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Extra info for Problems and Solutions on Quantum Mechanics: Major American Universities Ph. D. Qualifying Questions and Solutions
Example text
Lo) Solution: (a) As A sat isfirs :i quadratic equation it can t)(, rcprc~sc~itc~tl by a 2 x 2 matrix. c,clr~ntioliX":jX+2 = 0, X I = 1, X2 = 2. + C,! ) - (1 I JjLrl) = O . ,C is zero. ;~nnothi~v(:(:011111101i (ig(:l~sti~t,(:s. ) ' The eigenvalue equatiori 1068 then gives (L = 1, b = 0 for A = 1 and n = 0, b and eigenstates of A arc (h) (y). (c) Since A = A+, A is Hermitian and hence = 1 for X = 2. o possess only two energy eigensti~tc:~ tl(:ilotctl I I ) i~li(112). Tlic: systc:ln ;ilso inchides three other observablcs (t)csiclos t,hc c,ilc:rgy), known as P, I) aild R.
Give a brief discussion of why tlicre arc energy ballds in a crystallinc solid. Use the ideas of quantum ruechanics but do rlot att,em~,tt,o carry out any complicated calculatiolls. Yo11shollltl assllrnc that anyone reading rnc:cha~lic:sI)~ltdocs rlot ulldcrstar~~l your discussion undrrstands (~uallt~llm anything abolit the tl~eoryof solids. ( Wiscorrsrn) Solution: Solution: A crystal rnay l)e regarded as all infirlitc, ~)criotlic:array of j)otc:llti;tl wells, such as thc latticc structure givt:n in Problem 1065.
5 MeV, l, = r u - R = 33x10-'%1n (see pig. 58 x lo-"" 3x = 76 Problem9 and SoIutions ow Electromagnetism An electron with energy E = 1 PV is iticident upon a rectangular barrier of potential energy Vo = 2 eV (see Fig. 22). Allnut liow wide must tht. barrier be so t h a t the transmission prohability is where Vo is positive. If a particle wit,h mass rn is incident from the left with ilonrelativistic kinetic enprgy E, what is its p r ~ b a h i l i tfor , ~ trarlsmission I hrough the potential? For what values of E will t,his probabilib Ire unity?