Problem 2.13 (Engel)Determine in each case whether the function is an eigenfunction of the operator. If it is, what is the eigenvalue?A. function = e-iH3 x + 2 yL and operator = ∑2ê∑ x2Solution.∑2ÅÅÅÅÅÅÅÅÅ∑x2 e-iH3 x + 2 yL= = ∑ÅÅÅÅÅÅÅ∑x[ ∑ÅÅÅÅÅÅÅ∑xe-iH3 x + 2 yL]= ∑ÅÅÅÅÅÅÅ∑x[ -3i e-iH3 x + 2 yL]= H-3 iL2 e-iH3 x + 2 yL= -9 e-iH3 x + 2 yLThis result is a scalar multiple of the original function, so the function is an eigenfunction. The eigenvalue is -9.B. function = è!!!!!!!!!!!!!!!!x2+ y2 and operator = H1ê xL Hx2+ y2L H∑ê∑ xLSolution.H1êxL Hx2+ y2L ∑ÅÅÅÅÅÅÅÅ∑x è!!!!!!!!!!!!!!!!x2+ y2= x2+y2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅx ∑ÅÅÅÅÅÅÅÅ∑x Hx2+ y2L1ê2= x2+y2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅx H1ÅÅÅÅ2L Hx2+ y2L-1ê2 H2 xL= Hx2+ y2L1ê2This result is identical to the original function, so the function is an eigenfunction. The eigenvalue is 1.C. function = sin q cos q and operator = sin q dÅÅÅÅÅÅÅdq(sin q dÅÅÅÅÅÅÅdq) + 6 sin2qSolution.∂xHSin@xD Cos@xDLCos@xD2− Sin@xD2Expand@Sin@xD HCos@xD2− Sin@xD2LDCos@xD2Sin@xD − Sin@xD3∂xHCos@xD2Sin@xD − Sin@xD3LCos@xD3− 5 Cos@xD Sin@xD2Expand@Sin@xDHCos@xD3− 5 Cos@xD Sin@xD2LDCos@xD3Sin@xD − 5 Cos@xD Sin@xD36 Sin@xD2 Sin@xD Cos@xD6 Cos@xD Sin@xD3Combining the last two outputs, we obtainCos@xD3Sin@xD − 5 Cos@xD Sin@xD3+ 6 Cos@xD Sin@xD3Cos@xD3Sin@xD + Cos@xD Sin@xD3Factoring this result givesFactor@Cos@xD3Sin@xD + Cos@xD Sin@xD3DCos@xD Sin@xDHCos@xD2+ Sin@xD2LWhich simplifies toSimplify@Cos@xD Sin@xDHCos@xD2+ Sin@xD2LDCos@xD Sin@xDThis result is identical to the original function, so the function is an eigenfunction. The eigenvalue is
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