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Published byJasmin Pilot Modified about 1 year ago

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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

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In Hamilton’s formulation, the equations of motion were functions of r and p (or x and p x ) So we see that the momentum representation may be useful

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Recall k=p Now rather than putting vector symbols everywhere, I am going to jump to 1- dimension so let k x =p x I will let the wavefunction in momentum space be represented by (k x )

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Obviously

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Postulate 11 The Fourier transform (FT) converts functions from position representation to momentum representation of the conjugate momenta This means that FT(f(x))= (k x ), FT(f(y))= (k y ), etc. and vice-versa

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Recall

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A Mystery Solved Let’s Review: 1. We’ve seen new formulations of classical mechanics, in particular, H=f(x,p) 2. We’ve seen evidence that energy is quantized and learned that we cannot necessarily measure x and p simultaneously

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A Mystery Solved … 3. We learned that massless particles only have momentum which is related to the wavelength 4. Wavenumber is somehow related to the energy of the system Therefore…

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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?)

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Fourier Transform again We’ve just learned that I can switch between position representations and momentum representations using the Fourier transform

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From Schaum’s Math Handbook

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So

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An important result This is p x, the momentum in momentum representation This, then, represents momentum in POSITION representation!

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After all these years

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What about p x 2?

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What about the Hamiltonian? The Hamiltonian in position representation (and the left hand side of the Schroedinger Equation!)

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And Now for Something Completely Different (Not a Scotsman on a horse!)

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Maxwell’s Equations

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Now let’s assume just a wave propagating through space: = =0 Speed of propagation i.e c A “wave” equation i.e. any function that fits this differential equation is a wave traveling through space

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Let the wavefunction wave!

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Separating Stuff…

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So we have…

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We need Bohr and de Broglie We know from Bohr that mvr=h/2 And de Broglie says that 2 r= or r= 2 So mv 2 =h/2 =h/mv Mechanical energy is conserved under Bohr model so T+V= a constant =E

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Just a little more here Thus, E-V=T= ½ mv 2 And from this, 2m*(E-V)=m 2 v 2 mv = (2m*(E-V)) 1/2 Thus

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So we have…

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Finally

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Recall E=hf

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And what about H?

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The Time Dependent Schroedinger Equation: TDSE

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In 3 dimensions

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Postulate 12 The probability density, , is defined as absolute square of the wavefunction

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In momentum representation

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Probability current density = probability in time and space J = measure of the flow of probability from one place to another: “probability current density”

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Postulate 13 The probability current density, J, is defined as

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Huh? In E&M, J=I/A and has units of C/m 2 /s or (C/m 3 )*(m/s) which means *v Where =charge density v=average velocity of charges

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Before we go on, we have to define the expectation value of velocity

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A Faux Proof Take the difference between these two

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Now we need to recognize that

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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!

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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!

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Another E&M Analogy

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Time just keeps on slippin’ …

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Back to TDSE

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The other side E is the eigenvalue of H, the energy of the system. (r) is said to describe a “stationary state” This function (t) is just a phase factor for (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 ( ’)t which matters)

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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

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Problems with our brand new toy The Schroedinger Equation has two major problems 1. Not relativistic (see Dirac Equation) 2. Only describes systems which do not change! QFT—Quantum Field Theory covers this area

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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 | | 2 where V i is the interaction potential appropriate to the potential. This is only an approximation. Further approximations may be necessary

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