to reconsider it in the subspace of C0(−∞, ∞) containing all even functions, and show that it generates a strongly continuous semigroup It is interesting that our
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So d/dx x = 1 + x d/dx Since A B are operators rather than numbers, they don't necessarily commute If two operators A B commute, then AB = BA and their
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d dx ekx = kex where k may take on any value (2) The operator x d dx also has an infinite set of eigenfunctions xn
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1) OPERATORS: Mathematical notation signifying the nature of mathematics on a x d/dx do not commute; The result of x(d/dx)ψ will not be equal to the
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Prove that the product of two linear operators is a linear operator Answer: dx = 1 (b) [d2/dx2,x] Answer: [ d2 dx2 ,x ] = d dx d dx x − x d2 dx2 = d dx + d dx
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Another operation might be differentiation with respect to x, represented by the operator d/dx We shall represent operators by the symbol O (omega) in general,
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dx f(x) Operator: D Operates on: scalar functions Action: Differentiate with respect dx f(x)) = x d2f dx2 = (xD2)(f(x)) Therefore the commutator of D and xD is
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the commutation relation, DX = XD − i In particular D 2 dx (2 15) which may be taken as the definition of a Hermitian operator The basic opera- tors X and D
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Ob- viously, any f(x) = ekx with arbitrary k is an eigenfunction of the operator, with k the corresponding eigenvalue Going to the operator d 2 /dx 2 , again any
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to reconsider it in the subspace of C0(−∞, ∞) containing all even functions, and show that it generates a strongly continuous semigroup It is interesting that our
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