# Projection Operators

## Main Question or Discussion Point

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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Dr Transport
Gold Member
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

BvU
Homework Helper
2019 Award
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 lets 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
Staff Emeritus
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
stevendaryl
Staff Emeritus
hmm but lets 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.

DrClaude
Mentor
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 lets 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
$\langle m |(|m\rangle\langle m |m\rangle\langle m |) m\rangle = 1$