High School Projection Operator: What Math Operation Does it Represent?

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The expression |i><i| represents a projection operator in quantum mechanics, defined as \(\hat{P}=|i \rangle \langle i|\). This operator projects any vector |\psi\rangle into the direction of the unit vector |i\rangle, with the component of |\psi\rangle in that direction given by \(\psi_i=\langle i | \psi \rangle\). The completeness relation states that a complete set of orthonormal vectors |j\rangle can decompose any vector |\psi\rangle, expressed as \(|\psi \rangle=\sum_{j=1}^{\infty} |j \rangle \langle j|\psi \rangle\). The discussion emphasizes the utility of Dirac notation for simplifying operations in abstract Hilbert spaces, while also noting the connection to matrix representations. Overall, the projection operator serves as a fundamental concept in understanding vector decomposition in quantum mechanics.
Kevin McHugh
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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?
 
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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!):
$$|\psi \rangle=\hat{1} |\psi \rangle = \sum_{j=1}^{\infty} |j \rangle \langle j|\psi \rangle.$$
 
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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##.
 

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