
2 4
2 20
2π meE = -
4πε h
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
states that simultaneous measurements of position states that simultaneous measurements of position and momentum, of a particle can only be known with and momentum, of a particle can only be known with no better accuracy than Planck’s constant, h, divided no better accuracy than Planck’s constant, h, divided
by four times πby four times π h
p x 4π
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
Wave function is a probability density Wave function is a probability density function.function.
Heisenberg reasoned that probability Heisenberg reasoned that probability would be normally distributed, or Gaussian would be normally distributed, or Gaussian in nature given by expressionin nature given by expression
2
2
x - μ
2σ1f x; μ, σ = e
σ 2π
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
Heisenberg inferred that the Gaussian distribution of Heisenberg inferred that the Gaussian distribution of the position coordinate, q, would be expressed bythe position coordinate, q, would be expressed by
δq is the half-width of the Gaussian hump where the δq is the half-width of the Gaussian hump where the particle will be foundparticle will be found
uncertainty in position is given byuncertainty in position is given by
2
2
-q
2 δqψ q = Ce
δq = 2 q
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
Momentum distribution given byMomentum distribution given by
Substituting the function for Substituting the function for ψψ(q) gives(q) gives
Which through the magic of algebra can be rewritten Which through the magic of algebra can be rewritten asas
2π pq
h
-ψ p = e ψ q dq
i
2
2
-q2π pq2 δqh
-ψ p = C e e dq
i
22 22
2
2π p δq1 q 2π pδq + h2 δq h
-ψ p = C e e dq
i
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
Since second exponential term not dependent Since second exponential term not dependent on “q”, the expression can be rewritten ason “q”, the expression can be rewritten as
Letting Letting so so givesgives
22 2 2
2
2π p δq 1 q 2π pδq + h 2 δq h
-ψ p = Ce e dq
i
q 2π pδq
y = + δq h
i dq = δqdy
22 2
22
2π p δq1
yh 2
-ψ p = Ce e δqdy
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
This integral is symmetric so can be rewritten This integral is symmetric so can be rewritten asas
To solve this requires some math trickery first To solve this requires some math trickery first squaring both sides and selecting another squaring both sides and selecting another “dummy variable”“dummy variable”
22 2
22
2π p δq1
yh 2
0ψ p = 2Cδqe e dy
22 2
2 22
22π p δq
1 1y x2 h 2 2
0 0ψ p = 2Cδqe e dy e dx
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
Then rewriting it as a double integralThen rewriting it as a double integral
Combine exponential terms Combine exponential terms
22 2
2 22
22π p δq
1 1y x2 h 2 2
0 0ψ p = 2Cδqe e e dxdy
22 2
2 22
22π p δq
1y + x2 h 2
0 0ψ p = 2Cδqe e dxdy
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
because ybecause y22 + x + x22 = r = r22, the integral can be , the integral can be rewritten in polar coordinates asrewritten in polar coordinates as
Integrating over Integrating over θθ gives gives
22 2
22
22π p δq
12π r2 h 2
0 0ψ p = 2Cδqe e rdθdr
22 2
22
22π p δq
1r2 h 2
0ψ p = 2Cδqe 2π e rdr
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
Letting u = (1/2)rLetting u = (1/2)r22 and du = rdr gives and du = rdr gives
And solving givesAnd solving gives
Or Or
22 2
2
22π p δq
2 h -u
0ψ p = 2Cδqe 2π e du
22 2
2
22π p δq
2 hψ p = 2Cδqe 2π
22 2
2
2π p δq
hψ p = 2 2πC δq e
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
Comparison of the function for momentum Comparison of the function for momentum with the probability density function of normal with the probability density function of normal distribution givesdistribution gives
Which simplifies toWhich simplifies to
22
2 2
2π δq1 =
2σ h
h
σ = δp = 2π δq
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
Getting delta terms on left side givesGetting delta terms on left side gives
RememberingRemembering where where ΔΔq is q is standard deviation, the same applies for standard deviation, the same applies for δδpp, , givinggiving
hδpδq =
2π
δq = 2 q
h2 p 2 q =
2π
Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle
OrOr
hp q =
4π