3. hilbert space and vector spaces real vector spaces a real vector space is a set v with elements v...
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3. Hilbert Space and Vector Spaces
Real Vector SpacesA real vector space is a set V with elements v with the following properties:•You can add them•You can multiply them by real numbers:
– We define multiplication in either order the same: •The multiplication has associate and distributive properties:
1 2 1 2, V V v v v v
•We also define an inner product (dot product):•This dot product is linear in both arguments:
3A. Hilbert Space
,r V r V v vr rv v
1 2 1 2 1 2 1 2 1 2 1 2, ,r r r r r r r r r r r v v v v v v v v v
, ,V v w v w
1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2, , , , , , ,r r r r r r r r v v w v w v w v w w v w v w
•It is also positive definite: 2 , 0, zero only if 0 v v v v
•Consider wave functions at a specific time•Define the inner product of two wave functions:
Inner Product for Wave Functions
r
* 3, d r r r
•This inner product is linear in its second argument and anti-linear in the first:
•It is also positive definite:
* *1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2c , , , , , , ,c c c c c c c
2, 0, zero only if 0 . r
•Some other properties can be proven using just these two properties:
•Proof complicated in general:
•Schwartz inequality (homework):
*, ,
, , , ,
Hilbert Space , 1 •We are especially interested in normalized wave functions
Hilbert Space, denoted, is the set of wave functions satisfying ,
Hilbert space is a complex vector space with a positive definite inner product•You can add wave functions: •You can multiply them by complex numbers:•Let’s prove closure under addition:
1 2 1 2, H H
1 2 1 2 1 1 2 2 1 2 2 1, , , , ,
1 2 1 2 1 1 2 2 1 2 2 1, , , , ,
1 2 1 2 1 2 1 2 1 1 2 2, , 2 , 2 ,
1 2 1 2 1 1 2 2 1 2 1 2, 2 , 2 , ,
1 2 1 2 1 1 2 2, 2 , 2 ,
,c c H H
positive
Extended and Restricted Hilbert spaces 2 3, d r r•Hilbert space has only the restriction
•Sometimes, we wish to consider wave functions that do not fit this restriction, like eikr
•In this case, we are looking at a superspace of Hilbert space
•Sometimes, we want to consider more restrictive constraints– |(r)| < (r) continuous ’(r) exists, ’’(r) exists …
•In this case, we are looking at a subspace of Hilbert space
•These mathematical niceties won’t concerns us much
3B. Dirac Notation and Covectors
•A wave function and its Fourier transform contain equivalent information
•There are other ways to denote a wave function as well:
•We might want to consider cases not denoted just by a wave function– Spin– Multiple particles– Indefinite number of particles
•To keep our notation as general as possible, we will denote this information as an abstract vector, and clarify it by drawing a ket symbol around it:
r k
Vectors/Kets
0
n nn
c
r r
•A covector, such as f is a linear mappingfrom vectors to complex numbers•Linear means:
•Dirac drew a bra around a covector to signal that it’s a covector•When you put a bra with a ket, you get a bracket
– The extra bar is deleted for brevity
f
Covectors / Bras
f
1 1 2 2 1 1 2 2f c c c f c f
f f
Some examples of covectors/bras:•The position bra r|
•The momentum bra k|
•For any | , define | by
r r
3
3/22
ide
k rr
k k r
* 3, d r r r
• For any two covectors f1| and f2| and any two complex numbers c1 and c2, define the covector by
• Covectors form a vector space * as well
1 1 2 2 1 1 2 2c f c f c f c f
Covectors as a Vector Space
• In fact, there is a one-to one-correspondence between vectors and covectors
• Mathematically, this means the two vector spaces are the same, = *
• The process of turning a bra into its corresponding ket and vice-versa is called Hermitian Conjugation and is denoted by a dagger
1 1 2 2c f c f
† †
Sample ProblemWhat are the corresponding kets (wave
functions) for the position and momentum bras?
r r
3
3/22
ide
k rr
k k r
* 3, d r r r
3
* 33/2
2id
d e
k rk k
rr r r r
3/2
1
2ie
k r
k r
* 3d r r r r r r 3 r r r r
3 3
3 2 3 2, , ,2 2
i ie e
k r k r
r k k r r r r r k k k k
• Consider the Hermitian Conjugate of a linear combination:• This is defined by how it acts on an arbitrary ket:
Hermitian Conjugation and Complex Numbers
• Sums remain sums under Hermitian conjugation• But Hermitian conjugation takes the complex conjugate of complex
numbers
†
1 1 2 2c c
†
1 1 2 2 1 1 2 2 ,c c c c
* *1 1 2 2, ,c c
* *1 1 2 2c c
† * *1 1 2 2 1 1 2 2c c c c
† *c c
3C. Coordinate Bases
•In 3D space, it is common to write vectors in terms of components:
•In a complex vector space, we would similarly like to write
•It is complete if every vector can be written this way•It is independent if there is no non-trivial way to write the zero ket this way
•It is orthogonal if the inner product between any distinct pair vanishes
•It is orthonormal if:
ˆ ˆ ˆx y z r x y z
What’s a Coordinate Basis?
1 1 2 2 3 3c c c
1 1 2 2 3 3 1 2 30 0c c c c c c
0 ifi j i j
i j ij
• Any complete basis can be made into a complete, independent, orthonormal basis
• To make it independent, throw out redundant basis vectors:
Making an Orthonormal Basis
• To make it orthogonal, iteratively produce new vectors orthogonal to each other
1 1 2 20 n nc c c 1 1 2 2n nc c c
1 2, , , ,n
1 2, , , ,n
1 1 ,
1 22 2 1
1 1
1 3 2 33 3 1 2
1 1 2 2
1
1
ni n
n n ii i i
• To make it orthonormal, normalize them
1n n
n n
• Our basis is now:
1 2, , , ,n
• In general, given | and a basis, we wouldlike to find the complex numbers ci such that
• Assuming the basis is complete, this can always be done, but actually doing it may be difficult
• For an orthonormal basis, this is easy. Act on | with n|:
Pulling out the Coefficients
i ii
c
n n i ii
c i n ii
c i ini
c nc
n nc
Dimensionality of a Vector Space• The dimensionality of a vector space is the number N
of basis vectors in a complete, independent basis• Usually infinity for us
1 2, , , N
• Sometimes, it is more useful to work with bases that are labeled by real numbers rather than discrete integers
• Then a general ket can be written as
• An orthonormal basis would have the property:
• The coefficient functions are easily determined:
• Some examples in 3D of complete continuous bases:
• It is also common to simultaneously use mixed continuous and discrete bases
Continuous Bases
:
c d
c
orr k
, with , , iji i j
3D. Operators
•An operator is anything that maps vectors to vectors:– Usually, omit the parentheses
•I will (usually) denote operators by capital letters A•We will focus almost exclusively on linear operators:
– When I say “operator”, linear is implied
•Operators themselves form a vector space:– Technically, this space is *
V A V
Operators and Linear Operators
1 1 2 2 1 1 2 2A c c c A c A
1 1 2 2 1 1 2 2c A c A c A c A
A A
• The identity operator 1:– I will rarely bold the 1
• The position operators R = (X,Y,Z), which multiply by the coordinate:
• The momentum operators: P = (Px,Py,Pz) which take a derivative:
• The Hamiltonian operator:
• The Parity operator , which reflects the wave function through the origin:
• For any ket |1 and bra 2|, define |12| by
Examples of Operators
, , , ,X x y z x x y z
1
i P , , , ,xP x y z i x y zx
21
2H V
m P R
r r
1 2 1 2
• Let {|i} be a complete orthonormal basis• Consider the following expression:
• To figure out what this is, let it act on an arbitrary ket |
• We already know:
• Therefore:
• Since this is true for all |,
The completeness relation
i ii
i i i ii i
wherei i i ii
c c
i i i ii i
c
i ii
1
• We normally think of operators acting on vectors to produce vectors:
• Operators can also act on covectors to produce covectors– Logically this could be written as A(|), but instead we write this as
• This is defined by how we it acts on an arbitrary vector:
• Pretty much ignore parentheses
• We can just as easily define an operator by how it acts on covectors as on vectors
Operators acting on covectors (bras)
A
A
A A
• The product of two operators A and B is defined by
• From this property, you can prove the associative property for operators:
• Bottom line: Ignore the parentheses:
• You can also show the distributive law:
• Operator multiplication is usually not commutative
Multiplying Operators
AB A B
ABC
A B C A B C AB AC AB AC
A B C AC BC AC BC
AB C AB C A B C A BC A BC
AB BA
AB C A BC
3E. Commutators
•Usually, two operators do not commute:•Define the commutator of two operators A and B by
– Warning: avoid using [] as grouping when doing commutatorsLet’s work out the commutator of X and Px
•To do so, let X and Px act on an arbitrary wave function :
Definition of the Commutator:AB BA
,A B AB BA
xXP i Xx
i xx
xP X xP x i xx
i x
x
i i xx
, x x xX P XP P X i x i x ix x
i
, xX P i
Some Simple Commutators: , xX P i
•Generalize:•Some other simple ones:
•Some easy to prove identities to help you get more commutators:
,i j ijR P i , 0 ,i j i jR R P P
, ,A B B A
, , , and , , ,CA B C A C B C A B C A B A
, , , and , , ,AB C A B C A C B A BC B A C A B C
Sample ProblemThe angular momentum operators are defined as L = R P. Find the commutator of Lz with all components of R.
z y xL XP YP
,i j ijR P i , 0 ,i j i jR R P P , , ,AB C A B C A C B
, ,z y xL X XP YP X , ,y xXP X YP X
, , , ,y y x xX P X X X P Y P X Y X P 0 0 0Y i i Y
, ,z y xL Y XP YP Y , ,y xXP Y YP Y , 0yX P Y i X
, ,z y xL Z XP YP Z 0
3F. Hermitian Adjoints of Operators
•Let A be an arbitrary operator. Define the Hermitian adjoint A† by
where | = |†
•This is a linear operator:
Definition of the Adjoint of an Operator:
††A A
†††
1 1 2 2 1 1 2 2c c A A c c †* *1 1 2 2A c c
†* *1 1 2 2c A c A † †
1 1 2 2c A c A † †
1 1 2 2c A c A
•A very useful formula:
•From this, easy to show:
††A A ,A *, A
*A
*†A A
††A A
Conjugates of products of operators•What is (AB)†?
††AB AB †
A B † †B A † †B A
† † †AB B A
General Rules for Finding Adjoints of Expressions:•For sums or differences, treat each term separately•For things that are multiplied (bras, kets, operators), reverse the order•Replace A by A† for each operator•Replace bras kets, | |•Replace complex numbers c by their complex conjugates
Sample ProblemFind the Hermitian adjoint of the equation:
ji t
j jj
t e U
• Each term in the sum is treated separately• Reverse the order of everything multiplied• Adjoint everything
† ji t
j jj
t U e • Did we do it right?• Answer: both are correct• Warning: Don’t reverse the order of labels inside a ket
*
†
†
or
?
j
j
j
U
U
U
†Eric Carlson Eric Carlson Carlson Eric
†, , , , , ,s sn l m m n l m m
Hermitian and Unitary Operators
•A Hermitian operator A is one such that•A Unitary operator U is one such that
– Proving one of these is sufficient•Generally, easiest to prove by inserting an arbitrary ket and bra:
•For Hermitian, equivalent to
•Sufficient (and sometimes easier) to prove it with basis vectors:
†A A† †UU U U 1
† †
† †
A A A A
U U U U
1
*†
† †
i j j i
i j i j ij
A A A A
U U U U
1
*†A A A A
Examples of Hermitian Operators:
•The three position operators:
•The three momentum operators:
•The Hamiltonian
*†A A A A ? *
X X
? *
* 3 * 3x d x d r r r r r r * 3x d r r r? *
x xP P
*
?* 3 * 3i d i d
x x r r r r r r * 3i d
x
r r r
?
* * 30 i dx x
r r r r r * 3i dx
r r r
* x
xi dy dz
r r
21
2H V
m P R † † † †1
2H V
m P P R H
Examples of Unitary Operators
•Define the translation operators by howthey translate the position basis vectors:•Taking the Hermitian conjugate•We therefore have:
•Define the rotation operators by howthey rotate the position basis vectors:
– Where is a 33 matrix satisfying •We therefore have:
T a r r a
† †i j i j ijU U U U 1
†T r a r a
†T T r a a r r a r a 3 r a r a 3 r r r r
R r rR R
1T R R det 1 R
†R R r r r rR R R R 3 r rR
31
det r r
R 3 r r r r
3 r rR R
Sample ProblemIf | is normalized and U is unitary, show that U | is
normalized.2
1 †
2U †
U U †U U 1 1
If H is Hermition, show that |H | is real
*H †
H †H H
†H H
†U U 1
•U | is normalized
|H | is real
Sample Problem
3G. Matrix Notation
•Suppose we have an orthonormal basis {|1, |2, |3, …}•It is common to write vectors |v by simply listing their components
•Put the components vi together into a column matrix•A covector w| can also be written as a list of numbers
•Put the components together into a row matrix:•The inner product is then simply matrix multiplication
Matrices for Vectors and Covectors
,i i i ii
v v v 1
2
v
v v
i ii
w w ,i i i ii
w w w 1 2w w w
w v i j i ji j
w v i j ij i ii j i
w v w v
1 1 2 2w v w v w v
i i j ji j
w v
For an arbitrary operator A:
•Write A then as a square matrix:
•Any manipulation of operators, bras, and kets can be done with matrices
Matrices for Operators
i i j ji j
A A
AB
i i j ji j
A i ij ji j
ij i j
A
A A
11 12 1 1 1 2
21 22 2 1 2 2
A A A A
A A A A A
i ij j k kl li j k l
A B i ij j k kl li j k l
A B
i ij jl li j l
A B i lili l
AB ij jlilj
AB A B
•Often easier to understand if we think of bras, kets, operators as concrete matrices, rather than abstract objects•Even though our matrices are usually infinite dimensional, sometimes we can work with just a finite subspace, making them finite•In such cases, computers are great at manipulating matrices
Why Matrix Notation?
Sample ProblemFor two spin- ½ particles, the spin states
have four orthonormal basis states:
The operator S2, acting on these states, yields
Write S2 as a matrix in this basis.
, , ,
2 2 2 2
2 2 2
2 , 2 ,
S S
S S
1
2
3
4
•First, pick an order for your basis states:
2 2 2 2
2 2 2 22
2 2 2 2
2 2 2 2
S S S S
S S S SS
S S S S
S S S S
2
2 2
2 2
2
2 0 0 0
0 0
0 0
0 0 0 2
For a vector and its corresponding covector,
•So, if thought of as a matrix, | becomes | bychanging the column into a row and complex conjugating
•For an operator:
•For complex numbers, Hermitian conjugation means complex conjugation
•General rule: Turn rows columns and complex conjugate
Hermitian Conjugates in Matrix Notation
* *11 12 11 21
† * *21 22 12 22,
A A A A
A A A A A A
1 1
2 2
v v
v v v
1 2v v v * *
1 2v v
† †i jij
A A *
j iA *jiA
† *TA A
Announcements
9/15
ASSIGNMENTSDay Read HomeworkToday 3H, 3I 3.3Wednesday 4A, 4B, 4C 3.4, 3.5Friday 4D, 4E 3.6
All homework returned
On the web:•Chapter 3 Slides•Chapter 2 Homework Solutions
All the components of a vector, covector, or operator, depend on basis {|i}
What if we decide to change to a new orthonormal basis {|’i}?•Define the conversion matrix:
•This is a unitary matrix:
•Note that each column of V is components of |’i in the unprimed basis
Changing Bases (1)
, ,ij i j i i i iA A v v w w
1 1 1 2
2 1 2 2V
ij i jV
†
ijV V
†ik kj
k
V V *
k i k jk
*ki kj
k
V V i k k jk
Completeness
i j 1ij
† 1V V
We can now use V to convert a vector, covector, or operator
•Vectors:
•Covectors:
•Operators:
•Note: The primes don’t mean new objects, but the same old objects in a new basis
Changing Bases (2)
, ,ij i j i i i iA A v v w w ij i jV
i iv v i j jj
v *ji j
j
V v †v V v
i iw w j j i
j
w j jij
w V
†ij j
j
V v
w w V
ij i jA A i k k l l jk l
A *ki kl lj
k l
V A V †ik kl lj
k l
V A V†A V AV
3H. Eigenvectors and Eigenvalues
•An eigenvector of an operator A is a non-zero vector |v such that
where a is a complex number called an eigenvalue•If |v is an eigenvector, so is c|v
– We can normalize eigenvectors when we want to
•Eigenvectors of Hermitian operators are real:
•Eigenvectors of Unitary operators are complex numbers of magnitude one
Definition, and some properties
A v a v
v A v v a v a v v
real real
U v u v† *v U v u
† *v U U v v u u v
*v v u u v v1
* 1,i
u u
u e
• Hermitian operators can act equally well to the left or to the right on eigenvectors
• Let |v1 and |v2 be eigenvectors of A with distinct eigenvalues• Then |v1 and |v2 will be orthogonal to each other
The same property applies to Unitary operators U• Let |v1 and |v2 be eigenvectors of U with distinct eigenvalues• Recall: • We see that• Consider the expression:
Orthogonality of Eigenvectors
1 1 1
2 2 2
A v a v
A v a v
A v a v v A v a
2 1v A v 1 2 1a v v2 2 1a v v 2 1 2 1 0a a v v 2 1 0v v
1 1 1
2 2 2
U v u v
U v u v
*2 2 1u u
† *2 2 2v U v u
†2 1v U U v *
2 1 2 1u u v v2 1v v
*2 1 2 10 1v v u u * *
2 1 2 2 2 1v v u u u u *2 2 1 2 1u v v u u 2 1 0v v
Observables and Basis SetsSuppose that the eigenvectors of a Hermitian operator A form a complete set (mathematically, they span the space), so that we can always write:
•This can always be done in finite dimensional space, often in infinite•A Hermitian operator with this property is called an observable
Suppose we start with a complete set of eigenstates {|n}•Throw out any redundant ones•Construct an orthogonal basisset by the usual procedure:•Note that states with different eigenvalues don’t mix, so still eigenstates•Now make them orthonormalby the usual procedure:•Still eigenstates!•We end up with an orthonormal basis set which are eigenstates of A
n nn
c n n nA a
1
1
ni n
n n ii i i
1n n
n n
Labelling Basis States• When we choose a basis set that are eigenvalues of observable A, often
label them by their eigenvalues
• If we work in this bases, the matrix for A will be diagonal– Switching to this basis is called diagonalizing A
• Let B be another observablethat commutes with A, so
• If you act on any eigenstate of A with B, it is stillan eigenstate of A:
• This implies that B will be block diagonal• When you find eigenstates of B, they will exist just
within these subspaces with the same A eigenvalues• You can label them by both their eigenvalues under A and B
, , , ,n a n A a n a a n
,A B a n
1
1
2
2
0
0
a
aA
a
a
AB BA
,BA a n ,a B a n 11 12
21 22
33 34
43 44
0
0
b b
b bB
b b
b b
, , , , , , , , , , , ,a b n A a b n a a b n B a b n b a b n
Complete Sets of Commuting Observables• We can continue this procedure, until eventually our states are uniquely
labeled by their eigenvalues• The observables you are using are called a complete set of commuting
observables– They must all commute with each other
• Eigenstates can be labelled just by their eigenvalues
Why is this a good idea?• Oddly, making the requirements more stringent
can make them easier to find– Think of an animal whose name
ends in -ant • Often, the eigenvalues correspond to physically measurable quantities• In particular, the Hamiltonian is often chosen as one of the observables
, , ,A B Z
, , ,
, , , , , ,
, , , , , ,
, , , , , ,
a b z
A a b z a a b z
B a b z b a b z
Z a b z z a b z
that has a trunk
3H. Finding Eigenvalues and Eigenvectors
• Suppose an operator is given in matrix form• How can we find its eigenvectors and eigenvectors? • We want to solve: • Rewrite this as:
• This is N equations in N unknowns• Generally has a unique solution: |v = 0
– Unless det(A – 1) = 0 ! To find eigenvalues, solve det(A – 1) = 0 • This is an N’th order polynomial• It has exactly N complex roots, though
not necessarily distinct• For Hermitian matrices, these roots will be real
How to find Eigenvalues11 12
21 22
A A
A A A
A v v
0A v 1
11 12
21 22det 0
A A
A A
det 0A 1
To find the eigenvalues and orthonormal eigenvectors of operator A:•First find roots of det(A – 1) = 0
– This produces N eigenvalues {n}•Then, for each eigenvalue n:
– Solve the equation A|vn = n|vn– If you have any degenerate eigenvalues n, you may have to be
careful to keep them orthogonal– You then need to normalize them: vn|vn = 1
To change basis to the new basis vectors•Make the basis transformationmatrix V from the eigenvectors |vn•You can then use this to transformvectors, covectors or operators to the new basis •In particular, the operator A will be diagonaland will have the eigenvalues on the diagonal
– You don’t need to calculate it
Procedure for Eigenvalues and Eigenvectors
1 2V v v
†
†
v V v
w w V
B V BV
1
2
0
0
A
• If a matrix is already diagonal, theeigenvalues and eigenvectors are trivial
– The eigenvalues are the values on the diagonal– The eigenvectors are the unit vectors in this basis
• If a matrix is block diagonal, you can divide it into two(or more) smaller and easier problems
• If there is a common factor in an operator, take it out– Matrix without c has identical
normalized eigenvectors– Eigenvalues are identical except
multiplied by c
Shortcuts for Eigenvalues and Eigenvectors0
0A
1 0,
0 1
11
22 23
32 33
0 0
0
0
B
B B B
B B
22 231 11 2
32 33
,B B
B B BB B
11 12
21 22
cC cCC
cC cC
11 12
21 22
C Cc
C C
Sample Problem (1)For the S2 matrix found before,(1) Find the eigenvalues and orthonormal eigenvectors(2) Find the matrix V that relates the new basis to the old, and demonstrate explicitly that V†S2V is now diagonal(3) Write the state |2 = |+– in terms of the eigenvector basis.
2
2 22
2 2
2
2 0 0 0
0 0
0 0
0 0 0 2
S
2 2
2 0 0 0
0 1 1 0
0 1 1 0
0 0 0 2
S
• Take the common factor of 2 out.• Note that it is block diagonal• Two of the eigenvectors and
eigenvalues are now trivial• All that’s left is the middle
2 21 1 4 4
1 0
0 02 , , 2 ,
0 0
0 1
v v
Sample Problem (2)(1) Find the eigenvalues and orthonormal eigenvectors . . . 2
mid
1 1
1 1
S• Solve the equation det(S2 - 1) = 0
1 10
1 1
a a
b b
2 2
2 0 0 0
0 1 1 0
0 1 1 0
0 0 0 2
S
1 10 det
1 1
21 1 2 2 0 or 2
• For each eigenvalue, solve S2|v = |v
= 0:
= 2:
0
0
a b
a b
2
av
a
12
2 12
v
1 12
1 1
a a
b b
2
2
a b a
a b b
3
av
a
12
3 12
v
• The eigenvalues of S2 will be 2 times these, or 0 and 22
Sample Problem (3)
2 2
2 0 0 0
0 1 1 0
0 1 1 0
0 0 0 2
S
1 2 3 4, , ,v v v v
(1) Find the eigenvalues and orthonormal eigenvectors(2) Find the matrix V that relates the new basis to the old, . . .
1
0
0
0
0
0
0
1
12
12
0
0
12
12
0
0
21
2
23
24
2
0
2
2
1 2V v v
V
Sample Problem (4)
2 2
2 0 0 0
0 1 1 0
0 1 1 0
0 0 0 2
S
21
2
23
24
2
0
2
2
2 † 2V V S S
. . .and demonstrate explicitly that V†S2V is now diagonal
1 1 1 12 2 2 22
1 1 1 12 2 2 2
1 0 0 0 1 0 0 02 0 0 00 0 0 00 1 1 0
0 0 0 1 1 0 0 0
0 0 0 20 0 0 1 0 0 0 1
†A V AV
2
2 0 0 0
0 0 0 0
0 0 2 0
0 0 0 2
1 12 2
1 12 2
1 0 0 0
0 0
0 0
0 0 0 1
V
Sample Problem (5)
2 2
2 0 0 0
0 1 1 0
0 1 1 0
0 0 0 2
S
21
2
23
24
2
0
2
2
1 12 2
1 12 2
1 0 0 0
0 0
0 0
0 0 0 1
V
†v V v
†2 2V
1 12 2
1 12 2
1 0 0 0 00 0 1
0 0 0
00 0 0 1
12
12
0
0
1 12 32 2
v v
(3) Write the state |2 = |+– in terms of the eigenvector basis.