Free particle wavefunction

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Presentation transcript:

Free particle wavefunction If V=0 everywhere then solutions are (eta really hbar) but the exponentials are also eigenfunctions of the momentum operator can use to describe left and right traveling waves Can normalize in different ways; mostly we will skip details 1

1D Barriers (Square) Start with simplest potential (same as square well) V

1D Barriers E<V Solve by having wave function and derivative continuous at x=0 solve for A,B. As |y|2 gives probability or intensity |B|2=intensity of plane wave in -x direction |A|2=intensity of plane wave in +x direction

1D Barriers E<V While R=1 still have non-zero probability to be in region with E<V Electron with barrier V-E= 4 eV. What is approximate distance it “tunnels” into barrier? V A D B

1D Barriers E>V V A D B C X<0 region same. X>0 now “plane” wave Will have reflection and transmission at x=0. But “D” term unphysical and so D=0 V A D B C

1D Barriers E>V R T 1 E/V 1 Calculate Reflection and Transmission probabilities. Note flux is particle/second which is |y|2*velocity p=hbar k=mv “same” if E>V. different k Note R+T=1 R T 1 E/V 1

1D Barriers:Example 5 R -50 MeV 5 MeV neutron strikes a heavy nucleus with V= -50 MeV. What fraction are reflected? Ignore 3D and use simplest step potential. 5 R -50 MeV

1D Barriers Step E<V 0 a Different if E>V or E<V. We’ll do E<V. again solve by continuity of wavefunction and derivative As no “left” travelling wave after the barrier Incoming falling transmitted reflected 0 a

1D Barriers Step E<V x=0 x=a 4 equations A,B,C,F,G (which can be complex) A related to incident flux and is arbitrary. Physics in: Eliminate F,G,B

Example:1D Barriers Step 2 eV electron incident on 4 eV barrier with thickness: .1 fm or 1 fm With thicker barrier:

Example:1D Barriers Step Even simpler For larger V-E larger k More rapid decrease in wave function through classically forbidden region

Bound States and Tunneling State A will be bound with infinite lifetime. State B is bound but can decay to B B’+X (unbound) with lifetime which depends on barrier height and thickness. Also reaction B’+XBA can be analyzed using tunneling tunneling is ~probability for wavefunction to be outside well B’ V B A E=0

Alpha Decay Example: Th90  Ra88 + alpha Kinetic energy of the alpha = mass difference have V(r) be Coulomb repulsion outside of nucleus. But attractive (strong) force inside the nucleus. Model alpha decay as alpha particle “trapped” in nucleus which then tunnels its way through the Coulomb barrier super quick - assume square potential more accurate - 1/r and integrate

Alpha Decay 4pZ=large number Do better. Use tunneling probability for each dx from square well. Then integrate as V(r) is known, integral can be calculated. See Griffiths 8.2 A B C 4pZ=large number

Alpha Decay T is the transmission probability per “incident” alpha f=no. of alphas “striking” the barrier (inside the nucleus) per second = v/2R, If v=0.1c f=1021 Hz Depends strongly on alpha kinetic energy