Projection Operators

  • #1

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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Answers and Replies

  • #2
Dr Transport
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Writte it out in Dirac notation as you already have done and it pops out immediately.
 
  • #3
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
 
  • #4
BvU
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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
 
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  • #5
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?
 
  • #6
stevendaryl
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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##
 
  • #7
stevendaryl
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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##
 
  • #8
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!
 
  • #9
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.
 
  • #10
DrClaude
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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.
 
  • #11
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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##
 

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