Projection Operators: Explaining |m|2*|m|2 = |m|4

[Somali_Physicist]

Take a projection operatorP_m=|m><m|
However if the ket of m is a column matrix of m x 1 and its bra the complex conjugate with 1 x m length
therefore <m|m> = |m|²
since the m here is the same since projection operator is the same.
if A is a matrix
B = A
A*B=B*A

but P_m*P_m = P_m (Projection operator)
but this makes no sense as |m|²*|m|² = |m|⁴

Can anyone explain?

 

[Dr Transport]

Writte it out in Dirac notation as you already have done and it pops out immediately.

 

[Somali_Physicist]

Writte it out in Dirac notation as you already have done and it pops out immediately.

P_m*P_m=(|m><m|)(|m><m|) = |m><m|m><m|=|m><m| only if |m|^2 = 1

 

[BvU]

m here is the same the projection operator is the same

Same as what ?
Note that <m|m> = 1 so that indeed P²=P (and your |m| = 1)

Page 5 (page number 59) here

 

[Somali_Physicist]

Same as what ?
Note that <m|m> = I so that indeed P²=P (and your |m| = 1)

hmm but let's take a bra <m| is a row matrix of 2 elements in the dual vector space V*
and |m> is the corresponding column matrix of two elements in the vector of vector space V
now no matter which configuration i multiply these two things ( or which ever way they act on each other)
the corressponding matrix multiplication is the same value.

<m|m>=|m><m| = |m|^2
or is this system essentially assuming the magnitude is 1?

 

[stevendaryl]

Take a projection operatorP_m=|m><m|
However if the ket of m is a column matrix of m x 1 and its bra the complex conjugate with 1 x m length
therefore <m|m> = |m|²

Where are you getting that from? Let's take a simple example:

$|\psi\rangle = \left( \begin{array} \\ 1 \\ 0 \end{array} \right)$

(I don't want to use $|m\rangle$ because that's mixing up the number of components with the index of the component.)

Then $\langle \psi|\psi \rangle =\left( \begin{array} \\ 1 & 0 \end{array} \right) \left( \begin{array} \\ 1 \\ 0 \end{array} \right) = 1$

The projection operator $P_\psi = \left( \begin{array} \\ 1 \\ 0 \end{array} \right) \left( \begin{array} \\ 1 & 0 \end{array} \right) = \left( \begin{array} \\ 1 & 0 \\ 0 & 0 \end{array} \right)$

You can see that $(P_\psi)^2 = P_\psi$

 

[stevendaryl]

hmm but let's take a bra <m| is a row matrix of 2 elements in the dual vector space V*
and |m> is the corresponding column matrix of two elements in the vector of vector space V
now no matter which configuration i multiply these two things ( or which ever way they act on each other)
the corressponding matrix multiplication is the same value.

<m|m>=|m><m| = |m|^2
or is this system essentially assuming the magnitude is 1?

Yes. $|\psi\rangle \langle \psi|$ is a projection operator only if $\langle \psi|\psi \rangle = 1$

 

[Somali_Physicist]

Where are you getting that from? Let's take a simple example:

$|\psi\rangle = \left( \begin{array} \\ 1 \\ 0 \end{array} \right)$

(I don't want to use $|m\rangle$ because that's mixing up the number of components with the index of the component.)

Then $\langle \psi|\psi \rangle =\left( \begin{array} \\ 1 & 0 \end{array} \right) \left( \begin{array} \\ 1 \\ 0 \end{array} \right) = 1$

The projection operator $P_\psi = \left( \begin{array} \\ 1 \\ 0 \end{array} \right) \left( \begin{array} \\ 1 & 0 \end{array} \right) = \left( \begin{array} \\ 1 & 0 \\ 0 & 0 \end{array} \right)$

You can see that $(P_\psi)^2 = P_\psi$

Ahh thanks that explains it, i assumed commutativity!

 

[Somali_Physicist]

Yes. $|\psi\rangle \langle \psi|$ is a projection operator only if $\langle \psi|\psi \rangle = 1$

That also sorts my other conundrum.So for these transformations only can have 1s and 0s in it.The projection operator has the element at the Pm,m.My book used a value "n" at that position which i thought implied non 1 values.

 

[DrClaude]

That also sorts my other conundrum.So for these transformations only can have 1s and 0s in it.The projection operator has the element at the Pm,m.My book used a value "n" at that position which i thought implied non 1 values.

The matrix representation of a projection operator can have elements other than 1 and 0. Try construction the projection operator $\hat{P}_{\psi}$ for
$$
| \psi \rangle = \frac{1}{\sqrt{2}} \left( |+\rangle + |-\rangle \right)
$$
in the $\left\{ |+\rangle, |-\rangle \right\}$ basis.

 

[Mentz114]

hmm but let's take a bra <m| is a row matrix of 2 elements in the dual vector space V*
and |m> is the corresponding column matrix of two elements in the vector of vector space V
now no matter which configuration i multiply these two things ( or which ever way they act on each other)
the corressponding matrix multiplication is the same value.

<m|m>=|m><m| = |m|^2
or is this system essentially assuming the magnitude is 1?

Projectors are idempotent in this sense
$\langle m |( |m\rangle\langle m |) m\rangle = 1$ and inserting more projectors has no effect
$\langle m |(|m\rangle\langle m |m\rangle\langle m |) m\rangle = 1$