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##### A Cool Party Trick

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**Fundamentals of Schroedinger Notation**• Schroedinger notation is usually called “position representation” • However, Schroedinger notation can easily be transformed (as we will see) into “momentum representation” • If |a> represents a state, and |x> represents the unit vector in the x-direction then**In Hamilton’s formulation, the equations of motion were**functions of rand p (or x and px) • So we see that the momentum representation may be useful**Recall**• k=p • Now rather than putting vector symbols everywhere, I am going to jump to 1-dimension so let kx =px • I will let the wavefunction in momentum space be represented by f(kx)**The Fourier transform (FT) converts functions from position**representation to momentum representation of the conjugate momenta This means that FT(f(x))=f(kx), FT(f(y))=f(ky), etc. and vice-versa Postulate 11**A Mystery Solved**Let’s Review: • We’ve seen new formulations of classical mechanics, in particular, H=f(x,p) • We’ve seen evidence that energy is quantized and learned that we cannot necessarily measure x and p simultaneously**A Mystery Solved …**• We learned that massless particles only have momentum which is related to the wavelength • Wavenumber is somehow related to the energy of the system Therefore…**We make a choice**Since we don’t know if velocity is a particularly relevant number (i.e. massless particles) but we do know that wavelength (related to p) is very relevant and so we must choose to use Hamiltonian’s formulation of mechanics! (But what about uncertainty? i.e. that x and p cannot be measured simultaneously?)**Fourier Transform again**• We’ve just learned that I can switch between position representations and momentum representations using the Fourier transform**An important result**This, then, represents momentum in POSITION representation! This is px, the momentum in momentum representation**What about the Hamiltonian?**The Hamiltonian in position representation (and the left hand side of the Schroedinger Equation!)**And Now for Something Completely Different**(Not a Scotsman on a horse!)**Now let’s assume just a wave propagating through space:**r=s=0 A “wave” equation i.e. any function that fits this differential equation is a wave traveling through space Speed of propagation i.e c**We need Bohr and de Broglie**• We know from Bohr that • mvr=h/2p • And de Broglie says that 2pr=l or r=l/2p • So mvl/2p=h/2p • l =h/mv • Mechanical energy is conserved under Bohr model so • T+V= a constant =E**Just a little more here**• Thus, E-V=T= ½ mv2 • And from this, 2m*(E-V)=m2v2 • mv = (2m*(E-V))1/2 • Thus**Postulate 12**The probability density, r, is defined as absolute square of the wavefunction**Probability current density**• r = probability in time and space • J = measure of the flow of probability from one place to another: “probability current density”**Postulate 13**The probability current density, J, is defined as**Huh?**• In E&M, J=I/A and has units of C/m2/s or (C/m3)*(m/s) which means r*v • Where r=charge density • v=average velocity of charges**Before we go on, we have to define the expectation value of**velocity**A Faux Proof**Take the difference between these two**From E&M, the Continuity Equation**• The decrease of charge in a small volume must correspond to the flow of charge out through the surface (conservation of electric charge) • Hey, wait a minute!**Continuity Equation for Probability**• In this case, the decrease of the probability in one volume must be equal to the flow of probability out from the surface • Conservation of Probability!**The other side**• E is the eigenvalue of H, the energy of the system. • y(r) is said to describe a “stationary state” • This function c(t) is just a phase factor for Y(r,t) and does not contribute to the physics of the system. (unless you need to superimpose two different wavefunctions and then it is only the relative phase (w-w’)t which matters)**Transitions between 2 states**• In the last section, the unwritten assumption is that QM states are permanent; that is, a system in a given state will always remain in that state. • We know this cannot be true: radioactivity, among other phenomena, gives lie to this argument**Problems with our brand new toy**• The Schroedinger Equation has two major problems • Not relativistic (see Dirac Equation) • Only describes systems which do not change! QFT—Quantum Field Theory covers this area**Postulate 14**An initial state |i> can make a transition to a final state |f> by the emission or absorption of a quanta. The probability of this occurrence is proportional to |<f|Vi|i>|2 where Vi is the interaction potential appropriate to the potential. This is only an approximation. Further approximations may be necessary