Projection Operator: What Math Operation Does it Represent?

  • Context: High School 
  • Thread starter Thread starter Kevin McHugh
  • Start date Start date
  • Tags Tags
    Operator Projection
Click For Summary

Discussion Overview

The discussion revolves around the mathematical operation represented by the expression |i>

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants describe the projection operator $$\hat{P}=|i \rangle \langle i|$$ as a means to project a vector $$|\psi \rangle$$ into the direction of the unit vector $$|i \rangle$$.
  • Others emphasize the intuitive nature of Dirac's notation and its utility in expressing projections without needing to delve into matrix representations.
  • One participant mentions the relationship between the projection operator and the decomposition of arbitrary vectors using a complete set of orthonormal vectors.
  • There are inquiries about the expansion of the expression and its representation in matrix form, highlighting the distinction between bra and ket vectors.
  • Another participant points out the importance of recognizing the abstract nature of vectors in Hilbert space versus their realizations in function spaces or sequences.
  • Some participants discuss the isomorphism between different spaces, such as $$L^2$$ and $$\ell^2$$, and how this relates to the bra-ket formalism.

Areas of Agreement / Disagreement

Participants express various views on the nature and implications of the projection operator, with no consensus reached on certain aspects, particularly regarding the distinctions between abstract and realized vectors.

Contextual Notes

Participants note limitations in understanding the projection operator, including the dependence on definitions and the abstract nature of the Hilbert space versus its realizations.

Kevin McHugh
Messages
319
Reaction score
165
What mathematical operation does this expression represent"

|i><i|

I know the i's are unit vectors, and I know <i|i> is the dot product, but what operation is the ket-bra?
 
Physics news on Phys.org
The magic of Dirac's notation is that it is intuitive. Here you have an operator
$$\hat{P}=|i \rangle \langle i|.$$
To see what it does to a vector ##|\psi \rangle## just write it down:
$$\hat{P}=|i \rangle \langle i|\psi \rangle,$$
et voila, that's the right definition of the projector. It projects the vector ##|\psi \rangle## into the direction of ##|i \rangle##. This tells you that ##\psi_i=\langle i | \psi \rangle## is the component of ##|\psi \rangle## in direction of ##|i \rangle##. So it's precisely what a projection means!

If you have a complete set of orthonormal vectors ##|j \rangle##, you have
$$\sum_{j=1}^{\infty} |j \rangle \langle j |=\hat{1}.$$
This means the decomposition of an arbitrary vector is given by (again just write it down; Dirac's magic gives the correct expression!):
$$|\psi \rangle=\hat{1} |\psi \rangle = \sum_{j=1}^{\infty} |j \rangle \langle j|\psi \rangle.$$
 
  • Like
Likes   Reactions: Simon Phoenix, stevendaryl and DrClaude
Thank you Vanhees, is there an expansion for the expression? e.g. a1i + a2i + a3i ...
 
Kevin McHugh said:
Thank you Vanhees, is there an expansion for the expression? e.g. a1i + a2i + a3i ...

In matrix form the projector operation is the matrix resulting from column vector . row vector, where . is matrix multiplication. A ket is a column vector, a bra is a row vector, as you know if you do row vector . column vector you get a number, if you do column vector . row vector you get a matrix.
 
A bra vector is not a column vector but a vector in an abstract Hilbert space. Usually you organize the components of a vector with respect to a basis as a column vector and the matrix elements of an operator as a matrix. Of course, it's much more convenient to use the Dirac formalism since then you don't need to think about this organization much. For the operation of an operator on a vector you have
$$\hat{A} |\psi \rangle=\sum_{j,k=1}^{\infty} |j \rangle \langle j|\hat{A}|k \rangle \langle k|\psi \rangle=\sum_{j,k=1}^{\infty} |j \rangle A_{jk} \psi_k.$$
Now you organize the vector components as a column and the matrix elements as a matrix,
$$\tilde{A}=(A_{jk})=\begin{pmatrix} A_{11} & A_{12} &\ldots \\ A_{21} & A_{22} & \ldots \\ \vdots & \vdots & \vdots \end{pmatrix},$$
$$\tilde{\psi}=(\psi_k)=\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \end{pmatrix}.$$
Then you can have for the components of ##|\phi \rangle=\hat{A} |\psi \rangle##
$$\tilde{\phi}=\tilde{A} \tilde{\psi}$$
in the sense of the usual matrix-vector product. Note, however, that you have vectors and matrices with infinitely many components.
 
I was referring to the isomorphism between ##L^2(\infty)## and ##l^2(\infty)##.
 
It is very important to distinguish between the abstract vectors and various realizations. The Dirac bras and kets are in the abstract separable Hilbert space ##\mathcal{H}##. It can be realized as the function space ##L^2## of square integrable functions or the space ##\ell^2## of square-summable squences. These realizations are formally connected with the abstract Hilbert space of the bra-ket formalism by using (generalized) orthonormal bases. In the case of "wave mechanics" you use generalized position eigenstates as a generalized basis (note that these do not belong to the Hilbert space but to the dual space of the domain of the self-adjoint position operator). This leads to the one-to-one mapping between the abstract vector ##|\psi \rangle## to the wave function ##\psi(x)=\langle x|\psi \rangle##. If you use a discrete proper basis, e.g., the energy eigenfunctions of the harmonic oscillator. The map is to the sequence ##\psi_n=\langle n|\psi \rangle##.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 59 ·
2
Replies
59
Views
6K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K