for any pair of vectors u and v and any pair of complex numbers c 1 and c 2.

Transcription

1 Linear Operators in Dirac notation We define an operator  as a map that associates with each vector u belonging to the (finitedimensional) linear vector space V a vector w ; this is represented by  u = w. An operator is said to be linear if it obeys  [ c 1 u + c 2 v ] = c 1  u + c 2  v for any pair of vectors u and v and any pair of complex numbers c 1 and c 2. The linear operators themselves form a linear space in that the sum of two operators  and ˆB is defined by (  + ˆB ) u =  u + ˆB u and multiplication by a complex scalar is defined by the action of c  on any ket as follows: so that we have (câ) v = c(  v ). (1) Ĉ = c 1  + c 2 ˆB Ĉ v = c 1  v + c 2 ˆB v for all v. (2) We can also define the product of two operators Ĉ =  ˆB by for every v V. Ĉ u =  ( ˆB u ) (3) An operator is completely specified by e i  e j for all i and j, the n 2 elements. This is clear since we can figure out how  acts on arbitrary v. Let w =  v ; we have w i = e i w = e i  v = e i  v j e j. Now we use the fact that  is linear (and use distributivity) to obtain w i = e i  e j v j A ij v j. Therefore, we have the useful identification for the matrix element of the operator A ij = e i  e j. Observe that the matrix elements depend on which basis you use. Note also that an operator can act on bras as well as kets: v  yields another bra. Given  v = w how is w related to v? We write w = v  1

2 which defines the operator Â. How is this related to Â? One way of doing this is to find the (ij) th -matrix element of A. Let us introduce the notation 1 A e i = e i and e i A = e i. So we have ( A ) ij = e i A e j = e i e j = e j e i = e j A e i = A ji. The correspondence with the definition of a Hermitian conjugate of a matrix is evident.  is known as the adjoint or Hermitian conjugate of Â. You must remember this abstract definition of the adjoint of an operator: If  v = w then w = v Â. One can now define eigenvalues and eigenvectors of an operator as usual by  a j = λ j a j (4) Note that by using the properties of linear operators any multiple of a j is also an eigenvector with the same eigenvalue. We will typically use the normalized eigenvector. This allows standard definitions of Hermitian and Unitary matrices to be carried to the abstract operator in a finite-dimensional vector space. A Hermitian operator is defined by  = Â. The real cousin of a Hermitian matrix is a symmetric matrix. For real-valued A, we have A = A T the transpose for a symmetric matrix. Note that the eigenvalues of a Hermitian operator are real; in Dirac notation if  a j = λ j a j we have the corresponding bra version: a j  = a j λ j. Since  = Â, this implies a j  = a j λ j. Assuming normalized eigenvectors we have from the two boxed equations which clearly implies that λ j is real. a j  a j = a j ( a j ) = λ j a j a j = λ j = ( a j Â) a j = λ j a j a j = λ j (5) The orthogonality of eigenvectors corresponding to different eigenvalues follows simply. If  a j = λ j a j we have a i  a j = λ j a i a j. Since a i  = λ i a i we have a i  a j = λ i a i a j. Thus λ i a i a j = λ j a i a j and hence for i j the dot product must vanish. So in the case in which we have n distinct eigenvalues (non-degenerate) the n eigenvectors are orthogonal and they form a basis. It takes more effort to show that even in the case of degeneracy the set of eigenvectors is complete and we can construct an orthonormal basis from the set of eigenvectors. The completeness in the infinite-dimensional case of a Hilbert space is definitely harder to prove and we will assume the result. Define unitary operators: U a linear operator on V is said to be unitary if (i) UU = U U = I. 1 Note that ket e i need not have unit norm; all we know is that it is the result of acting on e i with A. 2

3 (ii) If v = U v and w = U w then the inner product v w = v w. This is clear since v = v U and w = U w so v w = v U U w = v w. Unitary matrices preserve the inner product and a fortiori the norm. You should think of unitary matrices as rotations in complex vector space. The real cousin of a unitary matrix is an orthogonal matrix OO T = I. Orthogonal matrices correspond to rotations in a real vector space. Here are some theorems: If A is a linear, self-adjoint matrix/operator then every eigenvalue of A is real. If U is unitary every eigenvalue of U has unit modulus. If an operator is unitary or Hermitian eigenvectors that correspond to distinct eigenvalues are orthogonal. If an eigenvalue has algebraic multiplicity k we can find k linearly independent eigenvectors that belong to that eigenvalue. If A is Hermitian then e ia is unitary. Thus e ia is unitary. We note that e ia = I + ia + i2 2! A2 + + in n! An. (e ia) ( i) 2 = I + ( i)a + A ( i)n A n = e ia. 2! n! ( e ia e ia) = e ia e ia = I. e A e B e A+B. Any Hermitian matrix can be diagonalized by a unitary similarity transformation: there exists a unitary U such that U 1 AU = D. A similarity transformation is defined as follows. Suppose A v = w. We use an ivertible transformation (a useful example is a rotation of the axes) S and obtain v = S v and w = S w by transforming v and w. Now we know that A acts on v to yield w. We wish to find  such that  v = w.  v = w  S 1 v = S 1 w. SÂS 1 v w. Thus if the vectors are transformed by S ( v = S v ) the operators are transformed according to  = SÂS 1 which is called a similarity transformation. If S is unitary the transformation is said 3

4 to be unitary. In general the characteristic polynomial defined by det(a λi) is invariant under a similarity transformnation. This can be seen using a very important result det(ab) = det(a) det(b). Note that this means that det(a 1 ) = 1/det(A) again showing the necessity of a non-vanishing determinant of A for A 1 to be well-defined. Now consider the charactersitic polynomial of SAS 1 given by det(sas 1 λi). We have det(sas 1 λi) = det(sas 1 λiss 1 ) = det(s[a λi]s 1 ) = det(s) det(a λi) det(s 1 ) = det(a λi). The second line follows since the product of the three matrices S, (A λi) S 1 yields SAS 1 λiss 1. Thus the characteristic polynomials are the same, in particular the detrminant and the trace are the same. Useful trick Consider a matrix in a basis in which it is diagonal. If the matrix is Hermitian or unitary this is always possible and so we will use this representation. Thus a 3 3 matrix A in the diagonal representation is a a a 3 where a 1, a 2, and a 3 are the eigenvalues of A. Note that the determinant of A is the product of the eigenvalues, a very useful result), a 1 a 2 a 3 in this case. What is the inverse of A in this representation? a 1 A = 0 a a 1 3 You can verify that this is the inverse by multiplying the two diagonal matrices. Clearly, if one of the eigenvalues vanishes then the determinant is zero and the inverse does not exist. This is a way of understanding this standard result. Recall the definition of the trace: it is the sum of the diagonal elements and this is the same in any basis. Note that if the eigenvalues of  are a i, the eigenvalues of  2 are a 2 i. Given  a i = a i a i we have  2 a i =  a i a i = a i  a i = a 2 i a i. Similarly, the eigenvalues of A n are a n i. What are the eigenvalues of ea. Recall that Compute e A a i by using the series. We have ( e A a i = e A = I + A A2 + 1 n! An. 1 + a i + a2 i 2 + an i n! 4 ) a i = e a i a i.

5 We have argued that the eigenvalues of e A are {e a i }. This is a useful result. For example if A is Hermitian its eigenvaklues are real and the eigenvalues of e ia are e ia j whose modulus is one which is expected for unitary operators. 5