Outer product problem of derac notation

[hasanhabibul]

why |a><b| expresses the projection...how can it be possible on matrix..if we multiply a ket a with a bra b ...we get a product of two matrix(one is a column matrix,an0ther is row matrix)..from where nothing can be realized very clearly..how this multiplication of matrix can give a projection..??

 

[strangerep]

why |a><b| expresses the projection...how can it be possible on matrix..if we multiply a ket a with a bra b ...we get a product of two matrix(one is a column matrix,an0ther is row matrix)..from where nothing can be realized very clearly..how this multiplication of matrix can give a projection..??

Think of the ket as a column vector, and the bra as a row vector.
So braket gives a scalar, but ketbra gives a square matrix.

 

[hasanhabibul]

I know ket bra is a square matrix...but what is the especiality of this squar matrix??how it expresses projection.

 

[malawi_glenn]

(|a><a|)*|b> = <a|b>|a>

It projects from the state |b> the part of it which is parallel to |a>

 

[hasanhabibul]

pleasez i know those..but i want the matrix format..if |A>=a
b
c
and <B|={d,e,f}. now perform a matrix multiplication this two matrix..and it will give u a result.and how this odd looking matrix express the projection..that was my question.

 

[malawi_glenn]

In what basis do you want to represent the operator |a><b| ?

(|a><a|)*|b> = <a|b>|a>

IS a matrix multiplying the ket |b> !

 

[jensa]

Look at it in a basis where |a> is a basis state
|a\rangle =\begin{pmatrix} 1\\0\\0\end{pmatrix}
then the matrix |a><a| will be
|a\rangle\langle a|=\begin{pmatrix} 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 0\end{pmatrix}
acting with this on any vector
|b\rangle=\begin{pmatrix} c\\ d\\ e\end{pmatrix}<br />
will give you just

(|a\rangle\langle a|)|b\rangle=\begin{pmatrix} c \\ 0 \\ 0 \end{pmatrix}=\langle a| b\rangle |a\rangle

Does this help?

 

[malawi_glenn]

u must choose basis of representation for your kets and bras as shown by jensa.