Simple angular momentum problem. QM

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Homework Help Overview

The discussion revolves around demonstrating that the angular momentum operator Lx is Hermitian within the context of quantum mechanics. Participants are exploring the properties of Hermitian operators and their implications for observables.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to show that the inner product < Lx f | f > equals < f | Lx f >, questioning the implications of the complex conjugate in this context. There is a discussion about the mathematical expressions involved and their interpretations.

Discussion Status

Some participants have provided insights into the definitions of Hermitian operators and the necessary steps to show that Lx is Hermitian. However, there remains uncertainty regarding the handling of complex conjugates and the specific forms of the expressions being evaluated.

Contextual Notes

There are indications of confusion regarding the mathematical representation of the operators and the assumptions about their properties. Participants are also grappling with the notation and the implications of the imaginary unit in the expressions.

armis
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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 realize 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
 
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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].
 
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>
 
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]
 

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