Simple angular momentum problem. QM

  • Thread starter armis
  • Start date
  • #1
103
0

Homework Statement



Show that Lx is a Hermitian operator

Homework Equations



Well, since Lx is an observable it must be represented by an Hermitian operator.

[tex] L = -i\cdot\hbar\cdotr\times\nabla [/tex]

If an operator is Hermitian, then it's equal to it's Hermitian conjugate

The Attempt at a Solution



I do realise what I have to do, however there are holes in my math.

So in order to show that the Lx operator is Hermitian I could show that:

[tex] < L_{x} f | f > = < f | L_{x} f > [/tex] is that correct?

If I assume that's correct and since [tex] < L_{x} f | f > = (< f | L_{x} f >)^* [/tex] I have a complex conjugate missing somewhere. Thus there is an error
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,306
48
You don't have a conjugate missing, the identity on the last line is true.
The point is that, by definition of Hermitian conjugate, if you want,
[tex]\langle f \mid L_x f \rangle = \langle L_x^\dagger f \mid f \rangle[/tex]
so if you show that it's also [itex]\langle L_x f \mid f \rangle[/itex] then effectively you are showing that [itex]L_x^\dagger = L_x[/itex].
 
  • #3
103
0
Thanks. That I understand but I probably don't understand something else or I don't really understand either, anyway to be more specific here is what I mean

[tex] \langle f \mid L_{x} f \rangle [/tex] which is

<f|-ih(y*d/dz - z*d/dy)f>. (I am sorry, Latex doesn't show up for this one. I guess I typed it wrong)

now if [tex] L_{x} [/tex] is Hermitian then as far as I can understand the above expression must be equal to

[tex] \langle L_{x} f \mid f \rangle [/tex] which is [tex] \langle f \mid L_{x}f \rangle^* [/tex]

and here is my problem, I get that [tex] \langle L_{x} f \mid f \rangle [/tex] is equal to [tex] \langle f \mid L_{x} f \rangle [/tex] conjugated and because of the i in the [tex] L_{x} [/tex] expression the two don't really match, do they? :confused: What am I missing?

Just to be absolutely clear I don't like the fact that [tex]
\langle L_{x} f \mid f \rangle [/tex] is <-ih(y*d/dz - z*d/dy)f|f> doesn't seem to match with <f|-ih(y*d/dz - z*d/dy)f>
 
  • #4
malawi_glenn
Science Advisor
Homework Helper
4,795
22
you can give a proof of this since the L_x operator is = y*p_z - z*p_y

now you proove that x_ip_j is hermitian. i.e show that

[tex] \int ( x_i\hat{p}_j \phi)^*\psi d^3r = \int \phi ^*x_i\hat{p}_j\psi d^3r [/tex]
 

Related Threads on Simple angular momentum problem. QM

  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
252
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
1K
Replies
8
Views
2K
Replies
14
Views
2K
Replies
4
Views
2K
Top