# Two quick ones

1. Apr 12, 2005

### broegger

Two quick ones :)

Hi, two questions:

1) How can I find the expectation value of the x-component of the angular momentum, $$\langle L_x \rangle$$, when I know $$\langle L^2 \rangle$$ and $$\langle L_z \rangle$$?

2) Say, I have a state $$|\Psi \rangle$$ and two operators A and B represented as matrices. Now, $$|\Psi \rangle$$ is given as a linear combination of the eigenstates of A and I want to express them as a linear combination of the eigenstates of the operator B instead. How do I do that?

Thanks :)

2. Apr 12, 2005

### vanesch

Staff Emeritus
I don't think you can ! I think you can if you are in an EIGENSTATE of L^2 and Lz, but not if you only know the expectation values, because I think I can make up two different wave functions with same expectation for L^2 and Lz and different expectation for Lx...

cheers,
Patrick.

3. Apr 12, 2005

### broegger

Oh, they ARE eigenstates of both L^2 and Lz. Sorry. What do I do? :)

4. Apr 12, 2005

### vanesch

Staff Emeritus
I think you will always find 0, no ? Because of the cylindrical symmetry of these states ?

The reason is that you can write Jx as 1/2 (J_+ + J_-) and these, acting on a state |j,m> will give you |j,m+1> and |j,m-1>, so
<j,m| Jx |j,m> = 0, no ?

cheers,
patrick.

5. Apr 12, 2005

### broegger

I don't quite get it. What does the |J>'s represent?

6. Apr 12, 2005

### vanesch

Staff Emeritus
Sorry, I should have written L. J stands in general for an angular momentum, L for an orbital angular momentum and S for a spin angular momentum. One usually uses the notation J if it doesn't matter (as is the case here) whether it is orbital or spin angular momentum one talks about.

I would think you are aware of J+ and J- (or L+ and L-), the ladder operators of angular momentum ?

cheers,
Patrick.

7. Apr 12, 2005

### vanesch

Staff Emeritus
I guess you mean that you work in a certain basis, in which you get to know the components of A (matrix) and B (matrix).
However, there is an ambiguity in the way you know psi: if you know its projections on the eigenvectors of A, you don't necessarily know its components in the basis in which A is given, because eigenvectors are only defined upto a complex multiplication factor ! So depending on how these eigenvectors were fixed your components will have an ambiguity (each of them individually) of a complex factor: just as well say that you don't know them (except if they happen to be 0).

Happily, you can hold the same reasoning for B :-) so I would say that a possible answer to the question will always be, that in a suitably scaled set of eigenvectors of B, psi will always take on the components {1,1,1,1,...1}

cheers,
Patrick.

8. Apr 12, 2005

### broegger

Sorry, I'm wasting your time. The situation in 1) is this:

An electron moving in a Coulomb-field from a proton, is in the following state (at time t=0)

$$|\phi,t=0> = \tfrac4{5}|100> + \tfrac{3i}5|211>$$​

where $$|nlm>$$ is the usual energy eigenstates of the hydrogenatom. They are also eigenstates of angularmomentum:

$$L^2|nlm>=l(l+1)\hslash^2|nlm>$$

$$L_z|nlm>=m\hslash|nlm>$$

The questions are now:

a) Calculate the expectation value <E> in the state $$|\phi,t=0>$$.

b) Calculate the expectation values for L^2 and L_z in the state $$|\phi,t=0>$$.

c) Calculate the expectation value for L_x in the state $$|\phi,t=0>$$.

It's c) I'm having trouble with. Hope this helps.

Last edited: Apr 12, 2005
9. Apr 12, 2005

### dextercioby

$$\langle \hat{L}_{x}\rangle_{|\psi\rangle} =:\langle\psi|\hat{L}_{x}|\psi\rangle$$

Okay?

Now,use the fact that

$$\hat{L}_{x}=\frac{1}{2}\left(\hat{L}_{+}+\hat{L}_{-}\right)$$

and the action of $\hat{\mbox{L}}_{+}$ and $\hat{\mbox{L}}_{-}$ on an arbitrary hydrogenoid wavefunction $|n,l,m\rangle$ (see textbook) and the orthonormalization of the states

$$\langle n,l,m|n',l',m'\rangle =\delta_{nn'}\delta_{ll'}\delta_{mm'}$$

Daniel.

10. Apr 13, 2005

### broegger

Yep, I with you now :) Thank you both.

11. Apr 14, 2005

### shyboy

I assume that eigenvectors of both A and B provide the complete basis. (?)

you have
|c>=|A><A|c>

you need to find

|c>=|B><B|c>

where |A> and |B> - eigenvectors

then

<b|c>=<B|A><A|c>;

12. Apr 14, 2005

### dextercioby

I don't follow.What and who are those A,B,a,b,,blah,blah,blah...?

Daniel.

13. Apr 14, 2005

### shyboy

that is an old linear algebra in QM notation. I assume that eigenvectors of each matrix forms an ortogonal basis.

You have vector in the basis of the A matrix

$$\vec\psi=a_1 \vec{a_1}+a_2 \vec{a_2}+....$$ (1)

you need to find

$$\vec\psi=b_1 \vec{ b_1}+b_2 \vec{ b_2}+....$$ (2)

so multiply (1) by $$\vec{b_1}, \vec{b_2}...$$ and so on.

$$b_1 =a_1 (\vec {a_1}\vec{ b_1})+a_2 (\vec {a_1}\vec{ b_1})+...$$

14. Apr 14, 2005

### dextercioby

Okay.What relevance does it have here?

Daniel.

15. Apr 14, 2005

### shyboy

well,

that was (1)

that was (2)

16. Apr 14, 2005

### dextercioby

Are u hijacking the thread?This was the context