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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 herma 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 hermb 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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 Fall '11
 Dr.DanielArenas
 mechanics

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