# B Projection operator

1. Sep 27, 2016

### Kevin McHugh

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?

2. Sep 27, 2016

### vanhees71

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!):
$$|\psi \rangle=\hat{1} |\psi \rangle = \sum_{j=1}^{\infty} |j \rangle \langle j|\psi \rangle.$$

3. Sep 27, 2016

### Staff: Mentor

4. Sep 27, 2016

### Kevin McHugh

Thank you Vanhees, is there an expansion for the expression? e.g. a1i + a2i + a3i .....

5. Sep 27, 2016

### ddd123

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.

6. Sep 28, 2016

### vanhees71

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.

7. Sep 28, 2016

### ddd123

I was referring to the isomorphism between $L^2(\infty)$ and $l^2(\infty)$.

8. Sep 28, 2016

### vanhees71

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 $\psi_n=\langle n|\psi \rangle$.