# Homework Help: Duality between generators of changes and observables in QM

1. Sep 5, 2007

### plmokn2

I've posted similar questions on a different forum previously, but since I'm feeling a bit guilty about going on at the one person there who answers, I'll post these here. Hope it's ok if I don't follow the template since they're more conceptual than a standard problem. Any help appreciated:

1. Why do we have this duality between generators of inifintesimal changes and observables in quantum mechanics? Is it a postulate in it's own right or derived from something else (is it even always the case?)? How do we know the observable from a particular generator corrisponds to the observable we attatch to it: is it just guessing based on classical mechanics?

2. Am I right in saying if you have a particle spin up in the z-direction and you measure the spin along a perpendicular axis you will get a result of + or - h(bar)/2 with a 50/50 probability of each?

Thanks.

2. Sep 5, 2007

yep.

3. Sep 5, 2007

### plmokn2

Thanks for your reply. In that case suppose I represent the particle by a spinor initially (psi+,0)Transposed what will the spinor look like once I've made the measurement of spin in the seond direction and got a definite result? Does the direction of the components of the spinor change so it's now written as (psi+,0)T or (0,psi-)T but with the spinor in a different direction or do the components of the spinor stay in the same direction?
Thanks again.

4. Sep 5, 2007

### genneth

I seem to believe that generators of infinitesimal changes (elements of a lie algebra) are usually hermitian. I also think this might be wrong...

Old age befuddles the brain... :D

Edit:

Indeed, I was wrong. A quick think reveals that the lie algebra corresponding to the group of unitary matrices are *anti*hermitian matrices.

Last edited: Sep 5, 2007
5. Sep 5, 2007

### olgranpappy

Actually, you might now recall, we usually stick in a factor of 'i' in order to make them Hermitian.

6. Sep 5, 2007

### dextercioby

I assume you won't appreciate a technical answer involving theorems and mathematical terms, so i'm giving you to read something nice written by the famous EP Wigner a while ago http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=300191&pageindex=10#page

7. Sep 5, 2007

### olgranpappy

The spinor you wrote down is the eigenvector (with eigenvalue plus hbar/2) of the matrix S_z in a particular basis. The one in which S_z is diagonal. Stick with that basis. Don't change to a different basis in which a different generator is diagonal... that seems to be part of what is confusing you.

If you measure the spin in a different direction (say, the x direction) and get a definite result (say, plus hbar/2) then you know that the state is now the eigenvector of the matrix S_x with eigenvalue plus hbar/2.

It's not either of the states you wrote down (where did a "psi-" come from?). You should be able to write down the eigenspinor of S_x with eigenvalue of plus hbar/2 in the basis you started with.

8. Sep 5, 2007

### plmokn2

Thanks, I'll have a read of that now.

I was being stupid, thanks for clearing it up. So then in this basis the new spinor would be (1,i)T or (i,1)T or (1,1)T or (1,-1)T depending on the result and the axis (all needing to be normalised). And if we were working in a different basis I'd be looking for eigenvectors of U^TSU where U is some unitary matrix?

Last edited by a moderator: May 3, 2017
9. Sep 5, 2007

### olgranpappy

yep.

10. Sep 5, 2007

### plmokn2

Cool, thanks.

Thanks, this makes a lot of sense. It does raise one question though: if we treat H as some general infinesimal time change generator, (pretending we don't know it's the Hamiltonian) all the results still work and so generators still give rise to conserved quantities. Then applying the results to time invariance we find there's some conserved quantity that corrisponds to the generator d/dt (ignoring constants). Analogy with classical mechanics or experiment then gives us that this conserved quantity is the total energy and we seem to have 'derived' the Schrodinger equation?
Is this right or have I managed to confuse myself?

Thanks

Last edited by a moderator: May 3, 2017
11. Sep 6, 2007

### dextercioby

Yes, it's perfectly true, this "deduction" of the SE is made by Sakurai in his brilliant book, see chapter 2.

12. Sep 7, 2007

Thanks