# Ladder Operators for a SHO

1. May 9, 2005

### zeta101

Hi,

Having trouble understanding something here, hoping someone can help...when dealing with a SHO, we can define two ladder operators a and a-dagger. The way I understand it is, applying a-dagger to an eigenstate of H (and that has, for instance, eigenvalue E) will give us a new eigenstate that has eigenvalue E+hw (h=h-bar). Similarly, applying a would have given eigenstate with eigenvalue E-hw, right?

The problem is, if you apply a and then a-dagger one after the other on some eigenstate, there should be no effect (one lowers the energy, one raises it), right?... so:

a a-dagger|phi> = |phi>

and surely:

a-dagger a|phi> = |phi>

so shouldn't the commutator between a and a-dagger be zero? (but I know that it is actually 1, so my reasoning is wrong, but why?)

Thanks
James

2. May 9, 2005

### Tom Mattson

Staff Emeritus
No, it doesn't work like that. The action of $a$ and $a^{\dagger}$ are given by:

$$a|n>=\sqrt{n}|n-1>$$
$$a^{\dagger}|n>=\sqrt{n+1}|n+1>$$

So we have:

$$aa^{\dagger}|n>=\sqrt{n+1}a|n+1>$$
$$aa^{\dagger}|n>=(n+1)|n>$$

and:

$$a^{\dagger}a|n>=\sqrt{n}a^{\dagger}|n-1>$$
$$a^{\dagger}a|n>=n|n>$$

So they are not equal.

Last edited: May 9, 2005
3. May 9, 2005

### dextercioby

I'm afraid that this shouldn't have been an issue.The defining commutation relation of the unit Heisenberg algebra of the SHO should have said it all.One needn't have looked for the standard basis in an irreductible space of weight "l",because that nonzero value of the commutator should have said it all...

Daniel.

4. May 9, 2005

### Tom Mattson

Staff Emeritus
It is an issue because zeta101 miscalculated $aa^{\dagger}|n>$ and $a^{\dagger}a|n>$. He made a mistake. It happens to everyone.

Actually, it did say it all. He knew from the commutator that he made a mistake. He just couldn't see what it was.

5. May 10, 2005

### dextercioby

No problem,i think we saw this "The problem is, if you apply a and then a-dagger one after the other on some eigenstate, there should be no effect (one lowers the energy, one raises it), right?" from different perspectives.

Daniel.

6. May 10, 2005

### zeta101

Thanks for the replys...i think my actual problem is between the mathematics and the actual physics. You two guys seems to think very mathematically...where my error came from was my interpretation of the *worded* definititon of the two ladder operators. Thats why I came up with the incorrect commutator (and i knew it was incorrect before I even posted, i have a text book and lecture notes and indeed the "defining commutation relation of the unit Heisenberg algebra of the SHO should have said it all" did say it all, I wanted to know what was wrong with the worded defintion). The issue was with what "picture" or worded definition to have in my head of what the two operators were doing.

...oh and dexter, you are assuming I have all your knowledge and thinking patterns. I'm sure you know it's bad to make unfounded assumptions. What is obvious to yourself may not be to others...:)

7. May 10, 2005

### vanesch

Staff Emeritus
What you have done is a very interesting and common mistake in quantum theory, so just as well learn from it. It is the following issue. The wordings (the creation operator puts yourself in a state E+hbar w...) are in fact correct ; however what you have forgotten - and what everybody forgets regularly ! - is that a "state" in Hilbert space is a RAY, and not one single representative of it.
So it is not because "applying operator X to state |b> gives you state |c>" that you can write X |b> = |c>. You can only write X |b> = a |c>, where a is a complex number. That is because the physical state |b> can be represented by any u |b> and the physical state |c> can be represented by any v |c>. If the physical action of the operator X is to map the physical state represented by |b> onto the physical state represented by |c> then the only requirement is that each element of the ray u|b> is mapped upon an element of the ray v |c>

This is an especially tricky issue when dealing with symmetries. A typical example is a rotation over 360 degrees. You'd figure that the operator R that rotates a system over 360 degrees must be mapping every state upon itself. In fact, it only needs to map every state upon itself times a complex number. For fermions, that complex number is -1.

I'm sure dexter will now beat me to death with projective group representations

cheers,
Patrick.

8. May 10, 2005

### dextercioby

This (the forum) is no place for lecturing .Do you know that most of the QM texts (some of them aim at graduates,even,like Newton,Sakurai and Merzbacher) don't treat Wigner's theorem,Bargmann's theorem and all the underlying truth behind the first postulate and symmetry groups...? :surprised

Daniel.

9. May 10, 2005

### vanesch

Staff Emeritus
Why not ?? You could use it to refine your pedagogical skills

cheers,
Patrick.

10. May 10, 2005

### dextercioby

Hmm,in my country "pedagogical skills" means that the teacher presents the course in front of the audience made up of students.He speaks and writes on the blackboard .I couldn't do the same here...

Daniel.

P.S.We're hijacking the thread. This is not GD. :tongue2:

11. May 10, 2005

### vanesch

Staff Emeritus
Which is also mine... (before I moved out)

Ah, puleeze ! Where is your sense of abstraction ?
What is important in "speaking" ? The words ? Or the pitch of the voice ?
What is important in writing on the blackboard: the fact that it is black, or the formulas ?
The words, and the formulas, can go here too :-)

I've already participated (and also organized) a few on-line courses. The problem is not the communications channel. It is the courage of the students to continue all the way :-)

cheers,
Patrick.

12. May 10, 2005

### dextercioby

1.I doubt you're Romanian.
2.In speaking:COHERENCE,the line of thought,heh,it's not that easy,especially when u don't have any lecture notes in front of you...
3.I find that "blackboard thing" entirely fascinating.
4.I guess you may figured out yet,i'm not able to lecture on projective representations of QM symmetry groups.Not yet.

Heck,i haven't even graduated.You'd be asking me to do something a professor would do.

Daniel.

P.S.I haven't been taught this chapter of QM not in Craiova last year,nor in Leuven this year.I've read it on my own.Thankfully,i found books.But i still can't lecture,i.e.write a coherent text on this subject.

13. May 10, 2005

### vanesch

Staff Emeritus
Indeed, sorry. I thought you were a Belgian native :-)
I should have known it ; the fact that you didn't know that Imec was not a guy's name :-))

cheers,
Patrick.