Fund Quantum Mechanics Lect & HW Solutions 31

Fund Quantum Mechanics Lect & HW Solutions 31 -...

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Unformatted text preview: 2.6. HERMITIAN OPERATORS 13 Answer: By definition of Inv, and then using [1, p. 43]: Inv sin(kx) = sin(−kx) = − sin(kx) Inv cos(kx) = cos(−kx) = cos(kx) So by definition, both are eigenfunctions, and with eigenvalues −1 and 1, respectively. 2.6 2.6.1 Hermitian Operators Solution herm-a Question: A matrix A is defined to convert any vector r = xˆ+ y into r2 = 2xˆ+ 4y . Verify ı ˆ ı ˆ that ˆ and are orthonormal eigenvectors of this matrix, with eigenvalues 2, respectively 4. ı ˆ Answer: Take x = 1, y = 0 to get that r = ˆ transforms into r2 = 2ˆ. Therefore ˆ is an ı ı ı eigenvector, and the eigenvalue is 2. The same way, take x = 0, y = 1 to get that transforms ˆ into 4ˆ, so is an eigenvector with eigenvalue 4. The vectors ˆ and are also orthogonal and ˆ ı ˆ of length 1, so they are orthonormal. In linear algebra, you would write the relationship r2 = Ar out as: x2 y2 = 20 04 x y = 2x 4y In short, vectors are represented by columns of numbers and matrices by square tables of numbers. 2.6.2 Solution herm-b Question: A matrix A is defined to convert any vector r = (x, y ) into the vector r2 = (x + y, x + y ). Verify that (cos 45◦ , sin 45◦ ) and (cos 45◦ , − sin 45◦ ) are orthonormal eigenvectors √ 1 of this matrix, with eigenvalues 2 respectively 0. Note: cos 45◦ = sin 45◦ = 2 2 √√ √ 1√ √ 1 1 Answer: For r = (cos 45◦ , sin 45◦ ) = ( 2 2, 2 2), x = y = 2 2 so r2 = ( 2, 2), and that √ √ √ 1 is twice r. For r = (cos 45◦ , − sin 45◦ ) = ( 1 2, − 2 2), x = −y = 1 2 so r2 = (0, 0), and 2 2 that is zero times r. The square length of r = (cos 45◦ , sin 45◦ ) is r · r, which is given by the sum of the square components: cos2 45◦ + sin2 45◦ . That is one, so the vector is of length one. The same for r = ...
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