Quantum Mechanics

("black" => opaque): from solid materials, liquids

The radiation from hot, solid materials looks samey (bright yellow) unlike every gas, however, had a spectral emission (think - neon lights.)

But!

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The light spectrum did depend on temperature, so what happened? Why is everything hot?

Max Plank => trying to model incoming light source from rays as basically all absorbed and not bounced back.

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Max Plank's Model 1 in this manner matched well with observations at long wavelengths (red hot). But, it predicted infinite brightness (it will just "keep bouncing") as wavelength => 0, which is wrong. This is the "ultraviolet catastrophie."

So, he made it better.

Max Plank's Model 2 is just Model 1, but an additional assumption that when Energy Transfers from e^- to EMWave, \(\delta E\) must be some constant * frequency of light.

So, to synthesize high frequencies, this cop out had the effect of supressing the infinite growth as \(\delta E\) would grow bigger and bigger to the point where all your energy would not go into the EMWave but to this transferring factor.

Which is like… Kind of a cop out. But it did fit medium frequencies better.

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Einstein > Light ! "wave"; instead, light are photon particles moving through space.

Impontant Knowledges::

Energy of each photon is equal to the plank constant (h) times the frequency (f). \(E = h*f\).

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\(\lambda * f = c\)

\(E_{photon} = h \times f\)

Instead of Hertz, however, the frequency of F could better be represented with \(\omega\), a unit of \(\frac{radians}{sec}\) that is derived as \(2 \pi f (\frac{radians}{s})\)

So to calculate energy with \(\omega\), simply use \(\bar{h} = \frac{h}{2\pi}\) and so \(E = \bar{h}\omega\)

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\(\Delta E \times \Delta t = \bar{h}\) => "uncertainty of energy times uncertanity in time is the reduced plank's contstant"

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Lifetime of the upper level => \(\Delta t\)

(Mean) lifetime of the "upper" energy level => \(\Delta t\). So, \(\Delta E = \frac{\bar{h}}{\Delta t}\).

If \(\Delta t\) is small, \(\Delta E\) is large.

As long as the units of two deltas end up as \(J \times s\), they would be related by the same way with \(\bar{h}\)

This \(\Delta P\) has an actual effect on our vision

THIS IS IMPORTANT, TOO! \(\Delta \vec{p} \times \Delta \vec{x} \approx \bar{h}\).

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Meaning, in the subatomic world, everything exists based on differening upper-energy-state-time based uncertainties.

"Diffraction through an apreture"

We could see a similar pattern in passing photons through a llit. \(Slit large, \Delta P_x small\) \(Slit small, \Delta P_x large\).

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This limits the width of the lens of a camera because of the uncertanity in momentum.

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Taking the angle, and dividing it by 3000, which is \(\frac{1}{60}\) degrees.

Even though Plank's constant is a tiny number, it effects how sharply you eyes could see b/c of this uncertainty.

There are three "flavor"s of Leptons, each with two variations — creating six different leptons.

Lepton => "small", but they are not actually that small as what their original namer had suggested.

• The Electron
  • Dirac's equations predicted the existance of a certain "positiron" which would be the oppostite of an electron. After self-determination (the "equation was too perfect to be wrong"), he set out hard to try to prove it.

Interactions in the small scale world occur through the creation and annihilation of particles.

Neutrinos interact only by weak interactions, which is (bar gravity, which is the weakest physical interaction) a very weak physical interaction.

Particle	Wat
Leptons	Fundimental one-half spin particles, experience strong interaction, have no quarks
Baryons	Componsite particles made of quarks + has 1/2, or 3/2, or 5/2 spin
Mesons	Composite particles made of quarks + has 0, 1, or any interger spin
Quarks	Fundimental strongly interacting particles that never appear singly
Hadrons	Bayrons and Mesons that strongnly interact
Nucleaons	Neutrons and protons that reside in the nucleai
Fermions	Leptons, quarks, and nucleans: all have 1/2 odd interger spin
Bosons	Force carriers, like mesons, have intergin spin

Positive pion decays into a positive muon, an antimuon, and a neutrino.

The negative pion decays into a negative muon and an antineutrino.

A pair of electrons could collide and form a pair of tou particles.

Three flavours of leptons cannot interchange or become each other, but they could interact.

Classical physics: objectc has a position X and velocity V that are completely defined

Quantum mechanics: wave function \(\psi\) \(\psi(\vec(x),t) = a+bi (\vec(x),t)\).

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The wave function yields the probablitiy density of finding the object at that location and at that time.

The momentum (\(\rho\)) of an object is associated with its wavelength (\(\lambda\)).

If \(\rho\) is completely defined, its position \(x\) is completely undefined

If \(x\) is completely defined, its position \(\rho\) is completely undefined

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Heisnberg Uncertainty, formally:

\(\Delta x \Delta \rho \approx h\). This is handwavy.

\(\rho_0 = \frca{h}{\lambda}\) to verify heisenberg

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Through this, we could find out that…

• \(\rho_0 = \frac{h}{\lambda} = \frac{\rho){}\)

Planck's Model:

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But this is only for solid. that it glows "hotter and hotter"

Gasses, however, looks different:

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Whereby at discrete wavelengths/energies, it glows; yet in the other states it does not.

This infers that there is only certain levels in which electron may exist, and hence these jumps occur when energy levels shift (?)

Planck's constant for \(\bar{h}\) had units of \(J/s\). This is also the units for angular momntum.

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Bohr's Model: completely classical with the addition that the angular momentum must equal an interger \(N\) times Planck's constant \(\bar{h}\). Meaning, \(angmom = N\bar{h}\).

So, the energy levels (angular momentums) in which electrons should exist should be an interger of planck's reduced constant.

Goal: Electron Energy in terms of an N and a contsant (hbar)

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Circular motion is a bound system (\(KE=\frac{1}{2}mv^2\)). Hence, as R increases, PE decreases.

\(PE=0 when R \to \infty\) \(PE_{electrical} = \frac{-k Q^2}{R}\);

\(|PE|=2KE\). \(E_{tot} = PE+KE = -KE\)

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Bohr Model: gets right — energy level of Hydrogen

For every N, we have a unique energy + a unique angular momentum.

Turns out, for every N, there are a variety of angular momentum that's possible

Bohr's circular orbit and definite position theories are definitely wrong.

There is, however, a maximum angular momentum for each N in units of H-bar.

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This equation is only useful if the particle is moving along on a potential.

As time increases, the potential does not change in time

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The function \(\phi\) solution to schodinger's equation in probably imaginary, but, multiplied by its complex conjugate, we could find the probability density that the particle will be found at (x).

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And given that its a probability density, we know a few things

\(\int_{-\infty}^{infty} \phi * \phi dx = 1\) because "if you look everywhere, you otta find it if it exist."

\(\phi(x)\) must be continuous.

The phi function must follow the property of being "bounded", meaning that \(\phi {x<0 \rightarrow 0, x>L => \rightarrow}\)

Solving for the differential equation that we proposed:

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But wait! B squared here resulted in an negative number. That's either imaginary (which actually works out because \(e^i = sin\), think talor), or we just use a different function:

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But! Remember that this function is bounded between 0 and L. Hence, we add multiple wave functions together like standing waves.

Continuity at X=L demand \(B*L = N*\pi\), whereby \(N=interger\). B !=0 because otherwise the particle would be just nonexistant anywhere.

For each N, a new standing wave is added. So the collapsed sum of the wave functions would be the probablitiy densities.

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On classical physics, the probability of oscellations should be constant (the particle could be at any point with equal probability.)

As N (location/"humps") increases, more and more smaller waves exist. So as \(N\rightarrow \infty\), there would be so many oscillations that it approaches the classical view

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As we know more certain its position (as L=>0, position is well known (between 0 and L smaller)), the energy goes to infinity, which makes momentum quite uncertain.

Visea versa.

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Outside of the box, the expotentials on the two sides would contain a stable velocity equaling to Q outside of the possible "existance" region.

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Wave function is not 0 outside the box! According to quantum physics, there is some probability of finding the object outside its physical boundary box.

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And this is "quantum tunneling"