How is a self-adjoint linear operator defined in quantum mechanics?

  • Thread starter Thread starter trosten
  • Start date Start date
  • Tags Tags
    Definition
Click For Summary
A self-adjoint linear operator A in quantum mechanics can be defined through the expression <u|A|u>, which represents the mean value of the observable associated with A for the state |u>. Two definitions arise from different contexts: one defines it as the squared norm ||A|u>||^2, while the other expresses it as the inner product <u|A|u>. The first definition relates to the length of the vector A|u>, while the second involves the projection of A|u> onto |u>. The discussion clarifies that these definitions are not equal as they pertain to different mathematical constructs, particularly when considering self-adjoint operators and their properties. Ultimately, the resolution highlights the nature of idempotent orthogonal projectors in quantum mechanics.
trosten
Messages
47
Reaction score
0
How do u define this <u|A|u>? where A is a selfadjoint linear operator.

I have one book that defines is as ||A|u>||^2=<u|A|u>
(equation 2.6 http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.ps)

and since ||A|u>||^2 is the length of the vector A|u> squared that is equal to (<u|A)(A|u>).

I have another book (j.j sakurai) that defines it as (<u|)(A|u>)=(<u|A)(|u>)=<u|A|u> but this is the projection of A|u> on |u>.

It seems to me that the two definitions arent equal !? Any ideas? :confused:
 
Physics news on Phys.org
Of course,they're not equal,as they involve different "animals".

\langle \psi|\hat{A}|\psi\rangle

is the mean/avrerage of the observable A described by the selfadjoint operator \hat{A} on the quantum state |\psi \rangle
,while the norm
|| \hat{A}|\psi\rangle ||^{2}

involves the matrix element of the SQUARE of the operator...

Daniel.

P.S.The operator is selfadjoint,that's why is the square...
 
Last edited:
I found the solution to the problem myself. If the operator is A=|n><n| then ofcourse AA = (|n><n|)(|n><n|) = |n><n|n><n| = |n><n| = A
Ofcourse repeated projections shouldn't change anything! silly me.
 
OKay,u should have told right from the beginning that u're interested in true and IDEMPOTENT ortogonal projectors... :wink:

Daniel.
 
haha yeah didnt think about that :smile:
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
37
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K