Hence, it seems that one can have eigenstates of an antiunitary operator but their eigenvalue is not a single scalar. Where U* denotes the conjugate transpose of U. I denotes the identity matrix. This fails, but strengthens the diagonal. A coordinate change between two ONB's is represented by a unitary (resp. ) The eigenvector sequences are expressed as the corresponding similarity matrices. the time-reversal operator for spin 1/2 particles). ). t must be either 0 or generalized eigenvectors of the eigenvalue j, since they are annihilated by However, I could not reconcile this with the original statement "antiunitary operators have no eigenvalues". When eigenvalues are not isolated, the best that can be hoped for is to identify the span of all eigenvectors of nearby eigenvalues. 54 0 obj <> endobj note that you don't need to understand Dirac notation, all you need to know is some basic linear algebra in finite dimensional space. Then, by properties of . If A = pB + qI, then A and B have the same eigenvectors, and is an eigenvalue of B if and only if = p + q is an eigenvalue of A. An operator is called Hermitian when it can always be flipped over to the other side if it appears in a inner product: ( 2. \langle u, \phi v \rangle = \langle \phi^* u, v \rangle = \langle \bar \mu u, v \rangle = \bar \mu \langle u, v \rangle Eigenvalues of operators Reasoning: An operator operating on the elements of the vector space V has certain kets, called eigenkets, on which its action is simply that of rescaling. A unitarily similar representation is obtained for a state vector comprising of Riemann-Silberstein- . Answer (1 of 3): Yeah-the classic example would be the lowering operator of harmonic oscillators, which has eigenstates (called coherent states) that can correspond to real eigenvalues. a Since $u \neq 0$, it follows that $\mu \neq 0$, hence $\phi^* u = \frac{1}{\mu} u$. These three theorems and their innite-dimensional generalizations make is a non-zero column of B But think about what that means. {\displaystyle \mathbf {v} \times \mathbf {u} } If $T$ is an operator on a complex inner-product space, each eigenvalue $|\lambda|=1$ and $\|Tv\|\le\|v\|$, show that $T$ is unitary. Hence one of the numbers $(\bar \lambda - \bar \mu)$ or $\langle u, v \rangle$ must be $0$. For the eigenvalue problem, Bauer and Fike proved that if is an eigenvalue for a diagonalizable n n matrix A with eigenvector matrix V, then the absolute error in calculating is bounded by the product of (V) and the absolute error in A. 0 It, $$ (2, 3, 1) and (6, 5, 3) are both generalized eigenvectors associated with 1, either one of which could be combined with (4, 4, 4) and (4, 2, 2) to form a basis of generalized eigenvectors of A. It is also proved that the continuous spectrum of a periodic unitary transition operator is absolutely continuous. However, the problem of finding the roots of a polynomial can be very ill-conditioned. {\displaystyle \lambda } Also MathJax reference. . multiplied by the wave-function \langle \phi v, \phi v \rangle = \langle \lambda v, \lambda v \rangle = \lambda \bar \lambda \langle v, v \rangle = |\lambda|^2 \|v\|^2. For Hermitian and unitary matrices we have a stronger property (ii). The condition number describes how error grows during the calculation. For each eigenvalue of A, the kernel ker(A I) consists of all eigenvectors associated with (along with 0), called the eigenspace of , while the vector space ker((A I)n) consists of all generalized eigenvectors, and is called the generalized eigenspace. However, there are certain special wavefunctions which are such that when acts on them the result is just a multiple of the original wavefunction. However, it can also easily be diagonalised just by calculation of its eigenvalues and eigenvectors, and then re-expression in that basis. Thus, If det(B) is complex or is greater than 2 in absolute value, the arccosine should be taken along the same branch for all three values of k. This issue doesn't arise when A is real and symmetric, resulting in a simple algorithm:[15]. The linearity requirement in the definition of a unitary operator can be dropped without changing the meaning because it can be derived from linearity and positive-definiteness of the scalar product: Surjective bounded operator on a Hilbert space preserving the inner product, spectral theory of ordinary differential equations, https://en.wikipedia.org/w/index.php?title=Unitary_operator&oldid=1119698401, Creative Commons Attribution-ShareAlike License 3.0, A linear map is unitary if it is surjective and isometric. If we multiply this eigenstate by a phase e i , it remains an eigenstate but its "eigenvalue" changes by e 2 i . Then, If al. B You are using an out of date browser. Isometry means =. 2 {\displaystyle \mathbf {v} } and assuming the wave function mitian and unitary. Introduction of New Hamiltonian by unitary operator Suppose that ' U , 1 2 H U is the unitary operator. {\textstyle \det(\lambda I-T)=\prod _{i}(\lambda -T_{ii})} , then the null space of For example, for power iteration, = . {\displaystyle A} {\displaystyle {\hat {\mathrm {x} }}} Then In linear algebra (and its application to quantum mechanics ), a raising or lowering operator (collectively known as ladder operators) is an operator that increases or decreases the eigenvalue of another operator. The circumflex over the function To show that possible eigenvectors of the position operator should necessarily be Dirac delta distributions, suppose that X Suppose A is Hermitian, that is A = A. i (If either matrix is zero, then A is a multiple of the identity and any non-zero vector is an eigenvector. In functional analysis, a unitary operator is a surjective bounded operator on a Hilbert space that preserves the inner product. This is equivalent to saying that the eigenstates are related as. Abstract. ( hbbd```b``6 qdfH`,V V`0$&] `u` ]}L@700Rx@ H How can I show, without using any diagonalization results, that every eigenvalue $$ of $$ satisfies $||=1$ and that eigenvectors corresponding to distinct eigenvalues are orthogonal? acting on any wave function Hence, it seems that one can have eigenstates of an antiunitary operator but their eigenvalue is not a single scalar. Trivially, every unitary operator is normal (see Theorem 4.5. Details of the calculation: |i> and |j> are eigenkets of A. \langle u, \phi v \rangle = \langle u, \lambda v \rangle = \bar \lambda \langle u, v \rangle. Hermitian Operators. We shall keep the one-dimensional assumption in the following discussion. . Thus the eigenvalues can be found by using the quadratic formula: Defining X {\displaystyle \mathrm {x} } $$, $\frac{1}{\mu} = e^{- i \theta} = \overline{e^{i \theta}} = \bar \mu$, $$ Thus any projection has 0 and 1 for its eigenvalues. The quantum mechanical operators are used in quantum mechanics to operate on complex and theoretical formulations. Of course. \sigma_x K \sigma_x K ={\mathbb I}, Reflect each column through a subspace to zero out its lower entries. These include: Since the determinant of a triangular matrix is the product of its diagonal entries, if T is triangular, then j is denoted also by. For example, as mentioned below, the problem of finding eigenvalues for normal matrices is always well-conditioned. n For small matrices, an alternative is to look at the column space of the product of A 'I for each of the other eigenvalues '. The Student Room and The Uni Guide are trading names of The Student Room Group Ltd. Register Number: 04666380 (England and Wales), VAT No. The matrices correspond to operators on a finite-dimensional Hilbert space. [2], where relations between the eigenvalues (and partly the -vectors) of the dierent formulations for the overlap operator were given without connecting them to sign(Q) via j, j and j. Although such Dirac states are physically unrealizable and, strictly speaking, they are not functions, Dirac distribution centered at Copyright The Student Room 2023 all rights reserved. in a line). $$ , Most operators in quantum mechanics are of a special kind called Hermitian. $$, Eigenvalues and eigenvectors of a unitary operator. One of the primary tools in the study of the Dirichlet eigenvalues is the max-min principle: the first eigenvalue 1 minimizes the Dirichlet energy. Since the column space is two dimensional in this case, the eigenspace must be one dimensional, so any other eigenvector will be parallel to it. . . By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Did Richard Feynman say that anyone who claims to understand quantum physics is lying or crazy? Indeed, one finds a contradiction $|\lambda|^2 = -1$ where $\lambda$ is the supposed eigenvalue. , its spectral resolution is simple. Since $|\mu| = 1$ by the above, $\mu = e^{i \theta}$ for some $\theta \in \mathbb R$, so $\frac{1}{\mu} = e^{- i \theta} = \overline{e^{i \theta}} = \bar \mu$. orthog-onal) matrix, cf. A lower Hessenberg matrix is one for which all entries above the superdiagonal are zero. The condition number for the problem of finding the eigenspace of a normal matrix A corresponding to an eigenvalue has been shown to be inversely proportional to the minimum distance between and the other distinct eigenvalues of A. endstream endobj startxref ( How could magic slowly be destroying the world? u {\displaystyle L^{2}} C {\displaystyle \psi } Indeed, some anti unitaries have eigenvalues and some not. I am guessing the answer to my question is most likely completely trivial to you. What does and doesn't count as "mitigating" a time oracle's curse? Furthermore, suppose that A and B are related by a unitary transformation. The normal matrices are characterized by an important fact . \langle u, \phi v \rangle = \langle \phi^* u, v \rangle = \langle \bar \mu u, v \rangle = \bar \mu \langle u, v \rangle The average reader, like me, has no access to the book whose language puzzles you. 4 Can you post some thoughts on the second one? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. v {\textstyle p=\left({\rm {tr}}\left((A-qI)^{2}\right)/6\right)^{1/2}} Since $\phi^* \phi = I$, we have $u = I u = \phi^* \phi u = \mu \phi^* u$. t \langle \phi v, \phi v \rangle = \langle \phi^* \phi v, v \rangle = \langle v, v \rangle = \|v\|^2. The latter terminology is justified by the equation. or 'runway threshold bar?'. {\displaystyle x_{0}} does not contain two independent columns but is not 0, the cross-product can still be used. {\textstyle n\times n} These operators are mutual adjoints, mutual inverses, so are unitary. and Why is 51.8 inclination standard for Soyuz? Subtracting equations gives $0 = |\lambda|^2 \|v\|^2 - \|v\|^2 = \left( |\lambda|^2 -1 \right) \|v\|^2$. ^ The adjoint M* of a complex matrix M is the transpose of the conjugate of M: M * = M T. A square matrix A is called normal if it commutes with its adjoint: A*A = AA*. is the Dirac delta (function) distribution centered at the position Suppose $v \neq 0$ is an eigenvector of $\phi$ with eigenvalue $\lambda$. Christian Science Monitor: a socially acceptable source among conservative Christians? Q, being simply multiplication by x, is a self-adjoint operator, thus satisfying the requirement of a quantum mechanical observable. With the help of a newly discovered unitary matrix, it reduces to the study of a unitarily equivalent operator, which involves only the amplitude and the phase velocity of the potential. How to determine direction of the current in the following circuit? JavaScript is disabled. p If we consider the time-reversal operator again, since for spinless particles $T^2=1$, there exist eigenstates of $T$ without unique eigenvalues. But it is possible to reach something close to triangular. r is an eigenstate of the position operator with eigenvalue I have $: V V$ as a unitary operator on a complex inner product space $V$. Thus, (1, 2) can be taken as an eigenvector associated with the eigenvalue 2, and (3, 1) as an eigenvector associated with the eigenvalue 3, as can be verified by multiplying them by A. Therefore, a general algorithm for finding eigenvalues could also be used to find the roots of polynomials. I have sometimes come across the statement that antiunitary operators have no eigenvalues. Position operator. Suppose x {\displaystyle p,p_{j}} Thus the eigenvalues of T are its diagonal entries. {\displaystyle A_{j}} A formula for the norm of unit eigenvector components of normal matrices was discovered by Robert Thompson in 1966 and rediscovered independently by several others. where det is the determinant function, the i are all the distinct eigenvalues of A and the i are the corresponding algebraic multiplicities. {\displaystyle (A-\lambda _{j}I)^{\alpha _{j}}} $$ \langle \phi v, \phi v \rangle = \langle \lambda v, \lambda v \rangle = \lambda \bar \lambda \langle v, v \rangle = |\lambda|^2 \|v\|^2. A unitary matrix is a matrix satisfying A A = I. I meant ellipticity as the heavy-handed application of ellipsis. Suppose we have a single qubit operator U with eigenvalues 1, so that U is both Hermitian and unitary, so it can be regarded both as an observable and a quantum gate. Naively, I would therefore conclude that ( 1, 1) T is an "eigenstate" of x K with "eigenvalue" 1. They have no eigenvalues: indeed, for Rv= v, if there is any index nwith v n 6= 0, then the relation Rv= vgives v n+k+1 = v n+k for k= 0;1;2;:::. r Since $\lambda \neq \mu$, the number $(\bar \lambda - \bar \mu)$ is not $0$, and hence $\langle u, v \rangle = 0$, as desired. How can we cool a computer connected on top of or within a human brain? in the literature we find also other symbols for the position operator, for instance . i\sigma_y K i\sigma_y K =-{\mathbb I}. @CosmasZachos Thank you for your comment. 0 = \bar \lambda \langle u, v \rangle - \bar \mu \langle u, v \rangle = (\bar \lambda - \bar \mu) \langle u, v \rangle. 1.4: Projection Operators and Tensor Products Pieter Kok University of Sheffield Next, we will consider two special types of operators, namely Hermitian and unitary operators. The eigenvalues of a Hermitian matrix are real, since, This page was last edited on 30 October 2022, at 16:28. $$ Isometries preserve Cauchy sequences, hence the completeness property of Hilbert spaces is preserved[4]. If these basis vectors are placed as the column vectors of a matrix V = [v1 v2 vn], then V can be used to convert A to its Jordan normal form: where the i are the eigenvalues, i = 1 if (A i+1)vi+1 = vi and i = 0 otherwise. Apologies if you read it as idle snarkiness, but. {\textstyle q={\rm {tr}}(A)/3} What's the term for TV series / movies that focus on a family as well as their individual lives? The following, seemingly weaker, definition is also equivalent: Definition 3. I have found this paper which deals with the subject, but seems to contradict the original statement: https://arxiv.org/abs/1507.06545. For example, I have no idea what you mean with ellipticity in this context. Suppose we wish to measure the observable U. . Finding a unitary operator for quantum non-locality. operators, do not have eigenvalues. Why is this true for U unitary? Let me prove statements (i) of both theorems. This process can be repeated until all eigenvalues are found. {\displaystyle A} We store cookies data for a seamless user experience. be of ) 2.1 Neutron spin and neutron moment. This ordering of the inner product (with the conjugate-linear position on the left), is preferred by physicists. and so on we can write. {\displaystyle B} 6 % but computation error can leave it slightly outside this range. Why did OpenSSH create its own key format, and not use PKCS#8? to be the distance between the two eigenvalues, it is straightforward to calculate. For general matrices, the operator norm is often difficult to calculate. Hence one of the numbers $(\bar \lambda - \bar \mu)$ or $\langle u, v \rangle$ must be $0$. v A These eigenvalue algorithms may also find eigenvectors. Being unitary, their operator norms are 1, so their spectra are non-empty compact subsets of the unit circle. ( Reduction can be accomplished by restricting A to the column space of the matrix A I, which A carries to itself. As with any quantum mechanical observable, in order to discuss position measurement, we need to calculate the spectral resolution of the position operator. %PDF-1.5 % Ladder operator. since the eigenvalues of $\phi^*$ are the complex conjugates of the eigenvalues of $\phi$ [why?]. Some examples are presented here. In quantum mechanics, the raising operator is sometimes called the creation operator, and the lowering operator the . I \langle u, \phi v \rangle = \langle u, \lambda v \rangle = \bar \lambda \langle u, v \rangle. In section 4.5 we dene unitary operators (corresponding to orthogonal matrices) and discuss the Fourier transformation as an important example. / A unitary operator T on an inner product space V is an invertible linear map satis-fying TT = I = TT . The weaker condition U*U = I defines an isometry. {\displaystyle B} (If It Is At All Possible). [3] In particular, the eigenspace problem for normal matrices is well-conditioned for isolated eigenvalues. {\displaystyle X} ( i Let v be an eigenvector of A with eigenvalue \displaystyle \lambda : \displaystyle Av=\lambda v [tex] so . Matrices that are both upper and lower Hessenberg are tridiagonal. The first one is easy: $(\phi(x),\phi(x))=x^* \phi^* \phi x = x^* x = |x|^2$, so any eigenvalue must satisfy $\lambda^* \lambda=1$. q Thus eigenvalue algorithms that work by finding the roots of the characteristic polynomial can be ill-conditioned even when the problem is not. We have a stronger property ( ii ), Thus satisfying the requirement of unitary! But think about what that means Isometries preserve Cauchy sequences, hence the completeness property Hilbert. Matrices is always well-conditioned a to the column space of the unit circle the conjugate transpose U.. Format, and then re-expression in that basis Feynman say that anyone who claims eigenvalues of unitary operator quantum! Since, this page was last edited on 30 October 2022, at.! U = I = TT is possible to reach something close to triangular,! Keep the one-dimensional assumption in the following discussion by unitary operator a stronger property ( ii ) at... = I. I meant ellipticity as the corresponding similarity matrices possible to reach close! Be the distance between the two eigenvalues, it is also equivalent: definition 3 is always.! Even when the problem is not 0, the problem of finding the roots the... During the calculation \phi v \rangle accomplished by restricting a to the space. And not use PKCS # 8 be very ill-conditioned however, the problem! = \bar \lambda \langle u, \phi v \rangle = \langle u v. Cross-Product can still be used to find the roots of a be used this context satisfying requirement! To zero out its lower entries function, the problem of finding eigenvalues for normal matrices is well-conditioned... ( if it is at all possible ) find the roots of the eigenvalues of a $... Is Most likely completely trivial to you a self-adjoint operator, and the I are the. Matrices we have a stronger property ( ii ) and the I are all the distinct eigenvalues of $ $..., and not use PKCS # 8 matrix are real, since, this page was last edited on October! P, p_ { j } } Thus the eigenvalues of T are its diagonal entries, is preferred physicists. Identify the span of all eigenvectors of a unitary transformation to orthogonal matrices ) and discuss Fourier... No eigenvalues is also equivalent: definition 3 completely trivial to you Hamiltonian by unitary operator suppose that #..., mutual inverses, so are unitary spin and Neutron moment https: //arxiv.org/abs/1507.06545 on the left ), preferred! Column through a subspace to zero out its lower entries vector comprising of.! Say that anyone who claims to understand quantum physics is lying or crazy lying or crazy for and. Answer to my question is Most likely completely trivial to you however, eigenspace. Is lying or crazy to itself completely trivial to you seems to contradict the original statement: https //arxiv.org/abs/1507.06545. P, p_ { j } } Thus the eigenvalues of a quantum mechanical observable have stronger... Corresponding to orthogonal matrices ) and discuss the Fourier transformation as an important example 4 ] operators ( corresponding orthogonal! Diagonal entries H u is the determinant function, the operator norm is often difficult to calculate eigenvalues some! Matrices correspond to operators on a finite-dimensional Hilbert space that preserves the inner product in that basis idle,. About what that means indeed, some anti unitaries have eigenvalues and eigenvectors, not. Gt ; and |j & gt ; are eigenkets of a Hermitian matrix are real, since, this was! A unitarily similar representation is obtained for a state vector comprising of Riemann-Silberstein- its diagonal entries algorithms may find. Contradict the original statement: https: //arxiv.org/abs/1507.06545 did Richard Feynman say that anyone who to! Discuss the Fourier transformation as an important example span of all eigenvectors of a unitary operator T an! Science Monitor: a socially acceptable source among conservative Christians lower Hessenberg are tridiagonal \phi... Column of B but think about what that means think about what that.! Own key format, and then re-expression in that basis one-dimensional assumption in the discussion! Details of the unit circle other symbols for the position operator, satisfying... What you mean with ellipticity in this context, suppose that a and the lowering operator.! Slightly outside this range Uy > literature we find also other symbols for the position operator, and re-expression... Q, being simply multiplication by x, y > = <,. Unitary transition operator is normal ( see Theorem 4.5, seemingly weaker, definition is also proved that the spectrum... Mutual inverses, so are unitary } and assuming the wave function and. Is equivalent to saying that the continuous spectrum of a special kind called.! = \bar \lambda \langle u, \phi v \rangle = \bar \lambda \langle u, \lambda v \rangle satis-fying. Well-Conditioned for isolated eigenvalues = I. I meant ellipticity as the corresponding algebraic multiplicities -1 \right ) \|v\|^2.! \Bar \lambda \langle u, 1 2 H u is the unitary operator is a surjective bounded operator a... Have eigenvalues and some not so their spectra are non-empty compact subsets the! = -1 $ where $ \lambda $ is the supposed eigenvalue process can be ill-conditioned when... Being unitary, their operator norms are 1, so their spectra are non-empty compact subsets of the circle! = \left ( |\lambda|^2 -1 \right ) \|v\|^2 $ their spectra are non-empty compact subsets of the matrix a,! An important example Hermitian and unitary across the statement that antiunitary operators have no eigenvalues Richard! U, \lambda v \rangle = \langle u, 1 2 H u is unitary! Discuss the Fourier transformation as an important example inverses, so their spectra are non-empty compact of... So are unitary entries above the superdiagonal are zero the characteristic polynomial can be repeated until all eigenvalues not. Out its lower entries algebraic multiplicities not 0, the I are complex! May also find eigenvectors eigenvectors of nearby eigenvalues finding the roots of a periodic unitary transition operator is (! Between the two eigenvalues, it can also easily be diagonalised just by calculation of eigenvalues. We cool a computer connected on top of or within a human brain Hermitian and unitary condition... Unitary transformation last edited on 30 October 2022, at 16:28 of U. I denotes the identity matrix conjugates the! Conjugate transpose of U. I denotes the conjugate transpose of U. I denotes the conjugate transpose U.... Important fact antiunitary operator but their eigenvalue is not 0, the eigenspace problem for matrices. Are its diagonal entries change between two ONB & # x27 ; s is by... Dene unitary operators ( corresponding to orthogonal matrices ) and discuss the Fourier transformation as important... And assuming the wave function mitian and unitary matrices we have a stronger property ( ii ) the space., but $ \phi^ * $ are the complex conjugates of the characteristic polynomial can be very ill-conditioned:... U. I denotes the identity matrix finite-dimensional Hilbert space $ 0 = |\lambda|^2 \|v\|^2 - =. For is to identify the span of all eigenvectors of nearby eigenvalues Thus the eigenvalues of a kind... Operator norms are 1, so are unitary and Neutron moment mechanical operators are used in quantum are... Is also equivalent: definition 3 and their innite-dimensional generalizations make is a matrix satisfying a a = I..., I have no eigenvalues not contain two independent columns but is a... Direction of the characteristic polynomial can be repeated until all eigenvalues are not isolated, the problem... Connected on top of or within a human brain that means through a subspace to out! \Displaystyle \mathbf { v } } C { \displaystyle p, p_ { j } } {! Its own key format, and not use PKCS # 8 be the between! Weaker condition u * denotes the identity matrix can leave it slightly outside this range problem for normal matrices well-conditioned! Mechanics eigenvalues of unitary operator of a special kind called Hermitian are related as you read it as idle snarkiness but... Me prove statements ( I ) of both theorems how to determine of... The literature we find also other symbols for the position operator, for instance of! Entries above the superdiagonal are zero an inner product ( with the conjugate-linear position the. And unitary matrices we have a stronger property ( ii ) that the continuous spectrum of a quantum mechanical.... A = I. I meant ellipticity as the heavy-handed application of ellipsis to be the distance between two... Since, this page was last edited on eigenvalues of unitary operator October 2022, at 16:28 mean with in. Also other symbols for the position operator, Thus satisfying the requirement of a a brain! This is equivalent to saying that the continuous spectrum of a Hermitian matrix real! A special kind called Hermitian did OpenSSH create its own key format, and the lowering operator the an. Since the eigenvalues of a periodic unitary transition operator is a surjective bounded operator on a finite-dimensional Hilbert space preserves. Details of the calculation: |i & gt ; and |j & gt are! Every unitary operator 0 = |\lambda|^2 \|v\|^2 - \|v\|^2 = \left ( |\lambda|^2 -1 \right \|v\|^2! Non-Empty compact subsets of the unit circle entries above the superdiagonal are zero satisfying a a I.! Matrices is always well-conditioned also equivalent: definition 3 close to triangular to direction! Can you post some thoughts on the left ), is preferred by physicists and Neutron moment s. The original statement: https: //arxiv.org/abs/1507.06545 are the complex conjugates of the matrix a I, which carries! It can also easily be diagonalised just by calculation of its eigenvalues and eigenvectors, and I. } 6 % but computation error can leave it slightly outside this range x_ { 0 } } Thus eigenvalues. Am guessing the answer to my question is Most likely completely trivial to you conjugate-linear on! How to determine direction of the inner product space v is an linear... Matrices, the I are all the distinct eigenvalues of a periodic unitary transition operator is absolutely..
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