Download Solution manual to Quantum Physics 3rd edition by:Stephen by Stephen Gasiorowicz PDF

By Stephen Gasiorowicz

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Additional resources for Solution manual to Quantum Physics 3rd edition by:Stephen Gasiorowicz

Example text

We get, very simply 〈n | x 2 | n〉 = (2n + 1) 2mω mω (2n + 1) 〈n | p2 | n〉 = 2 14. Given the results of problem 9, and of 10, we have (Δx)2 = (2n + 1) 2mω mω (Δp)2 = (2n + 1) 2 and therefore 1 ΔxΔp = (n + ) 2 15. The eigenstate in A|α> = α|α> may be written in the form | α 〉 = f (A + ) | 0〉 It follows from the result of problem 4 that the eigenvalue equation reads df (A + ) | 0〉 = αf (A + ) | 0〉 Af (A ) | 0〉 = + dA + The solution of df (x) = α f(x) is f(x) = C eαx so that αA + | α 〉 = Ce | 0〉 The constant C is determined by the normalization condition <α|α> = 1 This means that ∞ 1 (α *) n ⎛ d ⎞ αA + α *A αA + = 〈0 | e e | 0〉 = 〈0 | e | 0〉 ∑ 2 ⎝ dA + ⎠ C n!

We insert 〈l,m1 | L + | l,m2 〉 = (l − m2 )(l + m2 + 1)δm1 ,m2 +1 〈l,m1 | L − | l,m2 〉 = (l + m2 )(l − m2 + 1)δm1 ,m2 −1 4. Again we use Lx = 1 i (L+ + L − ); L y = (L− − L+ ) to work out 2 2 L2x = 1 (L + L− )(L + + L− ) = 4 + 1 = (L2+ + L2− + L2 − L2z + Lz + L2 − L2z − Lz ) 4 1 1 1 1 = L2+ + L2− + L2 − L2z 4 2 2 4 We calculate 〈l,m1 | L + | l,m 2 〉 = (l − m 2 )(l + m 2 + 1)〈l,m1 | L + | l,m 2 + 1〉 2 = 2 ((l − m2 )(l + m2 + 1)(l − m2 − 1)(l + m2 + 2)1/2δ m ,m +2 1 2 and 〈l,m1 | L − | l,m 2 〉 = 〈l,m 2 | L+ | l,m1 〉 * 2 2 which is easily obtained from the preceding result by interchanging m1 and m2.

If H is not hermitian, then all four eigenvalues are acceptable. 14. We have the equations (1) (2) Bu(1) a = b11ua + b12 ua (1) (2) Bu(2) a = b21ua + b22ua (2) Let us now introduce functions (v (1) a ,v a ) that satisfy the equations (1) (2) (2) Bv(1) a = b1v a ;Bv a = b2 v a . s. as b1(α u1 + β u2). We can now take the coefficients of u1 and u2 separately, and get the following equations α (b1 – b11) = βb21 β (b1 – b22) = αb12 The product of the two equations yields a quadratic equation for b1, whose solution is b11 + b22 (b11 − b22 ) ± + b12b21 2 4 2 b1 = We may choose the + sign for the b1 eigenvalue.

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