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Solution Manual for Modern Quantum Mechanics, 2nd Edition

Unlock the answers to every textbook problem with Solution Manual for Modern Quantum Mechanics, 2nd Edition, making studying more efficient and effective.

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2Chapter One1.AC{D, B}=ACDB+ACBD,A{C, B}D=ACBD+ABCD,C{D, A}B=CDAB+CADB, and{C, A}DB=CADB+ACDB. Therefore−AC{D, B}+A{C, B}D−C{D, A}B+{C, A}DB=−ACDB+ABCD−CDAB+ACDB=ABCD−CDAB= [AB, CD]In preparing this solution manual, I have realized that problems 2 and 3 in are misplacedin this chapter. They belong in Chapter Three. The Pauli matrices are not even defined inChapter One, nor is the math used in previous solution manual.– Jim Napolitano2.(a)Tr(X) =a0Tr(1) +￿￿Tr(σ￿)a￿= 2a0sinceTr(σ￿) = 0. AlsoTr(σkX) =a0Tr(σk) +￿￿Tr(σkσ￿)a￿=12￿￿Tr(σkσ￿+σ￿σk)a￿=￿￿δk￿Tr(1)a￿= 2ak. So,a0=12Tr(X) andak=12Tr(σkX).(b) Just do the algebra to finda0= (X11+X22)/2,a1= (X12+X21)/2,a2=i(−X21+X12)/2, anda3= (X11−X22)/2.3.Since det(σ·a) =−a2z−(a2x+a2y) =−|a|2, the cognoscenti realize that this problemreally has to do with rotation operators. From this result, and (3.2.44), we writedet￿exp￿±iσ·ˆnφ2￿￿= cos￿φ2￿±isin￿φ2￿and multiplying out determinants makes it clear that det(σ·a￿) = det(σ·a). Similarly, use(3.2.44) to explicitly write out the matrixσ·a￿and equate the elements to those ofσ·a.With ˆnin thez-direction, it is clear that we have just performed a rotation (of the spinvector) through the angleφ.4.(a)Tr(XY)≡￿a￿a|XY|a￿=￿a￿b￿a|X|b￿￿b|Y|a￿by inserting the identity operator.Then commute and reverse, soTr(XY) =￿b￿a￿b|Y|a￿￿a|X|b￿=￿b￿b|Y X|b￿=Tr(Y X).(b)XY|α￿=X[Y|α￿] is dual to￿α|(XY)†, butY|α￿ ≡|β￿is dual to￿α|Y†≡ ￿β|andX|β￿is dual to￿β|X†so thatX[Y|α￿] is dual to￿α|Y†X†. Therefore (XY)†=Y†X†.(c) exp[if(A)] =￿aexp[if(A)]|a￿￿a|=￿aexp[if(a)]|a￿￿a|(d)￿aψ∗a(x￿)ψa(x￿￿) =￿a￿x￿|a￿∗￿x￿￿|a￿=￿a￿x￿￿|a￿￿a|x￿￿=￿x￿￿|x￿￿=δ(x￿￿−x￿)5.For basis kets|ai￿, matrix elements ofX≡|α￿￿β|areXij=￿ai|α￿￿β|aj￿=￿ai|α￿￿aj|β￿∗.For spin-1/2 in the| ±z￿basis,￿+|Sz= ¯h/2￿= 1,￿−|Sz= ¯h/2￿= 0, and, using (1.4.17a),￿±|Sx= ¯h/2￿= 1/√2. Therefore|Sz= ¯h/2￿￿Sx= ¯h/2|.=1√2￿1100￿6.A[|i￿+|j￿] =ai|i￿+aj|j￿ ￿= [|i￿+|j￿] so in general it is not an eigenvector, unlessai=aj.That is,|i￿+|j￿is not an eigenvector ofAunless the eigenvalues are degenerate.

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Modern Quantum Mechanics by J. J. S...

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