# Quantum One: Lecture 14 1. 2 Ket-Bra Operators, Projection Operators, and Completeness Relations 3.

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Quantum One: Lecture 14 1

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Ket-Bra Operators, Projection Operators, and Completeness Relations 3

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Ket-Bra Operators In this lecture we introduce a class of operators whose action is, in a certain sense, self-evident. We refer to operators of this sort as ket-bra operators, because they are formed from an outer product of a ket and a bra, (opposite to the bra-ket associated with the inner product). Thus, if |φ 〉 and |χ 〉 are any two states in S then we can define an operator A = |φ 〉〈 χ| the action of which on any state gives the ket |φ 〉, multiplied by the complex number 〈 χ|ψ 〉. 5

Ket-Bra Operators In this lecture we introduce a class of operators whose action is, in a certain sense, self-evident. We refer to operators of this sort as ket-bra operators, because they are formed from an outer product of a ket and a bra, (opposite to the bra-ket associated with the inner product). Thus, if |φ 〉 and |χ 〉 are any two states in S then we can define an operator A = |φ 〉〈 χ| the action of which on any state gives the ket |φ 〉, multiplied by the complex number 〈 χ|ψ 〉. 6

Ket-Bra Operators In this lecture we introduce a class of operators whose action is, in a certain sense, self-evident. We refer to operators of this sort as ket-bra operators, because they are formed from an outer product of a ket and a bra, (opposite to the bra-ket associated with the inner product). Thus, if |φ 〉 and |χ 〉 are any two states in S then we can define an operator A = |φ 〉〈 χ| the action of which on any state gives the ket |φ 〉, multiplied by the complex number 〈 χ|ψ 〉. 7

Ket-Bra Operators In this lecture we introduce a class of operators whose action is, in a certain sense, self-evident. We refer to operators of this sort as ket-bra operators, because they are formed from an outer product of a ket and a bra, (opposite to the bra-ket associated with the inner product). Thus, if |φ 〉 and |χ 〉 are any two states in S then we can define an operator A = |φ 〉〈 χ| the action of which on any state gives the ket |φ 〉, multiplied by the complex number 〈 χ|ψ 〉. 8

Ket-Bra Operators In this lecture we introduce a class of operators whose action is, in a certain sense, self-evident. We refer to operators of this sort as ket-bra operators, because they are formed from an outer product of a ket and a bra, (opposite to the bra-ket associated with the inner product). Thus, if |φ 〉 and |χ 〉 are any two states in S then we can define an operator A = |φ 〉〈 χ| the action of which on any state gives the ket |φ 〉, multiplied by the complex number 〈 χ|ψ 〉. 9

Projection Operators: This ket-bra form is particularly useful for defining what are referred to as projection operators. An operator P is said to be a projection operator, or simply a projector, if it satisfies the idempotency condition Note that this implies that Pⁿ = P for all integers n ≥ 1. So P can act as many times as you want, but only the first time really counts. As an example, if |φ 〉 is square normalized to unity, so that 〈 φ|φ 〉 = 1, then the operator P φ = |φ 〉〈 φ| is the projector onto the direction |φ 〉, because 10

Projection Operators: This ket-bra form is particularly useful for defining what are referred to as projection operators. An operator P is said to be a projection operator, or simply a projector, if it satisfies the idempotency condition Note that this implies that Pⁿ = P for all integers n ≥ 1. So P can act as many times as you want, but only the first time really counts. As an example, if |φ 〉 is square normalized to unity, so that 〈 φ|φ 〉 = 1, then the operator P φ = |φ 〉〈 φ| is the projector onto the direction |φ 〉, because 11

Projection Operators: This ket-bra form is particularly useful for defining what are referred to as projection operators. An operator P is said to be a projection operator, or simply a projector, if it satisfies the idempotency condition Note that this implies that Pⁿ = P for all integers n ≥ 1. So P can act as many times as you want, but only the first time really counts. As an example, if |φ 〉 is square normalized to unity, so that 〈 φ|φ 〉 = 1, then the operator P φ = |φ 〉〈 φ| is the projector onto the direction |φ 〉, because 12

Projection Operators: This ket-bra form is particularly useful for defining what are referred to as projection operators. An operator P is said to be a projection operator, or simply a projector, if it satisfies the idempotency condition Note that this implies that Pⁿ = P for all integers n ≥ 1. So P can act as many times as you want, but only the first time really counts. As an example, if |φ 〉 is square normalized to unity, so that 〈 φ|φ 〉 = 1, then the operator P φ = |φ 〉〈 φ| is the projector onto the direction |φ 〉, because 13

Projection Operators: This ket-bra form is particularly useful for defining what are referred to as projection operators. An operator P is said to be a projection operator, or simply a projector, if it satisfies the idempotency condition Note that this implies that Pⁿ = P for all integers n ≥ 1. So P can act as many times as you want, but only the first time really counts. As an example, if |φ 〉 is square normalized to unity, so that 〈 φ|φ 〉 = 1, then the operator P φ = |φ 〉〈 φ| is the projector onto the direction |φ 〉, because 14

Projection Operators: This ket-bra form is particularly useful for defining what are referred to as projection operators. An operator P is said to be a projection operator, or simply a projector, if it satisfies the idempotency condition Note that this implies that Pⁿ = P for all integers n ≥ 1. So P can act as many times as you want, but only the first time really counts. As an example, if |φ 〉 is square normalized to unity, so that 〈 φ|φ 〉 = 1, then the operator P φ = |φ 〉〈 φ| is the projector onto the direction |φ 〉, because 15

Projection Operators: This ket-bra form is particularly useful for defining what are referred to as projection operators. An operator P is said to be a projection operator, or simply a projector, if it satisfies the idempotency condition Note that this implies that Pⁿ = P for all integers n ≥ 1. So P can act as many times as you want, but only the first time really counts. As an example, if |φ 〉 is square normalized to unity, so that 〈 φ|φ 〉 = 1, then the operator P φ = |φ 〉〈 φ| is the projector onto the direction |φ 〉, because 16

Projection Operators: This ket-bra form is particularly useful for defining what are referred to as projection operators. An operator P is said to be a projection operator, or simply a projector, if it satisfies the idempotency condition Note that this implies that Pⁿ = P for all integers n ≥ 1. So P can act as many times as you want, but only the first time really counts. As an example, if |φ 〉 is square normalized to unity, so that 〈 φ|φ 〉 = 1, then the operator P φ = |φ 〉〈 φ| is the projector onto the direction |φ 〉, because 17

Projection Operators: Thus, the action of on an arbitrary state |ψ 〉 is to take away those parts of |ψ 〉 not lying along |φ 〉, and to leave the part lying along the direction of |φ 〉 alone. As a simple extension, note that if the states {|i 〉 } form an orthonormal set so that 〈 i|j 〉 =δ_{ij, then we claim that the sum of orthogonal projectors is also a projection operator because 18

Projection Operators: Thus, the action of on an arbitrary state |ψ 〉 is to take away those parts of |ψ 〉 not lying along |φ 〉, and to leave the part lying along the direction of |φ 〉 alone. As a simple extension, note that if the states {|i 〉 } form an orthonormal set so that 〈 i|j 〉 =δ_{ij, then we claim that the sum of orthogonal projectors is also a projection operator because 19

Projection Operators: Thus, the action of on an arbitrary state |ψ 〉 is to take away those parts of |ψ 〉 not lying along |φ 〉, and to leave the part lying along the direction of |φ 〉 alone. As a simple extension, note that if the states {|i 〉 } form an orthonormal set so that 〈 i|j 〉 =δ_{ij, then we claim that the sum of orthogonal projectors then is also a projection operator because 20

Projection Operators: Thus, the action of on an arbitrary state |ψ 〉 is to take away those parts of |ψ 〉 not lying along |φ 〉, and to leave the part lying along the direction of |φ 〉 alone. As a simple extension, note that if the states {|i 〉 } form an orthonormal set so that 〈 i|j 〉 =δ_{ij, then we claim that the sum of orthogonal projectors then is also a projection operator because 21

Projection Operators: Thus, the action of on an arbitrary state |ψ 〉 is to take away those parts of |ψ 〉 not lying along |φ 〉, and to leave the part lying along the direction of |φ 〉 alone. As a simple extension, note that if the states {|i 〉 } form an orthonormal set so that 〈 i|j 〉 =δ_{ij, then we claim that the sum of orthogonal projectors then is also a projection operator because 22

Projection Operators: Comment: Projection operators always "project onto something". In this latter case, the operator P projects onto the subspace spanned by the vectors in this orthonormal set. 23

Projection Operators: Comment: Projection operators always "project onto something". In this latter case, the operator P projects onto the subspace spanned by the vectors in this orthonormal set. 24

Projection Operators: Comment: Projection operators always "project onto something". In this latter case, the operator P projects onto the subspace spanned by the vectors in this orthonormal set. 25

Projection Operators: Comment: Projection operators always "project onto something". In this latter case, the operator P projects onto the subspace spanned by the vectors in this orthonormal set. 26

Projection Operators: Comment: Projection operators always "project onto something". In this latter case, the operator P projects onto the subspace spanned by the vectors in this orthonormal set. 27

Projection Operators: Comment: Projection operators always "project onto something". In this latter case, the operator P projects onto the subspace spanned by the vectors in this orthonormal set. 28

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 29

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 30

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 31

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 32

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 33

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 34

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 35

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 36

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 37

Projection Operators: Let’s look at a geometrical example. Consider the vectors r in R³ and let us denote the unit vectors in this space as |x 〉,|y 〉, and |z 〉 so that r=|r 〉 =x|x 〉 +y|y 〉 +z|z 〉 wherex= 〈 x|r 〉 y= 〈 y|r 〉 z= 〈 z|r 〉 then the operatorP_{xy}=|x 〉〈 x| + |y 〉〈 y| projects |r 〉 onto the xy-plane, i.e., P_{xy}|r 〉 = [|x 〉〈 x| + |y 〉〈 y|](x|x 〉 +y|y 〉 +z|z 〉 ) = x|x 〉 +y|y 〉 is just the part of r lying in the xy-plane. The part of the vector lying outside this 2-D subspace has simply been removed. 38

Completeness: Notice in our original Hilbert space example above, that if the orthonormal states {|i 〉 } were complete, so that they actually formed an ONB for the space, then the subspace that they project upon would be the entire space S itself. Since, for such a basis, an expansion of the form exists for any state |ψ 〉 in the space, the action of the operator P on such a state is to just give back the state it acted on. We deduce that if the states {|i 〉 } form an orthonormal basis, then This relation, is referred to as the completeness relation for the states {|i 〉 }. 39

Completeness: Notice in our original Hilbert space example above, that if the orthonormal states {|i 〉 } were complete, so that they actually formed an ONB for the space, then the subspace that they project upon would be the entire space S itself. Since, for such a basis, an expansion of the form exists for any state |ψ 〉 in the space, the action of the operator P on such a state is to just give back the state it acted on. We deduce that if the states {|i 〉 } form an orthonormal basis, then This relation, is referred to as the completeness relation for the states {|i 〉 }. 40

Completeness: Notice in our original Hilbert space example above, that if the orthonormal states {|i 〉 } were complete, so that they actually formed an ONB for the space, then the subspace that they project upon would be the entire space S itself. Since, for such a basis, an expansion of the form exists for any state |ψ 〉 in the space, the action of the operator P on such a state is to just give back the state it acted on. We deduce that if the states {|i 〉 } form an orthonormal basis, then This relation, is referred to as the completeness relation for the states {|i 〉 }. 41

Completeness: Notice in our original Hilbert space example above, that if the orthonormal states {|i 〉 } were complete, so that they actually formed an ONB for the space, then the subspace that they project upon would be the entire space S itself. Since, for such a basis, an expansion of the form exists for any state |ψ 〉 in the space, the action of the operator P on such a state is to just give back the state it acted on. We deduce that if the states {|i 〉 } form an orthonormal basis, then This relation, is referred to as the completeness relation for the states {|i 〉 }. 42

Completeness: Notice in our original Hilbert space example above, that if the orthonormal states {|i 〉 } were complete, so that they actually formed an ONB for the space, then the subspace that they project upon would be the entire space S itself. Since, for such a basis, an expansion of the form exists for any state |ψ 〉 in the space, the action of the operator P on such a state is to just give back the state it acted on. We deduce that if the states {|i 〉 } form an orthonormal basis, then This relation, is referred to as the completeness relation for the states {|i 〉 }. 43

Completeness: Notice in our original Hilbert space example above, that if the orthonormal states {|i 〉 } were complete, so that they actually formed an ONB for the space, then the subspace that they project upon would be the entire space S itself. Since, for such a basis, an expansion of the form exists for any state |ψ 〉 in the space, the action of the operator P on such a state is to just give back the state it acted on. We deduce that if the states {|i 〉 } form an orthonormal basis, then This relation, is referred to as the completeness relation for the states {|i 〉 }. 44

Completeness: Notice in our original Hilbert space example above, that if the orthonormal states {|i 〉 } were complete, so that they actually formed an ONB for the space, then the subspace that they project upon would be the entire space S itself. Since, for such a basis, an expansion of the form exists for any state |ψ 〉 in the space, the action of the operator P on such a state is to just give back the state it acted on. We deduce that if the states {|i 〉 } form an orthonormal basis, then This relation, is referred to as the completeness relation for the states {|i 〉 }. 45

Completeness: These ideas are also extensible with some care to continuously- indexed states. If the states {|α 〉 } form an ONB for the space S then the operator is not a projection operator, since the state |α 〉 is not square-normalized to unity. Indeed, it is of infinite norm, so However, the integral of this operator over any region of the possible values taken on by the parameter α is a projector. That is, if we define then 46

Completeness: These ideas are also extensible with some care to continuously- indexed states. If the states {|α 〉 } form an ONB for the space S then the operator is not a projection operator, since the state |α 〉 is not square-normalized to unity. Indeed, it is of infinite norm, so However, the integral of this operator over any region of the possible values taken on by the parameter α is a projector. That is, if we define then 47

Completeness: These ideas are also extensible with some care to continuously- indexed states. If the states {|α 〉 } form an ONB for the space S then the operator is not a projection operator, since the state |α 〉 is not square-normalized to unity. Indeed, it is of infinite norm, so However, the integral of this operator over any region of the possible values taken on by the parameter α is a projector. That is, if we define then 48

Completeness: These ideas are also extensible with some care to continuously- indexed states. If the states {|α 〉 } form an ONB for the space S then the operator is not a projection operator, since the state |α 〉 is not square-normalized to unity. Indeed, it is of infinite norm, so However, the integral of this operator over any region of the possible values taken on by the parameter α is a projector. That is, if we define then 49

Completeness: These ideas are also extensible with some care to continuously- indexed states. If the states {|α 〉 } form an ONB for the space S then the operator is not a projection operator, since the state |α 〉 is not square-normalized to unity. Indeed, it is of infinite norm, so However, the integral of this operator over any region of the possible values taken on by the parameter α is a projector. That is, if we define then 50

Completeness: These ideas are also extensible with some care to continuously- indexed states. If the states {|α 〉 } form an ONB for the space S then the operator is not a projection operator, since the state |α 〉 is not square-normalized to unity. Indeed, it is of infinite norm, so However, the integral of this operator over any region of the possible values taken on by the parameter α is a projector. That is, if we define then 51

Completeness: These ideas are also extensible with some care to continuously- indexed states. If the states {|α 〉 } form an ONB for the space S then the operator is not a projection operator, since the state |α 〉 is not square-normalized to unity. Indeed, it is of infinite norm, so However, the integral of this operator over any region of the possible values taken on by the parameter α is a projector. That is, if we define then 52

Completeness: These ideas are also extensible with some care to continuously- indexed states. If the states {|α 〉 } form an ONB for the space S then the operator is not a projection operator, since the state |α 〉 is not square-normalized to unity. Indeed, it is of infinite norm, so However, the integral of this operator over any region of the possible values taken on by the parameter α is a projector. That is, if we define then 53

Completeness: We will refer to an operator such as as a projector density, since it's integral always gives a projector. The action of ρ on an arbitrary state gives ρ_{α}|ψ 〉 =|α 〉〈 α|ψ 〉 =ψ(α)|α 〉 the part of that state lying along the direction |α 〉, and removes the parts that don’t. 54

Completeness: We will refer to an operator such as as a projector density, since it's integral always gives a projector. The action of ρ on an arbitrary state gives ρ_{α}|ψ 〉 =|α 〉〈 α|ψ 〉 =ψ(α)|α 〉 the part of that state lying along the direction |α 〉, and removes the parts that don’t. 55

Completeness: We will refer to an operator such as as a projector density, since it's integral always gives a projector. The action of ρ on an arbitrary state gives ρ_{α}|ψ 〉 =|α 〉〈 α|ψ 〉 =ψ(α)|α 〉 the part of that state lying along the direction |α 〉, and removes the parts that don’t. 56

Completeness: We will refer to an operator such as as a projector density, since it's integral always gives a projector. The action of ρ on an arbitrary state gives ρ_{α}|ψ 〉 =|α 〉〈 α|ψ 〉 =ψ(α)|α 〉 the part of that state lying along the direction |α 〉, and removes the parts that don’t. 57

Completeness: As in the discrete case, if we now consider the projector which includes all the states in the basis, i.e., we project onto the entire space. Thus, since we can always write an arbitrary state in the form the action of the operator which has no restrictions on the values of α, is to reproduce whatever state it acts upon, i.e., This being true for all |ψ 〉, we deduce the completeness relation for continuously- indexed states 58

Completeness: As in the discrete case, if we now consider the projector which includes all the states in the basis, i.e., we project onto the entire space. Thus, since we can always write an arbitrary state in the form the action of the operator which has no restrictions on the values of α, is to reproduce whatever state it acts upon, i.e., This being true for all |ψ 〉, we deduce the completeness relation for continuously- indexed states 59

Completeness: As in the discrete case, if we now consider the projector which includes all the states in the basis, i.e., we project onto the entire space. Thus, since we can always write an arbitrary state in the form the action of the operator which has no restrictions on the values of α, is to reproduce whatever state it acts upon, i.e., This being true for all |ψ 〉, we deduce the completeness relation for continuously- indexed states 60

Completeness: As in the discrete case, if we now consider the projector which includes all the states in the basis, i.e., we project onto the entire space. Thus, since we can always write an arbitrary state in the form the action of the operator which has no restrictions on the values of α, is to reproduce whatever state it acts upon, i.e., This being true for all |ψ 〉, we deduce the completeness relation for continuously- indexed states 61

Completeness: As in the discrete case, if we now consider the projector which includes all the states in the basis, i.e., we project onto the entire space. Thus, since we can always write an arbitrary state in the form the action of the operator which has no restrictions on the values of α, is to reproduce whatever state it acts upon, i.e., This being true for all |ψ 〉, we deduce the completeness relation for a continuously-indexed ONB 62

In this lecture, we introduced a class of operators called ket-bra operators, whose action on arbitrary states is rather obvious. We used operators of this type to define another important class of operators called projection operators, which generally project an arbitrary state onto some subspace of S. Indeed, by considering complete sums of orthogonal projectors, we deduced the completeness relations for discrete and continuous ONBs, which provide a decomposition of the identity operator in terms of a “compete set of states”. As we will see in the next lecture, these completeness relations provide useful tools for generating representation dependent equations, from there representation counterparts. 63

In this lecture, we introduced a class of operators called ket-bra operators, whose action on arbitrary states is rather obvious. We used operators of this type to define another important class of operators called projection operators, which generally project an arbitrary state onto some subspace of S. Indeed, by considering complete sums of orthogonal projectors, we deduced the completeness relations for discrete and continuous ONBs, which provide a decomposition of the identity operator in terms of a “compete set of states”. As we will see in the next lecture, these completeness relations provide useful tools for generating representation dependent equations, from there representation counterparts. 64

In this lecture, we introduced a class of operators called ket-bra operators, whose action on arbitrary states is rather obvious. We used operators of this type to define another important class of operators called projection operators, which generally project an arbitrary state onto some subspace of S. Indeed, by considering complete sums of orthogonal projectors, we deduced the completeness relations for discrete and continuous ONBs, which provide a decomposition of the identity operator in terms of a “complete set of states”. As we will see in the next lecture, these completeness relations provide useful tools for generating representation dependent equations, from there representation counterparts. 65

In this lecture, we introduced a class of operators called ket-bra operators, whose action on arbitrary states is rather obvious. We used operators of this type to define another important class of operators called projection operators, which generally project an arbitrary state onto some subspace of S. Indeed, by considering complete sums of orthogonal projectors, we deduced the completeness relations for discrete and continuous ONBs, which provide a decomposition of the identity operator in terms of a “complete set of states”. As we will see in the next lecture, these completeness relations provide useful tools for generating representation dependent equations, from their representation independent counterparts. 66

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