20 Linear Algebra to be orthogonal to the k 1 vectors U and find that c = (U ,V )so i ik i k that k 1 u = V (U ,V ) U . (1.121) k k i k i Xi=1 The normalized vector then is uk Uk = . (1.122) (uk,uk) A basis is more convenient if its vectorsp are orthonormal.

1.11 Outer products From any two vectors f and g, we may make an outer-product operator A that maps any vector h into the vector f multiplied by the inner product (g, h) Ah= f (g, h)=(g, h) f. (1.123) The operator A is linear because for any vectors e, h and numbers z,w A (zh+ we)=(g, z h + we) f = z (g, h) f + w (g, e) f = zAh+ wAe. (1.124) If f, g, and h are vectors with components fi, gi, and hi in some basis, then the linear transformation is n n

(Ah)i = Aij hj = fi gj⇤ hj (1.125) Xj=1 Xj=1 and in that basis A is the matrix with entries

Aij = fi gj⇤. (1.126)

It is the outer product of the vectors f and g⇤. The outer product of g and f ⇤ is di↵erent, Bij = gi fj⇤. Example 1.20 (Outer Product). If in some basis the vectors f and g are i 2 f = and g = 1 (1.127) 3i 0 1 ✓ ◆ 3i @ A then their outer products are the matrices 2 2i 2 6i A = i 1 3i = (1.128) 3i 33i 9 ✓ ◆ ✓ ◆ 1.12 Dirac notation 21 and i 2i 3 B = 1 2 3i = 2 3i . (1.129) 0 1 0 1 3i 6i 9 @ A @ A

Example 1.21 (Dirac’s outer products). Dirac’s notation for outer products is neat. If the vectors f = f and g = g are | i | i a z f = b and g = (1.130) | i 0 1 | i w c ✓ ◆ @ A then their outer products are

az⇤ aw⇤ za⇤ zb⇤ zc⇤ f g = bz⇤ bw⇤ and g f = (1.131) | ih | 0 1 | ih | wa⇤ wb⇤ wc⇤ cz⇤ cw⇤ ✓ ◆ @ A as well as

aa⇤ ab⇤ ac⇤ zz⇤ zw⇤ f f = ba⇤ bb⇤ bc⇤ and g g = . (1.132) | ih | 0 1 | ih | wz⇤ ww⇤ ca⇤ cb⇤ cc⇤ ✓ ◆ @ A

1.12 Dirac notation Outer products are important in quantum mechanics, and so Dirac invented a notation for linear algebra that makes them easy to write. In his notation, a vector f is a ket f = f . The new thing in his notation is the bra g .The | i h | inner product of two vectors (g, f)isthebracket (g, f)= g f . A matrix h | i element (g, cf) of an operator c then is (g, cf)= g c f in which the bra h | | i and ket bracket the operator c. In Dirac notation, an outer product like (1.123) Ah =(g, h) f = f (g, h) reads A h = f g h , and the outer product A itself is A = f g . | i | ih | i | ih | The bra g is the adjoint of the ket g , and the ket f is the adjoint of h | | i | i the bra f h | g =(g )† and f =(f )†, so g †† = g and f †† = f . (1.133) h | | i | i h | h | h | | i | i The adjoint of an outer product is

( z f g )† = z⇤ g f . (1.134) | ih | | ih | 22 Linear Algebra In Dirac’s notation, the most general linear operator is an arbitrary linear combination of outer products

A = z k ` . (1.135) k` | ih | Xk` Its adjoint is

A† = z⇤ ` k . (1.136) k` | ih | Xk` The adjoint of a ket h = A f is | i | i

† ( h )† =(A f )† = zk` k ` f = zk⇤` f ` k = f A†. (1.137) | i | i | ih | i! h | ih | h | Xk` Xk` Before Dirac, bras were implicit in the definition of the inner product, but they did not appear explicitly; there was no simple way to write the bra g h | or the outer product f g . | ih | If the kets k form an orthonormal basis in an n-dimensional vector space, | i then we can expand an arbitrary ket in the space as

n f = c k . (1.138) | i k| i Xk=1 Since the basis vectors are orthonormal ` k = , we can identify the h | i `k coecients ck by forming the inner product n n ` f = c ` k = c = c . (1.139) h | i k h | i k `,k ` Xk=1 Xk=1 The original expasion (1.138) then must be

n n n n f = ck k = k f k = k k f = k k f . (1.140) | i | i h | i| i | ih | i | ih |! | i Xk=1 Xk=1 Xk=1 Xk=1 Since this equation must hold for every vector f in the space, it follows that | i the sum of outer products within the parentheses is the identity operator for the space n I = k k . (1.141) | ih | Xk=1 Every set of kets ↵ that forms an orthonormal basis ↵ ↵ = for the | ji h j| `i j` 1.12 Dirac notation 23 space gives us an equivalent representation of the identity operator n n I = ↵ ↵ = k k . (1.142) | jih j| | ih | Xj=1 Xk=1 These resolutions of the identity operator give every vector f in the space | i the expansions n n f = ↵ ↵ f = k k f . (1.143) | i | jih j| i | ih | i Xj=1 Xk=1 Example 1.22 (Linear operators represented as matrices). The equations (1.60–1.67) that relate linear operators to the matrices that represent them are much clearer in Dirac’s notation. If the kets B are n orthonormal basis | ki vectors, that is, if B B = , for a vector space S, then a linear operator h k| `i k` A acting on S maps the basis vector B into (1.60) | ii n n A B = B B A B = a B , (1.144) | ii | kih k| | ii ki | ki Xk=1 Xk=1 and the matrix that represents the linear operator A in the B basis is | ki a = B A B . If a unitary operator U maps these basis vectors into ki h k| | ii B = U B , then in this new basis the matrix that represents A as in | k0 i | ki (1.137) is

a0 = B0 A B0 = B U † AU B `i h `| | ii h `| | ii n n n n (1.145) = B U † B B A B B U B = u† a u h `| | jih j| | kih k| | ii `j jk ki Xj=1 Xk=1 Xj=1 Xk=1 or a0 = u† au in matrix notation. Example 1.23 (Inner-Product Rules). In Dirac’s notation, the rules (1.78— 1.81), of a positive-definite inner product are

f g = g f ⇤ h | i h | i f z1g1 + z2g2 = z1 f g1 + z2 f g2 h | i h | i h | i (1.146) z f + z f g = z⇤ f g + z⇤ f g h 1 1 2 2| i 1h 1| i 2h 2| i f f 0 and f f =0 f =0. h | i h | i () States in Dirac notation often are labeled or by their quantum numbers | i n, l, m , and one rarely sees plus signs or complex numbers or operators | i inside bras or kets. But one should. 24 Linear Algebra Example 1.24 (Gram Schmidt). In Dirac notation, the formula (1.121) for the kth orthogonal linear combination of the vectors V is | `i k 1 k 1 uk = Vk Ui Ui Vk = I Ui Ui Vk (1.147) | i | i | ih | i | ih |! | i Xi=1 Xi=1 and the formula (1.122) for the kth orthonormal linear combination of the vectors V is | `i uk Uk = | i . (1.148) | i u u h k| ki The vectors U are not unique; theyp vary with the order of the V . | ki | ki Vectors and linear operators are abstract. The numbers we compute with are inner products like g f and g A f . In terms of n orthonormal basis h | i h | | i vectors j with f = j f and g = g j , we can use the expansion (1.141) | i j h | i j⇤ h | i of the identity operator to write these inner products as n n g f = g I f = g j j f = g⇤f h | i h | | i h | ih | i j j Xj=1 Xj=1 n n (1.149) g A f = g IAI f = g j j A ` ` f = g⇤ A f h | | i h | | i h | ih | | ih | i j j` ` j,X`=1 j,X`=1 in which A = j A ` . We often gather the inner products f = ` f into j` h | | i ` h | i a column vector f with components f = ` f ` h | i 1 f f h | i 1 2 f f2 f = 0 h |. i 1 = 0 . 1 (1.150) . . B C B C B n f C B f C B C B n C @ h | i A @ A and the j A ` into a matrix A with matrix elements A = j A ` .Ifwe h | | i j` h | | i also line up the inner products g j = j g in a row vector that is the h | i h | i⇤ transpose of the complex conjugate of the column vector g

g† =(1 g ⇤, 2 g ⇤,..., n g ⇤)=(g⇤,g⇤,...,g⇤) (1.151) h | i h | i h | i 1 2 n then we can write inner products in matrix notation as g f = g f and as h | i † g A f = g Af. h | | i † One can compute the inner product g, f of two vectors f and g by doing h i the sum (1.149) of gj⇤fj over the index j only if one knows their components f and g which are their inner products f = j f and g = j g with j j j h | i j h | i the orthonormal states j of some basis. Thus an inner product implies | i 1.12 Dirac notation 25 the existence of an orthonormal basis and a representation of the identity operator n I = j j . (1.152) | ih | Xj=1 If we switch to a di↵erent basis, say from k ’s to ↵ ’s, then the com- | i | ki ponents of the column vectors change from f = k f to f = ↵ f , and k h | i k0 h k| i similarly those of the row vectors g† and of the matrix A change, but the bras, the kets, the linear operators, and the inner products g f and g A f h | i h | | i do not change because the identity operator is basis independent (1.142)

n n g f = g k k f = g ↵ ↵ f h | i h | ih | i h | kih k| i k=1 k=1 Xn X n (1.153) g A f = g k k A ` ` f = g ↵ ↵ A ↵ ↵ f . h | | i h | ih | | ih | i h | kih k| | `ih `| i k,X`=1 k,X`=1 Dirac’s outer products show how to change from one basis to another. The sum of outer products n U = ↵ k (1.154) | kih | Xk=1 maps the ket ` of the orthonormal basis we started with into ↵ | i | `i n n U ` = ↵ k ` = ↵ = ↵ . (1.155) | i | kih | i | ki k` | `i Xk=1 Xk=1

Example 1.25 (A Simple Change of Basis). If the ket ↵ of the new basis | ki is simply ↵ = k +1 with ↵ = k +1 1 then the operator that | ki | i | ki | i⌘|i maps the n kets k into the kets ↵ is | i | ki n n U = ↵ k = k +1 k . (1.156) | kih | | ih | Xk=1 Xk=1 The square U 2 of U also changes the basis; it sends k to k +2 .Theset ` | i | i of operators U for ` = 1, 2, . . . , n forms a group known as Zn. To compute the inner product (U, V ) of two vectors U and V , one needs the components Ui and Vi of these vectors in order to do the sum (1.89) of Ui⇤Vi over the index i. 26 Linear Algebra 1.13 Adjoints of operators In Dirac’s notation, the most general linear operator (1.158) on an n-dimensional vector space is a sum of outer products z k ` in which z is a complex num- | ih | ber and the kets k and ` are two of the n orthonormal kets that make up | i | i a basis for the space. The adjoint (1.134) of this basic linear operator is

(z k ` )† = z⇤ ` k . (1.157) | ih | | ih | Thus with z = k A ` , the most general linear operator on the space is h | | i n A = IAI = k k A ` ` (1.158) | ih | | ih | k,X`=1 and its adjoint A† is the operator IA†I n n A† = ` ` A† k k = ` k A ` ⇤ k . (1.159) | ih | | ih | | ih | | i h | k,X`=1 k,X`=1 It follows that ` A k = k A ` so that the matrix A† that represents A h | †| i h | | i⇤ k` † in this basis is T A† = ` A† k = k A ` ⇤ = A⇤ = A⇤ (1.160) `k h | | i h | | i `k k` in agreement with our definition (1.28) of the adjoint of a matrix as the T transpose of its complex conjugate, A† = A⇤ . We also have

g A†f = g A† f = f A g ⇤ = f Ag ⇤ = Ag f . (1.161) h | i h | | i h | | i h | i h | i Taking the adjoint of the adjoint is by (1.157)

† (z k ` )† =[z⇤ ` k ]† = z k ` (1.162) | ih | | ih | | ih | the same as doing nothingh at all.i This also follows from the matrix formula T T (1.160) because both (A⇤)⇤ = A and (A ) = A, and so

† T T A† = A⇤ ⇤ = A (1.163) the adjoint of the adjoint of⇣ a⌘ matrix is the original matrix. Before Dirac, the adjoint A† of a linear operator A was defined by

(g, A†f)=(Ag,f)=(f,Ag)⇤. (1.164)

This definition also implies that A†† = A since

(g, A††f)=(A†g, f)=(f,A†g)⇤ =(Af, g)⇤ =(g, Af). (1.165)

We also have (g, Af)=(g, A††f)=(A†g, f).