A QUANTUM LEAP IN MATH By Michael Smith. Introduction to Quantum Mechanics Niels Bohr Erwin Schrödinger Werner Heisenberg

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


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


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


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 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 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


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


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 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


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


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


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


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


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


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


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


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


OrOr

hp q =

4π

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A QUANTUM LEAP IN MATH By Michael Smith. Introduction to Quantum Mechanics Niels Bohr Erwin Schrödinger Werner Heisenberg