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

• B
• Somali_Physicist
In summary, the conversation discusses the concept of a projection operator and its properties in Dirac notation. It is explained that a projection operator is represented by the outer product of a column vector and its complex conjugate row vector. The conversation also highlights the fact that the projection operator is idempotent, meaning that multiplying it with itself results in the same operator. Furthermore, it is noted that the projection operator only has elements of 1 and 0, and this is related to the magnitude of the vector used in its construction.f

#### Somali_Physicist

Take a projection operatorPm=|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 Pm*Pm = Pm (Projection operator)
but this makes no sense as |m|2*|m|2 = |m|4

Can anyone explain?

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

Writte it out in Dirac notation as you already have done and it pops out immediately.
Pm*Pm=(|m><m|)(|m><m|) = |m><m|m><m|=|m><m| only if |m|^2 = 1

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)

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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?

Take a projection operatorPm=|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##

Somali_Physicist and BvU
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##

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!

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.

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.

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##