There’s one more useful characteristic of the eigenfuctions of aHermitian operator: they form a complete set. That means that any function in the abstract vector space containing the eigenfunctions of a Hermitian operator may be made up of a linear combination of those eigenfunctions.
Main Ideas of This Section
Hermitian operators may be applied to either member of an inner product and the result will be the same. Hermitian operators have real eigenvalues, and the nondegenerate eigenfunctions of a Hermitian operator are orthogo- nal and form a complete set.
Relevance to Quantum Mechanics
The discussion of the solutions to the Schr¨odinger equation inChapter 4 will show that every quantum observable (such as position, momentum, and energy) is associated with an operator, and the possible results of any measurement are given by the eigenvalues of that operator. Since the results of measurements must be real, operators associated with observables must be Hermitian. The eigenfunctions of Hermitian operators are (or can be combined to be) orthogonal, and the orthogonality of those eigenfunc- tions has a profound impact on our ability to construct solutions to the Schr¨odinger equation and to use those solutions to determine the probability of various measurement outcomes.
And since the inner products1|A,2|A, and3|Aare scalars (as they must be, since they represent A1, A2, and A3), you can move them to the other side of the basis kets|1,|2, and |3, and the expansion of the ket representingAbecomes
|A = |1 1|A + |2 2|A + |3 3|A. (2.39) This equation came about with the terms grouped as
|A = |1 ( )* +1|A
A1
+ |2 ( )* +2|A
A2
+ |3 ( )* +3|A
A3
,
but consider the alternative grouping
|A = |( )* +1 1|
P1
|A + |( )* +2 2|
P2
|A + |( )* +3 3|
P3
|A. (2.40)
As you can see from the labels underneath the curly braces, the terms|1 1|,
|2 2|, and|3 3|are the operatorsP1,P2, andP3. The general expression for a projection operator is
Pi= |i i|, (2.41)
in whichˆi is any normalized vector. This expression, with a ket standing to the left of a bra, may look a bit strange at first, but most operators look strange until you feed them something on which to operate. Feeding operatorP1the ket representing vectorAhelps you see what’s happening:
P1|A = |1 1|A =A1|1 (2.42) So applying the projection operator to|Aproduces the new ketA1|1. The magnitude of that new ket is the (scalar) projection of the ket that you feed into the operator (in this case,|A) onto the direction of the ket you use to define the operator (in this case,|1). But here’s an important step: that magnitude is then multiplied by the ket you use to define the operator. So the result of applying the projection operator to a ket is not just the (scalar) component (such asA1) of that ket along the direction ofˆ1, it’s a new ket in that direction. Put in terms of a vector in the Cartesian coordinate system, theP1projection operator doesn’t just give you the scalarAx, it gives you the vectorAxˆı.
In defining a projection operator, it’s necessary to use a ket representing a normalized vector (such asˆ1) within the operator; you can think of that vector as the “projector vector.” If the projector vector doesn’t have unit length, then its length contributes to the result of the inner product as well as the result
of the multiplication by the projector vector. To remove those contributions requires dividing by the square of the norm of the (non-normalized) projector vector.7
For completeness, the results of applying the three projection operatorsP1, P2, andP3to ket|Aare
P1|A = |1 1|A =A1|1 P2|A = |2 2|A =A2|2 P3|A = |3 3|A =A3|3.
(2.43)
If you sum the results of applying the projection operators for all of the basis kets in a three-dimensional space, the result is
P1|A +P2|A +P3|A =A1|1 +A2|2 +A3|3 = |A or
P1+P2+P3
|A = |A. Writing this for the general case in an N-dimensional space:
N n=1
Pn|A = |A. (2.44)
This means that the sum of the projection operators using all of the basis vectors equals the “identity operator”I. The identity operator is the Hermitian operator that produces a ket that is equal to the ket that is fed into the operator:
I|A = |A. (2.45)
This works for any ket, not only|A, just as multiplying any number by the number “1” produces the same number. The matrix representation (¯¯I) of the identity operator in three dimensions is
I¯¯=
⎛
⎝1 0 0
0 1 0
0 0 1
⎞
⎠. (2.46)
The relation
N n=1
Pn= N n=1
|n n| =I (2.47)
7That’s why you’ll see projection operators defined asPi=||ii|
i|2 in some texts.
is called the “completeness” or“closure” relation, since it holds true when applied to any ket in an N-dimensional space. That means that any ket in that space can be represented as the sum of N basis kets weighted by N components.
In other words, the basis vectorsnrepresented by the kets|n, and their dual brasn|inEq. 2.47form a complete set.
Like all operators, the projection operator in an N-dimensional space may be represented by an N×N matrix. You can find the elements of that matrix usingEq. 2.16:
Aij= i|Aj
. (2.16)
As explained inSection 2.2, before you can find the elements of the matrix representation of an operator, it’s necessary to decide which basis system you’d like to use (just as you need to decide on a basis system before finding the components of a vector).
One option is to use the basis system consisting of the eigenkets of the operator. As you may recall, in that basis the matrix representing an operator is diagonal, and each of the diagonal elements is an eigenvalue of the matrix.
Finding the eigenkets and eigenvalues of the projection operator is straight- forward. For projection operatorP1, for example, the eigenket equation is
P1|A =λ1|A, (2.48)
in which ket|Ais an eigenket ofP1with eigenvalueλ1. Inserting|1 1|for P1gives
|1 1|A =λ1|A.
To see if the basis ket|1is itself an eigenket ofP1, let|A = |1:
|1 1|1 =λ1|1.
But|1,|2, and|3form an orthonormal set, so1|1 =1, which means
|1(1)=λ1|1 1=λ1.
Hence|1 is indeed an eigenket ofP1, and the eigenvalue for this eigenket is one.
Having succeeded with|1as an eigenket ofP1, let’s try|2:
|1 1|2 =λ2|2.
But1|2 =0, so
|1(0)=λ2|1 0=λ2,
which means that|2is also an eigenket ofP1; in this case the eigenvalue is zero. Similar analysis applied to|3reveals that|3is also an eigenket ofP1; its eigenvalue is also zero.
So the eigenkets of operator P1are |1,|2, and|3, with eigenvalues of 1, 0, and 0, respectively. With these eigenkets in hand, the matrix elements (P1)ijcan be found by insertingP1intoEq. 2.16:
(P1)ij= i|P1j
. (2.49)
Settingi=1 andj=1 and usingP1= |1 1|gives(P1)11: (P1)11= 1|P1|1 = 1|1 1|1 =(1)(1)=1.
Likewise
(P1)12= 1|P1|2 = 1|1 1|2 =(1)(0)=0 (P1)21= 2|P1|1 = 2|1 1|1 =(0)(1)=0 (P1)13= 1|P1|3 = 1|1 1|3 =(1)(0)=0 (P1)31= 3|P1|1 = 3|1 1|1 =(0)(1)=0 (P1)23= 2|P1|3 = 2|1 1|3 =(0)(0)=0 (P1)32= 3|P1|2 = 3|1 1|2 =(0)(0)=0.
Thus the matrix representing operatorP1in the basis of its eigenkets|1,|2, and|3is
P¯¯1=
⎛
⎝1 0 0
0 0 0
0 0 0
⎞
⎠. (2.50)
As expected, in this basis the P1 matrix is diagonal with diagonal elements equal to the eigenvalues 1, 0, and 0.
A similar analysis for the projection operatorP2 = |2 2|shows that it has the same eigenkets (|1,|2, and|3) asP1, with eigenvalues of 0, 1, and 0. Its matrix representation is therefore
¯¯
P2=
⎛
⎝0 0 0
0 1 0
0 0 0
⎞
⎠. (2.51)
And projection operatorP3= |3 3|has the same eigenkets with eigenval- ues of 0, 0, and 1; its matrix representation is
P¯¯3=
⎛
⎝0 0 0
0 0 0
0 0 1
⎞
⎠. (2.52)
According to the completeness relation (Eq. 2.47), the matrix representations of the projection operatorsP1,P2, andP3should add up to the matrix of the identity operator, and they do:
⎛
⎝1 0 0
0 0 0
0 0 0
⎞
⎠+
⎛
⎝0 0 0
0 1 0
0 0 0
⎞
⎠+
⎛
⎝0 0 0
0 0 0
0 0 1
⎞
⎠=
⎛
⎝1 0 0
0 1 0
0 0 1
⎞
⎠= ¯¯I. (2.53)
An alternative method of finding the matrix elements of the projection operatorP1 is to use the outer product rule for matrix multiplication. That rule says that the outer product of a column vectorAand a row vectorBis
⎛
⎝A1
A2
A3
⎞
⎠ B1 B2 B3
=
⎛
⎝A1B1 A1B2 A1B3
A2B1 A2B2 A2B3
A3B1 A3B2 A3B3
⎞
⎠. (2.54)
Recall from Section 1.2 that you can expand basis vectors in their own
“standard” basis system, in which case each vector will have a single nonzero component (and that component will equal one if the basis is orthonormal).
So expanding the kets|1,|2, and|3in their own basis makes them and their corresponding bras look like this:
|1 =1|1 +0|2 +0|3 =
⎛
⎝1 0 0
⎞
⎠ 1| =
1 0 0
|2 =0|1 +1|2 +0|3 =
⎛
⎝0 1 0
⎞
⎠ 2| =
0 1 0
|3 =0|1 +0|2 +1|3 =
⎛
⎝0 0 1
⎞
⎠ 3| =
0 0 1 .
With the outer product definition ofEq. 2.54and these expressions for the basis kets and bras, the elements of the projection operatorsP1,P2, andP3can be found:
P1= |1 1| =
⎛
⎝1 0 0
⎞
⎠ 1 0 0
=
⎛
⎝(1)(1) (1)(0) (1)(0) (0)(1) (0)(0) (0)(0) (0)(1) (0)(0) (0)(0)
⎞
⎠=
⎛
⎝1 0 0
0 0 0
0 0 0
⎞
⎠
P2= |2 2| =
⎛
⎝0 1 0
⎞
⎠ 0 1 0
=
⎛
⎝(0)(0) (0)(1) (0)(0) (1)(0) (1)(1) (1)(0) (0)(0) (0)(1) (0)(0)
⎞
⎠=
⎛
⎝0 0 0
0 1 0
0 0 0
⎞
⎠
P3= |3 3| =
⎛
⎝0 0 1
⎞
⎠ 0 0 1
=
⎛
⎝(0)(0) (0)(0) (0)(1) (0)(0) (0)(0) (0)(1) (1)(0) (1)(0) (1)(1)
⎞
⎠=
⎛
⎝0 0 0
0 0 0
0 0 1
⎞
⎠.
You can see how to use the matrix outer product to find the elements of projector operators in other basis systems in the chapter-end problems and online solutions.
Main Ideas of This Section
The projection operator is a Hermitian operator that projects one vector onto the direction of another and forms a new vector in that direction;
operating on a vector with the projection operators for all of the basis vectors of that space reproduces the original vector. That means that the sum of the projection operators for all the basis vectors equals the identity operator; this is a form of the completeness relation. The matrix elements of a projector operator may be found by sandwiching the operator between the bras and kets of pairs of the basis vectors or by using the outer product of the ket and bra of each basis vector.
Relevance to Quantum Mechanics
As described inChapter 4, the projection operator is useful in determining the probability of measurement outcomes for a quantum observable by projecting the state of a system onto the eigenstates of the operator for that observable.