Hermitian operator- Problem

  • Thread starter Raz91
  • Start date
  • #1
21
0
Prove the equation

[itex]A\left|\psi\right\rangle = \left\langle A\right\rangle\left|\psi\right\rangle + \Delta A\left|\psi\bot\right\rangle[/itex]

where [itex] A[/itex] is a Hermitian operator and [itex] \left\langle\psi |\psi\bot\right\rangle = 0[/itex]


[itex]\left\langle A\right\rangle[/itex] = The expectation value of A.
[itex]\Delta A[/itex] = The standard deviation of A.


My attempt :

I tried to write [itex]\left|\psi\right\rangle[/itex] as a superposition of the eigenfuncion of the operator [itex] A [/itex] and used the fact that it's a Hermitian operator

[itex]A\left|\phi_{n}\right\rangle = \lambda_{n}\left|\phi_{n}\right\rangle[/itex] , [itex]\left|\psi\right\rangle = \sum a_{n}\left|\phi_{n}\right\rangle[/itex]

so that [itex]A\left|\psi\right\rangle = \sum a_{n}\lambda_{n}\left|\phi_{n}\right\rangle[/itex]

and [itex]\left\langle A\right\rangle = \sum |a_{n}|^{2}\lambda_{n} [/itex]


[itex]\Delta A = \sqrt{\left\langle A^{2}\right\rangle - \left\langle A\right\rangle ^{2}}[/itex]

and I wrote [itex]\left|\psi\bot\right\rangle[/itex] as [itex]\left|\psi\bot\right\rangle = \sum b_{n}\left|\phi_{n}\right\rangle[/itex]

[itex]\sum a^{*}_{n}b_{n} = 0[/itex]

I don't know how to go on from here...
any ideas?

thank you! :)
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,024
579
Hmm, using that the s-adj A has a purely discrete spectrum is not ok, it's not assumed in the problem. And I think you left out a square. I get

[tex] A\psi = \langle A \rangle \psi + (\Delta A)^2 \psi_{\perp} [/tex]

under the simplifying assumptions [itex] \psi_{\perp} \in D(A)[/itex] and [itex] \mbox{Ran}(A)\subset D(A) [/itex].
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
13,024
579
The first step I found to resolution is to calculate:

[tex]\langle \psi,(A-\langle A\rangle ) \psi\rangle [/tex]

What do you get and what do you conclude ?
 
  • #4
21
0
I didn't miss the squar.. this is the question.
you get wrong units if you put the squar there ...
 
  • #5
21
0
The first step I found to resolution is to calculate:

[tex]\langle \psi,(A-\langle A\rangle ) \psi\rangle [/tex]

What do you get and what do you conclude ?
this yields zero .. i didn't understand how it can help me

thank you
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
13,024
579
If it yields zero, then the vector in the rhs of the scalar product is perpendicular on psi, so it's psi perp times a non-zero constant.
 
  • #7
21
0
I don't understand something -
there are lots of functions that can be perpendicular to psi

e.g (euclidean space)- the vector z (cartesian coordinates) , x is prep to z and y is prep to z but also the superposition (x+y) is prep to z.

so by writing psi prep do u mean to the superposition of all the functions that prep to psi or it's enough to choose only one?
 
  • #8
dextercioby
Science Advisor
Homework Helper
Insights Author
13,024
579
Yes, Psi_perp is an arbitrary vector perpendicular to the psi already chosen. You get from ⟨ψ,(A−⟨A⟩)ψ⟩ = 0 that C psi_perp = A psi - <A> psi. All you need to show is that C is related to the standard deviation in square.
 

Related Threads on Hermitian operator- Problem

  • Last Post
Replies
8
Views
2K
Replies
10
Views
4K
Replies
5
Views
1K
  • Last Post
Replies
17
Views
4K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
1
Views
1K
Top